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.
We are 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.
Tuesday, September 27, 2011
Friday, September 23, 2011
Condensation Rate
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.
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.
Friday, September 16, 2011
Simulated Clouds, Part II
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.
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.
Friday, September 9, 2011
Simulated Clouds, Part I
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.
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.
Saturday, September 3, 2011
Latent Heat
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.
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.
Thursday, August 25, 2011
Condensation Point
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 = pwRw/pR = 0.62 pw/p,
where x is the concentration of water vapor, pw is its partial pressure, Rw 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.
xs = (T−250)2p/8000
Here xs 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.
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 = pwRw/pR = 0.62 pw/p,
where x is the concentration of water vapor, pw is its partial pressure, Rw 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.
xs = (T−250)2p/8000
Here xs 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.
Wednesday, August 17, 2011
Evaporation Rate
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)(xs − 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), xs 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 xs, 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%.
w = (0.007 + 0.005v)(xs − 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), xs 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 xs, 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%.
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