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Thermal Equillibrium

2012-01-22T20:12:00.000-08:00

An object is in thermal equilibrium when the amount of heat entering the object is equal to the amount of heat leaving it. In the case of a planetary system, consisting of a surface and atmosphere, the system will be in thermal equilibrium when the heat arriving from the Sun is, on average, equal to the heat the system radiates into space. The planetary system of our Circulating Cells program will be in thermal equilibrium when the short-wave radiation penetrating to the planet surface from the Sun is balanced by the long-wave radiation escaping into space from its surface, atmospheric gas, clouds, rain, and snow.

When our simulated system converges upon a state where the temperature of its surface and of its atmospheric layers fluctuates around some average value, our hope is that the heat penetrating to the surface from the Sun will be equal, on average, to the heat the system radiates into space. The heat penetrating from the Sun is, of course, the incoming short-wave radiation that is not reflected back into space by our simulated clouds. The heat radiated into space is the upwelling radiation at the top of our simulated atmosphere. We refer to the top of the atmosphere at its tropopause, and to the heat radiated into space as the total escaping power.

We instructed CC11 to print out the average penetrating power from the Sun and the total escaping power from the tropopause every ten hours, and plotted these values for the fourteen thousand hours of the experiment we performed in our previous post. For the final ten thousand hours, the temperature of the surface and the atmospheric layers fluctuate by ±1°C around their average values, as you can see here.

Over the final ten thousand hours, the average penetrating power is 290.0 W/m² and the average escaping power is 289.6 W/m². Given the size of the fluctuations in both quantities, and the errors introduced by certain simplifications in our simulation's calculations, we are well-satisfied with this agreement. Our simulation converges upon an equilibrium state that is also a state of thermal equilibrium.

Radiating Clouds

2012-01-18T10:00:00.000-08:00

The latest version of Circulating Cells implements the upwelling and downwelling radiation calculations we described in Up and Down Radiation. To run the program, download CC11 and follow the instructions at the top of the code. Clouds absorb and emit long-wave radiation as if they were black bodies. We now set the transparency fraction of our atmospheric gas to 0.5, so that it will be transparent to half the wavelengths in the long-wave spectrum and opaque otherwise. The planet surface can radiate heat directly into space at these transparent wavelengths, as it did in Simulated Planet Surface. But now we have clouds doing the same thing, while at the same time reflecting sunlight back into space.

We begin our simulation with the final state of Simulated Rain, which you will find in SR_1200hr. The initial surface air temperature is 292 K, and cloud depth is 1.5 mm. The following graph shows how air temperature and cloud depth vary in the first two thousand hours.

The following graph shows the first fourteen thousand hours. You will find the final state of the array in RC_14000hr. The average surface air temperature over the final ten thousand hours is 288 K, and the average cloud depth is 0.8 mm.

During the course of these fourteen thousand hours, the distribution of clouds in the atmosphere varies greatly. Sometimes there is a layer of clouds just above the surface of the sea. At other times there are clouds along much of the tropopause. For a view of the final state of the simulation, see here.

As we have discussed many times before, the absorption of long-wave radiation by the atmosphere gives rise to the greenhouse effect. The more opaque the atmosphere, the more heat must be radiated into space by the tropopause instead of the planet surface. In order to radiate more heat, the tropopause must be warmer. If the tropopause is warmer, the planet surface must be warmer too, in order to motivate convection to carry heat to the tropopause. When we change our atmospheric gas from 0% to 50% transparency, we expect the surface temperature drop. And indeed it does: by 4°C.

This cooling of 4°C is, however, far less than the cooling of 31°C we observed when we increased the transparency of our gas from 0% to 50% in the absence of simulated clouds. As we have already discussed, clouds and rain greatly reduce the sensitivity of surface temperature to changes in solar power. Now we find that they also greatly reduce the sensitivity of surface temperature to changes in the transparency of the atmospheric gas.

Up and Down Radiation

2012-01-13T14:42:00.000-08:00

We are going to add to our Circulating Cells simulation the absorption and emission of long-wave radiation by clouds. As we showed earlier, a liquid water depth of 100 μm absorbs over 99% of all long-wave radiation. Rain contains liquid water also, and ice is a good absorber of long-wave radiation too. We will add the equivalent depth of snow, rain, and cloud droplets for each cell, and so obtain the depth of water within the cell that acts to absorb long-wave radiation.

We note that the same addition of rain, snow, and cloud droplets does not apply to the transmission of short-wave radiation. Water is transparent to short-wave radiation, and clouds reflect it by refracting it through millions of microscopic droplets. But rain and snow contain thousands of times fewer drops and crystals for a given depth of water, so they are thousands of times less effective at refracting sunlight.

For simplicity, we will assume the water in a cell is either transparent or opaque to long-wave radiation, but not in-between. If the combined concentration of rain, snow, and cloud droplets in a cell is greater than wc_opaque, we will assume the entire gas cell is opaque to long-wave radiation. Otherwise, the cell will absorb long-wave radiation as if it were dry, as determined by our transparency_fraction. With our 300-kg cells, a concentration of 0.33 g/kg corresponds to 100 μm of water.

Now we are faced with the possibility of multiple layers of cloud, snow, and rain, all absorbing and emitting long-wave radiation in all directions. The first simplification we make is to assume each gas cell radiates only vertically upwards and downwards. Because our columns of cells are much the same as one another on average, the net effect of this simplification will be small. Even with this simplification, we see that a cloud can absorb radiation from a cloud below, and emit radiation back to that same cloud below, and upwards to a third cloud.

We will calculate the effect of long-wave radiation in the following way. We start at the surface and allow it to radiate as a black body. We allow this upward radiation to enter the first gas cell. We calculate how much is absorbed by the cell and how much keeps going. We calculate how much power the gas cell itself radiates upwards. We add this to the existing upward radiation. We move on to the cell above, and so on, until we get to the tropopause. At the tropopause, we assume the atmosphere above is transparent to long-wave radiation, so all upward-going radiation passes out into space.

We repeat the same process, going down. We start with the tropopause gas cell in each column and move down cell by cell until we arrive at the bottom, at which point all the downward-going radiation is absorbed by the surface. We first considered this kind of downward-going long-wave radiation in our Back Radiation post. It is distinct from the Solar radiation that penetrates the atmospheric clouds because it is radiation emitted by the clouds, rain, snow, and atmospheric gas themselves.

In any cell, the long-wave radiation going up is the upwelling radiation and the long-wave radiation going down is the downwelling radiation. At the tropopause, the upwelling radiation is the heat leaving our planetary system. It is our total escaping power. When our simulation converges to equilibrium, we should find that the average Solar power penetrating to the surface is equal to the average total escaping power.

Sinking Restored

2012-01-09T08:22:00.000-08:00

When we added rain and snow to our Circulating Cells program, we removed the slow descent of microscopic water droplets, saying that their movement would be insignificant compared with that caused by convection. This is indeed the case when an equilibrium with plenty of atmospheric convection is established.

Nevertheless, we have found in our recent tests, in which we are allowing the clouds to radiate heat directly into space, that clouds can form and sit directly upon the surface, where they block the Sun's light. The surface cools beneath these clouds, and the clouds themselves cool by radiating into space, and we have seen them sit there fore hundreds of hours. This is unrealistic, because in a hundred hours, a cloud will sink by at least a few hundred meters.
So we restored the sinking of cloud droplets to our simulation, at 3 mm/s, which is realistic for cloud droplets 20 μm in diameter.

When we restored the sinking, we noticed that our previous implementation had allowed the clouds to sink only when the cells containing them took part in a convection circulation. As a result, the clouds were sinking through our simulated atmosphere a hundred times slower than they should have been. The fast-sinking implemented by CC9 were in fact sinking at 3 mm/s instead of 300 mm/s, and the slow-sinking clouds were sinking at only 0.03 mm/s instead of 3 mm/s. Thus our fast-sinking clouds were a more realistic simulation of the manner in which actual clouds would sink, while our slow-sinking clouds were unrealistically slow. We run our sinking cloud experiments with a corrected version of CC9, and the new slow-sinking result looked like the former fast-sinking result.

When clouds sink at 3 mm/s, they can sit on the surface for a few hours, but after a hundred hours, they disappear. The droplets will sink 1000 m in a hundred hours, which is three times the height of a gas cells resting upon the surface. The slow descent of the droplets removes clouds that would otherwise freeze the surface, and therefore plays an important role in our simulation, despite the fact that convection, rain, and snow cause movements that are thousands of times faster.

Simulated Rain

2011-12-23T10:00:00.000-08:00

The latest version of our Circulating Cells program implements the simplified evaporation cycle we presented in our previous post. Freezing clouds turn into snowflakes and drift downwards. When the snowflakes pass through warm air, they melt and become raindrops. To run the program, download CC10 and follow the instructions at the top of the code.

We run the simulation starting with our cold-start state, CS_0hr. The program runs ten times slower than before. The water balancing calculations are more complex now that we have added precipitation, and we must perform them more often because rain and snow move quickly through the atmosphere. Nevertheless, a few hours running gives us six weeks of simulation time, and the atmosphere converges to the equilibrium state shown below, which you will find stored in SR_1200hr. The light gray cells are clouds of water droplets. The white cells are clouds of snowflakes. The dark gray cells are rain.

We see clouds of water droplets in the top row of cells. Here they are cooling by radiation. Our atmosphere is still opaque to long-wave radiation (transparency fraction is zero). Only the top cells can radiate into space. Their temperature is, however, well below 268 K (Tf_droplets), the temperature at which droplets are transformed into ice crystals. Snow forms within the clouds at 0.001 g/kg/s (freeze_rate_gps) and falls at 1 m/s (snow_speed_mps). When it sinks through a cell warmer than 278 K (Tm_ice), it melts at 0.01 g/kg/s (melt_rate_gps), forming rain. Rain falls at 5 m/s (rain_speed_mps).

The following graph shows how surface air temperature and average cloud depth vary with time from our cold start. The final cloud depth fluctuates by ±0.5 mm around an average value of 1.7 mm. The average temperature of the surface gas is 292 K, which is 19°C. Of the light that arrives from the Sun, 30% is reflected into space, giving our simulated planet an albedo of 0.3, which matches that of our own planet Earth.

With fast-sinking clouds, the cloud depth remained close to 2.9 mm and the surface temperature was −7°C. Evaporation from water at 19°C is roughly ten times faster than from from water at −7°C, but precipitation is so effective at removing water from the atmosphere, the sky is almost entirely clear. Indeed, the Sun shines directly upon our island half the time, heating its sandy surface up to 34°C.

Evaporation Cycle

2011-12-19T10:53:00.000-08:00

The following diagram presents the simplified cycle of evaporation and precipitation we propose to implement in Circulating Cells Version 10.1.

Evaporation takes place from the sea, as before. When a body of moist air rises, it cools, and microscopic droplets form by condensation. Clouds of such droplets that happen to descend from above will warm up, and some or all of their droplets will evaporate. A cloud of droplets whose temperature drops below some threshold T_f will be transformed into snowflakes by the Bergeron Process, warming the surrounding gas with latent heat of fusion. We choose T_f several degrees below the freezing point of water, so we can assume the freezing takes place rapidly.

In our simulation, snow will fall at an average of 1 m/s, which we base upon our own observations. We will implement snow fall in the same way we implemented sinking clouds. Snow that reaches the surface will melt and thus take its latent heat of fusion from the surface block. This melting at the surface is the simplest way we can think of to conserve the latent heat of fusion of the water involved in our evaporation cycle. We assume that our surface water itself never freezes, no matter how cold it gets.

Our clouds, meanwhile, will no longer sink. Real cloud droplets are of order ten microns in diameter and sink at a few millimeters per second. The contribution of such sinking to our new cycle would be negligible.

Snow that enters a gas cell at a temperature greater than T_m will melt, cooling the surrounding gas by absorbing its latent heat of fusion. The melted snowflakes become raindrops a few millimeters in diameter, and these fall at 5 m/s. With the simulation set up as we have it now, the cells are around 400 m high, so rain will take a minute or two to fall out of one cell into the next. We will choose Tm several degrees above the melting point of water so we can assume the melting takes place rapidly.

Our simplified evaporation cycle omits many interesting evaporation-related phenomena. When rain drops are carried up into cold air, for example, they form hail, which later falls to Earth. When liquid rain falls into sufficiently dry air, it evaporates and disappears altogether, giving rise to virga. When air rises at just the right speed along a mountain slope, water droplets join together to form rain drops, as in orographic precipitation. Our simulation will contain none of these interesting phenomena. But we believe it will capture the fundamental features of the Earth's evaporation cycle, and so allow us to investigate how this cycle influences the global surface temperature.

Rain

2011-12-12T10:00:00.000-08:00

Our simulated sky never clears. Clouds fill the atmosphere almost entirely. They are forever forming in air that rises from the sea, and forever sinking to the ground, but they never come falling out of the sky all at once in the big drops we know as rain.

The droplets in our clouds are tiny. Those in our slow-sinking clouds are only 10 μm in diameter and descend at 3 mm/s. Those in our fast-sinking clouds descend at 300 mm/s. The graph we present in Falling Droplets implies that these fast-sinking droplets are 100 μm in diameter. Rain falls to Earth at several meters per second, so the same graph tells us that rain drops are at least 500 μm in diameter. A drop 500 μm in diameter contains a hundred times as much water as a droplet of 100 μm and a hundred thousand times as much water as a droplet of 10 μm. Could it be that cloud droplets collide and coalesce in order to form rain drops? If so, how long does this take, and under what circumstances does it occur?

The Wikipedia page on rain describes convective precipitation and orographic precipitation. In both these forms of rain, a cloud moves up, and encounters rain drops descending from above. If the descending drops are 500 μm in diameter, and the air is moving up at 1 m/s, the drops will remain at the same altitude. The cloud moving up and past them carries microscopic droplets that can collide with the stationary drops, coalesce with them, and so enlarge them until they are heavy enough to fall out of the rising cloud and descend to the Earth as rain.

But further reading suggests that rain formed of coalescing droplets is rare. A far more potent source of rain drops are ice crystals. In Cloud Physics, we learn of the Bergeron Process, whereby ice crystals grow, sink, melt, and become rain drops. Large rain-drops are melted hail-stones. Small rain-drops are melted snow-flakes.

In our simulation, whenever the concentration of water vapor exceeds the saturation concentration, we assume the excess water condenses. It turns out, however, that the surface tension of liquid water makes it hard for water to condense into floating, microscopic droplets. If we provide a solid surface for the water to condense against, such as a blade of grass or a glass mirror, the water will condense when it reaches the saturation concentration, but in a body of air high above the ground, the only such surfaces would be dust particles, and these may be rare. Each one will serve as a catalyst for condensation until a droplet forms around it.

But the same is not true of ice crystals. In air saturated with water vapor and below the freezing point of water, an ice crystal can form on a grain of dust, and after that it will continue to grow. Water vapor deposits directly upon the surface of the crystal, thus changing state from gas to solid in one step, and the newly-created ice surface is an ideal foundation for further growth.

Now, suppose a cloud of microscopic water droplets rises until its temperature drops to −20°C. We might assume that the droplets will freeze. But pure water droplets resist freezing until they drop to −40°C. Ice crystals form in the midst of the cloud of super-cooled water droplets. As water vapor is deposited on the crystals, the concentration of water vapor in the air drops.

And here we encounter another curious physical phenomenon. The saturation concentration of water vapor with respect to an ice crystal turns out to be lower than the saturation concentration of water vapor with respect to super-cooled liquid water. Water vapor will deposit on the ice crystals until the concentration of water vapor drops to the saturation concentration of water vapor with respect to ice crystals. Because this concentration is below the saturation concentration with respect to super-cooled liquid water, the water droplets actually start to evaporate. The droplets evaporate, and their water is deposited onto larger and larger ice crystals.

Once the ice crystals are large enough, they start to fall, and they eventually fall into air that is warm enough to melt them. They turn into drops of water and fall to Earth as rain. It is this process that we will attempt to simulate in the next version of our Circulating Cells program.

Less Reflection

2011-11-28T08:19:00.000-08:00

With 350 W/m² arriving from the Sun, 75% of the surface covered by water, clouds sinking at 300 mm/s, and each 3 mm of cloud reflecting 63% of sunlight, our CC9 simulation converges upon a surface air temperature of −12°C. When we increase the Sun's power to 400 W/m², the temperature rises by a mere 0.5°C. Our simulated planet is kept cold by thick clouds that reflect the Sun's light back into space. Ice crystals drift down from the sky in some places, while elsewhere water evaporates from the frozen seas.

The surface of the Earth is at an average temperature well above the freezing point of water, and the Earth's sky is frequently clear of clouds. Our simulated sky never clears, and the surface is frozen. It never rains in our simulation, nor do our simulated clouds emit or absorb radiation. Perhaps these two omissions are responsible for our permanent clouds and frozen seas. Before we attempt to rectify them, however, let us consider the effect of decreasing the reflecting power of our simulated clouds.

We increased Lc_water from 3.0 mm to 6.0 mm, so that it now takes 6.0 mm of cloud water to reflect 63% of the Sun's light. With the reflecting power divided in half, we ran our simulation for eight thousand hours from the starting point CS_0hr. You will find the final state in LR_8000hr.

Compared to before, we now have more clouds in the sky. The following graph shows how cloud depth and surface air temperature vary with time.

Compared to before, we see the atmosphere reaches equilibrium in on third the time. The new temperature is higher and the cloud cover is thicker. The following table compares the state of the atmosphere for both types of clouds.

Our seas are now at −3°C. If they contain salt, they will not freeze. The air a few meters above our sandy island will be just below freezing. Our simulated world is still much colder than the Earth, and nobody standing on the island would ever see the Sun. We are, however, gratified to find that our simulation remains stable with such a large drop in cloud reflectance.

Negative Feedback

2011-11-22T20:01:00.000-08:00

With fast-sinking clouds, our circulating cells program reaches equilibrium in eight hundred hours of simulation time. With the 350 W/m² arriving continuously from the Sun, the surface air temperature settles to 261 K.

If we increase the power arriving from the sun, it seems reasonable to suppose that the surface temperature of our planet will rise. Indeed, before we added clouds to our simulation, we could use Stefan's Law to answer this question. The planet surface absorbed all the Sun's heat and the surface and tropopause radiated it all back into space, so if we increased Solar power by 4%, the absolute temperature of the surface and the tropopause would increase by 1%. But with clouds reflecting light from the Sun, we can no longer assume that all the Sun's heat will be absorbed, nor even that a constant fraction of it will be absorbed.

We ran our fast-sinking clouds simulation repeatedly from the same CS_0hr starting conditions, each time with a different Solar power. Each time we stopped the simulation after a thousand simulated hours, so that we could be sure it had reached equilibrium, and recorded the surface air temperature. We obtained the following graph.

Without clouds, a doubling of Solar power would cause the surface temperature to increase by roughly 20%. Here we increase Solar power from 200 W/m² to 400 W/m² and the surface temperature increases by only 1.5%. The following graph shows how the cloud depth increases with Solar power, thus decreasing the fraction of Solar power that penetrates to the surface.

The Sun's light, arriving at the surface, causes evaporation. This evaporation leads to clouds. But these same clouds reflect the Sun's light back into space. Thus one effect of Sunlight arriving at the surface is to reduce the amount of Sunlight arriving at the surface. The effect of clouds is an example of negative feedback. This negative feedback reduces the sensitivity of surface temperature to Solar power by more than a factor of ten.

Fast-Sinking Clouds

2011-11-15T14:18:00.000-08:00

In our previous post we allowed the clouds in our simulation to sink to the surface at 3 mm/s. We implemented this sinking by allowing an average of 0.001% of each cell's water droplets to drop down out of the cell every second (0.1% every 100 s). Today we repeat our experiment from the same starting point, but this time the gas cells lose 0.1% of their water droplets per second, which corresponds to droplets sinking at 300 mm/s. Here is the state of the atmosphere after thirteen thousand hours.

Although our screen shot is taken at thirteen thousand hours, the atmosphere converges to its equilibrium state in a mere eight hundred hours. Our previous simulation converged only after eight thousand hours. The following graph shows surface gas temperature and cloud depth versus time.

The following table compares the equilibrium state of the atmosphere at the end of our two experiments.

The faster-sinking clouds cause the surface to warm by 5.3 K. The Solar power penetrating to the surface increases by 13 W/m² because the clouds are slightly thinner. You may recall that our current simulation of clouds does not implement their absorption and emission of long-wave radiation, so we are working with transparency fraction set to 0.0, indicating an atmospheric gas that is opaque to long-wave radiation. The only place for this artificial atmosphere to radiate is at the tropopause. So we expect to see the tropopause radiating the same amount of heat that penetrates to the surface: the heat leaving the system must be equal to the heat entering. And indeed this is the case to within a couple of Watts per square meter.

We see that faster-sinking clouds cause the world to warm up, and this is in keeping with our expectation. The icy surface must warm up so that evaporation will keep up with the greater rate of return of water to the surface.

UPDATE: It turns out that our code was allowing clouds to sink only when they took part in a circulation, which resulted in them sinking roughly a hundred times slower than they should have, so our effective sinking rate here was more like 3 mm/s. When we correct our error, so that the clouds really do sink at 300 mm/s, the surface temperature warms by roughly 7 K. [07-JAN-12]

Slow-Sinking Clouds

2011-11-11T08:00:00.000-08:00

In Falling Droplets, we concluded that 10-μm water droplets will sink at 3 mm/s in air at pressure 100 kPa. The lowest atmospheric cells in our Circulating Cells program are at 100 kPa, and they are roughly 300 m high, so we see that it will take a hundred thousand seconds for a droplet to fall the height of the cell. Cells higher up are taller, but the gas within them is thinner. A droplet must fall more quickly through thinner air before its weight is matched by air resistance. For simplicity, we will assume that the time it takes a 10-μm droplet to fall the height of a cell is the same regardless of altitude.

As we found in Simulation Time, our program checks each gas cell every one hundred iterations on average, which corresponds to every 100 s. If it takes a droplet one hundred thousand seconds to fall the height of a cell, and the droplets in the cell are evenly distributed, 0.1% of the droplets will sink out of the cell every 100 s. If the cell rests upon a surface block, these droplets will return to the surface. We now have a way for water to leave the surface, by evaporation, and a way for water to return to the surface, by sinking. If the gas cell rests upon another gas cell, the droplets enter the cell below, where they may evaporate.

In Circulating Cells Version 9.1, we specify the sinking speed of droplets at 100 kPa with sinking_speed_mps in units of meter per second. We set sinking_speed_mps to 0.003 m/s and loaded CS_0hr into our array, which is the starting condition we used in Cold Start.

The planet warms quickly in the steady light of the Sun. After two hundred hours, the atmosphere is full of clouds and the planet starts to cool. The clouds sink towards the surface. After four thousand hours, they are thin enough that the sun starts to warm the surface. After eight thousand hours, this warming is stopped by the formation of new clouds. After a thirty thousand hours, the atmosphere settles to the steady state shown below, which you will find saved as a text array here.

The graph below plots temperature and cloud depth versus time for the first thirty thousand hours.

After thirty thousand hours, the sand and water surfaces are both at 260 K (−13°C), and the lower gas cells are at 255 K (−18°C). The average cloud depth is 3.2 mm and the average power reaching the surface is 120 W/m².

The combination of evaporation and sinking gives rise to an equilibrium in which the clouds allow just enough heat to reach the surface so that evaporation balances the return of water to the sea in the form of sinking droplets. This balance between evaporation and sinking controls the temperature of the planet surface. In our next post, we will increase the sinking speed by a factor of a hundred and see how this affects the surface temperature. We expect the surface temperature to go up, because only then will evaporation keep up with the increased loss by sinking.

UPDATE: It turns out that our code was allowing clouds to sink only when they took part in a circulation, which resulted in them sinking roughly a hundred times slower than they should have, so our effective sinking rate in this simulation was more like 0.03 mm/s. When we correct our error, so that the clouds really do sink at 3 mm/s, the surface temperature warms by roughly 7 K. [07-JAN-12]

Falling Droplets

2011-11-02T12:23:00.000-07:00

In Clouds Without Rain we saw immortal clouds circulating above a frozen planet, reflecting the Sun's heat back into space. We concluded that it is rain that saves our planet from freezing. It is rain that clears the skies so that the Sun's heat can reach us.

We must implement rain in our simulation, so that water vapor has some way of returning to the surface. Let us begin by considering how fast water drops fall through air. A falling drop accelerates until air resistance matches its weight. At that point, it continues to fall but it does not accelerate. It has reached its terminal velocity. In The Terminal Velocity of Fall for Water Droplets in Stagnant Air, Gunn et al. describe their apparatus for measuring the terminal velocity of water droplets, and present their measurements in graphs and tables. (The paper, published in 1948, is an enjoyable read that you can download here.)

We can calculate the terminal velocity of rigid, spherical objects using Stokes Law. Gunn et al. show that Stokes' Law applies well to water droplets of diameter less than 100 μm (one tenth of a millimeter). You may recall that the droplets in our simulated clouds are roughly 10 μm in diameter (one hundredth of a millimeter). For the droplets are small enough, surface tension is able to maintain a spherical shape in the face of air resistance.

But for droplets larger than 100 μm, Stokes Law over-estimates the terminal velocity. Larger droplets assume flattened shapes as they fall, and they are in constant motion, so that the air resistance they encounter is far greater than it would be for a rigid sphere. When the diameter exceeds 5 mm, the motion of the drop becomes so vigorous that the drop breaks into smaller drops.

The following graph shows the Gunn et al. measurements of terminal velocity, plotted against droplet diameter. We see that the maximum terminal velocity for the largest possible water droplets is around 10 m/s. For diameters less than 0.1 mm, Gunn et al. assure us we can use the terminal velocity given by Stoke's Law, so we also plot the terminal velocity calculated from Stokes' Law.

The 10-μm droplets in our simulated clouds will fall at a mere 3 mm/s through our gas cells. Given that our cells are a few hundred meters high, it will take a day or two for a cloud to fall from one cell to the cell below. Slow as this may be, the sinking of clouds does provide a way for water to move from one cell to another, and ultimately to return to the planet surface. In our next post, we will see how sinking clouds affect the result of our Cold Start simulation.

Cold Start

2011-10-19T07:42:00.000-07:00

In our previous post, we presented our simulation of clouds without rain. We started the atmospheric gas, the sandy island, and the watery sea, at a uniform 280 K (7°C). Water evaporated from the sea. The sand heated up in the sun. Hot air rose above the island and sucked moist air in from the sea. Clouds formed above the island, spread through the atmosphere, reflected the heat of the sun, and the world froze.

What if we start with a frozen world and a dry atmosphere? In our simulation of evaporation rate, no water will evaporate from a sea at 250 K (−23°C), so no clouds will form. We ran CC9, starting with the CS_0hr array, to find out what would happen. Our starting point is a uniform 250 K with no water vapor. We run with 350 W/m² continuous heat from the Sun.

After 20 hrs, the sandy island has warmed to 276 K (3°C). At 30 hrs, the average cloud depth is 0.03 mm, which is so thin that we don't bother plotting the clouds as white cells. But at 40 hrs we start to see the first thin clouds, and the average power arriving from the Sun drops to 335 W/m². At 50 hrs, the island reaches 283 K (10°C). From here on, it cools. At 100 hrs, the average cloud depth is 3.5 mm and only 120 W/m² is arriving from the Sun. The sea reaches 267 K (−6°C), which is the warmest it will ever get. By 200 hrs, cloud depth is 7.2 mm and power arriving from the Sun is only 40 W/m², as recored in CS_200hr.

We can see where the simulation is going to end up: a world kept frozen by immortal clouds. Regardless of our starting point, immortal clouds reflect the Sun's heat and cause the world to freeze.

Clouds Without Rain

2011-10-04T16:20:00.000-07:00

Today we present Circulating Cells Version 9 (CC9), and we use it to find out what would happen if we had clouds without rain. The simulation implements the following features of clouds.

(1) Evaporation from surface water, as in Evaporation Rate.
(2) Condensation in rising air, as in Condensation Point and Condensation Rate.
(3) Cooling and warming by latent heat of evaporation, as in Latent Heat.
(4) Reflection of incoming sunlight, as in Simulated Clouds, Part I.

The simulation does not yet implement the following features of clouds.

(5) Absorption and emission of long-wave radiation, as in Simulated Clouds, Part II.
(6) Cooling and warming by latent heat of fusion, as in Latent Heat.
(7) Rain and snow.

