Register to reply

Evaporative cooling power

by Angelos K
Tags: cooling, evaporative
Share this thread:
Angelos K
Jul21-08, 07:51 PM
P: 39
I wish to understand evaporative cooling. Apparently the principle consists in pumping the vapour above a bath, hence extracting it's most energetic corpuscles and achieving cooling.

The aim is to derive the cooling power. My textbook states:

[tex]\frac{dQ}{dt} = L\frac{dn}{dt} \propto p [/tex]

L being the latent heat per mol and p the gas pressure. I don't see why the proportionality holds. Sorry if I miss something simple, I'm battling a light cold ;-)

[From here on one can use the gas pressure formula

[tex] p = p_0 \exp(-\frac{L}{RT})[/tex]

to demonstrate, how with linearly decreasing temperature the cooling rate decreases exponentially]

Thanks in advance for any help!
Phys.Org News Partner Physics news on
Symphony of nanoplasmonic and optical resonators produces laser-like light emission
Do we live in a 2-D hologram? New Fermilab experiment will test the nature of the universe
Duality principle is 'safe and sound': Researchers clear up apparent violation of wave-particle duality
Jul22-08, 03:53 AM
P: 22,294
Evaporative cooling simply absorbs sensible heat in the air via the latent heat of evaporation of water. How much heat you absorb is the amount of water evaporated over a period of time (dn/dt) times the latent heat per unit of water (L). How much evaporation you get depends on the vapor pressure.

This equation you list has no information about air flow and the actual process of evaporation in it. Typically, you would get dn/dt directly from the absolute humidity of the air, but you also need to assume an efficiency factor in a real device (say, 75% for a decent cooling tower). You can easily diagram the process on a psychrometric chart.
Angelos K
Jul22-08, 09:11 AM
P: 39
Thanks! So my textbook seems to make several oversimplifying assumptions, right?

What I don't get is how he gets to the relation:

[tex]L\frac{dn}{dt} \propto p[/tex]

Jul22-08, 12:05 PM
P: 22,294
Evaporative cooling power

I don't know if I'd say oversimplifying assumptions - the math to actually calculate evaporative cooling via convection is pretty daunting and not very accurate in the real world anyway. It is best found via experimentation, giving a simple efficiency factor.

I'm not quite sure what is going on with that gas pressure formula (is that in the book too?), but I would think it should be proportional to the difference in partial pressure vs saturation pressure: ie, the relative humidity. In other words, as the relative humidity goes down, the number of moles evaporated goes up.
Angelos K
Jul22-08, 07:06 PM
P: 39
The gas pressure formula [exponential increase with temperature] is derived in the book from the Clausius-Clapeyron equation, by neglecting the volume of the fluid phase compared to that of the gaseous one AND making an ideal gas assumption for the gas.

I suppose the ideal gas behaviour to be a really rough approximation, even for helium, since the temperatures reached are around 1K

[This wikipedia article discusses the Clausius-Clapeyron equation and its less general form

[tex]\frac{dp}{dt} = \frac{Lp}{R{T}^2} [/tex]

used here]

Register to reply

Related Discussions
The typical formula for temperature drop in evaporative cooling models General Physics 3
Evaporative Heat Exchanger Materials & Chemical Engineering 1
Evaporative Cooling Introductory Physics Homework 5
Evaporative temperature Classical Physics 1
Evaporative cooling in a vacuum General Physics 3