# Homework Help: HVAC Humidity Question.

1. Nov 3, 2013

### SherlockOhms

1. The problem statement, all variables and given/known data
A pharmaceutical company operating a tabletting facility in Puerto Rico requires stringent environmental control set at 20°C and 40%RH in order to provide consistent high quality production. The HVAC system is designed to cope with outdoor tropical conditions of 30°C and 90% RH. The tabletting hall is 200$m^3$ and requires 5 changes per hour.

• How much water is removed from the air flow per hour?
What is the rating of each of the heat exchangers?

2. Relevant equations
Use of psychmoetric charts to find absolute humidity.

3. The attempt at a solution
So, to find the mass of water removed from the air flow per hour you find the difference in humidity between the starting and final conditions using the psychometric charts and multiply this found value by the mass flow of air into the room per hour. $Q$ can be found by multiplying the volume of the room, $V$ by the number of changes per hour, 5 to get the volume of air in per hour. Then dividing by the specific volume$\frac{1}{\rho}$, also found via the psychometric chart, we get the mass in per hour, $\dot{m}$. Multiplying $\dot{m}$ by the difference in humidity gives the mass of moisture removed per hour. Does this sound right? My real question is, having done this, how do I find the heat exchanger rating?

2. Nov 3, 2013

### Staff: Mentor

You're going to have to condense the water vapor out below 20 C to guarantee that, when the air goes back to 20 C, its relative humidity is 40%. What temperature do you have to condense the water out at? You start out with water vapor at 30 C, and you end up with saturated liquid water at the condensation temperature. Knowing the condensation temperature, what is the change in enthalpy per pound of water condensed? You also have to cool all the air down to this temperature.

3. Nov 3, 2013

### SherlockOhms

I'm a little lost, but I'll give this a go. So, the temperature at which the liquid will condense out at will be the saturation temperature for that given humidity, right? I'll list it below. Are you asking me for the change in enthalpy between the initial conditions (30C and 90%RH) and the condensation temperature (i.e. the saturation temp.)?

Initial Humidity: 0.0244
Final Humidity: 0.0058
Saturation Temp/Wet Bulb Temp for Final Humidity: 12.2 C
At Temp 30C and 90%RH, h = 92.45 kJ./kg.
At Temp 20C and 40%RH, h = 34.8 kJ/kg.
From the psychometric chart, how do I read off the enthalpy at the saturation temp (12.2C)?

Last edited: Nov 3, 2013
4. Nov 3, 2013

### Staff: Mentor

What units are you using for absolute humidity(mass per unit volume, or mass per unit mass)?

I think I would do this a little differently from how you did it in the original post. If there are 200 m^3 of air in the room at 20 C and 40% RH, I would first use this to calculate the amount of bone dry air and the amount of water vapor. The amount of bone dry air is what is constant in this problem. Therefore, the next question would be that, with this same amount of bone dry air, how much water would be present at 30C and 90% humidity. The difference would be the amount of water removed per charge.

As far as the second part is concerned, it would seem that there is more than one heat exchanger involved. Is there a flow diagram showing the process setup? The reason I'm asking is that some of the cooled air, either at 20 C or at 12.2 C could be used to precool the outside air. Also, it's necessary to reheat the saturated air coming out of the condenser from 12.2 C to 20 C. And, is there any incentive for using the cold condensate at 12.2 C, or is it just discarded?

Chet

5. Nov 4, 2013

### SherlockOhms

First of all, the absolute humidity readings are in units of kg/kg.

I see what you're saying. How would I calculate the mass of bone dry air present using the conditions (Temp and RH)? And on top of that, how would the amount of water vapour then be calculated? I'm a little slow to catch on to this method as the initial method I used was the way shown to us in class.