If we set the simulated atmosphere's transparency fraction to 0.0, we make the atmospheric gas opaque to long-wave radiation. No radiation escapes into space from the surface blocks nor from the rows of gas cells below the top row, regardless of the distribution of clouds within the atmosphere. The top row of cells, which is our simulated tropopause, does all the radiating of heat into space. Although this opaque atmosphere is not realistic, it does mask the fact that our clouds do not in themselves absorb or emit long-wave radiation, allowing us to proceed with a simulation that is at least self-consistent. Thus the copy of CC9 that you can download today has the transparency fraction set to 0.0 by default.

We ignore the warming of rising air by freezing water droplets, and the cooling of falling air by melting ice crystals. We will add ice crystals to our simulation later. For now, we trust that the error caused by our omission is not so great as to overturn the observations we make today.

By ignoring rain and snow, we are ignoring a feature of clouds that is so important to our climate that our simulation produces an entirely fantastic result. To watch the simulation in action, download CC9 and follow the instructions at the top of the code to run the program on your computer. Get the CWR_0hr array file and load it with the Load button. You will see an atmosphere at a uniform 280 K, and down at the bottom, an island of sand in a sea of water, also at 280 K. Press Run and the simulation will begin. The CC9 code runs in "Day" mode by default, with 350 W/m² arriving continuously from the sun.

Without rain and snow, any and all moisture that enters the atmosphere at the beginning of the simulation remains in the atmosphere for as long as the simulation runs. There is no means by which moisture can return to the surface of the planet. So long as the lower atmosphere is warm enough to absorb water vapor, however, the clouds can appear and disappear. The moisture they contain can either take the form of water vapor, as it will when the surrounding gas is warm, or it can take the form of water droplets, as it will when the surrounding gas is cold.

We represent clouds in our simulation with cells that are shaded white to gray. The thinnest clouds are white and the thickest are black. A cloud with 1 mm of water is white. A cloud with 12 mm of water is black. After 30 hours, the first white clouds appear over the island, the result of moist air from the sea being heated by the island and rising towards the tropopause. When it rises, it cools, and water vapor condenses to form the first clouds.

At 40 hours, the island reaches its peak temperature of around 311 K. After that, the clouds become more numerous. They reflect the Sun's light back into space. The surface begins to cool. After 150 hrs we end up with the following display.

There is fog over the sea and part of the island. There are thick clouds up in the tropopause. The average heat arriving at the surface from the Sun has dropped from 350 W/m² to only 50 W/m². The following graph shows how the atmosphere cools in the first 500 hrs.

Let us refer to the combined thickness of the clouds above a surface block as is its cloud cover. In our simulation, each 3 mm of cloud cover reflects 63% of incoming sunlight. If we press the Data button, a text window opens and here we will see a line of numbers printed every hour of simulation time. The first number is the time in hours, the second is the average cloud cover in millimeters. The third number is the average sunlight power penetrating to the surface through the cloud cover in Watt per square meter. After that we have four temperatures in Kelvin: average sand temperature, average water temperature, average surface gas temperature, and average tropopause temperature. We used these printed lines to obtain the data for the plot above.

After 3000 hrs, the tropopause has dropped to 158 K (−115°C) and the surface air is at 184 K (−85°C). The average cloud cover is 25 mm. Only 0.7 W/m² arrives at the surface. You can see this for yourself by loading CWR_3000hr into the simulation. Even at 158 K, the tropopause is still radiating 35 W/m², which is far more than the 0.7 W/m² reaching the planet surface. The tropopause will keep cooling until it reaches 60 K, at which point it will be radiating 0.7 W/m². Of course, at that point, nitrogen will condense into liquid.

If clouds remained aloft in the atmosphere indefinitely, the Earth would freeze. But in reality, clouds are forever falling towards the ground. They are made of droplets and crystals that are heavier than air. Rain and snow are what stop clouds from turning the Earth into a planet of frozen seas.

Summary to Date

2011-09-27T13:00:00.000-07:00

I have updated the Summary to Date, which now includes the following paragraphs describing where we are in our investigative journey, and what we are going to do next.

Once we were satisfied that our simulation handled convection properly, that we could relate the program iterations to the passage of time, and that all the heat entering the simulated system was accounted for by radiation from the top, we added blocks of either water or sand beneath the bottom gas cells, so as to simulate the planet surface. In Back Radiation we showed how the heat capacity and radiation produced by a semi-transparent atmosphere keeps the planet surface warm at night. In Island Inversion we see the surface of an island heating up ten times more than the surrounding ocean, while at night a layer of air a few hundred meters above the island is warmer, rather than cooler, than the air resting upon the island. Thus we see our simulation is consistent with our observations of surface cooling, including even temperature inversion.

Well-satisfied with our simulation of a dry atmosphere, we now turn to the simulation of a wet atmosphere, in which evaporation will cool the ocean and lead to the formation of clouds. To simulate cloud formation, we must have equations for the rate of evaporation from a water surface, the rate at which water vapor will condense out of rising air, the rate at which it will evaporate again in falling air, the cooling effect of evaporation upon the water surface, the warming effect of condensation upon the rising air, the amount of sunlight that will be reflected by existing clouds, and the amount of long-wave radiation that these same clouds will absorb and radiate. We obtained these relations in a series of posts Evaporation Rate to Consensation Rate. We have yet to consider the downward drift of water droplets that leads to their combining together and forming rain. But after so many posts of mathematics and empirical relations, we thought it was time to get back to the simulation, and so we will start our simulation of clouds without allowing rain, and perhaps we will see how important rain is for our climate.

I'm running CC9 right now, and will present it later this week, once I have made a reasonable effort to eliminate errors from my implementation of evaporation, condensation, and reflection. The clouds are going round right now, as gray-shaded cells, and the effect is entertaining. Ultimately, you may recall, our objective is to see how a change in the transparency of the dry atmosphere affects the surface temperature of the planet, so that we can determine the effect of CO2 doubling within a system dominated by the effect of cloud formation and rain.

Condensation Rate

2011-09-23T20:05:00.000-07:00

in Condensation Point we considered the temperature at which water vapor will begin to condense into water droplets, thus making a cloud. We did not consider how fast this condensation will take place. Consider air with 20 g/kg of water vapor (that's 20 g of water vapor mixed with each 1 kg of dry air to make 1.020 kg of moist air). This air rises rapidly, expands, and cools to a point where its saturation concentration of water vapor is only 10 g/kg. Does the excess 10 g/kg condense into droplets immediately, or does it take some time, in the same way that the original evaporation took time?

As we saw in Latent Heat, the evaporation of water requires 2.2 kJ of heat for each gram of evaporating water. Because we must put energy into the water to make it evaporate, evaporation takes place slowly. In the case of condensation, however, the exact opposite is the case: condensation liberates 2.2 kJ of heat for each gram of water that condenses. Condensation takes place much more quickly, but it cannot take place instantly. In order for condensation to take place, water vapor molecules must bump into one another and stick together. A dust particle helps accelerate the condensation process by providing a surface upon which water molecules can condense. Until such time as all condensation is complete, the water vapor concentration remains greater than the saturation concentration, and we say the water vapor is supersaturated.

The cloud chambers of early high energy physics experiments used supersaturated water vapor to detect charged sub-atomic particles. A cloud chamber consists of a piston with a glass top. We fill the piston with moist air and pull the piston down rapidly, so that the air cools by adiabatic expansion and becomes supersaturated. When a charged particle, such as a cosmic ray, passes through the chamber, water condenses into a trail along its path. Indeed, cosmic rays may play a part in promoting cloud formation in our atmosphere. The CLOUD experiment is an effort by high energy physicists to apply their experience with cloud chambers to the study of cosmic rays and cloud formation, especially cloud formation at high altitudes where the air is thin and the water vapor is scarce.

Even in a cloud chamber, however, supersaturated water vapor does not endure for long. A useful cloud chamber has a piston going up and down several times a second because the water vapor condenses on its own within a fraction of a second. In our Circulating Cells program, we will check the water vapor concentration of the cells every hundred seconds or so. For the purpose of our simulation, therefore, we will assume that condensation within a cell is complete within a hundred seconds. When we find a cell with 20 g/kg of water vapor and a saturation concentration of 10 g/kg, we will allow 10 g/kg to condense into droplets.

Not only do we expect to encounter moist air rising and cooling, we will also have cloudy air falling and warming. As it warms, the saturation concentration increases, so it is possible for some or all of the water in the droplets to evaporate again. Because evaporation rate is proportional to the surface area of water, the tiny droplets of a cloud will evaporate quickly. The 20-μm diameter droplets of a cloud provide 3000 cm² of surface area for each gram of water they contain. A 1-cm deep puddle, meanwhile, provides only 1 cm²/g. We expect cloud droplets to evaporate three thousand times more quickly than a 1-cm deep puddle. A 1-cm deep puddle will evaporate in less than ten thousand seconds, so a cloud will evaporate in less than thirty seconds. For the purpose of our simulation, therefore, we will assume that the evaporation of cloud droplets is complete within a hundred seconds.

Combining these two assumptions together, we see that whenever our simulation encounters a gas cell with water vapor concentration greater than the saturation concentration, we will remove the excess water vapor and turn it into cloud droplets. Conversely, whenever we have cloud droplets with water vapor concentration less than the saturation concentration, we will remove however many cloud droplets we can until the water vapor concentration is again equal to the saturation concentration.

Simulated Clouds, Part II

2011-09-16T10:12:00.000-07:00

In Part 1, we gauged the thickness of a cloud by how deep a layer of water it would make if we combined all its water droplets into a pool of the same area as the cloud. A thin cloud might contain 1 mm of water, while a thick storm cloud might contain 100 mm.

We also concluded that even the thinnest of clouds is opaque to long-wave radiation, and therefore a good radiator of its own heat. Meanwhile, clouds do not absorb short-wave radiation from the sun at all because water is transparent to sunlight. Instead, they reflect sunlight back into space. For the purpose of our Circulating Cells simulation, we decided that each 330 μm thickness of water will reflect 10% of sunlight. Perhaps that's too much reflection, perhaps it's too little. We can adjust the 10% reflection depth later if we need to.

Suppose we have a 1-mm cloud layer up near the tropopause, and a 10-mm cloud layer nearer the ground. The combined thickness of both clouds is 11 mm, from which we deduce that only 3% of sunlight will penetrate to the planet surface. This is a calculation we can perform easily in our simulation. We add the thickness of the clouds above each surface block, and apply our formula for reflection to obtain the fraction of sunlight arriving at the surface.

More complicated than the incoming sunlight is the absorption and radiation of heat by separate cloud layers. The surface radiates heat as if it were a black body, but our simulated atmospheric gas has a transparency fraction, which tells us the fraction of long-wave radiation passing through the gas. The rest of the radiation is absorbed. Suppose our transparency fraction is 60%, then 60% of the heat radiated by the surface will reach the bottom layer of cloud, where all of it is absorbed. The cloud itself radiates heat in proportion to the fourth power of its temperature, as if it were a black body, and of this heat 40% is absorbed immediately by the gas above, below, and even at the center of the cloud. The remaining 60% passes down to the surface and up to the upper layer of cloud. The upper layer of cloud absorbs all the radiation from below, and itself radiates heat in proportion to the fourth power of its temperature, as if it were a black body. Of the heat radiated by the upper cloud, 60% will pass back down to the bottom layer of cloud and out into space.

Thus we see that we have long-wave radiation flowing in both directions because of the clouds. If we had just one, thick, cloud layer, our calculation would be simpler. But we have fifteen rows of cells in our simulation, so we could have seven layers of cloud, each separated by a row of gas cells. Our way of handling this problem will be as follows.

For each column of cells, we start at the top and make our way down to the surface. When we encounter a cloud, we calculate how much heat it radiates downwards from its bottom surface. We proceed until we reach another cloud, and here we allow the downward heat to be absorbed at the top surface of the cloud. We continue to the bottom surface of the cloud, and keep going with the same procedure until we get to the surface. By this time we have added up the total cloud thickness and we can determine how much sunlight has reached the surface as well.

Now we start from the surface and go upwards. The surface radiates heat, and this is absorbed by the bottom surface of the lowest cloud. The top surface of this cloud radiates heat upwards. If there is another cloud above, its bottom surface will absorb the upward-going heat, but if there is no other cloud, the heat passes into space.

During this entire process, we keep track of the amount of heat that is added or subtracted from the surface and from each gas cell. Once we are done, we adjust their temperatures to account for the heat lost or gained.

Thus we see that our clouds will introduce new sources of radiation into space that are at a lower altitude than the tropopause that is currently doing all the radiating into space of our simulated atmosphere. On the other hand, the clouds obscure the hottest radiating surface of all, which is the ground.

Our calculation of up-welling and down-welling radiation might slow down our simulation a great deal. But we're not in any hurry, so we won't worry about the computation time.

Simulated Clouds, Part I

2011-09-09T15:25:00.000-07:00

When water condenses within a rising body of air, it forms a cloud of liquid droplets. A thickness of more than 20 μm of liquid water is opaque to long-wave radiation. In Clouds we showed that even a sparse cloud is a near-perfect absorber of long-wave radiation. By radiative symmetry, clouds are also near-perfect emitters of long-wave radiation. At the same time, we showed that clouds do not absorb short-wave radiation, such as sunlight. They either reflect it or allow it to pass through without absorption.

We will soon implement cloud formation in our Circulating Cells program. We must decide how to implement their absorption and emission of long-wave radiation, and their reflection of sunlight.

Looking at our graph of saturation concentration, we see that air with 50% humidity at 300 K contains around 25 g/kg of water. Suppose this air rises and a mere 1 g/kg of water vapor condenses. Our gas cells have mass 330 kg/m², so when 1 g/kg of water condenses, there will be 330 g of water over each square meter of the cell's base area. This 330 g, if spread over a square meter, has depth 330 μm. According to our absorption spectrum for water, 330 μm of liquid water is more than enough to absorb all long-wave radiation, but not enough to absorb even 1% of sunlight.

The condensed water forms a cloud of water droplets. Cloud droplets are typically twenty micrometers in diameter. Our 330 g/m² will form roughly a hundred billion such droplets. Sunlight passing vertically down through the cloud will encounter roughly thirty such droplets. Each drop will reflect and refract the light. We estimate that 10% of the descending sunlight will be reflected back out into space by such a cloud, while 90% will continue onwards. When 10 g/kg of water condenses, we will have 3.3 kg/m² of water vapor, and sunlight will encounter 300 droplets instead of 30. The fraction of light passing through the cloud will be 0.9¹⁰ = 35%, while 65% is reflected.

Thus we have a way of taking the concentration of condensed water in a gas cell, and calculating the fraction of light it will reflect back into space. We also have a simple way of handling the absorption and emission of long-wave radiation by clouds: any cloud in our simulation will be a both a perfect absorber and a perfect emitter of long-wave radiation.

Latent Heat

2011-09-03T20:54:00.000-07:00

In Evaporation Rate we considered the rate at which water evaporates from the sea, and in Condensation Point we considered the amount of water that will condense from humid air when it cools down. Today we consider the heat absorbed by evaporating water, and the heat liberated by condensing water vapor.

It takes 2.2 MJ of heat to evaporate one kilogram of water. This heat is called the latent heat of evaporation. Two million Joules is enough energy to raise a 100 kg load to the top of a two thousand meter mountain. It is the energy released by the explosion of a stick of dynamite, or the energy we obtain from eating two jelly donuts.

For the purpose of our simulation, let us suppose that only the top one meter of water supplies the heat of evaporation. The heat capacity of water is 4.2 kJ/kg, so our surface blocks of water will have heat capacity of 4.2 MJ/m². In an earlier example, we found that roughly 1.8 kg of water will evaporate every hour from each square meter of a lake at 290 K (14°C). The latent heat carried away by the evaporating water will come from the heat of the water it leaves behind, so the lake surface will cool by roughly 1°C/hr.

As we saw in Back Radiation, the lake absorbs heat from the sun during the day, and always radiates heat upwards. In Surface Cooling, Part I, we showed how a water surface heats up by less than 1°C during the day. A lake does not get hot enough with respect to the air above to cause significant convection. Thus heat loss by a water surface is dominated by radiation and evaporation.

When water vapor condenses from cooling, humid air, it releases its latent heat into the air around it. The volume occupied by the water vapor decreases by a factor of a thousand then it condenses, but at the same time its latent heat warms up the air, causing the air to expand. In Condensation and Convection we found that the expansion due to warming dominates the contraction due to condensation by almost an order of magnitude. A single gram of water vapor condensing out of kilogram of air causes the air volume to increase by 1%. When air expands, it becomes buoyant, so it will have latent heat of fusion. This is the heat required to melt ice, which is liberated when the water freezes. Water's latent heat of fusion is roughly 330 kJ/kg. If we have one gram of water freezing in 1 kg of air, the air will warm by roughly 0.3°C.

We can now implement in our Circulating Cells program the evaporation of water from the planet surface, its subsequent condensation into clouds of droplets in rising gas cells, and its eventual freezing into ice crystals. These clouds will, however, have a strong effect upon the manner in which the atmosphere radiates heat into space.

In Thick Clouds we saw how low, thick clouds block the sun's light from reaching the ground, thus causing it to cool down. In High Clouds we saw how thin, high clouds allow the sun's light to pass through, but block radiation by the planet surface, thus causing the surface to warm up.

Before we can implement clouds properly in our simulation, we must consider how to model their effect upon sunlight and radiation.

Condensation Point

2011-08-25T12:53:00.000-07:00

Suppose a body of moist air rises from the surface of the sea. The weight of air pressing down upon it decreases as it rises. Its pressure drops. It expands and cools adiabatically (see Adiabatic Balloons). So long as no condensation occurs, its temperature drops by 1°C for each 100 m that it ascends (see Tempearture, Pressure, and Altitude). Will the water vapor eventually condense into droplets?

If we know the initial concentration of water vapor in grams of water per kilogram of air, this concentration will remain constant as the air moves upwards. As the temperature of the air drops, however, perhaps there will come a time when the concentration of water vapor exceeds the saturation concentration, and at that point droplets will form. In Evaporation Rate we presented a graph of saturation concentration versus air temperature. But the data points of this graph correspond to measurements taken on the surface of the Earth, where air pressure is 100 kPa. What will the saturation concentration be at the lower pressures that apply to our rising body of air?

The pressure exerted by a gas is the force exerted by its molecules bouncing off whatever surface they encounter. The pressure of moist air is the sum of the pressures exerted by its nitrogen, oxygen, water, and all other molecules. The pressure exerted by the nitrogen molecules is the partial pressure of nitrogen. The pressure exerted by the water molecules is the partial pressure of water. It turns out that the maximum possible partial pressure of water molecules depends only upon temperature. This maximum is the saturation pressure of water at a particular temperature. Regardless of the other gases that might be mixed with the water vapor, the saturation pressure is always the same at a particular temperature. The graph below gives saturation pressure versus temperature, as indicated by an empirical formula we found here.

Water vapor is a gas just like nitrogen and oxygen. Its pressure depends only upon its volume and temperature. Water vapor in moist air will be at the same temperature as the air. It will share the same volume as the air. If the air pressure halves, so does the partial pressure of water vapor. For concentrations below a few percent, the concentration of water vapor at various air pressures is given by:

x = p_wR_w/pR = 0.62 p_w/p,

where x is the concentration of water vapor, p_w is its partial pressure, R_w is its specific gas constant, or 462 J/kgK, p is the air pressure, and R is the specific gas constant for air, or 287 J/kgK.

Consider moist air near the surface of the Earth at temperature 300 K, pressure 100 kPa, and water vapor partial pressure 2 kPa, which corresponds to concentration 12 g/kg. The water vapor pressure is roughly half the saturation pressure of 3.8 kPa shown on our graph. Suppose our moist air rises to 2000 m. Its cools to 280 K (1 K per 100 m) and its pressure drops to 80 kPa (adiabatic expansion of air). The concentration of water vapor remains 12 g/kg and its partial pressure drops slightly to 1.6 kPa. At 280 K, however, the saturation pressure has dropped all the way to 1 kPa, which corresponds to a concentration of only 7.6 g/kg. Thus each kilogram of air contains 4.4 g more water than it can hold as water vapor. This excess water must condense to form water droplets.

We would like a simple formula that will allow us to calculate the amount of water that must condense from moist, rising air in our Circulating Cells program. We combine the saturation concentration approximation of our Evaporation Rate post with the specific gas constants of air and water vapor, and with the assumption that the water vapor concentration is small, to arrive at the following approximation.

x_s = (T−250)²p/8000

Here x_s is the saturation concentration of water vapor in g/kg, p is the air pressure in kPa, and T is the temperature of the air in K. At 280 K and 80 kPa, this formula gives us 9 g/kg, which corresponds to a partial pressure of 1.2 kPa. This 1.2 kPa is close enough to the 1.0 kPa shown in the graph above.

The partial pressure of water vapor in moist air decreases as the air rises. But at the same time, the rising air cools by adiabatic expansion, and this cooling depresses the saturation pressure of water vapor so rapidly that condensation will eventually take place. With our approximate formula for saturation concentration with pressure and temperature, we will be able to simulate the condensation of water in moist, rising air.

Evaporation Rate

2011-08-17T11:00:00.000-07:00

We are preparing to add evaporation from surface water to our Circulating Cells program. Water will leave the surface and enter the atmosphere as water vapor. The rate at which water evaporates depends upon the humidity and movement of the air above. The following empirical equation, which we found here, tells us the approximate rate at which water will evaporate into air, assuming the air is at roughly the same temperature as the water.

w = (0.007 + 0.005v)(x_s − x),

where w is the evaporation rate in grams per second for each square meter of water surface (g/m²s), v is the velocity of the air in meter per second (m/s), x_s is the saturation concentration of water vapor in grams of water per kilogram of dry air (g/kg) for air at the same temperature as the water, and x is the actual concentration of water vapor in grams per kilogram of dry air (g/kg) in the air above the water.

In our simulation, we know the temperature of the surface water, and when we simulate a planet with a water surface using CC8, we find that the surface gas cells are within a few degrees of the temperature of the surface water. Our previous work on impetus for circulation suggests that the velocity of our gas cells is of order 4 m/s. When we implement evaporation, we will keep track of the water vapor concentration in each cell, so we will know x. What remains for us to determine is x_s, the saturation concentration of water vapor in air at the surface temperature.

The following graph shows measured values of saturation concentration in g/kg for a range of temperatures in Kelvin, based upon data we found here. To see the same plot in Centigrade, see here.

Also plotted on the graph is a parabolic approximation to the measured data, which is based upon two reference points: 0 g/kg at 250 K and 45 g/kg at 310 K. This approximation is good enough for our purposes, and will simplify our program. Thus our evaporation equation becomes:

w = [(T−250)²/80 − x] / 40

For example, if we have dry air over a lake at 290 K (14°C), water will evaporate at 0.5 g/m²s. In one hour, 1.8 kg of water will evaporate from each square meter. Our gas cells have mass 330 kg/m², so after an hour over the lake, the gas will acquire water vapor concentration 5 g/kg, which is well below the saturation concentration of 20 g/kg given by our approximation. Its relative humidity will be 25%.

Island Inversion

2011-08-10T10:00:00.000-07:00

We invite you to download Circulating Cells, Version 8.4 by clicking here so that you can watch it simulate air movement above an island in the sea. Start the program and set it to Cycle mode with Q_sun set to 700 W/m². Download this file and read it into the simulation with the Load button. You will see the an island of sand at the center of the surface blocks, surrounded by water on the left and right. The saved state of the cell corresponds to midnight, so you can start the simulation right away.

Just before dawn, at solar time 5.5 hr, you will see that the temperature of the air resting upon the sand of the island is slightly cooler than the air above. The following figure is a close-up of the air above the sand, in which ran the program with max_T set to 300 K and min_T set to 260 K in the source code so as to make the cool layer more obvious. Note that the orange-lined surface blocks are sand and the blue-lined ones are water.

Our simulation produces the inversion of atmospheric temperature that we discussed in Surface Cooling, Part IV. In that post, we proposed that the leaves of a forest would cool at night by radiation. Air in contact with the leaves would descend to the ground and drag warmer air down from above. We suggested that this process might cool the first ten to a hundred meters of air above the treetops, producing temperature inversion. Perhaps that is indeed what happens over a forest. But in our simulation, whenever the surface is cooler than the air resting upon it, we set the heat transfer by convection and conduction to zero. How, then, does inversion occur over our simulated island?

The sandy island cools by roughly 50°C to −12°C at night. At this temperature, it radiates only 260 W/m². Of this, 130 W/m² is absorbed by the bottom layer of the atmosphere, because we have transparency fraction set to 0.5. Meanwhile, the bottom layer of our simulated atmosphere radiates heat both night and day. At the start of the night, its temperature is around 300 K and it radiates 230 W/m² to the sand below, which is half the heat a black body at that same temperature would radiate. By radiation alone, we see that the cells above the sand are losing 100 W/m².

The mass of our cells is 330 kg/m², and their heat capacity at constant pressure is 1 kJ/kg, so 100 W/m² will cool them by roughly 1°C/hr. This is what we see during the night above our island. Inversion does not occur over the water because the water surface in our simulation has ten times the heat capacity as the sand, and so cools ten times less at night. The water continues to warm the lower atmosphere with its radiation during the night. Thus we see how back-radiation and a surface with low heat capacity work together to produce temperature inversion.

Back Radiation

2011-08-03T10:00:00.001-07:00

In our Rotating Greenhouse post, we use CC5 to simulate the alternation between day and night by varying the Solar power delivered to the gas cells resting upon our simulated planet surface. But we did not simulate the surface itself, nor did we distinguish between the temperature of the surface gas and the radiating temperature of the planet surface: we used the same temperature. But the CC8 program we introduced in our previous post does simulate the planet surface, so we can see how the temperature of the sand itself varies with day and night.

During the day, most of the heat passing into the surface gas cells does so by convection. But convection occurs only when the gas above is cooler than the surface below. In our discussion of atmospheric inversion we saw how the ground can be colder than the surface air at night, which can lead to a pocket of cold air sitting near the ground, with warmer air up above. In our simulation, we set the convection transfer to zero when the surface is colder than the surface gas.

At night, therefore, a sandy surface will radiate its heat into space, and receive no warmth from the sun. But it will receive warmth from the atmosphere, in the form of the back-radiation we described in our previous post. During the day, we found that our surface gas was radiating 226 W/m² down to the surface. This radiation will slow the cooling of the surface at night.

We ran our simulation with Cycle heating on a sandy planet surface, 700 W/m² solar heat during the day, daylight fraction set to 0.50, convection coefficient 20 W/m², and transparency fraction 0.50. You will find the equilibrium state of the cell array at midnight stored in a text file here. The following graph shows the average temperature of the sand blocks, the surface gas cells, and the tropopause gas cells during two complete day-night cycles. We plot the deviation of each temperature from its average value during the cycles, which is why we call the plots "anomalies".

The temperature of the surface sand varies by almost 50°C, dropping as low as −12°C just before dawn. The temperature of the air a hundred meters above the sand, at the center of the bottom row of gas cells, varies by 9°C, dropping as low as 23°C. The tropopause responds far less to the day-nigh cycle, with a variation of only 2°C. These results are consistent with our observations of the desert, which we discussed at length in our Surface Cooling posts.

When we turn off the back-radiation in our simulation, the temperature of the surface sand drops by another 50°C at night, in a manner reminiscent of the Moon. And so we conclude that our atmosphere, by means of back-radiation, keeps us warm at night.

Simulated Planet Surface

2011-07-28T10:00:00.000-07:00

Our Circulating Cells Version 8.3 simulates the planet surface with a row of blocks. You can download the latest code here. These blocks can be sand or water. If they are sand, they have low heat capacity and warm up quickly in the Sun's light. If they are water, they have high heat capacity and warm up slowly. By clicking upon one of the blocks, we change it from one type to the other.

In the long run, we will implement evaporation from the water blocks, but for now we concentrate upon the heat exchange between the surface and the atmosphere. In previous versions of our program, we allowed sunlight to pass directly into the surface gas cells. Now we allow sunlight to pass all the way through the entire atmosphere to be absorbed at a solid or liquid surface. Real water reflects 4% of short-wave radiation, and sand reflects something like 10%, but our simulated water and sand absorbs all short-wave radiation.

Both sand and water are near-perfect emitters of long-wave radiation, so we allow our surface blocks to radiate heat according to Stefan's Law. We also allow heat to leave the surface blocks by convection. We determine the convection heat loss by multiplying the temperature difference between the surface and the gas by a convection coefficient. This heat leaves the surface block and enters the gas cell above.

As we described in our previous post, our simulated atmosphere is partially-transparent, to an extent specified by the transparency fraction, τ. The gas above a surface block absorbs a fraction 1−τ of the block's radiation, and the remainder passes out into space. We add the absorbed heat to the gas cell above the surface block.

