What is the Relationship Between Evaporative Cooling and Gas Pressure?

In summary: Evaporative cooling is the process of absorbing sensible heat from the air by evaporation of water. It works by extracting the most energetic vaporized water particles, and as the temperature decreases, the cooling rate decreases exponentially. The equation used to calculate the cooling rate is derived from the Clausius-Clapeyron equation, which neglects the volume of the fluid phase compared to that of the gas. The ideal gas approximation is used to get an accurate calculation, even for very high temperatures.
  • #1
Angelos K
48
0
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!
 
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  • #2
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.
 
  • #3
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]
 
  • #4
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.
 
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  • #5
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

[http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation" discusses the Clausius-Clapeyron equation and its less general form

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

used here]
 
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1. What is evaporative cooling power?

Evaporative cooling power refers to the ability of a substance to cool its surroundings through the process of evaporation. It is a measure of how much heat energy is absorbed and converted into vapor during the evaporation process.

2. How does evaporative cooling work?

Evaporative cooling works by using the energy from the surrounding air to convert liquid water into water vapor. This process, known as evaporation, removes heat from the air and cools the surrounding area.

3. What factors affect the evaporative cooling power of a substance?

The evaporative cooling power of a substance is affected by factors such as the temperature and humidity of the air, the surface area of the substance, and the rate of air flow over the substance. Higher temperatures and lower humidity levels typically result in higher evaporative cooling power.

4. What are some common applications of evaporative cooling power?

Evaporative cooling power is commonly used in air conditioning systems, refrigerators, and cooling towers to lower the temperature of the surrounding air or water. It is also utilized in industrial processes such as food preservation and drying.

5. How is evaporative cooling power measured?

The evaporative cooling power of a substance is typically measured in terms of the amount of heat energy absorbed per unit of time (usually in watts). This can be calculated using the change in temperature and humidity of the air before and after the evaporation process.

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