Air changes per hour for room 1

[HawksHockey]

Ok, I've struggled with this one. I'm new at this, so here goes: I'm trying to determine a volumetric flow rate of air for a room which houses electrical equipment. I have the dimensions of the room at 15 x 27 x 20 and I have determined the cooling load at 250,000 BTU/hr. The room is underground, and will draw air at 90F and 90% RH. Here is the question, if I have accounted for 8 ACH (in my sensible and latent heat loads) but have 250,000 BTU/hr, therefore what is governing? From AHSRAE, Flow=Total Heat Load/(density of air x specific heat of air x 60 x temperature difference). With density at 0.07 and specific heat @ 0.24, and a temperature difference of 15 F, my flow rate comes up to 16,500 CFM? Could this be the volumteric flow rate through the 21 ton unit in order to cool the air and not the flow rate into the room (which I thought was 8 times the volume of the room)? I'm lost.

 

[gepman]

I am not sure I understand your information exactly but it appears that you are bringing 100% outside air. An electrical room should have very little if any latent load. Ventilation air requirements should be very little for an electrical room since the occupancy is very low. Therefore if your cooling load is 250,000 btu/hr from the electrical equipment then you should not have much outside air makeup to have any appreciable latent load.

Just bring in the minimum of outside air for ventilation requirements and recirculate all (almost) the air inside the room. If your 250,000 btu/hr includes the latent and sensible loads to bring the room down to whatever the design ambient in the room will be (I usually size 85 deg. F for worst case in an electrical room) then you will need a lot less then a 21 ton unit.

 

[ChrisConley]

250,000 Btu/hr with a 5 deg F temp rise will require a staggering 46300 cfm of outside to cool the space to 95F at your conditions.

It is much more pratical to consider mechanical cooling. At which point, you only need to bring in the minimum O/A and add that load to your 250,000 Btu/hr sensible load.

Is your minimum outdoor air 8 ACH? That seems high for an electrical room.

Given your high RH evaporative cooling doesn't look like an option.

 

[HawksHockey]

The electrical room is actually underground (mine) and we are drawing air from the drifts being supplied by the main shaft ventilation. However, you are right, there is very little latent load - primarily due to the room air changes, the vast majority of the load is sensible from the equipment in the room. The actual sensible load is 210,000 BTU/hr and the latent load is 40,000 BTU/hr. I can also live with 85 degree air at a higher RH, and I can reduce the latent load by reducing the outside air into the room, but do I not have to size the cooling for 210,000 BTU/hr (17.5 tons)?

 

[gepman]

I am unclear as to whether you are trying to cool this room by ventilation air only or by mechanical refrigeration (or a hybrid).

You definitely want to keep the RH lower than 90% in an electrical room. For ventilation only cooling, if you have high heat loads and high ambient air it requires a very high air flow.

If the sensible load in the room is 210,000 btu/hr you need to size for it. The heat load seems a little high for an electrical room. What is in it? Transformers and VFD's have relatively high heat loads for electrical equipment. FVNR starters, switchboards, etc. have relatively low heat loads.

 

[HawksHockey]

Sorry, maybe I should have called it a sub station. I've got the equipment list, it includes transformers (750 KVA, and 2 MVA transformer @ one each) which are the largest contributor to the loads at 25 W/KVA and 20 W/KVA respectively. So the load from the 2 MVA transformer looks to be 40,000 Watts or 136,600 BTU/HR.

 

[HawksHockey]

The more I look into this, the more confused I get. I was basing my calcs on an old version of the e-room and subs equipment list. I do not think that they are accurate. Therefore my loads may be out to lunch. I have a lot of work to do. Thanks for the help. It's appreciated.

 

[CountOlaf]

If the size of the transformers are correct (total of 2750 kVA), your equipment heat gain seems correct (58.75 kW, or 200 MBH, or 16.7 tons for the xfmrs alone). The 8 ACH doesn't seem too bad either (as a ventilation rule of thumb it could be 10 ACH), although this won't provide adequate cooling by itself of course. Perhaps you simply forgot to divide by 60 to get CFM? For instance, your room volume is 8100 CF. Exchanged 8 times per hour is 64,800 CF/HR which is only 1,080 CFM. This is less than 10% of the other CFM number you had (16,500 CFM), which would be a typical percentile for outside air without overpressurizing the space.

At any rate, as suggested, if you use non-mechanical cooling methods (all 90 deg air) and your upper limit is 104 deg F (40C), then the 14 deg temp difference will force the air flow full volume to be around 16,000 CFM as you calculated, assuming 250 MBH is correct. Is there a higher upper limit allowed by the electrical equipment, perhaps 50C? If so, your CFM without mechanical cooling will drop down to around 7,000 (temp diff of 32 deg F). WIth mechanical cooling, the delta T will go up and the air flow rate requirement will go down as well.