Now we come to a component in the heat exchange between the atmosphere and the surface that we have not simulated or calculated before. The atmosphere itself will radiate heat downwards towards the surface, and the surface, being a perfect absorber of such radiation, will absorb all of it. The heat radiated downward by the atmosphere is often called back-radiation or downward long-wave radiation. To calculate the back-radiation, we use the same equation we applied to the tropopause in our previous post. When the transparency fraction is 0.5, the gas radiates half as much heat as a black body at the same temperature.

We ran our simulation with Day heating of 350 W/m², all surface blocks made of sand, convection coefficient 20 W/m²K, and transparency fraction 0.50. The following figure shows the equilibrium state of the array. The surface blocks are color-coded for temperature in the same way as the gas cells, but they have an orange border to mark them as sand. A water cell has a blue border. You will find the equilibrium state of the array saved as a text file here.

As in our previous simulations, the temperature drop from the surface gas cells to the tropopause gas cells is very close to 50 K. The sandy surface settles to 303.0 K, at which temperature it radiates 480 W/m². Of this, 240 W/m² passes directly into space and 240 W/m² is absorbed by the gas above. The tropopause settles to 249.0 K and radiates 110 W/m² into space. The total radiation into space is 350 W/m², which is the amount that is arriving from the sun, so our planet is in thermal equilibrium.

The 110 W/m² radiated by the tropopause must pass up through the atmosphere by convection. The surface gas is at an average temperature of 298.5 K, radiating 226 W/m² back to the surface. Thus a net 14 W/m² passes from the surface to the atmosphere by radiation. The remaining 96 W/m² passes into the atmosphere by convection at the surface. The surface is on average 4.5 K warmer than the gas, for which we expect only 90 W/m² to flow by convection. But the gas cells warm by roughly 8 K while they sit upon the sand. When they first arrive from above, they are almost 10 K cooler, and 200 W/m² flows into them by convection. Just before they rise up, they are only 2 K cooler, at which point only 40 W/m² flows into them. When we get the program to print out the convection rate, we obtain an average of 96 W/m², so all the heat from the Sun is accounted for.

We note that almost all the heat flowing from the surface to the atmosphere is carried by convection. The heat radiated by the surface and absorbed by the atmosphere is nearly balanced by the radiation returned by this same atmosphere. The difference is only 14 W/m², compared to 96 W/m² passing into the gas by convection. In our next post, we will simulate night and day over a sandy desert and see if we come up with reasonable variations in temperature at the sandy surface, the air above the surface, and the air high up in the tropopause.

Simulated CO2 Doubling

2011-07-20T08:56:00.000-07:00

In Planetary Greenhouse we considered an atmosphere transparent to some long-wave radiation and opaque to others. In Circulating Cells Version 8.2, which you can download here, we simulate such an atmosphere with the new transparency fraction parameter.

For the moment, we ignore the temperature difference that must exist between the planet surface and the lower atmosphere (see Surface Cooling, Part VI). We assume that the solid or liquid surface beneath one of the bottom gas cells will be at the same temperature as the gas itself. Ever since our Earth Radiator post, we have assumed that the surface of a planet is a black body radiator. If a bottom cell is at 300 K, the surface below will be at 300 K also, and by Stefan's Law it will radiate 460 W/m². At 303 K, the radiated power increases to 480 W/m². Black-body radiation increases as the fourth power of the temperature, so a 1% increase in temperature causes a 4% increase in radiated power.

The absorption spectrum of the Earth's atmospheric layers varies in a complex and dramatic way with wavelength, as you can see here. What made our Total Escaping Power calculation so complicated was that we had to deal with the partial absorption of some wavelengths by each atmospheric layer, and therefore the partial emission of these same wavelengths by the same atmospheric layers. We want to avoid the complexity of partial absorption at a given wavelength, so we assume that the gas in our CC8 simulation has a spectrum that vacillates between perfect transparency to perfect opacity, and does so every fraction of a micron on the wavelength scale. As a result of this vacillation, a gas cell will absorb none of the radiation at a particular wavelength, or all of it. When we look at the fraction of black-body radiation that passes through a cell, this fraction is a constant property of the cell, regardless of its temperature or pressure. We call it the transparency fraction in the CC8 program, and here we will call it τ.

By the principle of radiative symmetry, each gas cells will radiate heat at the same wavelengths it absorbs. But because all the cells around it have the same absorption spectrum, none of the heat radiated by a cell will escape into space, except for the heat radiated by the cells in the top row, which we call our tropopause. These cells radiate directly into space. The power they radiate is the power that a black body would radiate, multiplied by 1−τ.

The incoming heat from the Sun, meanwhile, passes straight through our atmosphere, because we assume that the gas is perfectly transparent to short-wave radiation. The Sun's heat warms the planet surface, which for the moment we assume is something like sand. The sand heats up rapidly until it is losing heat by radiation and convection at the same rate it is gaining heat by absorption of the Sun's light. Some heat it radiates directly into space. The rest passes into the lower atmosphere and must be transported up to the tropopause by convection, where it is radiated into space.

Bottom Gas Cell Temperature = T_B
Planet Surface Temperature = T_S = T_B
Stefan's Constant = σ = 5.7 × 10⁻⁸ W/m²/K⁴
Emitted Surface Radiation = σ(T_S)⁴
Escaping Surface Radiation = τσ(T_S)⁴
Top Gas Cell Temperature = T_T
Emitted Tropopause Radiation = σ(1-τ)(T_T)⁴

We set τ=0.50 and ran CC8 with Day heating and the Sun's power set to 350 W/m². After ten million iterations we are confident that we have reached equilibrium, and we obtain this array. The average temperature of the bottom cells is 301.1 K (28°C) and of the top cells is 252.1 K (−21°C). The heat radiated by the surface is 468.5 W/m², of which 234.2 W/m² escapes directly into space. That leaves 115.7 W/m² of the Sun's heat to be transported up through the atmosphere. The heat radiated by the tropopause is 115.1 W/m², leaving 0.7 W/m² unaccounted for, which is well within the range of our rounding errors and the random fluctuations in our cell temperatures.

In our previous post, we ran our simulation for an opaque atmosphere, which corresponds to τ=0.00, and the surface temperature rose to 355 K (59°C). We see that τ=0.50 allows the surface to cool by 31°C to 28°C. In With 660 ppm CO2, we showed that doubling the CO2 concentration in the Earth's atmosphere will cause a 2% drop in the total escaping power. So now we set τ=0.49, so as to cause a 2% drop in the power escaping directly from our simulated planet surface into space. We arrive at a this array, in which the surface has warmed by 0.9°C to 302.0 K and the tropopause has warmed by 0.4°C to 252.5 K.

As a check, we run with τ=1.00, in which case the atmosphere is perfectly transparent and the surface radiates all its heat directly into space. The surface cools to 280 K (7°C). The atmosphere assumes the dry adiabatic lapse profile. But now we turn on the cell mixing by setting the mixing fraction to 0.2, and after a few hundred thousand iterations we see the entire atmosphere warm up to 280 K. This is the warm atmosphere we described in our original Greenhouse Effect post and again in Adiabatic Magic. When the atmosphere is not transporting heat to the tropopause, there is no greenhouse effect, and the atmosphere mixes until it arrives at a uniform temperature equal to the surface temperature.

We see that CC8 can simulate the effect of changing the concentration of a greenhouse gas like CO2, simply by making small changes to its transparency fraction. Once we have included evaporation, clouds, and rain into our simulation, we will be able to estimate the effect of changes in CO2 upon the average temperature of our planet surface.

Black-Body Tropopause

2011-07-13T18:45:00.000-07:00

Our CC7 program provides six different ways to heat the atmospheric cell array. In CC8, we eliminate most of these and replace them with only three: Day, Night, and Cycle. The others were useful in checking the performance of the simulation, but are no longer necessary. Because we now have a good understanding of the relationship between simulation time, impetus for circulation, and program iterations, we are now able to express the Sun's heat in W/m² instead of the less realistic K/iteration of our earlier programs. We represent the incoming solar power with Q_sun instead of the previous Q_heating. Furthermore, we can use Stefan's Law directly upon the top cells, as if they were black and the gas above them were transparent. Our program now contains a value for Stefan's Constant, which we set to 5.7×10⁻⁸.

For our convenience, we display the time of day in hours in the main window. Time 12.0 hr is noon, when the Sun is certain to shine, and midnight is 0.0 hr. We display the current solar power in W/m². Previously, we applied a sinusoidal variation in the Sun's power during the day, but now we simply turn the Sun's power on to Q_sun during daylight hours, and to zero during the night. As before, however, the Sun will shine for a fraction of the day given by day_fraction.

We run the program with Day heating and the Sun's power set to 350 W/m². We have ke_fraction at 0.0 and we un-check left_only. Our cells have mass 333 kg/m². Their specific heat capacity at constant pressure is 1 kJ/K. Thus the lower cells warm up at 0.001 K/s, which matches our previous Q_heating of 0.001 K/iteration. We allow the array to reach equilibrium, which takes a long time: five million iterations, or one hours on our lap-top. You will find the equilibrium state in Day_1.txt. You can load it into CC8 with the Load button. At equilibrium, the top row's average temperature is 280.0 K. Applying Stefan's Law, the top cells should radiate 350 W/m², which is what we expect, since that's what we are putting in. The surface cell average temperature is 335.4 K, giving us the 55-K drop from the surface to the tropopause. This drop is consistent with our previous results.

We run the program with Q_sun set to 700 W/m² and Cycle heating to simulate day and night. We have day_fraction set to 0.5. After a million iterations we see the temperature of the bottom and top rows varying by a few degrees during night and day, as we did in Rotating Greenhouse.

The equilibrium surface temperature of 59°C (335 K) is much hotter than the surface of the Earth (around 14°C), even though 350 W/m² is the average power of the Sun. The Earth is cooler because its surface and lower atmospheric layers are able to radiate almost half their heat directly into space, assuming there is no cloud cover (see Total Escaping Power and subsequent posts).

Our next step is to allow the surface cells to radiate directly into space, and we will see how the surface cools down as a result. We must implement the surface and tropopause radiation before we can model the effect of clouds.

Cells with Momentum

2011-06-30T10:09:00.000-07:00

We invite you to download Circulating Cells 7 by clicking here. Each cell now has vertical and horizontal kinetic energy in the manner we described in Cell Kinetic Energy.

The preservation of momentum after a circulation is controlled by the ke_fraction parameter. By default, this parameter is 0.0, and the simulation will run exactly as before, with the entire impetus for convection being transformed into viscous friction. But with ke_fraction set to 0.9, 90% of the impetus for convection will turn into kinetic energy.

In version 7.0, we have have no graphical illustration of the cell kinetic energy. Perhaps we can use short lines within the cells to indicate momentum in the future. But we can always can stop the simulation and save the array to a text file. The text file tells us the state of each cell. Text files written by 7.0 have a new format, as shown below. But please note that 7.0 can read in text files written by previous versions of the simulation.

row column marking temperature vertical_ke horizontal_ke

In the future, we will add moisture content and cloud concentration to this list of properties. For now, the cells remain dry. The kinetic energy is in units of J/kg, so if we want to know how fast the cell is moving, we take the absolute value of the kinetic energy, double it, and take the square root. If the energy is 32 J/kg, the speed is 8 m/s. The direction is given by the sign of the kinetic energy: positive is up or right.

We have been playing around with CC7, using the Left-Only and Planetary Greenhouse. We set ke_fraction to 0.9 and looked to see if a breeze developed along the lowest row of cells, from right to left. After several hundred thousand iterations, we saved the array to disk and find that all the cells along the bottom row are moving to the left, while all the cells along the top row are moving to the right. By marking cells, we can observe these movements as the simulation runs. We don't see a simple clockwise rotation: there is a lot of random movement on top of the rotation. In the middle rows, the cell movements are as random as they were in our original Left-Side Only simulation with no accounting for momentum. But along the surface and along the tropopause, we appear to have the prevailing breeze we were looking for.

What we have yet to determine is what value of ke_fraction is required to establish a steady breeze, and what fraction is realistic. We invite you to download the code and play with it yourself. We hope the implementation of kinetic energy is correct, but if not we hope you will point out any problems.

UPDATE: There is a flaw in the way we combine cell kinetic energy with impetus for circulation. When we subtract the kinetic energy of a cell from the impetus, this kinetic energy ends up disappearing from the array. We observed a similar program-induced loss of energy in Work by Circulation. Given that the effect of preserving cell momentum was not dramatic, we resolve to remove the kinetic energy calculation from our code, so that ke_fraction will remain zero.

Cell Kinetic Energy

2011-06-20T21:11:00.000-07:00

When a block of four cells rotates within our Circulating Cells simulation, the rotation does work, and we call this work the impetus for circulation. We express the impetus for circulation in units of energy per kilogram of gas in the four cells (J/kg). In our CC6 program, we take the impetus for circulation and add it back into the gas as heat. Our assumption is that the impetus is first used to accelerate the gas, and so turns into kinetic energy, but later is dissipated as viscous friction. At the end of our circulation, the gas is once again at rest.

But clearly the gas will not be at rest at the end of a circulation. Once it starts moving, it will tend to continue moving. In our previous post we showed that the cells coming to stop means that our simulation will never allow convection to produce a steady breeze. We would like our simulation to allow a cell to retain some of its kinetic energy after the circulation is complete, and thus allow this kinetic energy to influence subsequent movements of the same cell.

We propose that our upcoming CC7 program should handle kinetic energy in the following way. First, we give each cell two additional numbers that specify its kinetic energy per kilogram in the vertical and horizontal directions. When a cell is moving down, indicate its downward motion by giving its vertical kinetic energy a negative sign. When moving to the left, we give its horizontal kinetic energy a negative sign. The kinetic energy is in units of J/kg, just like the impetus for circulation. The use of a sign to indicate direction does not imply that the kinetic energy is really negative, because kinetic energy cannot be negative.

When four stationary cells rotate, we calculate the impetus for circulation just as we did in CC6. We rotate the cells if the impetus exceeds our impetus threshold. After that, we take a fraction of the impetus, given by the new ke_fraction parameter, and add it to the kinetic energy of each cell. In a clockwise rotation, the bottom-left cell acquires upward kinetic energy, the top-left cell acquires rightward kinetic energy, and so on. What is left of the impetus, we add into the cell temperature as viscous heat. If we set ke_fraction to zero, the simulation will run exactly as it did in CC6, because the entire impetus will turn into viscous heat, and no kinetic energy will be imparted to the cells.

When four cells with kinetic energy rotate, however, we add to the impetus for circulation whatever kinetic energy the cells might have in the direction they will be expected to move.

Suppose we have four cells, three of which are stationary, but the bottom-left one is already moving upwards with kinetic energy 40 J/kg. If the impetus for circulation due to buoyancy and expansion is 2 J/kg, we now add 10 J/kg to account for the fact that the bottom-left cell has 40 J/kg of kinetic energy that favors the rotation. The total impetus is 12 J/kg. Assuming our threshold is below 12 J/kg, the rotation takes place. If our ke_fraction is 0.5, each cell ends up with 6 J/kg in the direction of the rotation. The bottom-left cell ends up with 6 J/kg of upward, vertical kinetic energy, which is far less than the 40 J/kg it started with. Its kinetic energy was used to drive a circulation that might not have taken place at all, and in doing so, the it accelerated and heated three other cells. The bottom-left cell slowed down, but it is still moving up.

If the bottom-left cell also has kinetic energy in the horizontal direction, we ignore this fact, and assume that this horizontal energy will neither hinder nor help the rotation. When the rotation takes place, the kinetic energy of the bottom-left cell in the horizontal direction will remain unchanged.

This is what we plan to do in CC7. We welcome your comments before we proceed. The program is likely to slow down, so we are trying to figure out how to make the calculations faster. Not that any of us is in a hurry, of course.

Left-Side Only

2011-06-10T10:00:00.000-07:00

On a sunny day at the beach, the wind tends to blow towards the shore. The land warms up more than the water and warm air rises off the land. The air moving upwards sucks air sideways off the water to make the on-shore breeze. We wonder if our simulation will do something similar if we heat cells only on the left side of the array. The left-side would simulate land, and the right side would simulate water. We might see cells moving along the surface from the right, warming on the left, rising up to the top, and cooling as the move to the right again.

The figure below shows our Circulating Cells simulation program, Version 6.0, which you can download by clicking CC6. With the Left_Only box checked and Planetary Greenhouse heating, the surface cells on the left side receive twice the normal heat from the sun, while those on the right side receive none at all.

We started the simulation by loading the equilibrium state of the array with both sides receiving heat, which we have saved in PGH_Q001_M00. Following our recent discussion of enthalpy, we recognize this symmetric equilibrium state as the one in which all cells have the same enthalpy. Those at the top have more gravitational potential, but less internal heat and pressure energy, so that the sum of all three forms of energy is the same for all cells, or almost the same.

We checked the Left_Only box and increased Q_heating 0.01 K/s. You may point out that the unit of Q_heating should be Kelvin per iteration, not Kelvin per second, but we recall that one iteration corresponds to one second, so the two are equivalent. With Q_heating at 0.01 K/s, the left-side surface cells warm at 0.02 K/s and those on the right do not warm at all.

We ran the simulation for a million iterations and saved the cell array in PGH_Left_Only. The figure above shows the saved state of the cell array, after another ten thousand iterations. The temperature profile is consistent with a large-area circulation of air powered by heating on the left surface. The heated air rises to the tropopause and moves over to the right side as it radiates its heat into space. Once it has cooled, it descends to the right surface and moves along to the left.

We marked a few cells by clicking on them, and watched them go around. We invite you to do the same. The cells circulate in a clockwise direction. They rise to the tropopause on the left, but hardly ever rise to the tropopause on the right. Nevertheless, we don't see individual cells moving steadily in a clockwise direction across the width and height of the array. Often, cells rise on the extreme left and descend upon the center-left. Cells on the right rise up a little and fall again. They slowly drift to the left, but they do a lot of jumping around along the way.

When averaged over thousands of iterations, the combined movement of the cells is a large clockwise circulation, with a net movement of cells from right to left along the surface. But a simulated person standing on the center surface would not feel a steady breeze blowing from the right side. He would instead feel the wind changing every minute or two, and only by looking at the average wind speed would he be able to conclude that the net movement of air was on-shore.

Our simulation assumes that all momentum generated by circulation is dissipated as viscous heat at the end of each circulation. Thus each circulation affects only the temperature of the cells. No cell can build up momentum that encourages further circulation in the same direction.

We conclude that momentum is one of the driving forces behind the on-shore breezes. Buoyancy alone is not sufficient. If we want our simulation to produce steady winds, we must allow circulating cells to retain some of the momentum they gain during circulation, and we must allow this momentum to influence future circulations of the same cell.

Pressure Energy and Gravity

2011-06-02T10:00:00.000-07:00

In our Circulating Cells program, we assume each cell has a uniform pressure. We know perfectly well, however, that the weight of the gas in the cell causes the pressure at the bottom to be greater than the pressure at the top. If p is the average pressure of the cell, m is the mass of the cell per square meter of its base area, and g is gravity, the pressure at the top will be p−mg/2 and at the bottom will be p+mg/2. Today we consider how our assumption of uniform pressure affects our calculation of cell enthalpy.

Our diagram shows a cell in three consecutive states. In state 1, the cell has height h and average pressure p. Its center of mass is at altitude h/2. In state 2, the cell has warmed up and expanded, but its center of mass remains at the same altitude. As the cell expanded, it pushed upwards with pressure p−mg/2 and downwards with pressure p+mg/2. The top and bottom surfaces each moved outwards a distance δh/2. The top surface does work work (p−mg/2)δh/2 and the bottom surface does work (p+mg/2)δh/2. The total work done by the gas as it expands is the sum of these two quantities, which comes out to be pδh. The mg terms cancel. Thus we get the same value for the total work done that we would obtain if we assumed a uniform pressure p pushing outwards for a total distance δh.

In state 3 we allow the cell to rise up, so that its bottom surface is once again sitting on the ground at altitude zero. The altitude of its center of mass is now (h+δh)/2. As the cell moved up, its height remained constant. Its top surface pushed upwards with force p−mg/2 over a distance δh/2. Meanwhile, something pushed upwards upon its bottom surface with force p+mg/2, and did so over the same distance. The top surface does work (p−mg/2)δh, but the bottom surface absorbs work (p+mg/2)δh. The net work done is upon the cell is mgδh. This time, the terms in p cancel and we are left with a term in mg. If we were to assume the pressure in the cell was uniform, we would not arrive at this same result. We would instead conclude that no net work was done upon the surfaces of the cell to raise them up.

We recall that the enthalpy of a cell is the work required to replace it with an identical cell from a reservoir. Because of the pressure difference across a cell, we see that the work we must do to raise a cell to altitude z is mgz, so we our calculation of enthalpy must include not only the pressure energy of the cell and its internal heat energy, but also a term mgz. This term has a name: it is the gravitational potential energy, which we add to the internal and pressure energy of a stationary cell to obtain its enthalpy per kilogram, H.

H = CvT + RT + gz = CpT + gz

The gas above our cell rises a total distance of δh, which requires work pδh per square meter of base area. We could say that the gravitational potential of the gas above our cell has increased by pδh, and we would be correct, but this increase is already accounted for in our enthalpy equation. The increase in our cell's pressure energy is RδT, where δT is the amount by which we warmed the cell to make it taller. For an ideal gas, RδT = pδh. Thus the increase in the pressure energy of our cell is one and the same as the increase in the gravitational potential energy of the gas pressing down upon it from above.

The gravitational potential term in our enthalpy equation represents the effect of the pressure drop across a cell as we change its altitude. By including the gravitational potential term, we can continue our enthalpy calculations with the assumption that the cell pressure is uniform. But an increase in gravitational potential is not accompanied by any change in the appearance or state of our cell. Gravitational potential is a useful and elegant concept, but it has no physical existence of its own.

Impetus by Enthalpy

2011-05-26T21:12:00.000-07:00

In Impetus Dissected we showed that impetus for circulation is the sum of buoyancy work and expansion work. In terms of the diagram below, the sum of the buoyancy work and expansion work is Mgδh. The work available per kilogram of gas in the four cells is simply gδh/4. Today we show that we can obtain this same answer with a calculation based upon enthalpy.

If we ignore kinetic energy, the specific enthalpy of a gas (that's the enthalpy per kilogram) is CpT + gz, where z is the altitude of the cell. When we heat up cell A by δT we increase its specific enthalpy by ΔH1.

ΔH1 = δTCp+gδh/2

This is the energy required to bring about the change in the temperature of cell A, and includes the work required to raise up the cells above A as well as the work required to increase the internal energy of A, and the work required to raise A's own weight.

For an ideal gas cell, we have pV=MRT, and from this we deduce that δh = mRδT/p, where m is the mass per square meter of the cell's bottom surface. Thus we have the following expression for ΔH1.

ΔH1 = δT(Cp+mgR/2p)

When cell A rises, it cools by adiabatic expansion. It is now warmer by δT ' than cell B, and taller by δh'. Suppose we now allow A to cool by δT ', so that it becomes identical to cell B. Its enthalpy decreases by ΔH2, and the four cells and the gas above are in an identical state to the one they were in before we warmed cell A.

ΔH2 = δT 'Cp+gδh'/2

In buoyancy work we showed that δT ' = δT(1−Rmg/pCp). We also showed that δh' ≈ mRδT(1 + mgCv/pCp)/p. Using these relations, and the fact that mg≪p, we obtain the following expression for ΔH2 after several lines of simplification.

ΔH2 = δT(Cp−mgR/2p)

The expressions for ΔH1 and ΔH2 are identical, except the sign of the gravitational term is opposite. When subtract ΔH2 from ΔH1, we obtain the enthalpy per kilogram of cell A that has gone missing during the rotation.

ΔH1−ΔH2 = mgRδT/p = gδh

Enthalpy cannot go missing. Our expression for enthalpy ignores the kinetic energy of the cells when it should not. The kinetic energy of the cells after the rotation is equal to the enthalpy that is missing in our calculation. When we divide it among the mass of the four cells, the missing enthalpy is gδh/4, which is our impetus for circulation.

We performed the same enthalpy calculation for incompressible cells and obtained the same answer. We have arrived at the same expression for impetus by two different means, for both incompressible and compressible cells. We are now confident that our calculation is correct.

Impetus Dissected

2011-05-19T10:00:00.000-07:00

We return now to the unanswered questions we raised in Impetus for Circulation. We noted that the values our simulation program calculated for buoyancy work and expansion work differed by only a few percent. This equality made us suspect that we were in fact looking at the same source of energy in two different ways. If so, then it would be incorrect to add them together, because we would be counting the same energy twice. But how could they be the same source of energy, when we know that convection occurs in incompressible fluids like water? Surely there must be buoyancy work in water, even if there is no expansion work? Today we will answer these questions.

The following variation on our well-used buoyancy diagram shows four cells of incompressible fluid. The cells do not expand or compress as we change their pressure. As we showed earlier, water is almost, but not quite, and incompressible fluid.

When an incompressible cell rises, it does no expansion work, so it does not cool down. Unlike convection in a compressible gas, convection in an incompressible fluid requires no temperature gradient from bottom to top. If all the cells are at the same temperature, they can move around freely by exchanging places with one another. None of them are going to cool down or heat up as their pressure changes.

Thus we begin with all four incompressible cells at temperature T and warm up cell A by δT. Cell A gets taller by δh and becomes buoyant. Cell A rises and the block of four cells rotates clockwise. Cell B slides off cell A and onto cell C. In doing so, it drops by δh and does work Mgδh.

Meanwhile, the fluid above the four cells does not move at all. The left cells are taller by δh before and after the rotation. Thus the incompressible cells do not have to do any work raising the fluid above them. The total buoyancy work remains Mgδh. The expansion work is zero, so the total work done by the rotation is Mgδh.

Let us return to the gas-filled cells depicted here. The buoyancy work consists of two components. The first component is the work done by cell B as it slides off cell A. This work is Mgδh, just as it is for an incompressible fluid. The second component is the work required lift the gas above the block.

When cell A rises, the combined height of the two left-hand cells increases by a small fraction of δh (the fraction is mgCvδh/pCp). The increase is small, but the weight of the gas that must be lifted is great (the weight is p). The work required to raise the upper gas is significant (the work is MgδhCv/Cp). After subtracting this work from the work done by cell B, we are left with a net buoyancy work of only MgδhR/Cp (for air, R/Cp is 0.29).

Unlike an incompressible cell, however, a gas cell does work as it expands. In Expansion Work we found that the warm, rising cell does MgδhCv/Cp more work as it expands than is required to compress the cold, falling cell. This expansion work turns out to be exactly equal to the work required to raise up the gas above the rotating block. When we add the buoyancy and expansion work together, we find that the total work done by the rotating cells is Mgδh, which is the same as the work done by the incompressible cells.

We see that buoyancy work and expansion work are separate sources of energy after all. In an incompressible fluid, we have only buoyancy work. The expansion work is zero. In a gas, we have both, but their sum is the same. This sum is what we call the impetus for circulation. Given a certain δh, the impetus for circulation is the same for both gas cells and incompressible cells. When the warm cell is taller by δh, the impetus for circulation is Mgδh.

In CC5, our calculation of buoyancy work ignored the work required to raise the upper gas, and so produced a value for buoyancy work that was incorrect, but equal to the impetus for circulation. Our calculation of expansion work, meanwhile, used Cp to multiply δT instead of Cv, and so produced a value for expansion work that was incorrect, but equal to the impetus for circulation. By accident, therefore, CC5 calculated the correct impetus for circulation in two different ways, and this was the origin of our mystery.

In our next post we will confirm today's argument by considering the enthalpy of incompressible and gas-filled cells. After that, we will return to enhancing our simulation, being confident that we understand the forces involved in the circulation of our atmospheric cells.

Expansion Work

2011-05-11T10:00:00.000-07:00

Whenever we allow a block of four cells to rotate within our atmospheric simulation, we see a warm cell rise and expand while a cold cell falls and contracts. The expanding cell is larger, so it does more work than is required to compress the smaller, falling cell. In Impetus for Convection, we called the excess work done by the rising gas expansion work. Our CC5 program assumes the expansion work is equal to CpΔT, where Cp is the specific heat capacity of the gas at constant pressure, and ΔT is the drop in the average temperature of the rotating cells. Today we show that the correct value for expansion work is roughly one third less, and we find that its correct value is exactly equal to the work required to raise up the gas above the rotating block.

When a gas expands adiabatically, it does work pushing outwards, but no heat passes through its boundaries. By the First Law of Thermodynamics, the work done must be equal to the decrease in the internal heat of the gas. The internal heat of a kilogram of ideal gas is CvT, where Cv is its specific heat capacity at constant volume. A kilogram of gas that has cooled by ΔT through adiabatic expansion must have done CvΔT of work as it expanded. For air, Cv = 716 J/kgK, so this work would be 716 J/kg for a cooling of 1°C. If we multiply ΔT by Cp instead of Cv, we get a value 1003 J/kg, which is 313 J/kg too high.

