## Tuesday, June 21, 2011

### Cell Kinetic Energy

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.

1. Sorry, I have to point up that the circulating four cell aren’t conceptually acceptable because in contrast with the vectorial fields theory.

The four cells form a system without mass sources/sinks inside it and so the mass conservation requires that is ∇•(ρU) = 0, that means ρU = ∇Ψ. The vectorial field ρU admits the scalar potential Ψ, whose it is the gradient, and we know very well that its curl is ∇x(∇Ψ) = 0. That’s the vectorial field ρU is irrotational.

In this case we could eliminate the density ρ as the motion is very much subsonic and all claimed for the momentum ρU rest totally valid for the speed U.

Michele

2. Indeed: the sudden circulation of four cells creates angular momentum in one direction out of nothing, which violates the conservation of angular momentum. If we have a sum of random circulations, however, the net angular momentum might add up to zero, so we don't have a violation on the large scale. In our previous simulations, this appears to be the case.

But now we are proposing to set up a steady breeze along the bottom of our array, so that cells circulate around the entire array. They warm on the lower left and cool along the top. Thus we have angular momentum on large scale.

I agree that ∇•ρu = 0, where u is velocity and ρ is density. But I don't see how this implies that ρu = ∇Ψ, where Ψ is a scalar field. If we consider water spinning in a bucket, it moves in a circle, so that the curl of u, ∇xu, is not zero.

If I remember correctly, viscosity is necessary for the creation of rotation in a fluid in which mass is conserved.

3. Touché, my memory is playing tricks on me.
There is ∇•(ρU) = 0 if (ρU) = ∇x(A), that’s, (ρU) is the axe of the whirl of vector A, contained by a plan orthogonal to it. But, if I am not mistaking again, that means the air vertically rises while the surroundings horizontally rotate around it.

4. Dear Michele,

I'm trying to imagine horizontal rotation, like a hurricane, and whether the air in the center will be pushed up. But I'm not getting a clear picture of it.

Do you agree that we cannot get fluid to rotate without viscosity? If I have fluid in a bucket and I spin the bucket, the fluid will not spin unless viscous friction pushes it around. If I try to move it with a paddle, I will be unable to move the fluid unless viscosity disturbs the laminar flow. If I have a wing on a non-viscous fluid, it provides no lift. Only viscosity will move the stagnation point on the underside of the wing to the rear tip of the wing, giving us the rotation that creates lift.

Now, what if the simulation produces a large rotation of cells around the array? That's what we want it to do. Does viscosity make convection possible? I don't think I understand exactly how viscosity makes circulation possible.

Yours, Kevan

5. The hurricane is due to the Coriolis effect and to the fact of using a non-inertial reference frame.

The curl of the non-divergent vectorial field is, e.g., the magnetic field related to the electric field and vice versa.

Of course, all the other cases that you exposes require the viscosity but they are pertinent when/where a solid body exchanges momentum with a fluid. For the atmosphere this occurs only near the ground, within the boundary layer.

The large circulation is due to the continuity acting inside a bounded space. No bounds for the fluid, no circulating cells.

Michele