As for the heat exchangers, there's no PFD given to you as that's actually a sub question. You're asked to draw a PFD for the HVAC process. So, is there some set of charts or anything which you use to calculate the heat exchanger rating? I've been looking through our notes and can't find any reference to hear exchanger rating. Finally, how is it that you use the psychometric chart to read off the value of the enthalpy for the saturation temp? Because, if there's reheating required from the saturation temp. (12.2C) back to 20C, then the change in h is needed between these two temperatures. The way I'm reading them, following the diagonal line from the saturation temp to the enthalpy readings, gives the same enthalpy as the 20C and 40%RH condition. Not sure if you can visualise what I'm talking about or not though.

6. Nov 4, 2013

### Staff: Mentor

I don't have time to answer right now, but I'll get back with you later. But, I have had second thoughts about doing the first part by the method I discussed above, because it might be splitting hairs to get a tiny amount more accuracy. I think the method you used to get the mass of air and the mass of water condensed out is just fine. Get back with you later concerning the heat exchanger loads.

Chet

7. Nov 4, 2013

### SherlockOhms

Thanks for that. That and how to read the enthalpy at the saturation temp. from the psychometric charts would be great.

8. Nov 4, 2013

### Staff: Mentor

OK. I'm back. Sorry I took so long. Grandchildren visiting.

I hope you calculated the amounts (kgs) of water and air correctly in (a) the room at 20 C and 40% RH, and (b) the inlet air at 30 C and 90% humidity (containing the same mass of air on a dry basis as in (a). I've calculated these values too, so I can check you now. Also, how much water did you find is removed per charge?

Your value of 12.2 C is incorrect. The saturation vapor pressure of air at 20 C is 17.5 mm Hg, and, with 40% RH, the partial pressure of the water vapor in the room is only 7 mm Hg. The temperature at which the saturation vapor pressure of water is 7 mm Hg is 6 C, not 12.2 C.

As best I can tell, you are going to have 2 heat exchangers. One is a condenser to cool the initial air from 30 C to 6C, and condense out the liquid water. The other is a heater to reheat the 6C air coming out of the condenser back up to 20 C.

There is no phase change in the heater, and so it is a single stream. So it's easier to get the heat load. The mass fractions of air and water vapor in the stream passing through this heater are constant. At the inlet, the stream is at saturated with water at 6 C, so its RH is 100 %. I'm sure they must give the enthalpy per pound of this gas for 100% RH in your tables. Its absolute humidity is the same as the 20 C room air, but it is at 6C. The exit air is exactly the 20C room air.

Now for the condenser. The stream going into the condenser is the 30 C, 90% RH air. There are two streams coming out: the 6C saturated air, and the pure water removed at 6C. It is easy to get the enthalpies of the inlet and outlet air streams. But, part of the inlet air stream is water vapor that will eventually exit the condenser as the 6C condensate. You need to calculate the change in enthalpy in taking 1 kg of water vapor from 30 C to saturated liquid water at 6C, including the heat of condensation. You need to choose any convenient path to calculate this change. This is all necessary, unless your physhometric chart also includes mixtures of air saturated with water vapor and saturated liquid water in various proportions. That would make life much easier. Does it?

9. Nov 5, 2013

### SherlockOhms

No problem. Thanks for the reply.

Ok, I was reading the values of the wet bulb temp. as the condensation temp. instead of using the dew point temp. I've calculated the values for the amount of water removed/hr as follows:

abs humidity @30C and 90%RH = 0.024 kg/kg
abs humidity @ 20C and 40%RH = 0.0058 kg/kg

$\nu$ @ 20C and 40%RH = 0.837 m^3/kg. (Using this to calculate the mass flow rate of air in per hour. This is an assumption our lecturer told us to use)

$\dot{m}$ =( 200 m^3)(5 changes/hr)/(0.837 m^3/kg) = 1194.74 kg/hr.

1194.74 kg/hr(0.024 kg/kg - 0.0058 kg/kg) = 21.7 kg of water vapour removed per hour. Is this correct?

Also, my psychometric chart doesn't include the mixtures of air saturated with water vapor and saturated liquid water in various proportions.