Let us return to the four cells we considered in Buoyancy Work, and calculate the expansion work in terms of the same parameters. As before, we start with a block of four cells that has no tendency to circulate, and we warm the lower-left cell by δT. We use R for the specific gas constant. As always, we have Cp=R+Cv.

Before rotation, cell A is at T+δT. When it rises, it expands adiabatically from pressure p to p−mg. When mg≪p, the temperature drops by a factor of 1−Rmg/pCp. When cell B falls, it temperature begins at T(1−Rmg/pCp) and ends at T. A few lines of calculations will reveal that the drop in the average temperature of the four cells is simply δTRmg/4pCp. The expansion work, We, per kilogram of gas is:

We = δTRmgCv/4pCp

We have seen this quantity before. In Buoyancy Work we found that the work required to raise the gas above the block was δTRmgCv/4pCp. In our next post, we will explore the implications of this equality, and at last solve the mystery we presented in Impetus for Circulation.

Buoyancy Work

2011-05-06T10:00:00.000-07:00

When a block of four cells in our atmospheric simulation rotates by convection, the center of mass of the block drops. In Impetus for Convection, we called the work done by the descending block buoyancy work. When we calculated buoyancy work with our CC5 program, we considered only the movements of the four cells within the rotating block. Today we present a calculation of buoyancy work that takes account of the work required to move the cells above the block, and we find that our original calculation over-estimated buoyancy work by more than a factor of three.

We start with a block of four cells that has no tendency to circulate, warm up the lower-left cell, and consider the buoyancy work this single warm cell can generate for us. Here is the diagram we presented in our Buoyancy post. The cell on the lower left is warmer by δT than the adiabatic profile we derived earlier. We use R for the specific gas constant, Cv for the specific heat capacity at constant volume, and Cp for the specific heat capacity at constant pressure. As always, we have Cp=R+Cv for an ideal gas.

When a cell expands adiabatically from p to p−mg, its temperature drops from T to T(1−mg/p)^R/Cp. When mg≪p, this expression simplifies to T(1−Rmg/pCp), which is why the upper cells in our diagram have temperature T(1−Rmg/pCp).

As we heat cell A by δT, we raise its center of mass by δh/2 and the center of mass of cell B by δh. We also raise the gas above B by δh. Cells C and D do not move, nor does the gas above them. The weight of each cell is m per unit base area, and the weight of the gas above cell B is the pressure in cell B, which is p−mg. As we supply heat to A to warm it up, the work done by the expanding cell against gravity is W1.

W1 = mgδh + mgδh/2 + (p−mg)δh = mRδT(1 + mg/2p)

We now allow the four cells to rotate one step clockwise. Cells C and D move to the bottom. They are at temperature T. Cell B will be on the upper-right, with its temperature unchanged at T(1−Rmg/pCp). Cell A, having expanded adiabatically from pressure p to p−mg, will cool down, but will still be warmer than cell B by δT ' = δT(1−Rmg/pCp).

Suppose we remove heat from cell A until it is the same temperature as B. Once it has cooled by δT ', the block of four cells will be identical to the one we started with before we heated A by δT. As A cools by δT ', its center of mass drops by δh'.

δh' = mRδT '/(p−mg) = mRδT(1−Rmg/pCp)/(p−mg)

We note that mg≪p and this allows us to simplify the above expression.

δh' ≈ mRδT(1 + mgCv/pCp)/p

As cell A shrinks, its own center of mass drops by δh'/2 and the gas above drops by δh'. Now we can calculate the work done by gravity, W2, as the cell cools.

W2 = mgδh'/2 + (p−mg)δh' = δh'(p−mg/2) = mRδT(1 + mgCv/pCp)(1−mg/2p)

Once again we note that mg≪p, which allows us to simplify our expression.

W2 ≈ mRδT(1 + mgCv/pCp − mg/2p)

When we heated cell A by δT, we increased the gravitational potential energy of the cells and upper gas by and amount W1. When we allowed the cells to rotate, they produced buoyancy work, the size of which we would like to determine. When we allowed cell A to cool, we decreased the gravitational potential energy of the cells and upper gas by an amount W2, and in doing so we returned these gases to their original state. By conservation of energy, the difference between W1 and W2 must be equal to the buoyancy work.

W1−W2 = mRδT(Rmg/pCp)

The buoyancy work per kilogram of gas in the four cells, Wb, is therefore:

Wb = mgR²δT/4pCp

We interpret our expression for Wb as follows. When the cells rotate, cell B drops by δh, which does work mgδh = mgRδT/p. But at the same time, all the cells above A rise by δh'−δh>0, and this requires work mgRCvδT/pCp. The buoyancy work is the work done by B as it drops, minus the work required to raise the upper gas, divided by the mass of the four cells combined, which gives us the expression for Wb above. In our CC5 calculation, we ignored the work required to raise the upper gas, and so over-estimated the buoyancy work by a factor of Cp/R, which is 3.5 for air.

If cell B contains 300 kg/m² of air at 100 kPa and 300 K, and δT for cell A is 10°C, the buoyancy work is 6.2 J/kg, which is sufficient to accelerate the cells to 3.5 m/s. These cells would be 250 m high, so at 3.5 m/s they would circulate in one minute. Using our old calculation of buoyancy work, the circulation would take almost twice as long.

Buoyancy

2011-05-02T11:43:00.000-07:00

The following diagram shows four cells sitting on a flat surface, just as four cells in our atmospheric simulation might sit upon the ground. We mark the center of mass of each with a cross-shaded circle. The cells each have mass M and base area A. The mass per unit area is m = M/A. Pressure drops from p in the lower cells to p−mg in the upper cells, where mg≪p.

The temperature of cell D is T. Cells B, C, and D have the same enthalpy. If D were to expand adiabatically from p to p−mg, it would cool to the same temperature as B and C. When gas expands adiabatically, Tp^R/Cp remains constant, where Cp is the heat capacity of the gas at constant pressure and R is the gas constant. When pressure scales by (1−mg/p), temperature scales by (1−mg/p)^R/Cp. Because mg≪p, the temperature of the upper cells is very close to T(1−Rmg/pCp).

Cell A has been warmed by a small amount δT compared to cell D. Because it is warmer, it is taller than cell D by δh=mRδT/p. Cell B rises above cell C by a distance δh, and cell A itself rises by δh/2 compared to cell D

In our simulation program, we assume the temperature and pressure within each cell is uniform. Here we allow for variation in pressure from the top to bottom. The top is at p−mg/2 and the bottom is at p+mg/2, so that the total change from top to bottom is mg.

At the top of cell D, the pressure is p−mg/2, but the pressure within A at the same height is greater by mgδT/T. On the left side of our diagram is a graph of the excess pressure in cell A with altitude. Cell A pushes sideways into cell D with an average pressure mgδT/2T. Cell B pushes upon cell C with average pressure mgδT/T, which is twice as much. Cell B starts to push its way over the top of cell C, and in doing so places some of its own weight upon cell D. This additional weight causes the pressure at the base of D increases, and now cell D starts to push its way under cell A, forcing it upwards.

Thus cell A becomes buoyant, and the four cells will rotate clockwise to allow A to rise. Buoyancy manifests itself not only as an upward force upon the buoyant cell, but also as horizontal forces upon the cells above and beside the buoyant cell. The cells above slide out of the way, and the cells beside it push inwards. It is only after these horizontal movements have begun that the buoyant cell can rise.

Enthalpy

2011-04-25T15:12:00.000-07:00

The enthalpy of a volume of fluid is the energy it takes to replace this volume with an identical volume taken from a hypothetical infinite reservoir. Consider the following diagram. It shows a gas cell of volume V, pressure p, and temperature T at a height z. We show a pipe leading from the ground up to the base of the cell. A pressure valve allows gas to leave through another pipe when the cell pressure is greater than p. The cell is small enough that the pressure and temperature within it are uniform. We make the same assumption in our circulating cells simulation.

New gas entering at the lower-left corner of the cell must be at pressure p in order to force the old gas out of the pressure valve on the upper-right. Suppose the cross-section of the pipe is A and we push the gas in with a piston. The piston must move a total distance V/A to replace all the old gas. The front face of the piston must push with force pA. The work we do is pA×V/A = pV. The product pV is the pressure energy that we have mentioned before, but never explained in detail.

If we push the gas into the pipe at ground level, we have to do extra work to raise all the replacement gas up to the altitude of the cell. If the mass of the cell is m and gravity is g, the work we must do to raise the new gas is mgz. This is the gravitational energy component of enthalpy. When we use ρ for density, the gravitational energy becomes Vgzρ.

Our new gas must be at the same temperature as the old. We could choose 300 K as the temperature of the gas in our hypothetical reservoir. But we choose 0 K instead, because this eliminates a constant in our equations. The constant is of no use to us because we find that we are interested only in changes in enthalpy, never the absolute value of enthalpy.

Suppose our gas is ideal, with specific heat capacity at constant volume Cv. The heat required to warm up our replacement gas is mTCv. If we use Vρ for m the heat is TVρCv. The total energy required to replace the old gas in our cell with new gas from our hypothetical reservoir is the sum of the heat, gravitational energy, and pressure energy.

Enthalpy = H = pV + Vgzρ + TVρC_v

The specific enthalpy is the enthalpy per kilogram. We divide H by the mass Vρ to obtain the following, in which we use V to denote the volume per kilogram, which is the same as 1/ρ.

Specific Enthalpy = h = pV + gz + C_vT

For an ideal gas we have pV = RT, where R is the specific gas constant. We are already familiar with the heat capacity at constant pressure, Cp = R + Cv, so we arrive at the following expression for the specific enthalpy of an ideal gas.

h = (R + C_v)T + gz = C_pT + gz

When we establish convection in an atmosphere, the variation in temperature with altitude must be such that no work is required to move volumes of gas up and down. We established this principle in a previous post and showed with a dozen lines of calculations that the variation of temperature with altitude must be linear with slope −g/Cp. Now let us make the same point using the concept of specific enthalpy.

The enthalpy of a kilogram of gas is the energy required to replace it from our hypothetical infinite reservoir. The energy required to move a kilogram of gas from one place to another in our atmosphere is the difference in its enthalpy at the two locations. Our assumption that no work is required to move gas up and down means that the specific enthalpy of the gas in the atmosphere must be constant with altitude.

dh/dz = C_pdT/dz + g = 0 ⇒ dT/dz = −g/C_p

And so we see that the concept of enthalpy allows us to deduce the dry adiabatic lapse rate in two lines instead of twelve. Enthalpy is a useful tool for calculations. But we note that the pressure and internal energy calculation we performed in our previous post is the one that shows us the physical mechanisms by which enthalpy is conserved, and is therefore important for our understanding.

We will use enthalpy in our next post to help solve the mystery of why expansion work and buoyancy work are equal in our circulating cells simulation.

The BEST Project

2011-04-12T10:00:00.000-07:00

The Berkeley Earth Surface Temperature (BEST) has set out to re-calculate the global surface temperature trend. They describe the basis of their calculation here.

The first thing BEST addresses when they describe their calculation is the number of available weather stations, and how many of they should use to calculate a trend. We considered the same problem at length here and in brief here. According to BEST, there are around fifteen thousand weather stations reporting daily from 1970 to 2010. The Climatic Research Unit (CRU), the National Climatic Data Center (NCDC), and we ourselves based our calculations upon the GHCN station data. According to BEST, the GHCN data includes fewer than one in ten of the stations that recorded temperatures in 2000. The BEST team hope to use a greater fraction of the available stations.

BEST provided testimony to the US congress on 31st March. They have already applied their basic calculation to 2% of the fifteen thousand stations available in the period 1970 to 2010. They made no effort to correct for systematic errors like urban heating. And yet they arrive at a global surface temperature trend almost identical to that of CRU and NCDC. Here's what they have to say about this agreement.

The Berkeley Earth agreement with the prior analysis surprised us, since our preliminary results don’t yet address many of the known biases. When they do, it is possible that the corrections could bring our current agreement into disagreement. Why such close agreement between our uncorrected data and their adjusted data? One possibility is that the systematic corrections applied by the other groups are small. We don’t yet know.

We were just as surprised when we reproduced the CRU trend from the GHCN data by integrated derivatives. We used no reference grid. We used no corrections for systematic errors.

NASA estimates the urban heating effect in US weather stations to be roughly half a degree centigrade during the twentieth century (see our post here and NASA's paper here). But CRU claims that urban heating is either negligible or has been accounted for in their efforts. And NASA applies 0.5°C corrections and goes on to say that the corrected trend still shows a rise of 0.5°C in the twentieth century, and they claim this trend is accurate to ±0.1°C. We find all this confusing, and so do the people at BEST, which is why they are re-calculating the trend.

The GHCN inclusion criteria are another potential source of systematic bias in the trend. If we select stations from within the GHCN data set according to various criteria that appear to have nothing to do with temperature, we find that the trend alters in a significant way. The graph below shows the trend we obtain if we select from the GHCN data set only those stations that are reporting for at least 80% of the years between 1960-2000. We get a trend in which the 1930's are as warm as the 1990's.

We await with interest the BEST project's investigation of the effect of urban heating and station inclusion. We look forward to examining their trend-calculation algorithm. We have argued before that all such methods, with or without a reference grid, are pretty much equivalent, so we expect them to come up with something similar to the CRU and NCDC trends before they start to correct for systematic errors.

How BEST will correct for systematic errors in any meaningful or accurate way, we cannot say. If it was up to us, we would select long-lived stations in rural locations, and use the trends from those, even if there were only a dozen of them. But we have tried that already, and we get plots like the one above. The 1930s were hot, and it's just as hot now. But not especially hot.

Temperature, Pressure, and Altitude

2011-04-06T10:00:00.000-07:00

In Impetus for Circultation, we were surprised to find that "expansion work" and "buoyancy work" were equal. In any circulation of gas cells, the excess work done by the pressure of hot, expanding gases is equal to the excess work done by the weight of cold, falling gases. We found this equality mystifying. Today we demonstrate a linear relationship between altitude and temperature that may or may not help solve the mystery.

As we saw in Adiabatic Balloons, it takes work to raise gas up from the lower atmosphere. If the temperature of the atmosphere is uniform, the rising gas gets cold as it expands, and is therefore more dense than its surroundings. It tends to sink. At the same time, it takes work to pull gas down from the upper atmosphere. The falling gas gets hot, and is therefore less dense than its surroundings, so it tends to rise.

In Planetary Greenhouse Simulation, we saw what happens when we heat the atmosphere from the bottom, and cool it from the top. The lower atmosphere heats up until a particular temperature profile develops: a linear drop of 50 K from the bottom to the top. Once this profile is established, convection occurs freely. As gas rises, it expands and cools, but the gas around it is already just as cool, so the rising gas continues rising. As gas falls, it compresses and warms, but the gas around it is already just as warm, so the falling gas continues falling. The temperature profile that allows convection is the profile generated by the expansion and compression of circulating gases.

We determined that we could ignore mixing between cells in our simulation. When a volume of gas expands without mixing, it does so according to the equation for adiabatic expansion. In Atmospheric Pressure we calculated pressure as a function of altitude for an atmosphere at constant temperature. Today we calculate pressure as a function of altitude for an atmosphere whose temperature varies with pressure as it would for a dry, ideal, gas expanding adiabatically.

We see that the temperature of an atmosphere stirred by convection must drop linearly with altitude. Michele has proved this to us a number of times, in different ways. The differential form of the result is so simple I still find it astonishing.

dT/dz = − g/Cp

For dry air on Earth, we have g = 10 N/kg and Cp = 1 kJ/kgK, so we expect a drop of 10 K/km. Our simulation uses cells of dry air, and it shows a drop of around 50 K from bottom to top during convection. How high is the top? With a little more pencil-work, we arrive at the following equation for pressure as a function of altitude.

p = p₀ (T₀ − gz/Cp)^Cp/R / T₀^Cp/R (for dT/dx = −g/Cp)

The pressure at the bottom of our cell array is 100 kPa and at the top is 50 kPa, with the bottom temperature at around 300 K. So the height of our array must be around 5.4 km, giving the simulation an average temperature drop of close to 10 K/km.

In Atmospheric Pressure, we considered an atmosphere at uniform temperature, and arrived at the relation:

p = p₀e^−gz/RT (for T constant)

When our cell array is at a uniform temperature T = 250 K, its height will be 5.0 km. Thus we see that warming the atmosphere to initiate convection causes the top to rise by 400 m.

The −gz/Cp slope of the temperature profile is called the dry adiabatic lapse rate. The Earth's atmosphere is not dry. As moist air expands and cools, water condenses, which releases heat. Thus the wet adiabatic lapse rate is less than the dry adiabatic lapse rate. The lapse rate in the Earth's atmosphere appears to be 6 K/km.

To return to our mystery: we see that convection takes place when the temperature drops linearly with altitude. In an atmosphere stirred by convection, gravitational potential and atmospheric temperature are intimately related. A change in one will be matched by an equal and opposite change in the other. We have not solved the mystery, but I think we are getting closer.

Surface Cooling, Part VI

2011-03-30T10:00:00.000-07:00

We continue our study of Surface Cooling with the help of Circulating Cells, Version 5. We consider the transport of heat from a hot, sandy surface into the interior of a three-hundred meter cube of air. This three-hundred meter cube represents a single cell in the bottom row of our Rotating Greenhouse simulation. We refer to this cube as our super-cell. We are going to divide it into an array of sub-cells and heat the bottom row of sub-cells so as to induce convection within the super-cell.

We start by modifying our CC5 program. We set p_bottom to 100 kPa and p_top to 96.7 kPa. The total mass of our cell array is now 330 kg/m², which is the mass of our super-cell. The program will create an array of sub-cells with 30 columns and 15 rows. Each sub-cell will have mass 20 kg/m² and be 17 m high.

We set T_initial to 290 K, which is typical of a super-cell freshly-arrived at the surface our Rotating Greenhouse. We select Surface Heating, which warms the lowest sub-cells at a constant rate but allows no heat to escape from the array. According to our previous calculations, our super-cell will start to rise when it has warmed by a few degrees. Until then it will accumulate heat by convection of its own sub-cells.

When we start CC5, it calculates a value for the sub-cell impetus threshold using the procedure we describe in Impetus for Circulation. This value turns out to be 0.00003 K, a thousand times smaller than the value CC5 calculates for super-cells in our Rotating Greenhouse. As we showed in Simulation Time, each iteration of our simulation represents one second of planetary time. The heat arriving from the sun in the middle of the day can be as high as 1.4 kW/m². Let us suppose our sandy planet surface is receiving 800 W/m². The sub-cells have mass 20 kg/m² and heat capacity 1 kJ/kgK. We set Q_heating to 0.04 K so that the bottom sub-cells warm at 0.04 K/s. We set our mixing fraction to 0.10. Each time a sub-cell circulates, it will exchange one tenth of its volume with its neighbors.

We reset the sub-cell array and start running. The Figure below shows the simulation after about an hour of simulated time (four thousand iterations).

We record the average temperature of our sub-cell rows and obtain the following plots.

After half an hour, the bottom row of sub-cells has warmed by 20 K and the row above has warmed by 15 K. Meanwhile, the average temperature of all the sub-cells taken together has risen by 5 K. Once it has warmed by 5 K, the super-cell is likely to rise away from the surface, so further warming will be prevented.

The sub-cells above the surface warms by ten or twenty degrees during the day. This warming provides the impetus for the local convection we proposed in our previous post. At the end of the day, sub-cell convection stops, and super-cell convection brings cool air down from above.

And so our simulation confirms the process we described in Surface Cooling, Part III. After sunset, the air temperature in a sandy desert will cool by twenty degrees within a couple of hours. Furthermore, the temperature we experience standing on the sand at mid-day will be twenty degrees warmer than the temperature of the air three hundred meters up.

Rotating Greenhouse

2011-03-24T10:00:00.000-07:00

The Rotating Greenhouse configuration of Circulating Cells, Version 5 simulates the cycle of day and night that results from the rotation of a Planetary Greenhouse with respect to its sun. The day_length_hr parameter gives the length of the day-night cycle in hours. In Simulation Time, we concluded that one iteration of our simulation corresponded to one second of planetary time, so one hour is 3600 iterations and one twenty-four hour day is 86400 iterations. The sun will shine upon our Rotating Greenhouse for a fraction of the day given by day_fraction. You will find both parameters in CC5's configuration array.

As in the Planetary Greenhouse, the surface is something like sand. It heats up quickly and warms the bottom cells of the atmosphere. Now that the planet is rotating, the heat from the sun increases from zero at dawn to a maximum at mid-day, and decreases to zero again at sunset. We use a sinusoidal profile for the rise in solar heating during the day. For the entire night, no heat arrives at all. In the middle of the day, heat arrives from the sun at a rate π × Q_heating. When day_fraction is one half (the days and nights are of equal length), the average rate at which the bottom cells are warmed by the sun is Q_heating. But when the day is shorter, the average warming is less than Q_heating, and when the day is longer, the average warming is greater. For now, we leave the fraction at one half.

The top cells, meanwhile, radiates heat into space just as they did for the Planetary Greenhouse. Top cells cool by Q_heating per second when they are at temperature T_balance. Their cooling rate increases as the fourth power of their temperature.

We allow the Rotating Greenhouse simulation to run. The report line gives us the time in hours instead of the iteration counter. In the top-left of the report window we see the current rate of solar heating, in units of K/hr. After many days, the cycle of day and night reaches equilibrium, and we see the following each day.

During the day, the surface cells warm as they absorb the Sun's heat. Once they warm by a few degrees, they rise, as we describe in Impetus for Circulation. The average temperature of the surface rows rises by a few degrees during the day, but not more. Our simulation does not allow surface cells to cool. And yet the average temperature of the surface row drops by a few degrees at night. When we mark a few cells near the surface, and watch them move after night falls, we see the surface cells being replaced by cooler cells from above. It is air descending from above that cools the surface row. And so our simulation confirms the mechanism for surface cooling we described in Surface Cooling, Part III.

Our simulation does not, however, show the sudden ten-degree cooling we observe standing in the desert just after the sun sets. Understanding this sudden drop was the motivation behind our Surface Cooling series of posts. Next time, we will use CC5 in its Surface Heating configuration to simulate the transport of heat within one or our large surface cells, and so obtain an estimate of how much the first twenty meters of the atmosphere will warm up during the day.

PS. You will find the array data corresponding to the start of the day with Q = 0.001 K and no cell mixing in RGH_Q001

Impetus for Circulation

2011-03-17T10:00:00.000-07:00

Today we introduce Circulating Cells, Version 5. You can download the source code here. As usual, you will find instructions for running the program in the comments at the top. The source code is text file, so you can open it with any text editor. We have removed the Mix check box in CC5. You now turn off the cell mixing by setting the mixing fraction to zero.

Today's post is longer than usual. We present a mystery that we would like you to help us solve, but we also describe what happens to our simulation when we try to make sure that cell circulations take place quickly enough to be consistent with simulation time.

In Work by Circulation we showed that the rising cell does more work in the act of expanding than is required to compress the falling cell. This excess work will accelerate the cells so that they circulate. If the expanding cell cools by 3.032 K and the falling cell warms by only 3.000 K, the missing 0.032 K represents heat and pressure energy that has been converted into work. With heat capacity 1 kJ/kg, there will be 32 J of work available for each kilogram of gas in the rising cell. When divided between the four cells, there will be 8 J/kg. We call this the expansion work.

Convection occurs in water as well as air. Water at 20°C is 0.2% less dense than water at 4°C. A cell of water at 20°C will be buoyant within water at 4°C. We often think of water as an incompressible fluid. But it is slightly compressible. Water at 100 kPa is 0.004% less dense than water at 200 kPa. Water at 20°C rising from a depth of 10 m to the surface of a lake will do 6 mJ/kg of work by expansion. Meanwhile, water at 4°C falling from the surface to a depth 10 m will require 0.2% less work for its compression. The excess 12 μJ/kg is adequate to accelerate four equal cells to 2.5 mm/s. It is conceivable, therefore, that expansion work can power convection in water.

Nevertheless, there appears to be another source of work to power circulation. In a block of four cells, if we have a hot cell on the lower-left and a cold cell on the upper-right, the block's center of mass will descend when we rotate clockwise. The upper-left cell slides off the hot cell to rest upon the cold cell. As it slides, the top-left cell pushes the other three cells around to complete the circulation. The circulate routine of CC4 calculates how far the center of mass of a block of four cells will descend if we rotate the block by a quarter-turn. We describe this calculation in a comment here. If the block's center of mass drops by 0.8 m in gravity 10 N/kg, the weight of the block will provide 8 J/kg of work. We call this the buoyancy work.

In CC3 we rotated a block of cells only if the buoyancy work was positive. We ignored the effect of expansion work and buoyancy work upon the cells. Of the heat we put into the array at the bottom, only 85% emerged from the top by radiation. In CC4 we calculated the expansion work and divided it up between the four cells to represent viscous heating. At equilibrium, 100% of the heat we put in emerged from the top. Our simulation conserved heat, but it ignored buoyancy work. How can that be?

Furthermore, how large is the buoyancy work compared to the expansion work? If one is positive, is the other always positive? The circulate routine of CC5 includes code that calculates both the buoyancy work and expansion work, and reports them both. We ran our simulation with this code activated. In a wide variety of conditions, both evolving and convergent, the buoyancy work was always between 95.7% and 97.6% of the expansion work. To within the margin of error introduced by our simulation, buoyancy and expansion work are equal.

Why are they equal? Michele has proved to me several times that for adiabatic circulation, temperature will drop linearly with altitude. Could it be that this linear relationship causes buoyancy and expansion work to be equal? Or are buoyancy work and expansion work just two faces of the same process, so that we can count one or the other, but not both? If they are one and the same, then buoyancy work cannot exist without expansion work, which implies that convection would not occur in a perfect, incompressible fluid. But surely convection can occur in a perfect, incompressible fluid?

We would like to provide you with answers to these questions, but we don't have them yet. Perhaps, after a few more debates with Michele and anyone else who wants to take part, we will have answers. For now, however, we are going to assume that we cannot add buoyancy and expansion work together to obtain the total work available to accelerate our cells. We have a simulation that conserves heat at equilibrium, and we want to keep it that way. We will calculate and use the expansion work only, and add it back to the cells as viscous friction.

In the circulate routine of CC5, we calculate the drop in the combined temperature of the four cells in a block. We call this net loss our impetus for circulation. If the impetus is 0.032 K, our expansion work will be 8 J/kg when spread among the four cells. The cells could, in theory, reach a speed of 4.0 m/s. The cells in our atmospheric array are roughly 400 m high, so a circulation at 4 m/s will take place in 100 s. There are 450 cells in our 15×30 array, and our simulation picks a new block on every iteration. We expect one of the four cells in our block to be picked again within a hundred iterations. But one hundred iterations corresponds to 100 s, so we see 0.032 K is the minimum impetus required to make sure that our rotation takes place before we expect one of its cells to take part in another rotation.

Our CC5 program rotates a block of four cells only if its impetus for circulation is greater than an impetus threshold. With the impetus threshold set to 0.032 K, we ran CC5 to produce new versions of the graphs we presented in Simulation Time.

The impetus threshold has no effect upon the profiles we obtain with no mixing (M = 0.0). Nor do we see any effect with heavy mixing (M = 0.20). We do, however, see a slight change in the presence of mild mixing (M = 0.05). Mild mixing now raises the surface temperature by a few degrees.

The impetus threshold makes sure that rising cells are significantly warmer than falling cells. When a cell rises from row 7 to row 8, its temperature drops by a factor of 0.9879 (see here). If our impetus threshold is 0.03 K, warm cells about to rise from row 7 must be 0.03 K ÷ (1-0.9879) = 2.5 K warmer than cool cells about to arrive in the same row. Between the bottom rows, the required difference is 3.0 K, and between the top rows it is 1.7 K. With Q = 0.001 K, a cell freshly-arrived in the bottom row will warm by 3.0 K in three thousand iterations, which is almost an hour. Soon after, it will start to rise.

In our next post, we will see how well our program performs when we simulate night and day using CC5's new Rotating Greenhouse configuration.

PS. You will find the equilibrium array for Q = 0.001 K and M = 0.00 in PGH_Q001_M00. The array for Q = 0.001 K and M = 0.20 is PGH_Q001_M02. Saved arrays now have comments in them that tell us the conditions under which they were obtained.

Summary to Date

2011-03-13T16:46:00.000-07:00

On the side-bar, you will see a new link, Summary to Date. The new page gives a history of the posts on this site. It attempts to explain what we have achieved so far, what we are working on now, and what we hope to achieve in the future. I'll maintain the page, adding paragraphs every few months. With any luck, the page will allow newcomers to make sense of what's going on.

