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Air changes for heat removal - air compressor room 1

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breakingthescience

Mechanical
Jul 14, 2021
7
I have a 600 sq.ft. room, x 15 ft high that houses three 75HP air compressors. In lieu of ducting each compressor individually, I'd like to provide a single exhaust fan on the roof, and interlock it with a louver on the exterior wall.

Each compressor's internal air-cooled fan does 8,000 CFM.

When exhausting the room, instead of the compressor, I've seen manufacturers recommend to do twice the CFM of the compressor fan. The reason being that the airflow is sized to allow a ~30 degree rise, but you don't want a 30 degree rise in the room, so you therefore need to exhaust more.

My dilemma is that it would equate to 48,000 CFM for a 600 sq.ft. room. This is obviously ridiculous. Even 24,000 CFM seems crazy.

If I provide an exhaust fan that does 18,000 CFM, it would result in 120 air changes per hour. I would think that such a high airflow and exchange rate would throw any temperature rise formulas out the window (no pun intended). There just simply wouldn't be enough time for the room to rise in temperature, it would just become equal with the ambient makeup air.

I am looking for some thoughts on this and if anyone has an opinion on how many air changes are needed to essentially reject any amount of heat generated in a room.

Thanks!
 
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The back of my envelope says 35,000 CFM for a 20 degree F rise over intake air.
 
I think its actually closer to 25,000 CFM for 20F rise, but the question is more about whether that concept even holds true when the air is changing every 30 seconds. I'd think at a certain exchange rate, its pointless to calculate a temperature rise because it will just equal the intake air.

And how much temperature rise is too much when intake air is 95F outside air? I know I'd need to check the compressor manual, but I'd think the compressor won't like 115F air.
 
The temperature rise is from the air that passes through the fans in the compressors - if their intakes or exhausts or both are directly ducted then the room will see whatever temp it settles on from conduction with the ducts.

If the air is allowed to freely mix it will rise to some temperature close to the 20F rise.

Consider: the air coming in is at Temp A and the air leaving is at Temp B and in that temperature difference for the flow rate is the heat rejected by the compressors, so that will be the delta 20F or so as calculated above. Since the exhaust air has to be Temp B and the air from the compressor fans is freely mixed with the inlet air, then the room must be Temp A + 20F, meaning the exhaust from each compressor fan will be higher yet.

So, if you want to keep them in good shape, directly duct the outside air to the compressor fans. That ducting may required boost fans - they will add a bit of heat, but far less than the compressors will.

I can see why, to cool the room rather than duct the compressors, that the suggested flow seems ridiculously high.
 
Air-cooled air compressors do take a ridiculous amount of exhaust and it does catch people off guard. I would highly recommend ducting each individually with it's own fan. That way you might actually save operating costs because each fan will be less horsepower and will only run when it's respective compressor is running. This will also help the compressors be more efficient with their cooling because it eliminates any heated exhaust air from being pulled back into the compressor.
 
Thank you for your responses.

@nuuvox000 - I've been told that the compressors have no control output to control an external fan, so how do you suggest interlocking? Just a relay that turns the fan on if the compressor is pulling power?


I understand the heat gain is huge and I know the concept/formula of heat rise to CFM, etc. I get that the air entering and leaving the compressor will have a specific temperature rise, but my main question is about air changes. I have asked this question to many people outside of this forum and still get very similar answers to what I have received here. Everyone talks about the 1.08(CFM)dT formula but is ignoring the bigger question.

It only makes sense to me that at a certain air change rate in the room, the temperature rise idea is no longer applicable because there isn't enough time for any air in the room to rise in temperature. It is "instantly" flushed away. It is as if the compressors are in a room of infinite size, or simply placed outdoors if you want to think of it that way. You can't tell me that placing the compressors outside will raise the temperature of the outdoor environment by 20 degrees. If I have 120 ACH, or an air change every 30 seconds, I'd think the room would equalize with the temperature of the outdoor intake air, and any instantaneous heat gain from the compressors will be washed away in the wind tunnel. I used 120 ACH as an example because it is applicable to my scenario, but I'd think there is an air change rate that essentially becomes "diminishing returns" if you will. I don't know if that is 60 ACH, 120 ACH, 240 ACH, but at some point there becomes so much air exchange that any heat gain becomes negligible.


 
nuuvox000 said:
when it's respective compressor is running.

Brilliant part that I overlooked, as did the OP.

What is the duty cycle for the compressors?
 
MintJulep said:
Brilliant part that I overlooked, as did the OP. What is the duty cycle for the compressors?

I wouldn't say this was overlooked. I think it is irrelevant to the question. We are assuming a scenario with 100% load. Energy consumption is a separate conversation. You can also save energy by running the single room-exhaust fan on a VFD based on a temperature sensor, and only exhaust as much as needed with the varying makeup air conditions.

But let's get back on topic! All good conversations here, but I want to focus on the air change theory!

Thanks! :)
 
Unfortunately the actual wiring of controls is a weak point for me. But every time a put "interlock equipment with fan" on plans it just gets done by our wonderful controls installers ha ha.