I'm a little confused with the following part:
I can calculate the enthalpy of the air/water mixture at 30C and 90%RH to be 92.44 kJ/kg and the enthalpy of the saturated mixture at 6C and 100% RH to be 20.61 kJ/kg. So, if I , multiply 92.44 kJ/kg by the absolute humidity at 30C and 90%RH, do I get the enthalpy of the water vapour at these conditions? Then doing the same for the enthalpy value at 6C and 100%RH , I can calculate the change. Is this correct or have I gone wrong? Also, why is it that I'm trying to calculate this? Will it help me determine the heat exchanger rating? I've never actually heard of 'heat exchanger rating' before this problem, so apologies if I'm picking up on this a little slowly.

Last edited: Nov 5, 2013
10. Nov 5, 2013

### Staff: Mentor

Yes. I got 1200 kg/hr and 22.5 kg/hr.

The heat exchange rating is the same as the heat load of the heat exchange in kJ/hr.
There are two heat exchanges involved. First the easy one: You need to heat the saturated mixture at 6C and 100% RH back up to 20C, where its RH becomes 40%. From your calculations, the flow rate of this stream is 1194.74 kg/hr, and from your previous postings, its initial enthalpy per kg is 34.8 kJ/kg. So, the heat load of this heat exchanger is 1194.74(34.8-20.61)=16,700 kJ/hr = 16.7 MJ/hr.

Now for the other heat exchanger:

You have one stream entering: (1194.74+21.7)=1216.44 kg/hr at 30 C and 90% RH with specific enthalpy 92.44 kJ/kg=1216.44*92.44=112447 kJ/hr entering

You have two streams exiting:
1194.74 kg/hr at 6C and 100% RH with specific enthalpy 21.61 kJ/kg=1194.74*21.61=25818 kJ/hr leaving
21.7 kg/hr liquid water at 6C

To find the heat load for this heat exchanger, you need to know the specific enthalpy (kJ/kg) of the liquid water stream, so that we can multiply it by the 21.7 kg/hr to get the rate at which enthalpy is leaving in this stream. Even though the size of this stream is very small compared to the size of the air flow, the heat of condensation of the water is very high, and it could figure in the enthalpy balance. Using your phychometric tables, there has to be a way of getting the specific enthalpy of pure liquid water at 6 C. For example, if we knew the reference state for enthalpy (i.e., the conditions at which the specific enthalpy is taken to be zero for preparation of the table), we could do it. For example, one set of reference conditions might be: specific enthalpy of bone dry air at 20 C is taken to be zero, and specific enthalpy of pure liquid water at 20 C is taken to be zero. Please examine your psychometric table and find out the reference state that was used as a basis for preparing the table. This shouldn't be too have to find. It is usually on the first page of the tables, at the very top.

Without this information, we won't be able to figure out the heat load of the condenser.

Chet

11. Nov 6, 2013

### SherlockOhms

Thanks a million for that. Makes a lot more sense now. However, after looking at my psychometric chart I can't actually see any reference condition. I'll post a picture of it to clarify that. You see, it's not actually set of tables that we've been given. Only one handout of a psychometric chart in SI units.

The enthalpy is measured to be 0 kJ/kg at about 0C dry bulb temperature and 0%RH if that's needed? So, as well as the specific enthalpy of the water at 6C, is the latent heat of condensation also required to calculate the heat load of this condenser? Is it possible to find both these values from some chart/calculator on the internet?

12. Nov 6, 2013

### SherlockOhms

Here's the psychometric chart. It may not be fully legible, but you'll be able to make out that there's no reference condition at the top.

Last edited: Nov 6, 2013
13. Nov 6, 2013

### SherlockOhms

Actually, I may be completely wrong here but wouldn't it actually be like this:
The first heat exchanger is a condenser, used is the one to cool the incoming air from 30C and 90% RH to 6C and 100%RH. So, this incoming air would be at a rate of 1194.74 kg/hr (Calculated earlier, yes?), then the outflows from the condenser would be liquid water at 21.7 kg/hr and (1194.74 - 21.7) kg/hr of bone dry air. The way in which you're doing it is by saying that the incoming air is at (1194.74 + 21.7) kg/hr. But, isn't the 1194.74 kg/hr the total mass flow rate in (i.e. the mass flow rate of the air/water vapour mixture at 30C and 90%RH)?

