Evaporative cooling = refrigeration?

AI Thread Summary
Evaporative cooling is not considered refrigeration because it does not operate against a heat gradient in the same way traditional refrigeration does. While it cools the body by removing heat through the evaporation of sweat, it ultimately relies on the movement of water vapor to a cooler environment, where it can condense and release heat. The discussion highlights that evaporative cooling works against dry bulb temperature but not against wet bulb temperature, which reflects the true energy content of the air. Additionally, there is curiosity about naturally occurring refrigeration processes, as examples seem scarce compared to natural heat engines. The conversation concludes that while evaporative cooling is effective, it is more accurately described as a cooling mechanism rather than true refrigeration.
techmologist
Messages
305
Reaction score
12
I need convincing that evaporative cooling is refrigeration in the sense of a heat engine run backwards.

It has the outward appearance of refrigeration. Take perspiration, for example. Your body temperature drops even though the ambient air temperature is hotter. It requires work. You need a fan or a good breeze to blow the humid air away from you and surround you with dry air that can absorb more water. Or you have to walk around. Or at least convection must rise up from your head.

But is heat actually being delivered to a higher temperature? It seems like the water vapor is just being carried away until it is eventually in a cooler environment, where it can condense again and release the heat. So effectively, heat is just flowing from hot to cold.

Of course, this is ignoring the fact that the water vapor will take on the temperature of the ambient air, which is assumed to be hotter than your body. So maybe a better analogy is this. You cool off something by placing a block of ice on it, which absorbs some heat. Then you toss that block of ice in a furnace. Heat has been moved from cold to hot, but somehow that doesn't really seem like refrigeration. The last step was completely irrelevant. There is no cycle.

What am I missing?

Thanks.

Edit. Of course that second analogy sucks. What I really should say is that you cool something by placing a canister of compressed air up against it and letting it expand isothermally. Then toss the canister in a furnace (and duck).
 
Physics news on Phys.org
If your definition of "refrigeration" is that it works against a heat gradient, then no, evaporative cooling is not refrigeration. It may work against dry bulb temperature, but it doesn't go against wet bulb temperature, which is where the actual energy (enthalpy) of the air is measured.
 
Ah, so it isn't really refrigeration. Just cooling. I probably should have taken a hint from the name.

In that case, are there any examples of genuine refrigerators occurring in nature? Naturally occurring heat engines seem to be all over the place, but it's hard to think of a refrigerator. If there is one, it would probably have something to do with weather/climate.

I know there is a man-made refrigerator that has no moving parts besides an air compressor (Ranque-Hilsch refrigerator), so a refrigerator need not be as hi-tech as the Frigidaire in your kitchen. But even the Ranque-Hilsch, which isn't much more than a tube, still seems unlikely to occur naturally.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top