Simulation Time

2011-03-09T13:08:00.000-08:00

So far, we have measured simulation time in units of iterations. Today we relate iterations to time in seconds, and use this relationship to choose a heating rate better suited to a simulation of the Earth and the Sun.

Our simulation selects one block of four cells at random on each iteration. With thirty columns and fifteen rows in our array, a rising cell will find itself selected once in every hundred iterations. For circulation to occur, the rising cell must be one of the two lower cells in the block. Furthermore, the cell diagonally above it must be a falling cell. No more than half the cells are falling, so we see that a rising cell will move up by one row every four hundred iterations on average. A cell will take roughly six thousand iterations to rise from bottom to top.

The tropopause pressure in CC4 is 50 kPa, and the surface pressure is 100 kPa. Using the equation we presented in Atmospheric Pressure, we see our tropopause altitude is around 5 km. If one iteration corresponds to one second, cells will take six thousand seconds to rise five kilometers. Their average speed will be around 1 m/s. They will rise from the surface to the tropopause in one and a half hours. But if one iteration corresponds to ten seconds, rising cells will move at 0.1 m/s and take seventeen hours to reach the tropopause.

In Work by Circulation we presented an example circulation driven by a rising cell at 320 K and a falling cell at 300 K. This particular circulation generated 160 J of work for each kilogram of air in the rising cell. In our Planetary Greenhouse simulation, the average work produced by a single circulation is 60 J per kilogram of the rising cell. If this work turns into kinetic energy of the four rotating cells, each kilogram will receive 15 J. The cells will accelerate to 5 m/s. Our CC4 simulation assumes this kinetic energy is dissipated as viscous friction within the block. In that case, the average speed of the circulating cells will be less then their maximum possible speed, perhaps as low as 1 m/s. But the speed will certainly not be as low as 0.1 m/s.

We conclude that one iteration of our simulation is closer to one second of planetary time than it is to ten seconds. With one second per iteration, rising air will take roughly one and a half hours to travel from the surface to the tropopause. Falling air will take one and a half hours to descend.

As we showed in Solar Heat, the average power arriving from the sun per square meter of the Earth's surface is close to 350 W. Let's suppose the bottom cells of our simulation are warmed at an average rate of 350 W/m². With top pressure 50 KPa, bottom pressure 100 kPa, and 15 rows of cells, each row represents a drop of 3.3 kPa. With gravity 10 N/kg, the mass of each row must be 330 kg/m². The heat capacity of air at constant pressure is 1 kJ/kgK, so 350 W/m² will cause a cell to warm at 0.001 K/s, or 3.6 K/hr.

Thus we see that Q_heating of 0.001 K represents the Sun's effect upon the surface of the Earth far better than the 0.01 K that we have been using in our experiments until now.

With Q_heating set to 0.001 K instead of 0.01 K, the heat flow through our cell array will be ten times lower. The heat flow will be ten times smaller when compared to the heat capacity of the cell array. It will take ten times as many iterations for the simulation to converge. We find that it still takes six thousand iterations for a cell to rise up through the array, but now it takes six million iterations to approach a new equilibrium. Six million iterations represents six million seconds, or roughly two months. This corresponds well to the pace of the Earth's seasons. The shortest day in Boston, for example, is in December. But the coldest weather occurs two months later.

We ran our CC4 simulation in Planetary Greenhouse mode and allowed the temperature profile to reach equilibrium for various values of mixing fraction. We present our results below. In the legend, M is mixing_fraction and Q is Q_heating. Our CC4 program allows us to save and load previously-converged cell arrays. With the help of saved arrays, we were able to obtain convergence within two million iterations in most cases.

As we discussed in Circulation and Mixing, allowing cells to mix as they circulate can increase the temperature drop from the bottom to the top of the array. This is indeed the case when Q = 0.01 K, which corresponds to Solar heating of 3.5 kW/m², or ten times what we expect on Earth. But when we reduce Q to 0.001 K, mixing has the opposite effect. Michele's experiments show us that when Q = 0.001 K and M = 0.10, mixing alone is sufficient to transport all the heat through the array with a temperature drop of only 60 K. The temperature drop is less than that we would obtain without mixing. The temperature profile is, in Michele's words, hypo-adiabatic.

With M = 0.05, our simulation predicts a temperature of 297 K at the surface and 250 K at 6 km, which is very close to that of the Earth in middle latitudes. We observe steady circulation of cells from bottom to top.

When we increase M to 0.10, however, circulation slows down dramatically. If circulation slows down, mixing must slow down too, which means that our result for M = 0.10 presents us with an apparent contradiction. We see that mixing can occur only up to the point where it starts to slow down circulation. With Q = 0.001 K, we must have M ≤ 0.05. For M ≤ 0.05, the surface temperature is within one degree of its value for M = 0.00. Now that we have reduced Q to 0.001 K, it appears that we are better off running with the mixing turned off. We speed up execution, we avoid any self-contradictory mixing behavior, and we suffer no significant change to our atmospheric temperature profile.

PS. You can download the equilibrium state of the Planetary Greenhouse with M = 0.05 and Q = 0.001 K in PGH_M005_Q001.

Work By Circulation

2011-03-02T10:00:00.000-08:00

The CC3 program we presented in Planetary Greenhouse Simulation ignores the work done by cell circulation. Today we show how this omission caused the top cells to converge to a temperature 10 K lower than we expected. (Thanks to Michele for all his help in the comments.)

With Q_heating at 0.01 K, our simulation heats each cell in the bottom row by 0.01 K per iteration. With T_balance at 250 K each cell in the top row cools by 0.01 K per iteration. Our simulation assumes an ideal gas, so the heat required to warm or cool a cell by 0.01 K is always the same. At equilibrium, the heat entering must be equal to the heat leaving. We expect the top row converge to 250 K.

But the top row of CC3 converged to only 240 K. At 240 K, each top cell cools by only 0.0085 K per iteration. With circulation turned off, however, and only mixing to transport heat, the top row of CC3 did converge to exactly 250 K. We concluded that something was amiss with our circulation routine: it was making heat disappear.

In Work by Convection we showed how the expansion of warm, rising air does more work than is required to compress cool, falling air. The excess work manifests itself immediately as an upward force on the rising air. We could harness this force with a wind turbine. Otherwise the rising air will accelerate into a turbulent flow. Over time, viscous friction will transform this kinetic energy into heat. We discussed this phenomenon in Dissipation by Convection, and concluded that work by convection is what powers storms and winds.

Suppose we have a block of four cells in our simulation, at the bottom of the cell array. The lower-left cell is at 320 K while the upper-right is at 300 K. We rotate the block clockwise. The lower-left cell rises. Its pressure drops by 3.3 kPa, this being the weight per square meter of a single cell in CC3. If the cell expands adiabatically, its final temperature will be 320 K × 0.9903 = 316.90 K. Meanwhile, the upper-right cell falls. Its pressure rises by 3.3 kPa. Its final temperature will be 300 K ÷ 0.9903 = 302.94 K. One cell cooled by 3.10 K while the other warmed by 2.94 K. The two remaining cells, which do not change pressure, neither warm nor cool.

Both cells have the same mass, and the ideal gas inside has constant specific heat capacity C_p = 1.003 kJ/kg. One cell lost 3.10 kJ/kg while the other gained 2.94 kJ/kg. The difference is 160 J/kg. This 160 J/kg is the work done by the circulation. We did not account for this work in CC3. Each time a block rotated, the work by circulation disappeared.

Our CC4 program provides several new features. You can save the array temperatures to a text file, and load them also. You can track the movements and temperature of marked cells with the Tracking option. Within the code, we have greatly expanded the explanatory comments and we have simplified some of the routines and global variables to make future enhancements easier. But most important, CC4 accounts for the work done by circulation, as you can see for yourself in the circulate routine. The Viscosity option indicates whether or not we account for the dissipation of work by viscosity.

When CC4 rotates four cells, it calculate the work available just as we did above. We assume this work goes into accelerating the air within the cells and ends up dissipating as viscous friction within the four cells. The simulation adds the viscous heat to the four cells equally. The following graph shows the temperature profiles we obtained from CC4 after a million iterations for various values of mixing fraction (M) and heating rate (Q). Click on the graph for better resolution.

The top row converges to 250 K in each case. With no mixing, the bottom row is 50-K warmer than the top. With mixing fraction 0.1, the bottom is 58 K warmer than the top. With mixing fraction 0.4, it is 70 K warmer. The effect of mixing is almost exactly the same as it was for CC3. Also in the graph, you can see the profile that results when we turn off the circulation (NC) but set mixing fraction to 1.0. Heat is transported by mixing alone and the bottom is 62 K warmer than the top.

We conclude that the incorrect convergence of CC3 was indeed the result of our ignoring work by circulation. Indeed, we can obtain the CC3 results with CC4 simply by turning off CC4's viscous friction adjustment.

Circulation and Mixing, Part II

2011-02-22T12:10:00.001-08:00

The blue plot in the graph below shows the temperature of a single cell performing a complete circulation within our Planetary Greenhouse simulation. The cell exchanges no heat with its neighbors. We have mixing fraction zero, Q_heating at 0.01 K, and T_balance at 250 K. The cell warms up in the bottom row (the Sun heats the surface) and cools down in the top row (the tropopause radiates into space).

In any given row of the atmosphere, the cell is 20 K warmer on the way up than on the way down. Even though a rising cell may pass by a falling cell, they exchange no heat. The convection is adiabatic. The average temperature of the bottom row is roughly 50 K warmer than the average temperature of the top row.

Now suppose we allow the passing cells to exchange heat. The warm, rising cells cool down because they mix with the cool, falling cells. In order to reach the top, a cell will have to start off hotter than it would without mixing. Cells rise part-way, mix with their neighbors, fall, and mix with their neighbors again. The pink plot shows the variation in temperature, with mixing fraction 0.4, as a cell rises and falls during a hundred thousand iterations of the simulation. The average temperature of the bottom row is roughly 70 K warmer than the top row.

The following plot shows how bottom-row temperature varies with mixing fraction for Q_heating at 0.01 K. (Thanks to Peter Newnam for the data.) As we increase the mixing, circulation is further impeded, but but the mixing itself begins to transport sufficient heat to bring down the surface temperature.

Circulation alone is an efficient transporter of heat. The following plot shows that we see little change in the atmospheric temperature profile as we increase Q_heating by a factor of ten from 0.01 K to 0.1 K. The profile for 0.1 K is erratic, but on average is almost the same as that for 0.01 K.

When we allow mixing, we break up the continuous circulation that would otherwise transport heat from the surface to the tropopause. Instead of one continuous circulation we have a series of circulations carrying heat up in stages. Heat transport between stages takes place by mixing. The average temperature of the surface must increase because heat exchange by mixing requires an additional temperature drop over and above the adiabatic temperature drop produced by circulation.

To round off our explanation of the effect of mixing, imagine a continuous circulation from bottom to top consisting of two parallel columns of cells. One column is rising as the other falls. Now we suppose that each rising cell must be 10 K warmer than its side-by-side neighbor in the falling column, or else circulation will stop. Furthermore, we suppose that each time the circuit rotates by one cell, we allow some mixing between the side-by-side neighbors. The rising cell cools by 1 K and the falling cell warms by 1 K. Because of this mixing, a cell that has just descended will be 1 K warmer than without mixing. If it is to be 10 K cooler than its new side-by-side neighbor, this neighbor must be 1 K warmer than in circulation without mixing. The 10-K difference between the side-by-side neighbors in the column will be sustained only if the total temperature drop from bottom to top increases by 1 K multiplied by the number of cells from bottom to top. In our case, that would be 15 K.

Mixing raises the temperature difference necessary to transport heat through the atmosphere. The temperature difference is therefore greater than that dictated by adiabatic convection alone. In future posts we will enhance our simulation to allow cells within the array to radiate heat directly into space. This radiation, taking place before the cells reach the tropopause, will decrease the temperature drop between the bottom and the top. For now, however, we are satisfied that our program is effective in its simulation of mixing and circulation.

Circulation and Mixing, Part I

2011-02-14T18:45:00.000-08:00

We run CC3 program in Planetary Greenhouse mode with mixing_fraction = 0.1, Q_heating = 0.01 K, and T_balance = 250 K. The following graph shows the average temperature of the top and bottom rows versus iteration, which is the simulation's version of time. The entire array starts off at 250 K.

The top row represents our simulated planet's tropopause. We see the top row cooling during the first ten thousand iterations as it radiates its heat into space. At the same time, the bottom row warms up as it absorbs heat from the Sun. After the first ten thousand iterations, convection starts to transport this heat away from the bottom row and deliver heat to the top row. The pace of warming and cooling slows down. After a few hundred thousand iterations, the top row settles at 239 K while the bottom row settles at 297 K. When we watch the simulation, we see cells rising to the top, cooling, and sinking to the bottom in convection cycles. Now that thermal equilibrium has been reached, we know that the top row must be losing heat at the same rate that heat enters the bottom row.

Because we set T_balance to 250 K, we know the top row in our simulation will radiate as much heat as is arriving from the sun when its cells are all at 250 K. But the top cells radiate heat in proportion to the fourth power of their temperature. Furthermore, the temperature of the upper cells is not uniform. Cells arriving from below are warmer. Those that are warmer will radiate disproportionately more than those that are cooler. As a result, the average temperature of the top row can be less than 250 K. We will consider this effect in more detail later. Today we want to present the effect of mixing upon circulation. (UPDATE: The convergence to 240 K turns out to be due to our ignoring work produced by circulation, but this omission has little affect upon the magnitude of warming caused by mixing.)

We run our simulation three times, each with a different value of mixing fraction. After a million iterations, we plot the average row temperature versus the row number, which we call the temperature profile. In the legend, Q is Q_heating and M is mixing_fraction.

With no mixing, once a cell warms up in the bottom row, no heat passes in or out until it reaches the top. In the top row, it radiates its heat into space and cools down. Without mixing, the bottom row is 49 K warmer than the top. When we allow the cells to mix with their neighbors, each cell loses heat to its cooler neighbors and gains heat from its warmer neighbors. We see that when mixing_fraction is 0.1, the bottom row warms up until it is 58 K warmer than the top row. When mixing_fraction is 0.2, it is 62 K warmer.

We might think that the exchange of heat between atmospheric cells would encourage the overall transport of heat through the atmosphere, and so decrease the temperature difference between the bottom and the top. But quite the opposite is the case. In our next post, we will endeavor to explain why. At the moment, we don't have a compelling answer. Indeed, it may be that the warming caused by mixing is nothing more than a flaw in our simulation. Your thoughts on the subject will be most welcome.

Planetary Greenhouse Simulation

2011-02-10T11:27:00.000-08:00

Suppose we have a planet whose atmosphere is perfectly transparent to short-wave radiation from the Sun. The surface of the planet is made of sand, which warms up quickly. It warms until it loses heat by radiation and convection as fast as it receives heat from the Sun. Meanwhile, we suppose that the lower atmosphere is opaque to long-wave radiation so all the heat from the Sun ends up in the lower atmosphere. The lower atmosphere gets hot. It rises. Like all atmospheres, this one gets thinner as we ascend. Eventually it is so thin that it becomes transparent to long-wave radiation. This layer of the atmosphere is the tropopause. The tropopause radiates heat into space according to Stefan's Law. Air in the tropopause cools until it becomes so dense that it sinks towards the surface. And so atmospheric convection begins. The heat arriving from the Sun passes into the lower atmosphere, is carried up to the tropopause by convection, and is radiated into space.

We have described such a planet and atmosphere before, in Planetary Greenhouse. Today's CC3 program simulates the transportation of heat through such an atmosphere. You can download the program here. Running in its Planetary Greenhouse configuration, CC3 supplies heat to cells in the bottom row at a constant rate. Each bottom cell heats up by Q_heating K per iteration, regardless of its temperature. Meanwhile, a cell in the top row at temperature T_balance cools by Q_heating K per iteration. At other temperatures, the cooling rate of the upper cells increases as the fourth power of the cell temperature.

In the simulation, Q_heating represents the heat arriving from the Sun. The top cells represent the tropopause, and the heat they lose is the heat radiated by the tropopause into space. The heat radiated into space by the top cells will be equal to the heat arriving from the Sun when the temperature of all the top cells is T_balance. We start the simulation with all cells at 250 K, Q_heating = 0.01 K, and T_balance = 250 K. The top cells start to cool immediately. The lower cells start to heat up.

In CC3, we corrected our implementation of heat mixing in accordance with the suggestions made by Michele. You will find our new implementation in the heat routine. We added a reporting text window, as you can see in the screen shot above. You open the window with the Reporting button. The program calls report every reporting_interval iterations, or whenever you press the Report button. We instructed report to print the average temperature of the cell rows. We added a check procedure that checks for convergence of the simulation, as requested by Peter. In the comments below I'll provide some convergence code that you can paste into the check routine. In order to accommodate the planetary greenhouse temperature variations, we doubled the breadth of the color-coded temperature range. Red is now 320 K and blue is 220 K. The Configure button allows you to change the simulation parameters while the simulation is running. Be careful not to enter invalid numbers or else the program will abort with an error.

We're running the program now. It takes several hundred thousand iterations to reach equilibrium. In our next post, will present the vertical temperature profile under various conditions. We are curious to see how the temperature of the surface is affected by the amount of heat the atmosphere must transport, and by the amount of mixing we allow between cells.

Mixing Cells

2011-02-01T07:59:00.001-08:00

Our CC1 simulation allows cells to do work on one another so as to circulate by convection, but it does not allow them to pass heat to one another. The only place that heat could enter a cell was in one point along the bottom row. Today we present CC2, which allows heat to pass between cells so as to simulate mixing and conduction during atmospheric circulation. You can download the new program here.

As before, our simulation proceeds by selecting blocks of four cells at random. The simulation tests these block to see if buoyancy will cause them to rotate. In CC2, we allow heat to be exchanged between the cells of each block. We do this whether or not the cells rotate. The fraction of heat we allow to pass from each cell to each of its two neighbors is the mixing fraction. You can use CC2's new Configure button to set the mixing fraction for yourself. By default, we use 10%.

The new program provides an iteration counter, where the treatment of a block of four cells is one iteration. The display has several new buttons, as you can see below. The program starts running when you press Run. With accelerate checked, the program will update its display infrequently, so as to speed up the simulation. With the circulate and mix boxes checked, the simulation both circulates and mixes the cells. We have improved the marking of cells in this version. The marked cells have a black outline so you can see their temperature. We also increased the size of the display. Click on the following image to see the new size.

With "Point Temperature" we heat a single cell at the bottom. With "Surface Temperature" we heat all the cells along the bottom. You can set this temperature to which we heat them with the help of the Configure button.

Today we want to see how mixing affects the distribution of heat in the array with the same single-cell heating we used in our previous post. Without mixing, the cells settle down after half a million iterations to an adiabatic temperature profile, like this. With mixing the array looks pretty much the same after half a million iterations. The mixing has little effect upon the distribution of heat in the early stages of the simulation. But after 1.5 million iterations, we obtain the following array. The temperature at the top has risen from 250 K to 280 K. After two million iterations, the top of the array is at 290 K. After five million, it's above 295 K.

Half a million iterations is enough to establish an adiabatic temperature profile by convection. Five million iterations is enough to establish a uniform temperature by mixing.

We invite you to download CC2 and experiment with it yourself. For example, select surface heating to see what happens if you heat the entire bottom row of the array. Or try setting the heating temperature to 1000 K and turning off the circulation. You will see the mixing spreading heat through the array. Or run with the circulation turned on and the high heating temperature. You will see a plume of heat rising through the array.

The simulation appears to give reasonable results so far. We expected to arrive at a uniform temperature when we introduced mixing. Next time we will see what happens if we start to remove heat from the top of the array while we are inserting heat at the bottom. We wonder what type of temperature profile will evolve as heat flows up through the array.

Circulating Cells

2011-01-26T12:48:00.000-08:00

Today we present our new simulation program, Circulating Cells, Version 1, or CC1. The following picture shows the program's display shortly after we start it up. The array of cells represents a cross-section of the atmosphere. Each cell represents an equal mass of gas. The color of the cells indicates their absolute temperature, according to the legend at the top, with the exception of the black one. The black cell is black so we can watch it circulating around.

This version of the program heats one location on the bottom row. Any cell that enters this location we heat to 300 K. If the cell enters at a higher temperature, we let it keep its higher temperature. Thus we never remove heat from our array.

The hot cells are rising, and as they rise they cool down. They turn from red to green. Meanwhile, cold cells must descend to make way for the hot cells. The cells that descend warm up. They turn from blue to green. At the beginning of our simulation, all the cells are blue. They are at 250 K. If we press Reset, all the cells will return to 250 K and the simulation will start again.

In our simulation, convection occurs by rotating blocks of four cells. The program picks a block of four cells at random. It calculates how much a quarter-turn will raise or lower the block's center of gravity. If a rotation will lower the center of gravity, the program performs the rotation. Otherwise, it does not. When the program rotates a block, one cell rises, one falls, and two stay at the same height. The one that falls contracts and warms up. The one that rises expands and cools down. The two cells that move sideways do not change temperature. But they may rise when they move from the top of a cold cell to the top of a hot cell. Hot cells are taller than cold cells.

In the following picture we see the distribution of heat in the atmosphere after a few minutes of running. The black cell has moved. We can turn any cell into a black cell by clicking on it with the mouse. We can turn any black cell back into a normal cell by clicking it again.

The simulation is made possible by a fundamental approximation. We arrange the cells as if the cells in each row are all at the same height, even though some may be sitting on top of columns of air that are significantly warmer or cooler than the others. Without this approximation, we cannot arrange the cells in an array, and simulation becomes several orders of magnitude more complex and time-consuming.

In CC1, there is no way for heat to pass from one cell to another. There is no conduction, no mixing, and no radiation. There is no way for heat to leave the array. After an hour of running, we end up with the following distribution of heat, all arising from the single heated location at the center of the bottom row. There is one blue cell on the bottom remaining. It moves around a bit, but so far it has resisted being drawn into the warm location.

Our program is written in TclTk. We encourage you to download the program and run it for yourself. To download it, click here. The file is called CC_1.tcl. It won't take you more than five minutes to get the simulation going. The movement of the cells is fascinating. The pictures I have here cannot do justice to the evolving display. I have inserted detailed comments in the code, which I hope will make it clear how each routine works. There are also instructions on how to run the simulation on Linux, MacOS, and Windows.

Without self-regulation by clouds we estimate that high clouds will warm the Earth by 38°C, doubling CO2 will warm the Earth by 1.5°C, and thick clouds will cool the Earth by 96°C. It is our hope that Circulating Cells will evolve into a simulation of radiation, convection, precipitation, and, ultimately, self-regulation by clouds. We will then be able to estimate the effect of doubling CO2 concentration within a self-regulating climate.

That evolution may take another year or two, and maybe the task will prove too monumental for us. But I think it will be an enjoyable journey, no matter how it ends. So I hope you will accompany me.

Adiabatic Balloons

2011-01-17T13:00:00.000-08:00

In Atmospheric Convection we described how rising air must expand and cool, while falling air must compress and warm. We used the equation for the adiabatic expansion of diatomic gases to estimate the amount by which air will cool as it rises through the atmosphere.

In a debate that took place in the comments, one of our readers claimed that a temperature difference must always exist between the bottom and the top of an atmosphere. Air is always circulating to some degree, he said, and any air that rises must cool adiabatically. He asserted that the temperature as we ascend through the atmosphere will be related to the pressure by the equation of adiabatic expansion. In Adiabatic Magic, however, we showed that our reader's claim violated the Second Law of Thermodynamics. Spontaneous adiabatic circulation of air in the atmosphere is impossible.

If spontaneous, adiabatic circulation is impossible, there must be some physical process that stops it from happening. Although we have proved that some such process must exist, we have not yet described the process. We will attempt to do so today.

Consider the following diagram. We have a column of air 5 km high, enclosed in a box. The temperature of the air is uniformly equal to 250 K. Because of the weight of the air, the pressure at the bottom is twice the pressure at the top. We have a balloon of air at the top, and another at the bottom. Each balloon is insulating and reflecting, so no heat passes in or out. Each balloon is flexible, so the pressure of the gas inside is always equal to the pressure outside.

We attach a string to the upper balloon and start to pull it down. As it descends, the air around it pushes in upon it with greater pressure so as to compress its volume. The air inside undergoes adiabatic compression. Its pressure, p, and temperature, T, are such that the product p ^0.4T ^−1.4 remains constant.

We pull the balloon all the way down to the bottom of the column. Its pressure has doubled, so its temperature must rise from 250 K to 305 K. Our balloon has grown smaller, so the surrounding atmosphere must have expanded, and therefore cooled. But suppose our column is enormous compared to the balloon, so the cooling of the air outside the balloon is negligible. The air inside is at 305 K, and that outside remains at 250 K. The density of the air outside is 305/250 = 1.2 times greater than that of the air inside. Every kilogram of air in the balloon occupies the same volume as 1.2 kg of air outside the balloon.

By the principal of buoyancy, every kilogram of air in the balloon will experience an upward force equal to the weight of 0.2 kg of air. If we assume gravity is 10 m/s/s, we see that we must hold the balloon down with a force equal to 2 N/kg (two Newton per kilogram of air inside) or else it will rise, and it won't stop until it gets to the top again.

This buoyancy force was zero when we started pulling the balloon down, and at the end it was 2 N/kg. The average force would be close to 1 N/kg, which we apply over 5 km, so we must do roughly 5 kJ/kg of work to pull the balloon down.

Now suppose we pull the lower balloon up from the bottom at the same time. The net change in the volume of the rest of the column is is now zero, so we really can say that the outside air remains at 250 K. This other balloon, when it reaches the top of the column, will have expanded. It's pressure has halved, so its temperature must drop from 250 K to 205 K. It's density is 1.2 times greater than the air around it, so it will experience a negative buoyancy force of 2 N/kg. If we don't pull it up, it will sink all the way to the bottom again. In drawing it up, we must do 5 kJ/kg of work.

We must do 5 kJ of work to raise a single kilogram of air from the bottom to the top, and 5 kJ to lower another kilogram from the top to the bottom. Thus we see that the spontaneous circulation of air suggested by the Adiabatic Magic hypothesis is indeed impossible. It is buoyancy that stops the magic. We need a source of work to cause circulation, and no such source exists in our column of air at a uniform temperature.

One source of the necessary work is a heat engine in which we supply heat at a higher temperature to the bottom of the column and extract heat at a lower temperature from the top. Any time we have a source of heat at a higher temperature, and a place for the heat to flow to at a lower temperature, we can make a machine that does work. In the atmosphere, this machine is implemented, albeit inefficiently, by convection. We discussed work by convection earlier, but we we will return to it in future posts.

UPDATE: I originally made the column 10 km high, but Michele pointed out that I had my math wrong. The column should be 5 km high. Indeed, it turns out that the heigh of the column for which we have half the pressure at the top is a function of temperature only, which did not occur to me until I tried to duplicate Michele's calculation. See comments for details.

Surface Cooling, Part V

2011-01-15T16:13:00.000-08:00

If you go back and look at Surface Cooling, Part III, you will see that we have improved and abbreviated our explanation of the rapid cooling that occurs in the desert when the sun goes down. It was my father who pointed out a simple mechanism by which this cooling must occur. As we already noted, the sand will cool down quickly when the sun sets. As the day-time convection of air above the desert slows to a halt, it brings cool air down from above, and this air will not be warmed by hot sand as it was during the day. It remains cool, which causes the temperature to drop by ten or twenty degrees.

Mercury Bulb, Part II

2011-01-11T08:08:00.000-08:00

Suppose we expose our mercury bulb thermometer to radiation from the Sun and the ground, as shown below. We have short-wave radiation arriving from the Sun far away, long-wave radiation arriving from the ground just below, and the bulb radiating as well. If we assume the air below the bulb is transparent, the heat the bulb receives from the ground is equal to the heat it would receive from a hemispherical shell of the same material.

The ground and the bulb are opaque to short-wave and long-wave radiation. The emissivity of mercury is roughly 0.10 for short-wave and long-wave, while that of sand and dirt is 0.90 for both short-wave and long-wave.

In the following calculations, we use the letter ε for emissivity, and a subscript to indicate ground or bulb. The radius of the bulb is r. We calculate the net absorption of radiation by the bulb, in Watts. We assume that this extra heat will cause a slight increase in the bulb's temperature over the temperature of the air, leading to conduction of the excess heat away from the bulb. We apply the equation for steady-state conduction we derived last time, and so arrive at an equation for the change in the bulb temperature in terms of power arriving from the sun, ground temperature, and air temperature.

