Hello. First, this thread was in "homework", but I thought this probably wasn`t a "homework" because is a lot more conceptual , and maybe deeper, than a simple exersise. Not more difficult though. I apologize if someone feels bad because I copied the thread here.

I have some problems analysing the results of a simulation of Rayleigh-Benard problem. I`m confused with a energy balance. Since it`s more a general question, I`ll describe first the situation. Please, be patient, I`m going to describe the problem carefully in order to help the understanding :

The configuration:
In a vertical rectangle (width 2, heigh 0.5), you have air inside. It´s a bidimensional problem. Side walls are adiabatic (insulated), always.
On t=0, air is at temperature T=0 (non dimensional temperature), and top wall is at temperature T=0 too. At the same time, on bottom wall is imposed a temperature T=1, that remains. So the bottom wall temperature is constant. Air temperature, upper wall temperature, and side walls are not. Rayleigh number is enough to make convection possible.
So as time elapses, there will be heat transpor, and in some time, it will be done by convection (creating some regular patterns of flux), and the system will reach an steady state (where convection exists)

The situation:
When I plot Nusselt number on top wall, and on bottom wall, I see they converge to a value , i.e they have the same numerical value, as the system goes to a steady state. Nusselt number is defined as (H/k)*(q/(T1-T0), that`s , the heigh, divided by the fluids thermal conductivity, multiplied by the heat flux, divided by the temperature difference betweeen walls (dimensional temperature).

The question:
If Nusselt numbers (at bottom and top wall) are the same, that should imply that the heat fluxes are identical too. So all the energy that enters by the bottom wall, leaves by the upper surface (top wall). Now, where does the energy necessary to maintain convection (the velocity of the fluid and the patterns) come from? In other way, to make the fluid get some velocity, and get regular patterns, you need work. But where the system gets it, if all the energy that`s supplied, leaves.

Thank for your time

Alejandro

This is an interesting question, because it highlights the difference between energy and availability of energy. You are correct that it takes work to drive convection. Why does it take work? Because without a constant energy input, the convection will stop. Why will it stop? Because of viscous drag. Where does viscous drag put the energy? It turns it into heat. Therefore convection will take an energy input and convert it into heat.

But the energy input here IS heat, so convection is converting heat into heat ... what's wrong here? The energy is entering the system at a high temperature T=1 and exiting at a lower temperature (which you don't know until you have solved the problem, but it will be 0<T_exit<1). Heat at a high temperature carries less entropy into the system, and heat a low temperature carries more entropy out of the system. Therefore entropy is being created inside the system, which is just what you would expect, since viscous drag is converting the kinetic energy of the convective flow into heat.

Where does the energy in the convective flow come from? Since the heat enters the system at a high T and low S, there is some AVAILABILITY, defined as the amount of work that can be extracted by and ideal heat engine before dumping the waste heat at the lowest accessible temperature. The process of convection acts like a heat engine (it IS a heat engine) converting the available part of the incoming heat into work, which is where the kinetic energy of the convective flows comes from. Friction then takes this energy and converts it back into heat, basically wasting all the available energy. When the heat exits at T_exit, it has no availability left, because it has reached the lowest temperature that exists within the system.

This situation is no different from putting hot steam into a steam engine, using that engine to turn a wheel where friction converts all the work back into heat, and then letting the engine's waste heat plus the friction generated heat exit from the system. Since the system is in steady state, this exiting heat must equal the entering heat.

As an interesting side topic, this is how the earth works. Sunlight puts energy into the earth at a high temperature, the available part of this energy drives atmospheric and oceanic convection, and also all the chemistry of life via photosynthesis, while the unavailable part remains as heat. These processes result in waste heat, which (together with the unavailable part of the incoming heat) exit the system. Since the earth releases heat into space at the same rate as it receives it from the sun, the temperature of the earth stays steady. (A slight imbalance between these flows leads to global warming or global cooling.)

So you see that the fact that the same amount of heat enters and exits does not mean that it can't do anything interesting while it is in transit through the system. It is the availability that is used up, not the heat.

Hope this helps!

--Stuart Anderson

Thanks!, it really helped . I`ve got only one more question, that I think would get things totally clear:

I you put a shaft in the cavity, in such a way that convection could turn it (i.e. use it actually as a heat engine), and supposing that the same patterns are formed. In other words, you manage to get work through a shaft from the system. Then, am I right when thinking that the heat imput now will be equal to the work in the shaft plus the heat that leaves, and in this case, heat that enters is not the same that heat that leaves?
In other words, does my mistake was considering a work inside the system to make a global balance of fluxes outside the system?

Alejandro

Yes, you are exactly right. Convection could turn a shaft which would export work outside the system, and in that case, heat input = work output + heat output, just exactly like any other heat engine.

I think you've got it totally clear now.

Good luck in all your studies!

--Stuart Anderson