# Cooling with Evaporation

StoningTree
Homework Statement:
A room should be cooled by 3 Kelvin using evaporation. The temperature of the water is 15 ° C. Furthermore, the volume of the room is 180m ^ 3, at a temperature of 25 ° C and a humidity of 50%. Hom much water is needed?
Relevant Equations:
x=0.622*j*ps/(p-j*ps)
h = cpa t + x [cpw t + hwe
First i calculated the latent heat of vaporation: h_vap(15°C)= 2465.4 kJ/kg
In the next step i calculated how much water is already in the air:
x=0.622*j*ps/(p-j*ps)=0.00988 kg/kg
where:
j=50%
ps=0.03171 bar
p=1.01325 bar

Now that i know that i can calculate the enthalpy of humid air:
h_hum = cpa t + x [cpw t + hwe] = 50.3 kJ/kg
where:
cpa = 1.006 kj/(kg*C)
cpw = 1.86 kj/(kg*C)
hwe = 2500 kJ/kg
t = 25

With 25° air temperature and 50% humidity the density is at around: 1.177 kg/m^3
--> h_hum=59.2 kJ/m^3

Now if i want to cool the room from 25°C to 23°C m=3*180*59.2/2465.4 = 12.9 kg water is required?

Mentor
This does not seem to me correct in many respects. Let's start over.

Based on the ideal gas law, what is the number of moles of moist air at 25 C for a volume of 180 m^3 and a total pressure of 1 bar? What is the mole fraction water vapor in the air initially? How many moles of water vapor are there initially? What is the mass of water vapor in the air initially? What is the mass of bone dry air?

The problem statement is actually very ambiguous. Is the added water mixed with the moist room air, or are they kept separate by a barrier? Are they both supposed to come to the same final temperature? Have you provided an exact verbatim statement of the problem?

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etotheipi
StoningTree
It's more like a thought experiment, so we can say that the air and the water never touches/are separated.

I'm not sure how i take the humidity in account when using the ideal Gas Law:
p*V = n*Ru*T
n=p*V/Ru*T (With 180m^3, 25°C and 1.01325 bar) = 7356.6044 mol ?

The mole fraction of water vapour i calculated like this:
x=partial pressure/pressure= 1583.5 Pa/101325 Pa = 0.0156
Partial pressure at 25°C and 50% humidity is 1583.5 Pa

Moles of water vapor initially:
n*x = 7356.6044*0.0156 = 114.863 mol??

Mass of water vapor in the air initially:
c(H2O) = x * rho(air)*M(H2O) = 11.508 g/m^3

rho(air) is the density of the air: rho(air) = 101325 Pa/(Ru...*288.15 K) = 41.739 mol/m^3
Ru = 8.31446261815324 kg m^2 / s^2 mol K
M(H2O)= 18.01534 g /mol

Mass of dry air:
28.9644 g /mol

Mentor
It's more like a thought experiment, so we can say that the air and the water never touches/are separated.

I'm not sure how i take the humidity in account when using the ideal Gas Law:
p*V = n*Ru*T
n=p*V/Ru*T (With 180m^3, 25°C and 1.01325 bar) = 7356.6044 mol ?

The mole fraction of water vapour i calculated like this:
x=partial pressure/pressure= 1583.5 Pa/101325 Pa = 0.0156
Partial pressure at 25°C and 50% humidity is 1583.5 Pa

Moles of water vapor initially:
n*x = 7356.6044*0.0156 = 114.863 mol??

Mass of water vapor in the air initially:
c(H2O) = x * rho(air)*M(H2O) = 11.508 g/m^3

rho(air) is the density of the air: rho(air) = 101325 Pa/(Ru...*288.15 K) = 41.739 mol/m^3
Ru = 8.31446261815324 kg m^2 / s^2 mol K
M(H2O)= 18.01534 g /mol

Mass of dry air:
28.9644 g /mol
My questions here were based on assuming that the air and water were going to be mixed. Apparently, this is not the case.

This raises a few more questions. Is the air in a constant-volume room, hermetically sealed, so that the mass and volume of air are constant? Or, is there somehow a lid on the room so that the volume can change, and hold the pressure constant?

Is the water in a container where its volume is held constant? If so, is there a certain fraction of head space to start with? Is water the only component in the container? Or,, does the volume of the container change so that water and vapor are always at equilibrium?

Did you make up this problem, or is it out of some book?

