Calculating Ventilation Cooling Load
Calculating Ventilation Cooling Load
(OP)
I'm working a job that needs to cool supply air for a small laboratory that requires a high rate of fresh air exchange. They need the environment maintained at a DBT of 65ºF (humidity will be controlled via high pressure nozzle system and will fluctuate based on user-input). I was hoping to run my calculations by someone in the HVAC field.
Application:
Air Flow- 300 CFM
Target Room DBT- 65ºF
ASHRAE Data:
Max Dry Bulb / Mean Coincident Wet Bulb Temp- 97.5 / 76.1ºF (37.85%RH)
Max Wet Bulb / Mean Coincident Dry Bulb Temp- 80.24 / 90.86ºF (63.46%RH)
Psychrometric Data:
DBT: 90.86ºF
WBT: 80.24ºF
RH: 63.46%
Humidity Ratio: 0.0199 lb/lb
Spec Volume: 14.3173 cu.ft./lb
Enthalpy: 43.7257 Btu/lb
Load Calculation
I know what my target dry bulb temperature is (65ºF), but how do I know what the resulting humidity level will be after the cooling coil? Should I just assume all the moisture will be pulled out of the air?
Here's my initial calculation based on the change in enthalpy (assuming 1% RH after cooling). I'm not sure the assumptions are correct, but is the formula correct?
DBT: 65ºF
WBT: 41.15ºF
RH: 1.00%
Enthalpy: 15.7411 Btu/lb
300 CFM = 18,000 cu.ft/hr
18,000 cu.ft./hr / 14.3675 cu.ft/lb = 1,252.83 lbs/hr
43.7257 - 15.7411 = 27.9846 Btu/lb (delta enthalpy)
27.9846 Btu/lb x 1,252.83 lbs/hr = 35,059.95 Btu/hr
35,059.95 / 12,000 = 2.92 tons
Thanks in advance!
Application:
Air Flow- 300 CFM
Target Room DBT- 65ºF
ASHRAE Data:
Max Dry Bulb / Mean Coincident Wet Bulb Temp- 97.5 / 76.1ºF (37.85%RH)
Max Wet Bulb / Mean Coincident Dry Bulb Temp- 80.24 / 90.86ºF (63.46%RH)
Psychrometric Data:
DBT: 90.86ºF
WBT: 80.24ºF
RH: 63.46%
Humidity Ratio: 0.0199 lb/lb
Spec Volume: 14.3173 cu.ft./lb
Enthalpy: 43.7257 Btu/lb
Load Calculation
I know what my target dry bulb temperature is (65ºF), but how do I know what the resulting humidity level will be after the cooling coil? Should I just assume all the moisture will be pulled out of the air?
Here's my initial calculation based on the change in enthalpy (assuming 1% RH after cooling). I'm not sure the assumptions are correct, but is the formula correct?
DBT: 65ºF
WBT: 41.15ºF
RH: 1.00%
Enthalpy: 15.7411 Btu/lb
300 CFM = 18,000 cu.ft/hr
18,000 cu.ft./hr / 14.3675 cu.ft/lb = 1,252.83 lbs/hr
43.7257 - 15.7411 = 27.9846 Btu/lb (delta enthalpy)
27.9846 Btu/lb x 1,252.83 lbs/hr = 35,059.95 Btu/hr
35,059.95 / 12,000 = 2.92 tons
Thanks in advance!





RE: Calculating Ventilation Cooling Load
RE: Calculating Ventilation Cooling Load
Stick this in Google:
cooling coil psychrometric chart
You'll get lots and lots of resources regarding the process.
You also must consider the air conditioning load from the equipment, people, and lights in the room. 65F supply air temperature is not likely to maintain 65F room air temperature.
Best to you,
Goober Dave
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RE: Calculating Ventilation Cooling Load
It seems I may be a bit lost. I've oriented myself to the psychrometric chart, but most of the formulae [Q = 4.5 x CFM x (delta h), or Btu/hr = CFM x Density Factor x (T1-T2) + CFM x 0.69143 x (G1 - G2)] require that you already know the coil leaving air conditions, including humidity. How do I determine coil leaving humidity? Is this determined by my target room humidity?
If I say my target humidity is 70% RH for the lab, then here's the calculation I was able to come up with:
Q(coil load) = 4.5 x CFM x (delta enthalpy)
Q = 4.5 x 300 x (43.7257 - 25.6181)
Q = 4.5 x 300 x 18.1076
Q = 24,445.26 Btu/hr = 2.04 tons
or, using the second formula:
Btu/hr (total) = 300 x (1.08 + [(70-97.52) x 0.024)/10]) x (97.52 - 65) + 300 x 0.69143 x (7000 x 0.01444993 - 7000 x 0.00919207)
Btu/hr = 300 x 1.013952 x 32.52 + 300 x 0.69143 x 36.80502
Btu/hr = 9,892 + 7,634
Btu/hr = 17,526 = 1.46 tons
Something doesn't add up there. Could anyone walk me through this load calculation? I would be forever in your debt.
Thanks in advance.
RE: Calculating Ventilation Cooling Load
>>>These fora should not be used to bypass your own in-depth research on the issues that affect you, nor is it intended to be a substitute for appropriate professional assistance within your field or geographical region.
You should hire an Engineer who does this.
RE: Calculating Ventilation Cooling Load
Ten air changes is high but not huge. One point to consider: Your lab has people, equipment, lights, walls, a roof, infiltration, and other what-not that will be adding or taking heat from the space. Supply air of 65F will never make your room 65F. We don't make the supply air conditions equal to the desired room conditions. We choose supply air conditions to meet the load in the room.
Best to you,
Goober Dave
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