When the airflow rate becomes "negligible" is up to you. Pick a temperature in the room that you don't want it to rise above and pick an outdoor air temperature. Temperature rise formulas would not go "out the window" at any airflow rate. You could calculate this easily if you assume all of the air will be well-mixed between the incoming outdoor air and the compressor exhaust air (this would be your worst case scenario).

As far as an air change every 30 seconds; yes it is a lot of air changes but you will have 8,000 CFM from each of the compressors when running at full load at 30 degrees rise (I think you said 30). That's a lot of hot air being dumped into a small room.

You'll have much better efficiency using separate 8,000 CFM fans for each compressor than one large fan on a VFD (assuming the fan can even turn down to 1/3 of the full CFM)
 
In the comparison to free-air it's as if the atmosphere has been ducted into and out of the compressor cooling fan. Outside isn't closed as the atmosphere can radiate heat into space, which it generally does quite well. It also isn't infinite, it's just big enough that a few Hp isn't going to be noticeable vs the 1kW/m^2 (or so) of solar radiation it normally encounters.

The air-changes is for a control volume for which there is a fixed volume, a fixed inlet, and a fixed outlet; this is your situation. The heat has to be rejected through the outlet and the amount is based on the difference in temperature with the inlet and mass flow rate.

What air-changes does is limit the room exhaust temperature which, since the exhaust is the temperature of the mixed room air, means that is the temperature of the room.

You say you have three 8000 CFM blowers pushing a delta-T of 30F. They are going to do a good job of mixing so, while the process is not "instantaneous," it is very rapid. To prevent mixing, you need to use ducting which, compared to the size of the compressor cooling section, provides how many ACH? I'd bet it's between 100 and 1000.
 
Ok, back on topic.

All the HP goes to heating the space.

3 * 75 = 225 HP --> 573007 BTU/hr

(my envelope last night had 4 compressors in it for some reason)

Ventilation cooling:

BTU/hr = 1.08 * cfm * ΔT

You have given the desired ΔT of 20 degrees F.

Solving for cfm = 26528

Not so very far from 3 * 8000 = 24000. So maybe, just maybe the guys that selected the fan for the compressor had the same thinking.
 
3DDave said:
The air-changes is for a control volume for which there is a fixed volume, a fixed inlet, and a fixed outlet; this is your situation. The heat has to be rejected through the outlet and the amount is based on the difference in temperature with the inlet and mass flow rate.

This helped me look at it differently. My thoughts about the number of air changes is like saying, "8000 CFM across this little radiator?? That's so much airflow across a tiny surface, the temperature rise will be negligible" Wrong. There is still a rate of heat exchange associated with a volume of air. So, if my room is viewed as a heat exchanger instead, the 1.08 * cfm * ΔT still applies in calculating the temperature rise between the inlet and outlet. Not sure why that was so hard for me to see before...

So, if you have 95F air entering the room, and its rising to 115F as it leaves, your compressor cooling section theoretically starts seeing 115F air entering, and 135F leaving?

I appreciate everyone's participation in this conversation!
 
Close - if, as you said, it puts 30F across the heat exchanger on the compressor - the compressor heat exchanger exhaust air will be 145F for a room at 115F. The compressor air will initially exhaust at 95F + delta 30 F, 125F. But not all of that will leave the room, some will go out and some will recirculate so that now it's 100F + delta 30F, then 105F + delta 30F, and so forth until the room reaches equilibrium, which for some large ACH could be 115F into the heat exchanger and 145F out. By then the room exhaust carries enough heat to match what the compressors are rejecting.

I had a similar problem to solve for an electrical test equipment rack - the original system did the ACH concept and destroyed the test gear. I put in a baffle to prevent recirculation so that every instrument was getting its own access to outside air rather than taking in heated air from items lower in the rack. The system was about home refrigerator size and could dump up to 10 kW as heat. As produced the max exhaust temp was limited to the peak temp of any of the instruments as they warmed incoming air in parallel rather than the original serial system.
 
That is a fairly decent size room, but Very small rooms are a problem because the air will just recirculate and cause the compressor to overheat. I had a project where the owner had stuffed 400hp of rotary screw units in a room about the size of a 40 foot connex box.

We ended up designing a utility set fan to pull through the duct work connected to the compressors the duct work was too small but it was the largest we could fit in the room and it had to pull about 3” of static to make it work. The majority of the wall in the front of the room had to be one large Louver to get enough into the room.



 
75hp is not the heat dissipated. It is the nominal motor shaft power. Heat dissipated comes from motor inefficiency, and compression heat. Nominal power also could mean actual power demand on shaft is 60hp etc. It doesn't seem the manufacturer actually provided heat to be removed. They provided the air flow required instead. So unless you get better information from manufacturer, do what they say. Unless you know actual cooling load generated, go by their airflow numbers.

The overall fan system needs to be designed to assume all compressors running in summer unless this is an N+1 setup and a control ensures only 2 compressors are running at any given time. You could implement controls that modulate room air exchange based on a maximum room temp, and a dT to ambient. You also want to do that ot not freeze the room in winter. Two important temps missing are the maximum room temp the manufacturer allows, and the maximum ambient temperature at the local climate.
 
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