In Arizona over Thanksgiving, we observed the ground to be at around 320 K at mid-day and the air at 300 K. In Solar Heat, we showed that the power arriving from the sun is 1,400 W for each square meter we hold up perpendicular to its rays. A thermometer exposed to the sun and the ground under these conditions will read 1.5°C warmer than the air. At night, we observed the air and the ground to drop to around 280 K. A thermometer free to radiate its heat into space under these conditions will read 1.5°C colder than the air.

Suppose we house our thermometer in a box with holes on the sides to let air through, but a solid roof and floor. We make the box out of a good insulator, like wood. Air can move freely through the box, but radiation from the ground and the sun are blocked. The outside of the box may get hot during the day or cold at night, but the inside surface will settle to the air temperature, along with the thermometer bulb. The radiation the bulb receives from the box will be exactly equal to the radiation it emits towards the box.

During the day, the ground warms the air. In order for heat to pass from the ground to the air, there must be a sharp temperature gradient near the surface, because the conductivity of air is so poor. If we place our thermometer one meter above the surface, it will be out of this heating layer, and so measure the temperature of the main body of air moving along the ground. When the wind blows, the thermometer will be more effective at measuring air temperature, because the movement of air around the bulb greatly facilitates the transfer of heat to and from the bulb. But once again, the wind near the ground can be erratic, or shielded by obstacles, so placing the thermometer a meter above the ground would make sure that it experiences the temperature of the main body of moving air.

We conclude that a thermometer mounted in a perforated box one meter above the ground will give us an accurate measurement of air temperature, despite the effects of radiation, conduction, and convection.

Mercury Bulb, Part I

2011-01-03T07:03:00.001-08:00

When the temperature recorded by a thermometer two meters above the desert drops by 10°C in an hour, what does this drop mean? Does it mean that the air has cooled down, or does it mean that the radiating bodies around it have cooled down? When we feel cold walking in the desert at night, is this because the air is cold, or is it because the sand all around us is no longer radiating heat that would otherwise warm our bodies? So far, we have assumed that it is the air that cooled down. In any case, we feel it is time to consider the role played by conduction, radiation, and convection in determining the temperature recorded by a mercury bulb thermometer.

Let us start with conduction. We will simplify the conduction question by considering steady-state heat flow out of a spherical mercury bulb with thin walls. By "steady-state" we mean that no element in our problem is either warming up or cooling down. The heat conducted away from the bulb is constant. The advantage of this simplification is that it has a simple analytical solution, as we show in the following diagram.

Meanwhile, the heat capacity of the bulb allows us to relate the heat flow out of the bulb to the rate at which the temperature of the bulb must be falling, as we show below. In our continuing equations, we use T and r for the temperature and radius of the bulb.

The following graph plots the temperature of mercury bulbs of various radii versus time after a sudden 10°C drop in temperature. We see that a bulb of radius 2 mm will cool to within 0.5°C of the new air temperature within a few minutes.

Most mercury bulbs are a few millimeters in diameter, so it appears that mercury thermometers can measure air temperature accurately provided that the changes in air temperature occur on a time scale of ten minutes or longer. In our next post we will consider whether radiation from the ground, the sun, and the bulb itself will disturb the thermometer's measurement of ambient air temperature.

Surface Cooling, Part IV

2010-12-30T13:29:00.001-08:00

In our previous post, we concluded that air above the desert cools down by 10°C in the first hour after sunset because cold air descends slowly from above as the day's convection cycle comes to a stop. This explanation is not one of the several you will find on the web if you search for "Why does the desert get so cold at night?" These other explanations talk about dry air allowing heat to radiate away into space. But our calculations in Day and Night showed that the rate at which the Earth and its atmosphere radiate heat is barely adequate to cool the air by half a degree in ten hours, let alone ten degrees in one hour.

One of our readers points us to another night-time cooling phenomenon called "night inversion". According to this description, the air immediately above a forest can be 10°C cooler at night than the air 100 m up. In our explanation of the desert cooling, the cold air above the desert was still warmer than the air higher up, but in this case, the air above the forest is actually colder than the air higher up. We will conclude our Surface Cooling series by offering an explanation for nigh inversion also.

The first 100 m of the atmosphere has heat capacity 100 kJ/K/m². The forest is a good black body, so it will radiate of order 100 W/m² directly into space through a clear night sky. A leaf, being roughly 1 mm thick, and consisting mostly of water, has heat capacity roughly 4 kJ/K/m². When it radiates 100 W/m² it will tend to cool by more than 1°C every minute. It is the heat capacity of the air around it that will stop the leaf from freezing.

The forest leaves will set up a convection cycle in which air is cooled by the upper leaves, which are the ones radiating into space, and warmed by the surface of the Earth below. The air immediately above the forest will be sucked into this cycle, and the air immediately above that also, so that the very cold air of the tree-tops mixes slowly with the air ten to a hundred meters up, creating a layer of air that is colder and more dense than the air above. The convection cycle transports roughly 100 W/m². Suppose that the mixing we describe causes 30 W/m² to be removed from the heat capacity of the first 100 m of air. In that case, the average temperature of the first 100 m will drop by 5°C in six hours.

So we see that radiation by leaves held ten or twenty meters above the ground can cause the first hundred meters of the atmosphere to cool by several degrees compared to the air above, and so create a static, cold layer of air above the forest. If there is any wind, of course, the leaves will no longer need their convection cycle to keep them warm: the wind will warm them, and blow away any cold air as well. The night must be still for such inversion to take place, and air is more often still in a bowl or valley.

Our Surface Cooling posts have introduced us to the heat capacity of the atmosphere, and shown us that traditional explanations for cooling and warming may be unreliable. But we have no proof that it is an exchange of warm air for cold air that causes the sudden drop in temperature at sunset in the desert. Nor do we have any proof that a cold convection cycle in a forest causes night inversion. Both explanations are merely hypotheses. Any help devising experiments to test either hypothesis would be much appreciated. Our ideas so far, which involve helium balloons and battery-powered thermometers, appear impractical, not to mention expensive.

Surface Cooling, Part III

2010-12-21T06:32:00.001-08:00

UPDATE: The original version of this post included an argument by Second Law of Thermodynamics that, although correct, did not point out the actual physical mechanism by which desert air cools at sunset. I cut out this argument and added a description of the physical mechanism as suggested by my father.

The air above the desert cools by 20°C at night. But the heat capacity of the air above the desert is so great that it is impossible for it to cool so quickly by radiation into space or by conduction to the ground. We ended our previous post with the suggestion that the drop in temperature over the desert at night is caused not by the cooling of air, but by the replacement of warm air by cold air.

In Biggles Flies South, the author tells us that flying over the desert during the day is exhausting and uncomfortable because the air is so "bumpy". The best time to fly is the early morning, when the air is still and there is daylight to see by. Our understanding of atmospheric convection suggests that the air above the desert will be a succession of convection cycles, something like the drawing below, where small features of the surface end up dictating the boundaries of the cycles, and the cycles are angled sideways by a prevailing wind.

All this convection is caused by the sun heating the desert sand until it warms the air and causes the air to rise. But when the sun goes down, the sand cools off quickly. The circulation will eventually come to a stop, but it's momentum will keep it going for a while. When it is near to stopping, air is descending slowly from above, and comes to a stop. The sand is no longer hot, and no longer warms the air.

When the sun goes down over the desert, the day-time convection of the atmosphere slows to a halt. The slowing cycle brings cold air down from above. This cold air is not warmed by the hot sand as it would be during the day. The temperature drops by ten or twenty degrees in an hour. For the rest of the night, the air is still and cools slowly by radiating into space. Perhaps the temperature will drop by another one or two degrees by dawn.

Surface Cooling, Part II

2010-12-15T06:20:00.000-08:00

In our Surface Cooling, Part I, we saw that the air above the ocean cools by less than 1°C at night. The air above the desert in Arizona, however, cools by 20°C at night. In Night and Day we saw that the heat radiated into space by the Earth is barely sufficient to cool the atmosphere by 1°C at night, let alone 20°C, so we are left wondering how the air above the desert can cool down so fast.

The table below presents the thermal properties of some common surface materials. We will make use of this table in future posts. Values for composite materials like wet soil are approximate. We picked values that are representative from resources like this, this, and this. Today we will use only the properties of dry sand. In the first column we have the thermal conductivity in Watt per meter per Kelvin (W/mK = W/m°C). A layer of material with thermal conductivity α, thickness d, temperature T_b at the bottom and T_t at the top, will conduct α(T_b−T_t)/d for each square meter of its surface area. A 10-cm thick layer of dry sand that is 20°C cooler on top than bottom will conduct 0.20 W/mK × 20 °C / 0.1 m = 40 W/m².

           Conductivity  Density   Capacity 
Material     (W/mK)     (kg/m^3)  (J/kgK)
Air           0.024        1.2       1000
Asphalt       0.75        1300        920
Copper         390        8900        420
Ice            2.2         910       2000
Mercury        8.7       14000        140
Sand, Dry     0.20        1600        830
Sand, Wet      4.0        2000       2500
Sandstone      1.7        2500        800
Soil, Dry      0.4        1300       1400
Soil, Wet      4.0        1500       2500
Snow, Light   0.10         100        420
Steel           43        7800        460
Water         0.58        1000       4200
Wood, Oak     0.17         770       1700

The second column gives the density in kilogram per cubic meter (kg/m³). A 10-cm layer of dry sand has mass 0.1 m × 1700 kg = 170 kg/m². The third column gives the specific heat capacity in Joule per kilogram per kelvin ( J/kgK = J/kg°C). A layer of material with specific heat capacity C, density ρ, and thickness d will require Cρd Joules of heat per square meter of surface area to raise its temperature by 1°C. A 10-cm thick layer of dry sand requires 800 J/kgK × 1700 kg × 0.1 m = 136 kJ/m² to warm up by 1°C. Alternatively, we must remove 136 kJ/m² to cool it by 1°C. If this layer loses loses 80 W/m², it will take half an hour to cool down by 1°C.

Our SC1 program simulates the cooling of the surface of the Earth at night. You can download it and run it yourself, by following the instructions in the code. The program works by dividing the surface material into thin layers and calculating the temperature of each layer in a sequence of small time steps. We assume the material is at a uniform temperature when it begins to cool, and that the heat loss from the upper surface is a constant 200 W/². In earlier versions of the code, we calculated the surface loss as a combination of radiation into space and convection with the surface air, but our results were pretty much the same, so we reverted to the assumption of constant loss to keep things simple.

The following graph shows what our simulation program tells us happens to dry sand at various depths in millimeters when it starts to cool from the surface at 200 W/m². We see the surface cools by 13°C in 1000s while the sand 40 mm down cools by roughly 0.1°C.

Materials with greater conductivity cool more slowly because heat rises from below to keep the surface warm. If we apply our simulation program to sandstone or asphalt, we find the surface cools down ten times more slowly than the surface of dry sand. So it is above dry sand or soil that we expect to see the greatest drop in surface temperature at night.

Our calculations show that a sandy surface can cool by 20°C at night. They show that one meter down, the heat of the day will hardly affect the temperature of the sand. But they still don't explain why the temperature above the desert sand cools by 20°C at night. The surface of the sand cools down because it loses heat to the air and to outer space. If anything, the cooling sand warms the air. We already know that the heat radiated by the air itself is not sufficient to cause such cooling. Indeed, our calculations show that it is impossible for air above the desert to cool by 20°C at night. And yet we know that a thermometer 2 m above the surface of the desert can register a drop of 20°C.

We conclude that the air above the sand at night cannot be the same air above the sand during the day, and indeed this conclusion leads us to a most satisfactory explanation for cold desert nights.

Surface Cooling, Part I

2010-12-08T07:38:00.000-08:00

In Day and Night we found that the heat capacity of the atmosphere keeps the Earth warm at night. There are ten tons of air above every square meter of the Earth, and this ten tons has heat capacity 10 MJ/°C. When the atmosphere radiates 200 W/m² into space, which is typical of a clear sky, its average temperature drops by less than 1°C in twelve hours.

The heat capacity of water, meanwhile, keeps a pond warm at night. In a 1-m deep pond, there is a ton of water beneath every square meter of surface, and this ton has heat capacity 4 MJ/°C. If the water surface radiates 80 W/m² into space, which is typical when the sky is clear, this radiation will cool the pond by less than 1°C in twelve hours. The water surface will, of course, be losing more heat by atmospheric convection and by evaporation. But in the ocean, where the top ten meters of water are well-mixed by waves, even a surface loss of 800 W/m² will cool the water by less than 1°C at night.

Here is a twenty-day recording of air and water temperature from an ocean buoy, which we obtained from the National Data Buoy Center. The data is here.

We see that the air temperature varies by 5°C from one week to the next, but by less than 1°C from day to night. The water itself, meanwhile, is hardly affected by the weekly variations, and not at all by the daily variations.

Why, then, does the temperature in Arizona drop by 20°C at night? The surface of Arizona is sand and rocks. These solid, opaque materials transport heat within themselves by conduction alone. During the day, their surfaces get hot. During the night, they cool down. In our next post, we will present the thermal properties of some common surface materials, along with a computer program to calculate how fast these surface materials will cool at night. We will find out whether the thermal properties of sand can explain why the desert is hot in the day and cold at night.

Day and Night

2010-12-01T10:02:00.001-08:00

In a previous post, we considered the dramatic change in the temperature of the moon as it revolves around the Earth. In the middle of its night, the moon cools to −190 °C. In the middle of its day, it warms to 100 °C. I was in Phoenix, Arizona last week. The sky was clear all day and all night, and the air was dry. In the middle of the night, the temperature dropped to 0°C. In the middle of the day, it rose to 20°C. The day-night variation on the moon is 290°C, but on the Earth it is only 20°C.

The moon's days are one month long. At night, the surface rocks have time to cool down, losing all the heat they gained during the day. They get so cold they draw heat out of the moon's interior, and this flow of heat from the depths of the moon is what stops the moon's surface from dropping below −190°C. During the day, the rocks heat up until they start to radiate almost as much heat as is arriving from the Sun. But some heat flows into the depths of the moon to make up for the heat drawn out during the previous night.

The moon has no atmosphere, so the only thing stopping it from cooling down to absolute zero at night is the heat capacity and thermal conductivity of its surface rocks. On Earth, we have the heat capacity of the atmosphere to keep the surface warm during the night. Above each square meter of the Earth's surface is 10,000 kg of air (see Atmospheric Weight). The heat capacity of dry air is is roughly 1 kJ/K/kg. So the air above each square meter of the Earth has heat capacity 10 MJ. The Earth and its atmosphere radiate something like 250 W/m² into space. Suppose this 250 W/m² were extracted uniformly from the entire column of air above each square meter of the surface. It would take 10 MJ ÷ 250 W = 10 hrs to cool the air by 1°C.

We see that the heat capacity of the atmosphere is significant in reducing the temperature changes caused by the setting of the sun or the formation of thick clouds. Indeed, the temperature changes we observe from day to night on Earth are more than ten times greater than our simple heat capacity calculation suggests. Perhaps if we consider the individual 3-km layers of the atmosphere, we will come up with a different answer.

As day turns to night, the Sun no longer warms the Earth, but the Earth and its atmosphere continue to radiate heat into space. The following table gives the escaping power in Watts per square meter from each 3-km layer of the Earth's atmosphere up to the tropopause, as calculated by our TEP2 program under conditions we describe here.

--------------------------------------------------
      Name    Temp       BB     Layer   Escaping  
--------------------------------------------------
   Surface    290.0    401.1    388.8     80.0 
      0-km    280.0    348.5    262.6     46.1
      3-km    270.0    301.3    184.8     39.7
      6-km    250.0    221.5    111.5     44.6
      9-km    230.0    158.7     50.7     26.4
     12-km    220.0    132.8     22.2     10.1
     15-km    220.0    132.8     14.4      5.5 
--------------------------------------------------
Total:                                   252.5
--------------------------------------------------

The surface of the Earth radiates 390 W/m². Of this, 80 W/m² escapes directly into space. The rest of it is absorbed by the atmosphere. But the atmosphere radiates heat towards the surface also. The 0-km layer (the first 3 km of the atmosphere) radiates 260 W/m² upwards and downwards. We may have thought only of the upward component in the past, but the downward component exists also, and is equal to the upward component.

Let us ignore the heat passing up through the atmosphere by radiation and convection, and assume that the Earth's surface starts to lose 80 W/m² when the sun sets. Suppose the surface itself is water 1 m deep. The heat capacity of water is roughly 4 kJ/K/kg and its density is 1,000 kg/m³. Beneath each square meter is fluid with heat capacity 4 MJ/K. This water will cool by only 1 °C in 10 hrs.

The 0-km layer radiates 46 W/m² directly into space. The mass of air in this first 3-km layer is a little less than 3,000 kg, with heat capacity roughly 3 MJ/K. At a loss of 46 W/m², this layer will cool down by only 0.5°C in ten hours.

The changes in temperature we observe at the surface of the Earth from day to night are much greater than we would expect from a consideration of the heat capacity of the atmosphere and total escaping power alone. What is going on when the sun sets, that makes the Earth's surface cool down so fast? We will hope to answer this question in our next post.

Adiabatic Magic

2010-11-23T07:38:00.000-08:00

A while back, a visitor here tried to persuade us that heat flow through an atmosphere was not required to make it warmer at the bottom than the top. He was convinced that adiabatic circulation would always take place within the atmosphere, regardless of whether or not heat was flowing, and this circulation would always make the bottom warmer than the top. At Science of Doom, meanwhile, several authors try to convince us that the surface of Venus will be super-hot even if it is not being heated by the Sun. They describe some kind of adiabatic process, which we will call adiabatic magic, by which any atmosphere will always be warmer at the bottom than the top, even with no heat flow. Indeed, one author goes so far as to say that heat flow from bottom to top will actually decrease the temperature difference between the lower and upper atmosphere.

We are well aware that air taking part in atmospheric convection will cool down as it rises, and warm up as it falls, but this circulation is powered by heat absorbed from the planet and subsequently radiated into space by the atmosphere. It is not spontaneous or self-generated. The circulation requires that the planet surface be warmed by the sun and that the atmosphere be capable of radiating heat into space at higher altitudes. With no heat flow, there will be no circulation, and no temperature difference between top and the bottom of the atmosphere. In other words: there is no such thing as adiabatic magic, as we will now demonstrate.

The diagram below shows how we can use adiabatic magic to create a machine that is 100% efficient at turning heat into work. Such a machine is also known as a perpetual motion machine of the second kind.

In the center of the diagram is an air column. The action of gravity is indicated by the double-headed arrow with the letter g. The column is high enough that the weight of the air it contains causes the pressure to be greater at the bottom than the top. By adiabatic magic, the bottom of the column is hotter than the top, so T₁ > T₂. We exploit this temperature difference with a heat engine. Metal veins absorb heat from the air at base of the column. A metal rod, well-clad with thermal insulation, carries this heat to a heat engine. A heat engine is something like a steam turbine, which exploits the flow of heat from hot to cold in order to generate work. Heat flows out of the engine through another metal rod to top of the column. The heat flowing into the engine is q₁ and the heat flowing out is q₂. The work produced by the heat engine is W. We express the heat flow and work output in Watts.

The First Law of Thermodynamics requires that W = q₁ − q₂. The Second Law of Thermodynamics requires that the entropy of the heat going in must be less than or equal to the entropy of the heat going out. With the absolute temperature scale, this means that q₁ / T₁ ≤ q₂ / T₂. Subject to these constraints, we see that our heat engine can produce work by exploiting the temperature difference between the bottom and the top of the air column.

Of course, we are taking more heat out of the bottom of the column than we are restoring to the top, so the total heat of the column will tend to decrease. We're not sure what the adiabatic magic theory has to say about an air column as it cools, so we are going to avoid this issue by adding heat to the column from a large reservoir. We add exactly as much heat as the column is losing by the action of the heat engine. The heat we add is Q, so Q = q₁ − q₂ = W.

So now let us consider the air column and the heat engine acting together. They constitute a perpetual motion machine. We see that we have Q entering the machine and W leaving. The machine turns heat into work with 100% efficiency. It is a perpetual motion machine of the second kind.

The adiabatic magic idea is one of a class of impossible ideas: those that propose the spontaneous separation of heat into hot and cold. The above diagram could be adapted to disprove any such idea with equal facility. If someone claimed they had invented a new material that always became hotter at one end than the other, even with no heat flow from one end to the other, we could disprove his claim with the above diagram.

Because adiabatic magic is impossible, we see that a transparent atmosphere can be no warmer at the bottom than at the top. Transparent gas cannot radiate heat into space (see Radiative Symmetry). In the case of Venus, if nothing heats the surface, the gas at the surface will be no warmer than the gas at the altitude to which sunlight does penetrate.

Of course, our atmosphere does radiate heat into space, so the lower air is warmer than the upper air, in order to transport that heat. And Venus's atmosphere gets hotter all the way to the surface, which means it is transporting heat away from the surface and radiating it into space.

Condensation and Convection

2010-11-16T17:15:00.000-08:00

One of our readers has been asking me about the role of condensation in accelerating convection, as described in papers like this one. So let us dicsuss the effect of evaporation and condensation in convection.

Consider a kilogram of air that has just been warmed at the surface of the Earth, and is now rising towards the tropopause as part of atmospheric convection. This kilogram has temperature 300 K, pressure 100 kPa, and volume 860 l (liters). Let us suppose this air contains 10 l of water vapor. This water vapor has mass 10 g and would occupy 10 ml (milliliters) if it condensed into liquid.

Our rising volume of air expands and cools. Its pressure drops to 50 kPa and its temperature to 250 K. It expands to 1400 l. At this temperature, almost all of our water must condense into liquid. Before condensation, the water occupies 14 l. When it condenses it will turn into microscopic water droplets that occupy only 14 ml. Of course, these droplets will freeze afterwards, but let's ignore that for now, and consider what happens when the water first condenses.

Most likely, the condensation will occur gradually as the air expands, but we will imagine that it happens after the expansion is complete, so we can calculate the net effect of condensation upon the pressure of our gas. At the end of the expansion, our air contains supersaturated water vapor, like the gas in a cloud chamber.

Consider the first gram of water that condenses. Let us assume, for the time being, that this condensation happens instantly, so that our two cubic meters of air has no time to expand or contract during the process. That is to say: we assume the condensation takes place at constant volume.

Before condensation, this one gram occupied 1.4 l in our volume of 1400 l. After condensation, the water droplets occupy a combined volume of only 1 ml. The remaining air and water vapor expand into our 1400 l. As it expands, its pressure decreases by 0.1%.

When water condenses, it gives up its latent heat of evaporation, which is 2 MJ/kg. When the first gram of water condenses, it gives up 2 kJ of heat to our 1400 l of air. The heat capacity of air at constant volume is roughly 700 kJ/kgK, so our kilogram of air warms up by around 3 K. It's pressure increases by 3 K ÷ 250 K = 1.2%.

We see that the first gram of water condensation causes a net increase in pressure of 1.2% - 0.1% = 1.1% in our kilogram of air. Suppose the air all around is dry. It has not been warmed by condensation. It's pressure has not increased. Our two cubic meters pushes outwards with its greater pressure and expands until it's pressure is the same as that of the surrounding air. It will expand by 1.1%. Its 1 kg of mass will occupy the same volume as 1.011 kg of surrounding dry air. It will experience an upward buoyancy force of 11 g weight, or 0.11 N. This force will cause it to accelerate upwards at 0.11 m/s². Within less than a minute, it will be traveling upwards at a few meters per second. At that point, other forces will come into play to slow it down. We note that we allowed only 1 g of our water vapor to condense. If all 10 g condensed at once, our kilogram of air would experience ten times more lift.

We see that condensation does indeed cause air to rise more quickly than surrounding dry air.

Moist air will expand more than dry air as it rises towards the tropopause. Evaporation, on the other hand, causes air to rise more slowly. Air at the surface of a dry desert heats up fast in contact with the hot sand. It rises quickly. We get dust devils and sand storms. But air on the surface of the ocean does not heat up quickly. Water evaporates from the ocean, cooling the ocean surface and the air, so that the air lingers above the ocean, accumulating more and more water vapor. Eventually, the air will rise, if only because it becomes saturated with water vapor. But we see that water rising from the ocean is not as hot compared to the air above it as is air rising from the desert. The initial rise will be slow compared to the rise over a desert. But if the water starts to condense out of the moist air, and the moist air is surrounded by dry air, the moist air will begin to rise more quickly, sucking more moist air up beneath it.

Thus we expect low-altitude storms over a desert and high-altitude storms over an ocean.

PS. Another writer concludes that the upward force on a cubic yard of air due to condensation will be close to a thousand pounds. He says, "I’m not intending to do the calculations any further here, because it is basic knowledge." Had he performed the calculations, he would have discovered that the effect of water vapor contraction is opposite to his claims (the contraction tends to make the air more dense and therefore sink, while he claimed that the contraction decreases pressure and therefore creates lift) and that his forces were off by four orders of magnitude (the net lift is of order 0.1 pounds per cubic yard, not 700 pounds per cubic yard).

PPS: The debate on this subject continues over at Watt's Up With That?.

PPS. Look in the comments for an explanation of evaporation and condensation on a molecular level.

Venus

2010-11-08T07:00:00.000-08:00

UPDATE: Initial version of this post used diatomic gas equation for the compression of CO2. Have now corrected this oversight, and find that our estimate of Venus's surface temperature is much improved.

One of our readers suggested we consider the atmosphere of Venus. What a good idea. Let's see how well we can estimate Venus's surface temperature using our understanding of atmospheric convection and the greenhouse effect.

According to Wikipedia, Venus's tropopause is at an altitude of 65 km with temperature 243 K (−30°C) and pressure 10 kPa. The surface pressure, meanwhile, is almost a thousand times greater: 9,300 kPa, which is ninety-three times the surface pressure on Earth. Venus's atmosphere is made up almost entirely of CO2, but also contains 150 ppm of SO2 (sulfur dioxide), and this SO2 condenses into liquid droplets so that the atmosphere below the tropopause is filled with pale yellow clouds.

Venus reflects 90% of incident sunlight (it's Bond Albedo is 0.90). The remaining 10% is absorbed. We're not sure what fraction of the Sun's light reaches the surface of Venus, but our guess is 1%. The SO2 clouds refract and reflect light like our own clouds, but SO2 is a pale yellow liquid, not a clear liquid, and will absorb sunlight eventually.

According to Schriver et al., even a 0.5-μm film of SO2 ice will absorb over 20% of long-wave radiation, so a 10-μm droplet of SO2 liquid will absorb it all. The clouds of Venus are near-perfect absorbers of long-wave radiation, and near-perfect radiators too, just like our own water clouds. Unlike the Earth, however, Venus is always entirely covered with clouds. Neither the planet surface nor the lower atmosphere has any opportunity to radiate heat directly into space. Venus radiates heat directly into space only from its cloud-tops and upper atmosphere.

Although most of the sun's heat is reflected back into space, 10% is absorbed, and we estimate that around 1% reaches the surface of the planet. This heat will raise the temperature of the surface until it forces convection. Here is the convection diagram from our Atmospheric Convection post.

Gas warms at the surface. It rises, expands, radiates heat into space, falls, contracts, and warms again. As the gas expands, it gets cooler. As it contracts, it gets warmer. In our simplistic analysis, we assumed that the expansion and contraction were adiabatic, meaning they took place without any heat entering or leaving the gas. In reality, the gas radiates heat to nearby gas, absorbs heat from nearby gas, mixes with nearby gas, and generates heat through viscous friction. But our adiabatic assumption allowed us to estimate the temperature changes using the equation of adiabatic expansion and contraction. For an ideal diatomic gas, such as N2 or O2, p^−0.4T^1.4 remains constant during adiabatic changes, where p is pressure and T is temperature. For CO2, however, the equation is p^−0.3T^1.3 at 300 K and p^−0.2T^1.2 at 1300 K (see here for thermodynamic properties of CO2).

The Earth's tropopause is at altitude 15 km. According to our typical conditions, the tropopause is at 220 K with pressure 22 kPa, while the surface pressure is 100 kPa. Adiabatic compression implies that air descending from the tropopause to the surface will heat up to 340 K. But the Earth's surface is at only 280 K. Air descending 10 km from the tropopause to the surface appears to lose 20% of its heat while contracting to a third of its original volume.

The atmosphere of Venus will heat up as it descends from the tropopause to the surface, and it will be hottest if it does not lose any heat on the way down. The pressure rises from 10 kPa in the tropopause to 9,300 kPa at the surface. The temperature starts at 240 K. Adiabatic compression of an ideal diatomic gas would heat the gas to 1700 K. If we use p^−0.25T^1.25 as an approximation for CO2, we estimate a final temperature of 941 K.