 

[HawksHockey]

Yes, if I understand correctly - I will be utilizing some sort of cooling. At this point I was looking at a water/air cooling coil as I have access to chilled water. So if I use 100% outside air (which is conditioned air to 90 F) I should be able to cool the room incoming air to a respectable value (between 75 and 80) at between 6 and 10 ACH, and exhaust a controlled amount of "heated air" from the room.

 

[ChrisConley]

This isn't working in my head. I'm going to restate one more time.

HH, you have two loads:

1) Electrical equipment load
2) Outside air load

If you use mechanical cooling you will need to have a fan coil or similar device to cool the 250 MBH load you have from your electrical equipment.

If you introduce O/A (at 90F) you will need to cool this air to the acceptable room temperature.

If 90F is lower than the acceptable room temperature, you can use O/A to provide some, or all, of your cooling. The factor that will determine how much O/A you are using is what is an acceptabe air volume rate.

Where does the 6-10ACH come from? What is the acceptable room temp?

 

[HawksHockey]

Once again thanks for the help. The acceptable room temperature is 80F. Your two points are valids Electrical load and 2) outside air load. I have calculated all heat loads (ie. Latent and Sensible). The total cooling load is in the neighbourhood of 250,000 BTU/hr. ACH of 6-10 was chosen for human occupancy (if required).

 

[gepman]

I assume that you need all of the air changes because it is an underground mine. If you don't need all these air changes I would cut back on them and reduce the load somewhat. Otherwise you have it right, you have your equipment (sensible) load and your outside air load (latent and sensible).

 

[quark]

You shouldn't, first, fix up number of air changes per hour. Calculate the flowrate based upon the sensible load and check whether you can manage dehumidification as well.

Now, you have the volumetric flowrate with you. If this is less than the no. of ACPH you require, you can increase the flowrate. However, you may have to either reheat the conditioned air or treat part of the air and then mix with untreated air. In this case, volume flowrate through coil and volume flowrate through room can differ.

If volumetric flowrate by cooling load calculation is more than volumetric flowrate obtained by no. of ACPH, then the flowrate through coil and that of room should be equal.

 

[HawksHockey]

Thanks for the help everyone, all the information was extremely helpful and has me going in the proper direction...Thanks

 

[gepman]

I guess it all depends on what you consider the definition of air changes per hour to be. Is it calculated on the basis of outside (fresh) air or total air (outside and recirculated) air through the coil?

 

[HawksHockey]

That is the question I've come to as well. ASHRAE recommends certain air changes per hour, but is this fresh treated air (ACH) or is this total volumetric flow rate to (through) the room (ACH). I have a hard time beleiving that 250,000 BTHU/HR of load can be dissipated with 8 or 10 air changes per hour at a temperature that is 15 F cooler than outside air! This is what I've been struggling with.

 

[CountOlaf]

IMO, forget about the ACH. The recommended number of ACH for ventilation is probably only valid if you were exhausting the electrical room, and you were making up the exhaust with 100% outside air (or transfer air) and this air temp was enough to keep the equipment below whatever upper limit it coud handle--in other words, satisfy the load.

Since you want to keep the space at 80 or something, we've established you can't do it with ventilation alone, therefore you need mechanical cooling. Since it's an unoccupied space, you don't require ANY ouside air. Therefore, you can recirculate all of the air and cool it to satsify the load and the CFM requirement will come from this analysis alone, not ACH. Now, as I suggested, if you want to bring in 10% or so of outside air, this will help pressurize the space and may even provide (coincidentally) around 6 to 10 ACH of fresh air.

 

[ChrisConley]

Where does ASHRAE recommend ACH in an electrical room? Their ventilation standard (62-2007) is primarily based on cfm/person + cfm/ft2.

If their is a recommended ACH (from some source) then it is definetly outside air only.

 

[RossABQ]

If the room is down in a mine, I would be leery of not having some true OSA ventilation (not air from outside the room, still mine air), due to potential for methane or other gases that are typically present downhole. Any accumulation in an electrical room could be trouble. I'm sure there is also a reuqirement for monitoring and routine maintenance of the equipment, that requires personnel to be in there at times.

 

[gepman]

I would assume the the outside or ventilation air requirements would come from MSHA CFR 30, Chapter 57, Subpart G.

http://www.msha.gov/30cfr/57.0.htm

I agree with ChrisConley that ASHRAE requirements for electrical rooms don't require 8-10 ACH of outside air.

Obviously minimizing the outside air will minimize your cooling load since the outside air is warmer than the temperature setpoint of the room.