Mentor
m=3*180*59.2/2465.4 = 12.9 kg water is required?
59.2 is the enthalpy of the air. To change the temperature you need the specific heat times dT (or beginning minus final enthalpy).

Also, I'll add that the method and level of precision required may depend on whether this is a homework question or practical problem. For example, if the mass of water vapor already in the air is only 1% of the mass of air, you might safely ignore it for a practical problem, but in a thermo class the professor may want to see it. Same goes for whether the system is open or truly closed.

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Mentor
Let's suppose, for discussion purposes, that you had the air and the cooling water water in two separate containers of constant volume (i.e., the room and the water container). To cause the air in the room to cool down, at time zero, you bring the two containers into contact with one another, and allow the air and water to equilibrate. For this system, the first law of thermodynamics tells us that, under these circumstances, the change in internal energy U, rather than the change in enthalpy, is zero for the combined system.

Let's focus on the water container, and let's assume that, at the final instant that the containers reach equilibrium, the very last bit of water evaporates. It will then be a saturated vapor at 22 C, with a pressure of 0.0264 bars and a specific volume of 51.45 m^3/kg. So, if the mass of liquid water in the container were m, the volume of the container would be 51.45m cubic meters. We can backtrack to determine what the contents of the container was in the initial state at 15 C. The specific volume of liquid water and water vapor at 15 C are, respectively 0.001 and 77.93 m^3/kg. So if x is the mass fraction of liquid water initially, to have the same volume of water in the container as in the final state, we would have $$0.001x+77.93(1−x)=51.45$$This would tell us that x = 0.34, and the initial amount of liquid water would be 0.34 m. This would be the amount of liquid water that evaporated. The other 0.66 m kg of water in the container was already vapor.

If U was the internal energy of the water in the container, we know from the first law of thermodynamics that
$$ΔU=Q$$From the steam tables, the specific internal energies of liquid water and water vapor at 15 C are 62.99 and 2396 kJ/kg; and at 22 C, the specific internal energy of water vapor is 2406 kJ/kg. So the change in internal energy for the water in the container would be $$\Delta U=m[2406-(0.66)(2396)-(0.34)(62.99)]=803.2m\ Joules.$$So, per kg of water evaporated, the amount of heat required would be ##(803.2m)/(0.34m)=2362 J/kg##.

Mentor
@Chestermiller I'm having trouble seeing how that relates to the problem the OP is trying to solve. Can you explain what that was for?

Mentor
@Chestermiller I'm having trouble seeing how that relates to the problem the OP is trying to solve. Can you explain what that was for?
In doing this, he needs to be precise about the water part of the system that he is determining the heat for. This has to include whether the water is at constant pressure or constant volume, or whatever. Otherwise, the work and the change in internal energy and enthalpy will be different, and the heat will be different. My suggestion was to start with a closed container of water (and water vapor) that just finishes evaporating when the water equilibrates with the air at the final temperature.

Mentor
In doing this, he needs to be precise about the water part of the system that he is determining the heat for. This has to include whether the water is at constant pressure or constant volume, or whatever. Otherwise, the work and the change in internal energy and enthalpy will be different, and the heat will be different. My suggestion was to start with a closed container of water (and water vapor) that just finishes evaporating when the water equilibrates with the air at the final temperature.
Er...sorry, I just saw the part where he said the water and air are to be kept separated. That's a really odd constraint.

The method in the OP works fine for a standard evaporative cooling scenario, it just contains a small math error. But if the OP really means they should be separate, I'll have to put some more thought into it...

Mentor
Er...sorry, I just saw the part where he said the water and air are to be kept separated. That's a really odd constraint.

The method in the OP works fine for a standard evaporative cooling scenario, it just contains a small math error. But if the OP really means they should be separate, I'll have to put some more thought into it...
Yes. I thought a much more interesting problem would be if they were mixed.

Mentor
It's more like a thought experiment, so we can say that the air and the water never touches/are separated.

I'm not sure how i take the humidity in account when using the ideal Gas Law:
p*V = n*Ru*T
n=p*V/Ru*T (With 180m^3, 25°C and 1.01325 bar) = 7356.6044 mol ?
This is correct. The molar heat capacity of air at constant volume (neglecting the tiny amount of water vapor) is 2.5R = 20.8 J/mole. So the heat load to lower the air temperature 3 C will be (7356)(20.8)(3)=459000 J = 459 kJ. This is much less than you estimated.