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Hello. HVAC is not my area of expertise. My background is in industrial/agricultural engineering. However, 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

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

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Originally Posted by olywafox
Hello. HVAC is not my area of expertise. My background is in industrial/agricultural engineering. However, 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

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

There are problems here. For one, your target RH of 1% puts your dew point at -57 deg F. To drive the air down low enough to get that the air has to be at or below -57 deg F. Now do you mean 41.15 RH? Which would be obtainable at only 40.9 deg dew point.

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If you have 2.9 tons to throw it and the entering air is 300 cfm at 90/80 100% OA you can achieve about 47/46.57 off your coil. That gives you a DP of 46.2 so with a room air temp of 65 and DP of 46.2 your RH is 51% (theoretically).

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Originally Posted by vangoghsear
There are problems here. For one, your target RH of 1% puts your dew point at -57 deg F. To drive the air down low enough to get that the air has to be at or below -57 deg F. Now do you mean 41.15 RH? Which would be obtainable at only 40.9 deg dew point.
Are you referencing dew point in relation to the required coil temperature? I.e., with a coil temp of 40.9ºF, the coil leaving humidity will be 41.15% RH?

The target RH for the lab is more like 70%. Most of the formulae I've come across [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

However, something doesn't add up there. Could anyone help clarify this? I don't want to utilize 2.92 tons, if it isn't necessary.

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Your BTU calculations seem correct for your heat load. if you want to figure out how much DE-humidification is needed you need to calculate the difference in grains per lb. calculate the difference in grains then multiply by the amount of lbs. per cf of air that you are moving per hour and that will give you your total amount of grains per hour needed to be removed.

Divide that number by 7,000 (7,000 grains = 1lb. of water) and that will give you how many pounds of water that needs to be removed(8.33 lbs = 1 gallon). you can then size your unit appropriately. be careful when looking at condensing units when sizing. a 3 ton unit will not give you 3 tons of sensible cooling. might be 2.5 tons sensible and .5 tons latent.

I hope that helps.

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The problem you are having is that the point you are aiming at is the final condition after the air reheats or mixes in the room; the unit doesn't go straight there. It rides the system curve over by the wetbulb curve on the psych chart, then reheats as it mixes with the room air, following the dew point line directly right to the final room condition.

Using your example to get 70% RH in the lab and 65 deg F DB you have to look directly left to the wet bulb curve following the dew point line not the wet bulb line to the system curve closest to your starting temp. You are out of range of that curve (which is why I said theoretical in my example this needs to be a unit capable of high OA levels), but if we extend the curve down to your desired point and look straight to the left of your plotted final conditions following the dew point line on the psych chart, your coil needs to leave at approx 55 deg, so your delta enthalpy at these conditions 90/80 down to the coil leaving temp 55/54.51 is really 43.7257 - 22.8795 = 20.8462

Q = 4.5 x 300 x 20.8462
Q = 28142.37 Btu/hr = 2.345 tons

Hope that makes more sense now.

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Also, and I alluded to this already, you can't expect a standard DX unit to handle more than about 25% OA. If the mixed air temperature is above 80 Deg F for an extended period of time, your standard unit will be constantly faulting going out on high head until it eventually burns out the compressor. They make units capable of high OA but unless this is one of them, don't do it. A chilled water coil doesn't care, as long as it is part of a larger system, but DX is a problem.

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Thanks for the help, MCCRefTech & Vangoghsear.

I wonder if you could elaborate on the issues with high OA%? What would I be looking for if I wanted to find a DX unit capable of handling a high OA% (100%)? I am assuming that a chiller system will have a significantly higher installed cost. So, a DX unit would be preferable here.

Thanks!

9. Originally Posted by olywafox
Thanks for the help, MCCRefTech & Vangoghsear.

I wonder if you could elaborate on the issues with high OA%? What would I be looking for if I wanted to find a DX unit capable of handling a high OA% (100%)? I am assuming that a chiller system will have a significantly higher installed cost. So, a DX unit would be preferable here.

