checonbill
Electrical
Could someone let me know what would be a reasonable air change for a small compressor room 15 x 15 x 12 with a 150 hp compressor I need to drop the temperature in the room it is apx 140 degrees when the outside temp is 90.
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Air compressor room air changes
- Thread starter checonbill
- Start date
- Status
checonbill (Electrical)
I have a similar compressor room. 10x20x12. I am removing 1500cfm which gives me 2 air changes per min.
Even that is marginal on a hot day with the sun beating on the compressor room.
I am presuming that your compressor is venting it's heat into the room.
B.E.
I have a similar compressor room. 10x20x12. I am removing 1500cfm which gives me 2 air changes per min.
Even that is marginal on a hot day with the sun beating on the compressor room.
I am presuming that your compressor is venting it's heat into the room.
B.E.
trashcanman
Mechanical
As per ASHRAE Handbook of Fundamentals, for a motor driven machine, both in the conditioned space,
heat gain = (Hp/Eff.)*load factor*2545 = BTUH
Load factor is the fraction of the load delivered (how much does it run?).
Assuming 80% efficiency and 50% load factor (runs half the time),
BTUH = (150/.8)*.5*2545 = 238594 BTUH.
Assuming room is to be kept at 95,
CFM = 238594/((95-90)*1.08) = 44184 CFM. I would use a roof mounted axial fan. They move lots of air at low static pressure.
Of course, this neglects solar heat gain. As the ventilated room temperature approaches ambient air temp, the cfm required goes to infinity. The only way to get it to ambient (without refrigeration) is to remove the roof.
Hope this helps.
heat gain = (Hp/Eff.)*load factor*2545 = BTUH
Load factor is the fraction of the load delivered (how much does it run?).
Assuming 80% efficiency and 50% load factor (runs half the time),
BTUH = (150/.8)*.5*2545 = 238594 BTUH.
Assuming room is to be kept at 95,
CFM = 238594/((95-90)*1.08) = 44184 CFM. I would use a roof mounted axial fan. They move lots of air at low static pressure.
Of course, this neglects solar heat gain. As the ventilated room temperature approaches ambient air temp, the cfm required goes to infinity. The only way to get it to ambient (without refrigeration) is to remove the roof.
Hope this helps.
Our numbers are a little off, but I agree with trashcanman's calcualtions. I usually figure a laod factor of 1, so my numbers are double for the same delta-T. It seems liek every industrial compressor room I have ever been in have all units running full-bore.
In lieu of specific manufacturer data, here is what I have used.
1 HP = 746 Watts
Assumme 80% efficiency, so each compressor HP = 933 Watts
Add 1/10 HP per compressor HP for dryers or 93 Watts
This totals 1,026 Watts per compressor HP or roughly 3,500 BTUH per compressor HP.
(CFM/HP) = (BTUH/HP)/1.08/Delta-T
Your Delta-T is your allowable or desirable room temperature minus the outside air temp. In other words, if you want to keep the room at 105, and your outside air is 95, then your Delta-T is 10.
Using 10 deg Delta-T, your required CFM/HP is 3,500/1.08/10, or 324 cfm/HP
With a 150HP compressor, you will need 48,600 CFM. This is obviously a truckload for such a small room. You may not have enough free wall/roof space for fans and louvers. Direct vent the compressor heat outside if you can. Be careful, however, because most air-cooled compressor fans can handle very little static. Consult with the manufacturer.
In lieu of specific manufacturer data, here is what I have used.
1 HP = 746 Watts
Assumme 80% efficiency, so each compressor HP = 933 Watts
Add 1/10 HP per compressor HP for dryers or 93 Watts
This totals 1,026 Watts per compressor HP or roughly 3,500 BTUH per compressor HP.
(CFM/HP) = (BTUH/HP)/1.08/Delta-T
Your Delta-T is your allowable or desirable room temperature minus the outside air temp. In other words, if you want to keep the room at 105, and your outside air is 95, then your Delta-T is 10.
Using 10 deg Delta-T, your required CFM/HP is 3,500/1.08/10, or 324 cfm/HP
With a 150HP compressor, you will need 48,600 CFM. This is obviously a truckload for such a small room. You may not have enough free wall/roof space for fans and louvers. Direct vent the compressor heat outside if you can. Be careful, however, because most air-cooled compressor fans can handle very little static. Consult with the manufacturer.
There may be an error in here somewhere. If a 150 HP motor/compressor has an efficieny of 80%, that means that 80% or 120 hp goes into useful work of compressing the gas. The remaining 20% shows up in heat from the motor windings, bearings, inefficiency, etc. The heat in the room (igonoring solar, vents, conduction, and other sources and sinks) is then 20% of 150 hp or 30 hp which equates to about 1272 btu/min. An airflow of 14,553 ft^3/min is required for a 5 deg temp difference at a load factor of 1. It gives 5.4 air changes per minute.
Regards
Dave
Regards
Dave
checonbill (Electrical)
This is why I asked if the compressor was venting it's heat into the room.
Without the heat load of the inter cooler and the dryer, 2 to 3 air changes per min should cover it with capacity to spare, if not then as CESSNA1 says you will need 5 to 6 air changes. The calculation imok2 (Mechanical) did on your related post looks closest if you do not have an inter cooler venting in the room.
This is why I asked if the compressor was venting it's heat into the room.
Without the heat load of the inter cooler and the dryer, 2 to 3 air changes per min should cover it with capacity to spare, if not then as CESSNA1 says you will need 5 to 6 air changes. The calculation imok2 (Mechanical) did on your related post looks closest if you do not have an inter cooler venting in the room.
- Status
Similar threads