Then, wouldn't the incoming stream to the second HE(the heater used to raise the temp of the bone dry air) be at (1194.74 - 21.7) kg/hr of bone dry air seeing as the 21.7 kg of water vapour has been removed earlier by the condenser?

So, essentially, where we're differing is you're saying that the mass flow rate into the room every hour is 1194.74 kg of bone dry air, excluding the water vapour. And then, you're adding the 21.7 kg of water removed to get a total mass flow rate in.

While, what I'm saying is the mass flow rate calculated earlier, of 1194.74 kg/hr, is the total mass flow rate of the air/water vapour and the 21.7 kg/hr is the water vapour component of this. Thus, the flow rate of the bone dry air by itself is 1194.74 - 21.7 kg/hr and not 1194.74 + 21.7 kg/hr.

See what I'm saying? I could be completely incorrect though. Apologies for getting a bit sidetracked.

Last edited: Nov 6, 2013
14. Nov 6, 2013

### Staff: Mentor

Well, actually, you calculated the 1194 for the 20 C air at 40% RH, which had most of the water vapor removed. That's why your instructor allowed you to us the specific volume of dry air. To get the flow rate of the air with the 30 C 90% humidity air (which enters with much more water in it), you have to add back the amount of water that condensed out.

Getting back to our calculation of the condenser heat exchanger, it is clear from the psychometric chart you sent that the datum for zero enthalpy of water for the chart is liquid water at 0 C. Note that, at 0 C on the chart, the enthalpy at 100% RH is 10 kJ/kg. This is consistent with the heat of vaporization of water at 0 C. I looked up the heat of vaporization of water at 0C, and it was about 2500 kJ/kg. From your chart, the absolute humidity of this air at 100% RH and 0 C is about 0.004 kg/kg. So, if the datum of zero enthalpy for the water were 0 C, the specific enthalpy of the air at these conditions should be (2500)(0.004)=10 kJ/kg. This confirms that the datum for zero enthalpy of water for the chart is liquid water at 0 C.

Next, lets calculate the rate at which enthalpy is exiting the condenser in the liquid stream. The heat capacity of water is 4.186 kJ/kg. So the rate is 21.7(4.186)(6 - 0) = 545 kJ/hr. We can now do the heat balance on the condenser.

Q = 25858 + 545 - 112447=-86000 kJ/hr = -86 MJ/hr

This is the rate at which heat is being removed from the gas stream (86 MJ/hr) by the condenser.

The heat of condensation/vaporization is inherently included in the enthalpies of the streams, and does not have to be accounted for separately. I know that I may have implied that this might not be the case in a previous post, but, clearly, separate accounting of the latent heat is not necessary (mia culpa).

The next thing to think about is how you might be able to save energy by using the cold gas at 6 C to precool the 30 C and 90% RH air, while, at the same time, reheating the 6C air.

Another thing to think about is how you might go about using first principles to construct a physchometric chart like this one from scratch.

Chet

15. Nov 6, 2013

### SherlockOhms

I got it. So, would I actually have been able to use the specific volume of the air at 30C and 90%RH and then find the dry air by subtracting the amount of water that was removed?

Anyway, thanks for that. Why is it that energy in the water stream was calculated using a 0C as it's initial temperature? I can see why it's final temp. is 6C but when is is 0C?

I'll have a think about it. I guess it'll reduce the energy required by the heater to heat the cool gas to the final temp as well as reducing the amount of energy required cool the initial air. I'll give it some thought. Thanks for that.

16. Nov 6, 2013

### Staff: Mentor

It would be necessary to know the volume of the stream to do this. But, this is unnecessary, because you know the amount of dry air from the 20 C calculation, and you know the absolute humidity. In fact, this is what you actually used to do the 30C stream.
0C is not an initial temperature. The enthalpy of all the streams is calculated relative to 0C. So, to be consistent with how the other streams are referenced, you need to get the enthalpy relative to 0C.

17. Nov 6, 2013

### SherlockOhms

Got it! Thanks a million for all the help.