According to Wikipedia, the surface temperature of Venus is actually 740 K. Carbon dioxide falling 65 km from the tropopause to the surface appears to lose 20% of its heat while being compressed into less than a hundredth of its original volume.

Despite the 20% difference between our calculations and our observations, we see that atmospheric convection makes the surface of Venus incredibly hot while leaving the surface of the Earth delightfully temperate.

If we recall our post on work by convection and another on atmospheric dissipation, we see that a strong convection cycle causes powerful weather. Venus's convection cycle produces an order of magnitude more contraction and expansion than the Earth's. We expect Venus's weather to be an order of magnitude more extreme. And indeed it is.

Self-Regulation by Clouds

2010-10-31T09:00:00.000-07:00

On a typical spring day at latitude 30°N, our TEP2 program tells us that the formation of high clouds will tend to warm the Earth's surface by 38°C, while the formation of thick clouds will tend to cool the Earth's surface by 96°C.

Over the past few weeks here in Boston, the warmest days have been those with high, thin clouds, and the coolest have been those with low, thick clouds. But the difference between the warm days and the cold days is only 10°C. Despite alternating clouds, rain, and sunshine, the temperature at a particular location anywhere in the world is almost always within 10°C of its monthly average at mid-day, and these monthly averages vary by less than 1°C from one year to the next.

How can it be that the Earth's surface temperature remains so stable when clouds have such tremendous power to heat and cool the surface? Some kind of self-regulation of cloud formation and evaporation must be taking place. Indeed, at first glance it seems to us that warmth will promote evaporation from the oceans and lead to the formation of the low, thick clouds that cool the Earth, while cold will suppress evaporation, encourage atmospheric convection, and lead to the formation of the high, thin clouds that warm the Earth. So each type of cloud tends to promote the formation of the other, which is a basis for self-regulation.

An example of self-regulation is the action of a thermostat in a house's heating system. When the heater turns on, the house warms up. If the heater stays on, the house will get too hot. The thermostat turns off the heater when the house is warm enough. The house cools down. If the heater stays off, the house will get too cold. The thermostat turns on the heater when the house has cooled enough. The difference between the turn-on and turn-off temperatures is the hysteresis. The temperature half-way between is the set-point.

The heating system in our house can heat the inside to 30°C when the outside is at 0°C. With the heater off, the house will eventually cool to 0°C. But our house is always close to 20°C inside. That's because the set-point of our thermostat is 20°C and its hysteresis is only 1°C.

High clouds at latitude 30°N in spring would heat the Earth to 55°C if the day were long enough and no change to the high clouds took place. Thick clouds would cool the Earth to −79°C. But the mid-day temperature remains within ±5°C of 17°C. The set-point of cloud self-regulation is roughly 17°C and its hysteresis is 10°C. The self-regulation reduces a maximum possible variation of 134°C to an observed variation of 10°C. That's a factor of thirteen reduction through self-regulation.

Our TEP2 program tells us that if we double the CO2 concentration in the atmosphere, the Earth's surface will tend to warm up by 1.5°C. Given the strong self-regulation of warming and cooling by clouds, how much warming will we actually observe if we double the CO2 concentration? It may be that the self-regulation of clouds is entirely independent of the heat retained by extra CO2, so that doubling the CO2 will increase the average temperature of the Earth's surface by a full 1.5°C.

But cloud self-regulation is not independent of the heat absorbed by extra CO2. According to our calculations, doubling the CO2 concentration of the atmosphere causes the atmosphere to retain an additional 5.0 W/m² of heat. According to our typical atmospheric conditions, air close to the surface contains 1% water vapor, which amounts to roughly 10 g/m³. When this water condenses into water droplets, thus forming a cloud, the water vapor gives up 20 kJ/m³ of latent heat. If this condensation takes place over 1000 s (about fifteen minutes) in a column 1000 m high, it generates 20 kW/m², which is four thousand times the heat retained by our extra CO2. The heat retained by extra CO2 and the heat produced by condensation are exactly the same type of heat. They are indistinguishable. We think it likely, therefore, that cloud self-regulation will reduce the effect of extra CO2 in the same way that it reduces the effect of cloud variation: by a factor of thirteen. In that case, doubling CO2 concentration will warm up the world by only 0.1°C.

In the presence self-regulation, however, we cannot assume even that CO2 doubling will have a net warming effect. Our calculations show that the extra CO2 causes the heat to be retained mostly between altitudes 3 km and 6 km. In an earlier post we showed how turning on an electric heater in the same room as your home's thermostat will cause the the rest of the house to cool down. It could be that the heat retained by CO2, by virtue of being concentrated in the middle-troposphere, affects cloud self-regulation in such a way as to cause a slight cooling of the surface of the Earth below. Such an outcome seems unlikely, but it is possible.

Another possibility suggested by documents like this one is that the 1.5°C warming from CO2 doubling will in fact be amplified by the self-regulation of clouds, so that we see a 3°C or even 4°C warming due to CO2 doubling. We can think of no examples of such behavior by self-regulating systems, nor do we understand the arguments presented by climatologists who propose such amplification.

In future posts we hope to explore the sources of self-regulation by clouds, and perhaps construct a simple model to investigate how the average temperature of the self-regulated system will be affected by such things as CO2 concentration and the availability of water. Before that, however, we will test our climate model by applying it to day-night variations, and to the planet Venus.

Thick Clouds

2010-10-26T06:03:00.000-07:00

We have been using our computer program, TEP2 to estimate how much the Earth's surface will tend to warm up if we make certain permanent changes to its atmosphere. In With 660 ppm CO2 we found that doubling the atmosphere's CO2 concentration from 330 ppm to 660 ppm tends to warm the Earth's surface by 1.5°C. In High Clouds we found that the formation of high, thin clouds in a clear sky will tend to warm the Earth's surface by 38°C.

Today we consider the effect of thick clouds. We start with the Earth on a typical clear spring day at latitude 30°N, as represented by our EA1.txt absorption spectra. Our TEP2 program tells us that the total escaping power is 252.5 W/m². We allow a thick cloud to form from altitude 3 km to 12 km. In Clouds we argued that a cloud this thick will reflect 90% of the Sun's radiation and absorb 100% of the Earth's radiation. We simulate such a cloud by setting the absorption of our 3-km, 6-km, and 9-km layers to 1.0 for wavelengths 5 μm to 60 μm to produce EA4. We apply TEP2 to EA4 and obtain the following output.

------------------------------------------------
      Name    Temp       BB     Layer   Escaping
------------------------------------------------
   Surface    290.0    401.1    388.8      0.0
      0-km    280.0    348.5    262.6      0.0
      3-km    270.0    301.3    292.0      0.0
      6-km    250.0    221.5    214.0      0.0
      9-km    230.0    158.7    152.3    120.8
     12-km    220.0    132.8     22.2     10.1
     15-km    220.0    132.8     14.4      5.5
------------------------------------------------
Total:                                   136.5
------------------------------------------------

This output is exactly the same as the one we obtained for high clouds. Both the high clouds and the thick clouds have the same effect upon the total escaping heat. The difference between high clouds and thick clouds is that thick clouds block 90% of the Sun's heat from reaching the Earth, while high clouds block only 10%.

As we did in High Clouds, let us suppose that the Earth radiates as much heat as it absorbs when its surface and atmosphere are as described by EA1. The total power escaping the earth is 252.5 W/m², so the total power arriving from the Sun must be 252.5 W/m² also. In the conditions described by EA4, 90% of the heat arriving from the Sun is reflected back into space, so only 25.2 W/m² is absorbed by the Earth. But the total power escaping the Earth is 136.5 W/m². When these large clouds form, the Earth will start to radiate 111 W/m² more than it absorbs. It will cool down. In our TEP2 program, we decrease the temperature of the Earth and its atmosphere until we arrive at the following output.

------------------------------------------------
      Name    Temp       BB     Layer   Escaping
------------------------------------------------
   Surface    194.3     80.8     76.0      0.0
      0-km    187.6     70.2     59.0      0.0
      3-km    180.9     60.7     56.4      0.0
      6-km    167.5     44.6     40.8      0.0
      9-km    154.1     32.0     28.7     22.3
     12-km    147.4     26.8      4.5      2.9
     15-km    147.4     26.8      2.1      1.1
------------------------------------------------
Total:                                    26.3
------------------------------------------------

Here we see the total escaping heat is 26.3 W/m², which is pretty close to the 25.2 W/m² arriving from the Sun through the thick clouds. The Earth's surface has cooled by 96°C.

According to our simulation, the formation of a 9-km thick layer of cloud at altitude 3 km has a cooling effect sixty-four times stronger than the warming effect of doubling the atmosphere's CO2 concentration.

UPDATE: When we look at discussions like this one about clouds, we see that climatologists recognize the power of clouds to cool and warm the Earth. But we don't see them giving us values in Centigrade in the same way that they do for a hypothetical doubling of CO2 concentration.

High Clouds

2010-10-18T08:33:00.000-07:00

Sitting on a plane over the Atlantic Ocean last month, I was at 13,000 m looking down on a continuous layer of thin, high cloud. Through occasional openings, the ocean below appeared brightly-lit despite the clouds above it. Today we estimate the effect of such a cloud layer upon total escaping power using our TEP2 program.

Let us place a 1-km thick layer of cloud in our 9-km atmospheric layer. This layer extends from altitude 9 km to 12 km. In our previous post we showed that a 1-km thick layer of cloud absorbs 100% of incident long-wave radiation, but only 10% of incident short-wave radiation. In our atmospheric absorption spectra we simulate such a cloud by setting the absorption of our 9-km layer to 1.0 for wavelengths 5 μm to 60 μm. We start with our EA1 spectra (typical, clear, spring day at 30°N) and produce EA3. We submit EA3 to TEP2 and obtain the following table of escaping power in W/m². We add the final column by hand, which gives the change in escaping power caused by the introduction of the cloud.

-------------------------------------------------------
      Name    Temp       BB     Layer   Escaping  Change
-------------------------------------------------------
   Surface    290.0    401.1    388.8      0.0   -80.0
      0-km    280.0    348.5    262.6      0.0   -46.1
      3-km    270.0    301.3    184.8      0.0   -39.7
      6-km    250.0    221.5    111.5      0.0   -44.6
      9-km    230.0    158.7    152.3    120.8    94.4
     12-km    220.0    132.8     22.2     10.1    0.0
     15-km    220.0    132.8     14.4      5.5    0.0
-------------------------------------------------------
Total:                                   136.5 -116.0
-------------------------------------------------------

No radiation escapes into space from the Earth's surface or from the first three layers of the atmosphere. A total of 210.4 W/m² is absorbed by the lower surface of the cloud layer. The cloud layer itself, however, is a fine radiator of heat, and radiates 120.8 W/m² into space.

We note that the black-body radiation for our 9-km layer is 158.7 W/m² (BB column) while the heat that TEP2 calculates by numerical integration is only 152.3 W/m² (Layer column). Our spectra extend to 60 μm, but the cool cloud (−43°C) radiates some power in the range 60-100 μm. We will accept this 4% error and move on.

The cloud causes the total escaping power to drop by 116 W/m², or 46%. We now ask how much warmer the Earth's surface will have to get in order to make up a 46% drop in total escaping power caused by the formation of a 1-km layer of cloud at altitude 10 km. The 1-km cloud absorbs some of the heat arriving from the Sun, and we must take this into account somehow when we perform our estimate.

Let us suppose that the Earth radiates as much heat as it absorbs when its surface and atmosphere are as described by EA1. Because the Earth and its atmosphere radiate a total of of 252.5 W/m², we suppose that a like amount of heat arrives from the Sun and lands upon the Earth's surface. When we place a 1-km layer of cloud in the way, we estimate that 10% of the Sun's heat will be reflected back out into space. When the cloud layer forms, the heat arriving from the Sun decreases to 227 W/m² and the heat escaping from the Earth decreases to 136 W/m².

By experimenting with TEP2, we find that a 13% increase in the absolute temperature of the Earth and its atmospheric layers will produce the following result.

------------------------------------------------
      Name    Temp       BB     Layer   Escaping
------------------------------------------------
   Surface    327.7    653.9    629.9      0.0
      0-km    316.4    568.3    413.4      0.0
      3-km    305.1    491.3    284.7      0.0
      6-km    282.5    361.1    169.3      0.0
      9-km    259.9    258.7    250.4    201.4
     12-km    248.6    216.6     34.6     14.8
     15-km    248.6    216.6     23.5      8.7
------------------------------------------------
Total:                                   224.9
------------------------------------------------

The total escaping power is now 225 W/m², which is close enough to the 227 W/m² arriving from the Sun. When a cloud forms at altitude 10 km, the heat escaping from the Earth will balance that arriving from the Sun only if the Earth's surface warms by 38°C. If, instead of a cloud forming at 10 km, we double the CO2 concentration from 330 ppm to 660 ppm, the same calculation suggests the Earth will warm by only 1.5°C.

According to our simulation, the formation of a 1-km thick layer of cloud at altitude 10 km has twenty-five times the warming effect as doubling the atmosphere's CO2 concentration.

UPDATE: Michele points out that high clouds will contain microscopic ice particles instead of water droplets. Ice, like water, is also opaque to long-wave radiation and transparent to short-wave radiation.

Clouds

2010-10-14T06:10:00.001-07:00

Clouds contain microscopic droplets of water. The droplets are around 10 μm in diameter and there are hundreds of them in every cubic centimeter (see here). We discussed the absorption of long-wave radiation by water in an earlier post. We presented a graph of water's absorption length versus radiation wavelength. The graph shows that the absorption length for 5 μm to 60 μm radiation is less than 20 μm. A 100-μm layer of water will absorb 99% of long-wave radiation.

Let us consider the fate of a single long-wave photon as it enters a cloud. The photon may pass all the way through the first centimeter of the cloud without touching a single droplet, or it might strike a droplet. If we assume our droplets have cross-section 100 μm², the chance of our photon striking any particular droplet while passing through the first cubic centimeter is one in a million. The chance of the photon striking any one of a hundred droplets in the same cubic centimeter is a hundred times greater, or one in ten thousand. In every 100 m of cloud, the expected number of droplets our photon will strike is one (that's one in ten thousand per centimeter times ten thousand centimeters).

When the photon strikes a droplet, it can reflect off the surface or it can enter the liquid. According to this graph from Querry et al., the reflectance of water to long-wave radiation is 2% when the photon approaches at 10° to the perpendicular, 10% at 60°, 20% at 70° and 80% at 87°. The chance of a long-wave photon reflecting from a random spot on the surface of a droplet is around 10%. We also note that most reflection takes place when the photon strikes the droplet at a small angle to the surface, so the photon will continue into the cloud after reflection.

If a photon enters a 10-μm droplet, it will pass through something like 10 μm of water and strike the opposite, inner surface. Here it might reflect back into the droplet. But let's keep things simple: if the photon strikes a droplet, it will most likely enter the droplet, pass through 10 μm of water and keep going.

With 100 droplets in each cubic centimeter, our photon must pass through one droplet on average for each hundred meters of cloud. In 1 km of cloud, our photon will pass through ten droplets. In order to pass through the cloud, it will have to pass through roughly 100 μm of water. But we know that 100 μm of water absorbs 99% of all long-wave radiation. So a 1-km cloud is opaque to long-wave radiation.

Clouds have a different effect upon short-wave radiation, such as sunlight. Short-wave radiation reflects off the air-water surface far more readily than long-wave radiation, and once inside, it is more likely to reflect off the internal surface as well (internal reflection within droplets is what causes rainbows). But on the other hand, the absorption length of visible light in water is tens of meters. The water in a thick cloud will not absorb short-wave radiation. Instead, short-wave radiation changes direction as it encounters water droplets. It can change direction by reflecting off the outer or inner surface of a droplet. It can change direction by refraction as it enters or leaves a droplet. The reason thick clouds are dark is because sunlight is being turned around and re-directed back into space by the cloud's water droplets, not because sunlight is being absorbed by the clouds.

A short-wave photon passing through 1 km of cloud will encounter only ten droplets. These ten droplets are enough to absorb long-wave radiation, but they are not enough to guarantee that the short-wave photon gets re-directed back out of the cloud. With the help of some pencil drawings, we estimate that it would require encounters with one hundred droplets to guarantee that the photon came back out of the cloud on the same side that it went in. A 1-km thick cloud might reflect 10% of the short-wave radiation incident upon it, while a 10-km thick cloud will reflect 90%.

A 1-km layer of cloud acts like a black-body absorber of long-wave radiation. By radiative symmetry, this same 1-km layer must also be a perfect black-body radiator of long-wave radiation. We estimate that this same 1-km of cloud will allow 90% of short-wave radiation to pass through, while reflecting 10%. A 10-km layer of cloud, meanwhile, will allow only 10% of short-wave radiation to pass through, while reflecting 90%.

Heat radiated by the Earth will be absorbed by clouds. These same clouds will radiate their own heat out into space. Sunlight will be reflected off their top surfaces. In our next post, we will consider the effect of a 1-km layer of cloud at altitude 10 km upon the Earth's total escaping power and upon the amount of heat arriving from the Sun.

UPDATE: Cirrus clouds appear to be examples of our 1-km cloud layer. According to NASA, cirrus clouds are opaque to long-wave radiation, but largely transparent to short-wave radiation.

UPDATE: We see here confirmation of our estimates: clouds reflect from 10% to over 90% of short-wave radiation (albedo is 0.1 to over 0.9).

Absorption By All Higher Layers

2010-10-07T18:33:00.000-07:00

Our Total Escaping Power program calculates the total power radiated by the Earth and its atmosphere given the absorption spectra of a sequence of atmospheric layers. The program does not dictate the thickness of these layers, nor even do the layers have to have the same thickness, but we did assume that radiation passing through one layer will be certain to pass through all higher layers. If we want to experiment with greater water vapor content in an upper layer, or if we want to include a layer of clouds, we will need a program that does not rely upon this assumption.

We modified our TEP1 program to produce TEP2. The new program applies the absorption of all higher layers to upward radiation. The result is a decrease in total escaping power. Here is the output of the new program when we apply it to our EA1 spectra, which assume 330 ppm CO2. We have added by hand a final column giving the change in escaping power caused by applying the absorption of all upper layers.

---------------------------------------------------------
      Name    Temp       BB     Layer   Escaping  Change
---------------------------------------------------------
   Surface    290.0    401.1    388.8     80.0    -9.8
      0-km    280.0    348.5    262.6     46.1   -12.8
      3-km    270.0    301.3    184.8     39.7   -10.0
      6-km    250.0    221.5    111.5     44.6    -6.8
      9-km    230.0    158.7     50.7     26.4    -3.0
     12-km    220.0    132.8     22.2     10.1    -1.1
     15-km    220.0    132.8     14.4      5.5    -0.0
---------------------------------------------------------
Total:                                   252.5   -43.5
---------------------------------------------------------

In our discussion of TEP1, we claimed that our assumption of upper transparency would cause less than a 10% error in our calculation of total escaping power. It turns out that including the absorption of upper layers causes a 15% drop. Let's apply TEP1 to our EA2 spectra, which assume 660 ppm CO2.

------------------------------------------------
      Name    Temp       BB     Layer   Escaping
------------------------------------------------
   Surface    290.0    401.1    388.8     78.8
      0-km    280.0    348.5    264.0     44.2
      3-km    270.0    301.3    188.6     38.2
      6-km    250.0    221.5    115.2     43.9
      9-km    230.0    158.7     53.4     26.4
     12-km    220.0    132.8     24.2     10.3
     15-km    220.0    132.8     16.3      5.7
------------------------------------------------
Total:                                   247.5
------------------------------------------------

We see a 5 W/m² decrease in total escaping power due to doubling the CO2 concentration from 330 ppm to 660 ppm. The drop is 2.0%, slightly less than the 2.2% calculated by TEP1. If we repeat our calculation of CO2 doubling temperature using TEP2, we arrive at a value of 1.5°C, slightly less than the 1.6°C calculated by TEP1.

The enhancement of our model makes only a slight change in our CO2 doubling temperature, but it allows us to include upper layers that are more absorbent than lower layers. We can look forward to including layers of clouds and layers of excess humidity, and seeing how they effect the total escaping power.

With 660 ppm CO2

2010-10-06T08:06:00.000-07:00

In our In our previous post we extended our atmospheric absorption spectra to cover wavelengths 5 μm to 60 μm. These spectra assume CO2 concentration 330 ppm, which is typical for the Earth today, and temperature and water vapor concentration typical for a 30°N on a clear spring day. We plotted the extended spectra and applied them as input to our total escaping power program. The program says that the total power radiated by the Earth's surface and its atmosphere below the tropopause is 296.0 W.

Today we make one change to our atmospheric conditions: we double the CO2 concentration to 660 ppm. We leave water vapor concentration, temperature, and pressure the same. With the help of the Spectral Calculator, we obtain new absorption spectra and plot them.

We feed our new spectra into our total escaping power program, and obtain the following output. The final column we added by hand. It gives us the change in total escaping power when we raise CO2 concentration from 330 ppm to 660 ppm.

-------------------------------------------------------
      Name    Temp       BB     Layer   Escaping Change
-------------------------------------------------------
   Surface    290.0    401.1    388.8     88.2   -1.6
      0-km    280.0    348.5    264.0     55.6   -3.3
      3-km    270.0    301.3    188.6     47.8   -1.9
      6-km    250.0    221.5    115.2     51.0   -0.4
      9-km    230.0    158.7     53.4     29.7   +0.3
     12-km    220.0    132.8     24.2     11.5   +0.3
     15-km    220.0    132.8     16.3      5.7   +0.2
-------------------------------------------------------
Total:                                   289.4   -6.6
-------------------------------------------------------

The total escaping power drops by 6.6 W/m². The power escaping from the Earth's surface and the first three 3-km layers of the atmosphere decreases by 7.2 W/m², but the power radiated by the top three layers increases by 0.8 W/m². In the lower layers, more CO2 causes more radiation to be absorbed on its way into space. In the upper layers, more CO2 allows the atmosphere to radiate its heat more efficiently into space. We see that the increased absorption far outweighs the increased radiation, and the result is a 2.2% reduction in the heat escaping the Earth.

According to Stephan's Law, the amount of power radiated by a black body increases as the fourth power of its temperature. For small changes in temperature, the amount of power radiated by any body will increase as the fourth power of its temperature. A 1% increase in temperature will cause a 4% increase in radiated power. In order to increase total escaping power by 2.2%, we can increase the temperature of the Earth's surface and all the atmospheric layers by one quarter of 2.2%, or 0.55%. We increased all our layer temperatures by 0.55% and applied our total escaping power program to the resulting atmospheric layers. We obtain the following output. Note that the new temperatures are recorded in the first column.

------------------------------------------------
      Name    Temp       BB     Layer   Escaping
------------------------------------------------
   Surface    291.6    410.0    397.5     90.6
      0-km    281.5    356.1    269.3     57.0
      3-km    271.5    308.1    192.3     48.8
      6-km    251.4    226.5    117.4     52.0
      9-km    231.3    162.3     54.4     30.2
     12-km    221.2    135.8     24.7     11.7
     15-km    221.2    135.8     16.7      5.9
------------------------------------------------
Total:                                   296.1
------------------------------------------------

We see that there is no net change in total radiated power when we combine a doubling of CO2 concentration with a 1.6°C warming of the Earth's surface. That's not to say that the Earth will actually warm up by 1.6°C if we double the CO2 concentration, because the climate might be a chaotic system that defies such simple logic, or a system with feedback that acts in quite the opposite manner to our expectations. Nevertheless, in the absence of such complex behavior, the doubling the CO2 concentration will tend to warm the Earth by 1.6°C. This 1.6°C is our estimate of the Earth's CO2 doubling temperature.

Our analysis considered only the heat radiated by the troposphere, which is the region of the atmosphere below the tropopause. We ignored the effect of increased CO2 upon the stratosphere, which is the region above the tropopause. Even though there is no net heat flow between the troposphere and the stratosphere, an overall warming or cooling of the stratosphere will affect the temperature of the troposphere, and therefore change its total radiated power. Furthermore, we ignored the effect of increased CO2 upon radiation arriving from the Sun. We will consider these effects in future posts. In our next post, however, we will try doubling the water vapor concentration in various atmospheric layers and re-calculating total escaping power. By this means, we will be able to determine whether the effect of CO2 upon total escaping power is significant compared that of water vapor.

UPDATE: See slightly improved calculation of CO2 warming in next post, where we arrive at 1.5°C instead of 1.6°C as the size of the effect.

Extended Atmospheric Absorption Spectra

2010-09-29T07:22:00.001-07:00

In our previous post we concluded that we needed to extend our existing absorption spectra to cover the range 5 μm to 60 μm. So we returned to the Spectral Calculator and, entered our gas properties for each atmospheric layer, and obtained the absorption spectra from 30 μm to 60 μm. We added these to our existing spectra and obtained the following graph (click for a larger version).

As a point of detail, each spectrum we receive from the Spectral Calculator contain tens of thousands of values. We condense the spectra using a program that calculates the average absorption in each 0.1-μm window. The result is a spectrum with a point every tenth of a micron. You will find the original spectra in our data archive, in the EA_1 folder. You will find the condensed spectra here. The condensed file can act as input to our climate models and is convenient for generating plots like the one above.

We apply our total escaping power program to our extended spectra and obtain the following values for power per square meter. For an explanation of the various columns, see here.

------------------------------------------------
      Name    Temp       BB     Layer   Escaping
------------------------------------------------
   Surface    290.0    401.1    388.8     89.8
      0-km    280.0    348.5    262.6     58.9
      3-km    270.0    301.3    184.8     49.7
      6-km    250.0    221.5    111.5     51.4
      9-km    230.0    158.7     50.7     29.4
     12-km    220.0    132.8     22.2     11.2
     15-km    220.0    132.8     14.4      5.5
------------------------------------------------
Total:                                   296.0
------------------------------------------------

Let's look at the first line of numbers, which apply to the surface of the Earth. We treat the surface as a black body, as we justify elsewhere. Using Stefan's Law we find that a black body at 290 K (that's 17°C) should radiate 401.1 W/m². Our calculator adds the power radiated in each 0.1-μm window from 5 μm to 60 μm, in a process we call numerical integration, and arrives at 388.8 W/m². That's closer to 401.1 W/m² than the 349.0 W/m² we obtained when we our calculation extended only to 30 μm. But we still have a difference of 12.3 W/m².

We experiment with the black-body radiation spectrum using this program. When we integrate in 0.1-μm steps all the way from 1 μm to 300 μm, we get a total power of 401.0 W/m², which is almost exactly what we obtain from Stefan's Law. We use our program to determine the power radiated by a black body at various wavelengths.

--------------------------
Range (um)  Power (W/m^2)
--------------------------
0.1-1          0.0
1-5            3.8
5-60         388.8
60-100         6.4
100-200        1.8
200-300        0.2
--------------------------
Total        401.0
--------------------------

We have a total of 12.2 W outside the 5-60 μm range of our absorption spectra. In the future, perhaps we will extend our spectra to include this missing 3%, but for now we are content with our omission. Furthermore, we are satisfied that our numerical integration in 0.1-μm steps, which is the basis of our climate model's calculations, is accurate to better than 0.1%.

Our next step is to generate a new set of spectra for the atmosphere with 660 ppm CO2, and see how the total escaping power changes. Our guess is that the total escaping power will decrease, but we are not sure by how much.

Total Escaping Power

2010-09-22T07:54:00.000-07:00

Our current approach to modeling the Earth's greenhouse effect is to divide the Earth's atmosphere in to 3-km layers. So far, we have considered how each of the first seven layers absorbs and emits long-wave radiation. Our study of these seven layers has been motivated by a desire to calculate the total heat radiated into space by the Earth and its atmosphere, and ultimately to estimate the strength of the relationship between the Earth's surface temperature and the concentration of CO2 in its atmosphere.

In previous posts, we simplified our discussion by assuming that each atmospheric layer was either perfectly transparent or absolutely opaque at each wavelength. This assumption, combined with an examination of the absorption spectra, allowed us to say that all radiation passing through one layer would be sure to pass through all higher layers as well. Furthermore, our assumption allowed us to ignore the fact that most upward radiation is not vertical, but at some angle to the vertical. Radiation traveling at 60° to the vertical will pass through 6 km of air as it ascends through a 3-km layer, and so is more likely to be absorbed by the layer. But if we assume that the 3-km layer is either perfectly transparent or absolutely opaque, we conclude that radiation of a particular wavelength will either pass entirely through the layer or be entirely absorbed by the layer, regardless of its direction of travel.

