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# Cooling calculation3

## Cooling calculation

(OP)
Could anyone tell me the calculation to work out the cooling load for a chiller on an AHU. The flow rate I've been given is 9m3s and a delta T of 10 degC from 25 down to 15 degC. I know with water we use Q=m.c.dT where m is the flow rate in l/hr and we normally use '1' for 'c' assuming it is water. By the way Q is the load in k/cals/hr. We then convert back to kW by dividing by 860.

I'm not sure if the same formula applies and if so, do I need to convert the 9m3s to m3h?

### RE: Cooling calculation

(OP)
I should have mentioned that the figures I posted relate to air flow for an AHU, which is why I'm struggling as my previous experience is only with chilled water systems.

### RE: Cooling calculation

Well, the same formula will work for the sensible load.  You would need the specific heat value for air, in the appropriate unit system, and the density of air to work on a mass flow basis.

But there is that latent load to consider, for which you will need more information that you have given.

### RE: Cooling calculation

Chilled water at 25C? typically it is in the range of 12-14C return/6-8C supply.

### RE: Cooling calculation

(OP)
Thanks for the help.

The air tempature is from 25 down to 16, not the water. As you say we would then supply water to the AHU at 6 or 7 degC.

Are there any rules of thumb we can use to calculate approx cooling loads?

### RE: Cooling calculation

Q = 4.5 x cfm x (delta h)

Q is in Btu/hr
cfm is air flow in cubic feet per minute
Delta h is the difference in enthalpy of entering air and leaving air.

HVAC68

### RE: Cooling calculation

You would need the entering and leaving air temperature and humidity conditions in addition to the airflow.

Total heat in Btu/hr = CFM x 1.1 x (T1-T2) + CFM x 0.69143 x (G1 - G2)

Where T1 = °F entering drybulb temperature
T2 = °F leaving dry bulb temperature
G1 = Grains Entering = 7000 x humidity entering in lb moisture / lb dry air
G2 = Grains leaving = 7000 x humidity leaving in lb moisture / lb dry air

The sensible heat Btu/hr = CFM x 1.1 x (T1-T2)

and the Latent Heat Btu/hr = CFM x 0.69143 x (G1-G2)

### RE: Cooling calculation

Pete,
Liliput has gievn you some good formulas, but being industrial, do you have a psychrometric chart? and do you understand how one works?
In addition, you work with the metric system which most of us here the US sweep under the rug. I can understand that you sweep our IP system under the rug just as well.

Why don't you post all your data, your system application, especially your geographic location, and we will pull together here and size this thing for you.

Best

### RE: Cooling calculation

Lilliput has separated the sensible and latent components and given the formula.  The formula given by me earlier gives a combined one and the result would have taken care of both the sensible and latent portions.  Either way, the result should be the same.

In any case, since you have the air quantity with you, the information that's required is

(a)  Difference in enthalpy

OR

(b) Difference in dry bulb temperature and difference in specific humidity.

HVAC68

### RE: Cooling calculation

(OP)
Hi all, Thanks for the support. Sorry about the delay in coming back. As it was the long bank holiday weekend we were away since last week.

I've work in the process cooling sector (primarily with water chillers) for the last 16 years, so I've no experience in sizing A/C applications. I've looked over my college notes from a long time ago, but can't remember how to use the psychrometric / PH chart.

The spec has changed slightly to 7m3s from 25 degC down to 16 degC. The site location is Killenny in Ireland. Unfortunately this is the only information I have recieved. I'll see if I can get the consultant to supply further info.

### RE: Cooling calculation

I don't see much latent load reduction at 160air temperature. Considering the heat load as purely sensible, you require about 1.1*14818*16.2 = 264057btu/hr or 22TR

### RE: Cooling calculation

Using the metric system and sweeping the IP system under the rug . . . . .

Assuming air on is 25oCdb and 50%RH, enthalpy: h2 = 52.91 kJ/kg

Assuming air off is 16oCdb and 15.5oCwb, enthalpy: h1 is 43.42 kJ/kg

Mass flow rate: m = 7 m3/s x 1.2kg/m3 = 8.4 kg/s

Specific heat of air: Cp = 1.01 kJ/kg.C

Total cooling = m (h2 - h1) = 8.4 x (52.91 - 43.42) = 79.7kW

Sensible cooling = m Cp (t2 - t1) = 8.4 x 1.01 x (25 - 16) = 76.4kW

As Quark said, cooling is nearly all sensible.

If selecting the coil also remember to add fan kW and any duct heat gains which will increase the coil cooling load.

Note there are online psychrometric calculator such as http://www.linric.com/webpsysi.htm which can be useful or have a look at http://www.engineeringtoolbox.com/cooling-dehumidifying-air-d_695.html

### RE: Cooling calculation

(OP)
Thank you all for your help. It is very much appreciated.

To prevent me annoying you all again, could anyone explain how to work out the following:

1) The enthalpy values.

2) Do I always convert the m3/s (x 1.2) to kg/s for the marcoh's calculation.

3) Do I always use 1.01 for the specific heat of air.

4) Do I simply add the kW rating of the fan motor to the cooling capacity.

I owe you all a pint.

### RE: Cooling calculation

1. Enthalpy values can be read from a psychrometric chart. FAQ403-1255 can be of good use to you for understanding the process. Alternatively, you can download this Psychrometrics spreadsheet for obtaining properties of state points.

2. You will get heat load in kW when you use the formula mCpdT using the units of kg/s for mass flow rate, kJ/kg0C for specific heat and 0C for temperature difference. So, kg/sxkJ/kg0Cx0C gives you kJ/s or kW. 1.2kg/m3 is the density of standard air.

3. Generally, yes.

4. Yes, if you have your motor and fan in the air path. Otherwise check for past threads that discussed this topic.

### RE: Cooling calculation

If the cooling coil has been properly sized, the entering air condition would account for return fan heat gain and the leaving coil condition would account for the supply fan heat gain.
The cooling coil sensible load should include the following:
-room sensible heat load (exterior envelope, people sensible, lights and equipment sensibleL
- return fan heat gain
- return duct heat gain
- supply fan heat gain
- supply duct heat gain

You should read up on psychrometrics to understand the air process as the return air pics up return duct heat gain, return fan heat then mix with outdoor air to produce the cooiling coil condition. Then you have to work backward from the design room condition to select the cooling coil CFM and leaving air condition such rhat the cooled air will pick up the supply fan heat gain, supply duct heat gain and room heat gain to arrive at the room design conditions.

The cooling coil latent load should include the following:

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