Thanks!
Playing with dynamite.
http://efficientcomfort.net/asp/Psyc...ilCalc_Web.asp
This is a load calc site.
Concerned about a lab that specs. 70%RH. Usually labs spec. <50%RH. You load is mostly latent during evening hours. Most a/cs max at 70/30 sensible latent ratios.
The Ultra-Aire 1200SD is 40/60 sensible/latent. A couple of these units could work. Your extreme is 15,800 btus/hr of latent. To provide 65^F,50%RH requires a 5 ton unit with reheat.
You need serious help with the design specs.
Regards TB

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Originally Posted by teddy bear
Playing with dynamite.
http://efficientcomfort.net/asp/Psyc...ilCalc_Web.asp
This is a load calc site.
Concerned about a lab that specs. 70%RH. Usually labs spec. <50%RH. You load is mostly latent during evening hours. Most a/cs max at 70/30 sensible latent ratios.
The Ultra-Aire 1200SD is 40/60 sensible/latent. A couple of these units could work. Your extreme is 15,800 btus/hr of latent. To provide 65^F,50%RH requires a 5 ton unit with reheat.
You need serious help with the design specs.
Regards TB
I like that online Psych calculator, Bear.

I agree he is playing with dynamite.

I agree 70% RH is probably wrong (I was just using his numbers to show the process). My experience also tells me RH should be closer to 50% for a lab.

The thing with 100% OA is that conditions can change so drastically. A unit with only 25% OA still has 75% RA coming back to it to blend at room air conditions, but a 100% OA unit can see all the extremes that weather can throw at it. So the unit needs to be able to follow those changes adjust its capacity without throwing its output conditions out of whack or burning or freezing itself.

A DX 100% OA unit for a lab should have modulating hot gas reheat, digital scroll compressor(s), a modulating form of main heat, which can be heat pump with backup electric, or gas. Hot water or steam can be used but would require a special form of non-freeze coil like a face and bypass coil or a pump for constant water movement.

There are cheaper ways to do this. You can get a bypass box or bypass curb for a roof top unit and modulating air flow to provide up to double the air flow so that the unit always has false RA returning to it. How this works is say you need 300 CFM at 100% OA so the unit runs at 600 CFM. 50% of this (300 cfm) bypasses straight back into the unit so if it is 90/80 outside and the unit is putting off 65/50% (after reheat) the mixed air becomes approximately 76/67 (within the range of a standard DX unit with reheat).

You really should consult an HVAC engineer for this project. Labs are often critical spaces for temps, humidity and air quality.

11. Originally Posted by vangoghsear
I like that online Psych calculator, Bear.

The thing with 100% OA is that conditions can change so drastically. A unit with only 25% OA still has 75% RA coming back to it to blend at room air conditions, but a 100% OA unit can see all the extremes that weather can throw at it. So the unit needs to be able to follow those changes adjust its capacity without throwing its output conditions out of whack or burning or freezing itself.

A DX 100% OA unit for a lab should have modulating hot gas reheat, digital scroll compressor(s), a modulating form of main heat, which can be heat pump with backup electric, or gas. Hot water or steam can be used but would require a special form of non-freeze coil like a face and bypass coil or a pump for constant water movement.

There are cheaper ways to do this. You can get a bypass box or bypass curb for a roof top unit and modulating air flow to provide up to double the air flow so that the unit always has false RA returning to it. How this works is say you need 300 CFM at 100% OA so the unit runs at 600 CFM. 50% of this (300 cfm) bypasses straight back into the unit so if it is 90/80 outside and the unit is putting off 65/50% (after reheat) the mixed air becomes approximately 76/67 (within the range of a standard DX unit with reheat).

You really should consult an HVAC engineer for this project. Labs are often critical spaces for temps, humidity and air quality.
I agree with the concept of partial outside air to a conventional a/c. At full cooling load conditions most a/c can be set up to handle up to 25% OA. Adding supplemental dehumidification, like the Ultra-Aire ducted dehumidifiers of adequate capacity to maintain the desired %RH in the space. A properly setup a/c with a 200 pints per day Ultra-Aire may handle 300 cfm of make-up air. We need more info like space latent load and the size of the a/c.
Regards TB

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