Today we present our first layered-atmosphere computer model. The model does not assume that each layer is either transparent or opaque. When we calculate the power escaping into space from layer A at temperature T between wavelengths 10.0 μm and 10.1 μm, we proceed as follows. We start with the black-body radiation power density at wavelength 10 μm and temperature T. This power density is almost constant in the narrow range 10.0 μm to 10.1 μm, so we multiply the power density at 10.0 μm by 0.1 μm and obtain a good estimate of the power radiated between 10.0 μm and 10.1 μm. Now we multiply this power by the fractional absorption of layer A at 10.0 μm. The principle of radiative symmetry dictates that the efficiency with which a body emits radiation at a particular wavelength, as a fraction of the power emitted by a black body, is exactly equal to the efficiency with which it absorbs that same wavelength. So we now have the power radiated by layer A between wavelengths 10.0 μm and 10.1 μm. By assumption, whatever is not absorbed by the layer above A will escape into space, so we multiply the power radiated by A by the fractional transmission at 10.0 μm (that's one minus the fractional absorption at 10.0 μm) of the layer immediately above, and so arrive at the power between 10.0-10.1 μm that escapes from layer A into space.

To obtain the total power escaping into space from each layer, we add up the power escaping in all the 0.1-μm bands from 5 μm to 30 μm, which is the extent of our absorption spectra. We will include the surface of the Earth as its own layer. As we have in the past, we will assume the Earth's surface is a black-body radiator. Thus our program will calculate for us the total power escaping from the Earth into space, and allow us to see how much of this power is radiated by the surface and by each atmospheric layer.

Although we have dropped the transparent-opaque simplification, we retain two of its corollaries. First, we ignore the fact that most upward radiation is not vertical. Second, we assume that any radiation that manages to pass through the layer immediately above will be sure to pass through all remaining layers. These assumptions will effect our estimates, but we believe the effect will be less than 10%. In the future we may enhance our model to remove these assumptions, but for now we will start with a simple model.

We wrote our computer model in TclTk. You will find it here. The program takes as its input this file containing the names of the layers, their temperatures, and their absorption spectra. It produces the following output.

------------------------------------------------
      Name    Temp       BB     Layer   Escaping
------------------------------------------------
   Surface    290.0    401.1    349.0     89.7
      0-km    280.0    348.5    225.3     58.8
      3-km    270.0    301.3    150.0     49.8
      6-km    250.0    221.5     81.5     44.4
      9-km    230.0    158.7     31.2     15.4
     12-km    220.0    132.8     16.3      5.8
     15-km    220.0    132.8     13.1      4.2
------------------------------------------------
Total:                                   268.2
------------------------------------------------

The "Layer" column is the power radiated upwards by each square meter between 5 μm and 30 μm. The "Escaping" column is the power that escapes. We used 290 K for the surface temperature, which is 17°C. The total power escaping into space is 268 W/m². We don't have radiation from the 18-km layer, because we use this layer only as an absorber for radiation from the 15-km layer. The atmosphere above 15 km plays little or no part in atmospheric convection because it is above the tropopause.

We see immediately that the power escaping from the 15-km layer is only 1.5% of the total escaping power, and twenty times less than the power escaping from the Earth's surface. The top three layers of the atmosphere combined contribute less than 10% to the total escaping power. These are the layers in which radiation is dominated by CO2. In the lower layers, radiation is dominated by water vapor.

Before we proceed with exercising our model, however, we must correct a deficiency in our existing spectra. Our spectra extend from 5 μm to 30 μm, but it turns out that a significant portion of the heat radiated by the Earth's atmospheric layers is radiated in the range 30 μm to 60 μm, as you can see in the graph below. (In case you are interested, we obtained the curves of this graph using this program.)

The "BB" column in our table gives the power a black body would radiate, as calculated using Stefan's Law. Our calculation assumes the Earth's surface is a black body, so when we see 401 W/m² in the "BB" column, we expect to see the same value in the "Layer" column. But we don't, and the reason we don't is because our calculation adds up only the power radiated by each layer in the range 5 μm to 30 μm. A black body at 290 K radiates only 349 W/m² in this range. The remaining 52 W/m² it radiates in the range 30 μm to 60 μm.

We must go back to the Spectral Calculator and extend our spectra to 60 μm. While we are there, we will obtain the spectra for 660 ppm CO2 and 165 ppm CO2 as well.

CO2 Absorption Band

2010-09-15T04:55:00.000-07:00

In our previous post, we presented this graph of absorption by the Earth's atmosphere between 18 km and 21 km. This absorption is due entirely to CO2. In our calculations we set water vapor concentration to zero. Indeed, the absorption in the range 13-17 μm is due almost entirely to CO2 starting at an altitude of 9 km. The following plot gives us a close-up of the CO2 absorption in 3-km layers.

The four graphs are for 9, 12, 15, and 18 km, but we have named them after their atmospheric pressures, 410, 300, 220, and 160 mbar respectively. These pressures do not, in fact, correspond exactly to the pressures we observe in the atmosphere, because our pressure calculation ignores the drop in temperature as we ascend, but we used these pressure values when we calculated the spectra.

What we see between the absorption at 160 mbar and 300 mbar is the effect of doubling the density of CO2 molecules. In both cases, CO2 concentration is 330 ppm, which means 330 out of every 1,000,000 molecules are CO2. But the number of molecules per cubic meter increases approximately with the pressure, so the 300-mbar plot represents twice as much CO2 per cubic meter as the 160-mbar plot.

The absorption at 160 mbar for wavelength 14 μm is 21%, rising to 99% at 15 μm. The absorption at 300 mbar for wavelength 14 μm is 42%, rising to 99% at 15 μm. Twice as much CO2 in a 3-km layer leads to twice as much absorption of 14-μm radiation, but no significant increase in absorption of 15-μm radiation. We are particularly interested in the effect of doubling CO2 concentration, so let us try to get a better idea of how much more absorbing power the atmosphere has at 300 mbar than 160 mbar. Suppose a lower layer of the atmosphere emits uniformly in the band 13-17 μm, so that there is equal power in each 0.5-μm division of our graph. We can count squares under each graph and divide by the total number of squares to get the fraction of power absorbed by CO2 at each pressure. We come up with 43% at 160 mbar and 56% at 300 mbar.

Of course, the lower layers of the atmosphere do not radiate uniformly in the 13-17 μm band. Their radiation in this band is dominated by their own CO2, in accordance with radiative symmetry. The 300-mbar layer will receive radiation from the 410-mbar layer below it, and this radiation will have a power distribution that matches the 410-mbar layer's own absorption spectrum. At 14 μm, for example, the 410-mbar layer will radiate only 60% as much power as it does at 15 μm. In our spreadsheet we multiply the radiation spectrum of the 410-mbar layer by the absorption spectrum of the 300-mbar layer, and find that the 300-mbar layer absorbs 74% of the radiation arriving from the 410-mbar layer below. Meanwhile, the 160-mbar layer absorbs 70% of radiation from the 220-mbar layer below it.

The greenhouse effect causes a convection cycle in the Earth's atmosphere, which transports heat from the Earth's surface to the tropopause. All layers of the Earth's atmosphere up to the tropopause radiate heat out into space through the thinner, upper layers. As our previous posts have shown, radiation by the lower layers is dominated by water vapor, while radiation by the upper layers is dominated by CO2. Our calculations suggest that doubling the atmosphere's CO2 concentration will cause only a slight increase in the absorption and emission of radiation by the upper layers. In future posts, we will try to estimate how much this slight increase in absorption will raise the surface temperature of the Earth.

The Earth's Tropopause

2010-09-01T06:05:00.000-07:00

In our previous post, we noted that the temperature of the atmosphere does not drop by more than a few degrees Centigrade between altitude 12 km and 18 km. We concluded that altitude 15 km corresponds roughly to the atmosphere's tropopause, and therefore marks the top of its convection cycle. Today we consider if there is any significant heat transfer between the tropopause and the upper atmosphere.

The following graph shows the absorption of long-wave radiation by the seventh 3-km layer of the Earth's atmosphere, from altitude 18 km to 21 km. This line is the same as the 18-km line in our Earth's Atmosphere post. The absorption is typical of a clear day in April, at latitude 30°N. We assume water vapor content is zero, but CO2 concentration remains 330 ppm. We assume air pressure is 160 mbar and temperature is 210 K, or −63°C.

In the seventh layer, absorption by CO2 remains strong. As before, we say our 3-km layer is opaque at a particular wavelength when it absorbs more than 63% of that wavelength, and is transparent otherwise. The seventh 3-km layer is opaque from 14.3 to 15.6 μm, while the sixth layer was opaque to 14.3 to 15.7 μm. The seventh layer will absorb roughly 93% of the heat radiated upwards by the CO2 in the sixth layer.

We note that the actual pressure in the Earth's atmosphere at 18 km is closer to 130 mbar. We calculated atmospheric pressure using our constant-temperature model of atmospheric pressure, and this model starts to be increasingly inaccurate in the cold upper layers of the atmosphere.

When we chose our temperatures, we were looking at this plot and rounding to the nearest 10 K. If you look at the plot yourself, you will see that the minimum temperature varies with latitude, and that the minimum spreads over a large range. We see this variation with latitude and the extent of the minimum in plots like this one also. In the case of 30°N, the temperature minimum extends from roughly 12 km to 22 km. The tropopause is the altitude at which the temperature stops dropping with pressure as it would in adiabatic expansion. Thus the tropopause is not at the altitude of minimum temperature, but rather at the altitude where temperature stops dropping adiabatically with pressure.

We go back to this plot and estimate the tropopause altitude as 9 km at 90°N (the north pole), 12 km at 30°N, and 13 km at 0°N (the equator). Why is the tropopause higher in the tropics? Our model of the greenhouse effect and the atmospheric layers must provide us with an answer to this question.

We look at the water vapor concentration in this plot, we see that we have 10 ppm water vapor concentration up to roughly 9 km at 0°N, 12 km at 30°N, and 13 km at 0°N. As we saw in the fifth layer, water vapor concentration of 10 ppm marks the altitude at which the atmosphere becomes transparent to wavelengths 5 to 9 μm, thus allowing the atmosphere below to radiate these wavelengths into space. Between 12 km and 21 km there is no significant change in the absorption spectrum of the atmosphere. A large body of air rising from 12 km to 21 km will be unable to radiate heat into space as it rises. If it cannot radiate heat, it will not sink, and so we will have no convection. Thus we expect the convection cycle to stop at the altitude where water vapor concentration drops to around 10 ppm, and this is indeed what we observe in the Earth's atmosphere.

There is no significant drop in temperature for at least 10 km above the tropopause. Without a temperature gradient, there can be no heat transport by conduction. There is no significant change in the absorption spectrum as we ascend these 10 km, so each layer of the atmosphere will receive as much heat by radiation from the layers above and below as it radiates back to those same layers. Thus there can be no significant heat transport by radiation either. We have already concluded that convection stops at the tropopause, so we see that there is no significant heat exchange between the tropopause and the atmosphere above it.

The Sixth 3-km Layer

2010-08-25T08:23:00.000-07:00

The following graph shows the absorption of long-wave radiation by the sixth 3-km layer of the Earth's atmosphere, from altitude 15 km to 18 km. This line is the same as the 15-km line in our Earth's Atmosphere post. The absorption is typical of a clear day in April, at latitude 30° North. Water vapor content is only 1 ppm, which is one tenth thousandth the concentration in the surface layer. But CO2 concentration is 330 ppm, which is the same as in the surface layer. Air pressure is 220 mbar, which is 22% of the pressure in the surface layer. What is striking about the sixth layer is that the temperature is 220 K, or −53°C, which is the same as in the fifth layer.

In the sixth layer, absorption by water vapor is nowhere more than 5%. But absorption by CO2 remains strong. As before, we say our 3-km layer is opaque at a particular wavelength when it absorbs more than 63% of that wavelength, and is transparent otherwise. The sixth 3-km layer is opaque from 14.3 to 15.7 μm and transparent otherwise. The fifth layer, immediately below, is opaque to 14.2 to 15.8 μm. It radiates in these wavelengths also. The sixth layer absorbs 90% of the heat radiated upwards by the fifth layer, but the remaining 10% will pass out into space.

The fifth and sixth layers are at the same temperature. We did not make up the temperatures, pressures, and concentrations we use in our 3-km atmospheric layers. We obtained them from actual measurements of the Earth's atmosphere.

We have talked at length about the tropopause that exists at the top of the convection cycle in an atmospheric greenhouse. In our simple examples, the tropopause marks the top of the convection cycle, and occurs when the atmosphere above becomes so thin as to allow heat to radiate directly into space. The Earth's atmosphere is not so simple as in our examples. Above 12 km, the Earth's atmosphere is transparent to most long-wave radiation, but remains opaque in the narrow band that is absorbed by CO2. At a high enough altitude, this CO2 will become thin enough that it becomes transparent. But the tropopause of the Earth's atmosphere appears to occur at a lower altitude. It occurs when the atmosphere is transparent to over 90% of long-wave radiation.

In later posts, we will see how absorption by ozone in the stratosphere effects our greenhouse convection cycle. In the meantime, we will accept that there is very little change in temperature from altitude 12 km to 18 km, and therefore these altitudes mark the top of the Earth's greenhouse convection cycle.

First Differences

2010-08-23T06:20:00.001-07:00

Over at Climate Audit, Hu McCulloch as a guest post The First Differences Method that describes a simple way to calculate the global surface trend from station data. This method turns out to be same as the Integrated Derivative Method we describe in our Climate Analysis essay. I said as much in the comments of Hu's post, and pointed to our Continuous Stations plot, which we generated with our Integrated Derivative Method.

The Fifth 3-km Layer

2010-08-22T09:05:00.000-07:00

The following graph shows the absorption of long-wave radiation by the fifth 3-km layer of the Earth's atmosphere, from altitude 12 km to 15 km. This line is the same as the 12-km line in our Earth's Atmosphere post. The absorption is typical of a clear day in April, at latitude 30° North. Water vapor content is 10 ppm, which is one tenth the concentration present in the fourth 3-km layer, just below. Air pressure is 300 mbar (that's 30% of the pressure at sea level), temperature is 220 K (that's −53°C), and CO2 concentration is still 330 ppm.

As before, we say our 3-km layer is transparent at a particular wavelength when it absorbs less than 63% of that wavelength, and opaque otherwise. The fifth 3-km layer is transparent from 5.0 to 14.2 μm and from 15.8 to 30 μm.

Between the fourth and fifth layer of the Earth's atmosphere, water vapor concentration drops from 100 ppm to 10 ppm. Absorption by water vapor almost disappears. So far as our transparent-opaque simplification is concerned, the absorption by water vapor does indeed disappear. The atmosphere becomes transparent to all long-wave radiation except for 14.2 to 15.8 μm. These wavelengths are absorbed by CO2. Unlike water vapor concentration, CO2 concentration remains constant as we ascend from 0 km to 12 km.

The concentration of C02 may be constant, but the amount of CO2 in each cubic meter of air is still decreasing. The density of the air decreases as we ascend. Thus the absorption by CO2 in the fifth layer is less intense than in the fourth layer. The CO2 in the fourth 3-km layer absorbs radiation from 14.1 to 15.9 μm. By radiative symmetry, these are the same wavelengths that the fourth layer itself radiates up to the fifth 3-km layer. Of this radiation, only the narrow ranges 14.1 to 14.2 μm and 15.8 to 15.9 μm will pass through the fifth layer and out into space. The remaining 90% will be absorbed by CO2 in the fifth layer.

The Fourth 3-km Layer

2010-08-16T07:46:00.000-07:00

The following graph shows the absorption of long-wave radiation by the fourth 3-km layer of the Earth's atmosphere, from altitude 9 km to 12 km. This line is the same as the 9-km line in our Earth's Atmosphere post. The absorption is typical of a clear day in April, at latitude 30° North. Water vapor content is only 100 ppm (that's 0.01%), which is one tenth the concentration present in the third 3-km layer, just below. Air pressure is 410 mbar and temperature is 230 K (that's −43°C). We have CO2 present in 330 ppm, just as in the earlier layers.

As before, we simplify our analysis of the absorption spectrum by saying that our 3-km layer is transparent when it absorbs less than 63% of a wavelength and opaque otherwise. The fourth 3-km layer is transparent from 5.0 to 14.1 μm, with the exception of a couple of narrow bands around 6 μm which we will ignore for the sake of simplicity. The fourth layer is again transparent from 15.9 to roughly 28 μm.

The biggest change between the third and fourth layers is the new transparency from 5.4 to 7.3 μm. In the third 3-km layer, water vapor was still absorbing at these wavelengths, but now the water vapor is so sparse that it becomes transparent. Radiation by water vapor in the third 3-km layer at these wavelengths will pass through the fourth and higher layers, and so out into space. We also see an end to the absorption by water vapor for the range 23 to 28 μm.

Another thing that is striking about the fourth 3-km layer is how clearly the absorption by CO2 stands out and remains strong. Let us apply our opaque-transparent simplification to the absorption of the third and fourth layers to estimate how CO2 affects radiation into space by the third 3-km layer. In the third layer, CO2 was opaque from 13.8 to 16.2 μm. In this layer it is opaque from 14.1 to 15.9 μm.

To the first approximation (we split the atmosphere into layers and we assume the atmosphere is either opaque or transparent but never half-way between), 80% of the upward-going radiation emitted by the CO2 in the third layer will be absorbed by the CO2 in the fourth layer, and 100% of the downward-going radiation emitted by the CO2 in the fourth layer will be absorbed by the CO2 in the third layer.

This Time It's Different

2010-08-08T07:24:00.000-07:00

In The Right and the Climate, an opinion piece for the New York Times, Ross Douthat discusses the abandonment of cap-and-trade legislation by the US senate. He argues that Democrats must share the blame for the failure of cap-and-trade because they have exhausted the population's patience for doomsday stories. "The Seventies were a great decade for apocalyptic enthusiasms, and none was more potent than the fear that human population growth had outstripped the earth’s carrying capacity," he says. "The catastrophes never materialized, and global living standards soared." But the global warming doomsday story is different, Douthat says, because, "History, however, rarely repeats itself exactly — and conservatives who treat global warming as just another scare story are almost certainly mistaken."

People like Paul Ehrlich, who predicted global famine, and Rachel Carson, who claimed that DDT was killing humans, both believed in what they were saying. It turns out that they were wrong in their most famous claims, but they were doing their best to understand the dangers presented by lower infant mortality and widespread use of new chemicals. They believed in their cause and they believed in what they told us.

When Douthat says, "the evidence that carbon emissions are altering the planet’s ecology is too convincing to ignore," he seems to think that the evidence for global famine and chemical poisoning were unconvincing at the time that Ehrlich and Carson made their claims. But that's not the case. The evidence was convincing. A lot of people were convinced, including the US government under John Kennedy. But "convincing" is not the same as "correct". It turns out that their arguments were convincing but not correct.

Douthat is an example of the many well-educated people who have chosen to believe in anthropogenic global warming despite the history of environmental scare-mongering. The internet bubble was caused by people saying "this time it's different: revenue counts, not profit." The housing bubble was caused by people saying, "this time it's different: prices will never go down." The anthropomorphic global warming bubble has been caused by people saying, "this time it's different: the evidence is too compelling."

Here's my prediction: this time it's not different.

The Third 3-km Layer

2010-08-04T07:37:00.000-07:00

The following graph shows the absorption of long-wave radiation by the third 3-km layer of the Earth's atmosphere, from altitude 6 km to 9 km. The graph is typical of a clear day in April, at latitude 30° North. Water vapor content is 1,000 ppm (that's 0.1%), CO2 is 330 ppm, air pressure is 550 mbar, and air temperature is 250 K (that's −23°C). This line is the same as the 6-km line in our Earth's Atmosphere post.

As before, we say our 3-km layer is transparent at a particular wavelength when it absorbs less than 63% of that wavelength. The third 3-km layer is transparent from 5.0 to 5.4 μm, 7.3 to 13.8 μm, and 16.3 to 23.1 μm.

The transparency in the range 16.3 to 23.1 μm is due to the drop in water vapor concentration and pressure that occurs as we ascend through the atmosphere. Absorption in this band is due mostly to water vapor dimers instead of bonds within the water molecule. Absorption by dimers is proportional to the square of the pressure. Continued absorption in 5.4 to 7.3 μm is due to bonds within water vapor molecules, and does not drop with pressure as quickly as absorption by dimers. Continued absorption in the range 13.8 to 16.3 μm range is mostly due to CO2.

Hockey Stick Defenders

2010-07-28T15:07:00.000-07:00

I had a look through this defense of the Hockey Stick over at Realclimate. The author is Tamino. Here he writes at length defending Michael Mann's hockey stick graph against Steve McIntyre, Andrew Monfort, and others. He insults their integrity and character. I'm not surprised. Tamino called me an idiot and blocked me from commenting on his website when I asked, "Has there been any statistically significant warming in the past twenty years?"

Steve McIntyre has his own answer to Tamino's post. He concentrates upon how Tamino mis-quotes him.

But here's my answer to Tamino. The central argument against the Hockey Stick, as put forward by Steve McIntyre and Ross McKitrick, and confirmed by the Wegman Repot for the US congress, is that Michael Mann's data analysis generates the hockey stick from random proxy data. We explain how his analysis produces the hockey stick here, and I first posted about the effect here.

Nowhere in Tamino's discussion do I see any mention of the fact that random proxy data combined with CRU's global surface trend produces a hockey stick. Given that random input produces the hockey stick, it's pretty clear that the graph has no statistical significance. When I said as much in a comment at Realclimate, my comment was blocked. That's the third comment of mine that Realclimate has blocked.

UPDATE: In addition to the temperature hockey stick, there is also a CO2 hockey stick that you get by splicing modern CO2 measurements onto the end of the ice core measurements, like this. We talk about such splicing in the Ice Cores and Carbon Dioxide sections here. In Questioning the CO2 Hockey Stick, Jaworowski et al. challenge the absolute accuracy of the ice core CO2 measurements.

UPDATE: There's a new paper out in the Annals of Applied Science, A Statistical Analysis of Multiple Temperature Proxies: Are Reconstructions of Surface Temperature Over the Last 1000 Years Reliable? by McShane et al. You will find a copy of the paper here. The paper concludes that Michael Mann's method produces a hockey stick graph from random data, and that the hockey stick shape has no statistical significance.

UPDATE: In the new Realclimate discussion about the McShane paper, I again attempted to post a comment stating that Mann's method generated the hockey stick from random data. That's my third attempt to make the same statement. Comments posted after mine have now appeared. I assume my comment was blocked. So far as I can tell, nowhere in this latest discussion does any comment or the author mention the fact that Mann's method produces the hockey stick from random data. [24-AUG-10]

The Second 3-km Layer

2010-07-20T14:09:00.001-07:00

The following graph shows the absorption of long-wave radiation by the second 3-km layer of the Earth's atmosphere, from altitude 3 km to 6 km. The graph is typical of a clear day in April, at latitude 30° North. Water vapor content is 3,000 ppm (that's 0.3%), CO2 is 330 ppm, air pressure is 740 mbar, and air temperature is 270 K (that's −3°C). This line is the same as the 3-km line in our Earth's Atmosphere post.

In our previous post, we concluded that the first 3-km layer was transparent only between 8 μm and 13 μm. By transparent we mean that less than 63% of radiation is absorbed. We settled upon this definition of transparent in Upper Gas. Our definition is an approximation, and is to some extent arbitrary. We could have chosen 10% absorption or 30% absorption. But 63% absorption is what occurs after one absorption length in a medium, so it is a convenient value. Given our definition of transparent, we will say the layer is opaque, whenever it is not transparent. So our definition of transparent, which we will use later in a computer program that models the absorption and radiation of the layers, will be binary, and therefore simple to implement, without any significant loss in accuracy.

With our simple, binary, definition of transparent, we see that the second 3-km layer is transparent from 7.6 to 13.4 μm and for a small range from 17.7 to 18.8 μm. The emerging transparency around 18 μm is due to the drop in water vapor concentration and pressure that occurs as we ascend through the atmosphere. For wavelengths around 18 μm, absorption by water vapor is dominated by absorption in water dimers. As we saw in earlier, absorption by dimers is proportional to the square of the water vapor pressure, and drops rapidly with altitude.

The first 3-km layer is opaque to wavelengths 7.6 to 8.0 μm, 13.0 to 13.4 μm, and 17.7 to 18.8 μm, but the second and higher layers are transparent to these wavelengths. By the principle of radiative symmetry, the first 3-km layer will radiate in the ranges for which it is opaque. Wavelengths in these three ranges will be radiated by the first 3-km layer and pass out through the upper layers and into space.

The Surface Layer

2010-07-17T14:18:00.000-07:00

The following graph shows the absorption of long-wave radiation by the first three kilometers of the Earth's atmosphere. The graph is typical of a clear day in April, at latitude 30° North. Water vapor content is 10,000 ppm (that's 1%), CO2 is 330 ppm, air pressure is 1000 mbar, and air temperature is 280 K (that's 7°C). This line is the same as the 0-km line from our Earth's Atmosphere post.

The 5 μm to 30 μm range includes over 95% of the heat radiated by the Earth's surface (look at the brown power density line in this graph). We see that the first 3 km of the atmosphere absorbs almost all radiation from 13 μm to 30 μm, and from 5 μm to 8 μm. According to the Spectral Calculator, the absorption between 5 μm and 8 μm is due to water vapor, between 13 μm and 17 μm is due to CO2, and above 17 μm is due to the water vapor again.

Most radiation between 8 μm and 13 μm passes through the first 3 km of the atmosphere. The higher layers, being thinner and more sparsely supplied with water vapor, absorb even less of this radiation, as you can see in our compound graph. To the first approximation, the first 3 km layer of the atmosphere is transparent from 8 to 13 μm and opaque to other long-wave radiation. The 8 to 13 μm radiation emitted by the Earth's surface will pass right through the atmosphere and into space, provided there are no clouds. As we declared in our recent review, we are ignoring clouds for the time being.

In our up-coming posts, we will consider each 3-km layer of the atmosphere up to altitude 18 km in turn. We will approximate each layer as being transparent to some wavelengths and opaque to others. Once we have considered all the layers, we will write a computer program to calculate the radiation of heat from the various layers, and so obtain an estimate of the Earth's greenhouse effect without clouds.

Heat Transport and Lapse Rate

2010-07-09T01:38:00.001-07:00

Over at RealClimate we find a post entitled A Simple Recipe for GHE by someone called Rasmus. The post is an attempt to explain the greenhouse effect, and to convince us that a doubling the CO2 concentration will raise the Earth's surface temperature by a few degrees Centigrade. We have already claimed that all explanations of the greenhouse effect we encountered on the web violated at least one law of physics or thermodynamics. The manner in which this new explanation violates the laws of thermodynamics is interesting enough to merit a short discussion.

The author agrees with our conclusion that the Earth's atmosphere radiates its heat at an altitude where it becomes transparent. And so he brings the atmospheric temperature lapse rate to our attention in his section (iv). The lapse rate is the drop in temperature with altitude. In Atmospheric Convection we found that the drop in temperature with altitude is the result of the adiabatic expansion of gas during during the transport of heat by convection to the upper atmosphere. If there were no greenhouse gases in the atmosphere, there would be no radiation of heat by the upper atmosphere, and no heat to transport. With no heat to transport, there would be no convection, no adiabatic expansion, and therefore no atmospheric lapse rate. To suggest that a convection cycle can exist in a viscous fluid without the transport of heat and without any source of physical work, such as a paddle, is to believe in a perpetual motion machine. In the comments of Motl on CO2 Sensitivity you will see us debating this same issue with a reader. The reader claims that convection will take place even in a transparent atmosphere without any machines to move the air and without any upward heat transport. Rasmus appears to share our reader's belief, because he makes no mention of the relationship between heat transport and atmospheric lapse rate. He refers to hydrostatic balance, but such balance in no way requires a drop in temperature.

When we ignore the dependence of the atmospheric lapse rate upon heat transport, we conclude that the atmosphere keeps cooling as we go up, so that there is no layer in the atmosphere where the lapse stops. In short, Rasmus's understanding of the lapse rate implies that there is no tropopause, even though the existence of the tropopause in the Earth's atmosphere is well-established. In the first comment following the post, we see a reader asking some cogent questions about the tropopause, and Rasmus answering that the tropopause is above the highest altitude of heat radiation by the atmosphere. But this cannot be true, because without vertical heat flow, the temperature of the atmosphere would not be dropping. The tropopause must mark the upper limit of the atmospheric heat transport away from the Earth.

Rasmus's argument requires that we set aside the law of conservation of energy, which is the First Law of Thermodynamics. We prefer explanations that adhere to established laws, so we feel justified in dismissing his post.

PS. Originally spelled "lapse" without the "e", but corrected spelling in title and text after Chuck pointed out my error, see comments.