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# Manual Cooling Coil Sizing Calculator | Estimating Cooling Coil Load

In designing an Air Conditioning system, determining the heat load of a conditioned space will be the first most important step. And once it is determined then the next immediate step is to calculate the required supply air and determine cooling coil sizing accordingly. A wrongly estimated cooling coil load will have a drastic effect on the efficiency of the entire air conditioning system. In this article am going to explain how to determine cooling coil size with a simple case example.

Cooling Coil Sizing Calculation - Case Example:
Consider a space to be conditioned is having below heat loads, outdoor conditions in summer and comfort requirements as below
• Summer sensible heat gain: 10 kW.
• Latent heat gain: 3 kW.
• Internal design air condition: 20ÂºC, 50% sat.
• Summer external design air condition: 28ÂºC, 50% sat.
• Occupancy minimum outdoor air requirement is 325 l/s. - (Changes as per the type of room - Office space, Auditorium, Lobby etc)Â
• Acceptable supply temperature in summer: 14ÂºC.
Step1:

Supply air volume and temperature - considering the cooling caseÂ

The sensible heat is given by:

Qs =Â  m cp Î”t
10 = m x 1.026 Ã— (20-14)
m = 1.626 kg/s

cp = Specific heat capacity of air in kJ/kg K
Î”tÂ  = Temperature differential between supply air and room airÂ
mÂ  =Â  mass flow rate of air in kg/s

The latent gain is given by:

Ql = m hfg Î”g
3 = 1.626 Ã— 2450 Ã— Î”g
Î”g = (gr â€“ gs) = 0.000753 kg/kg

mÂ  =Â  mass flow rate of air in kg/s
hfg = Latent heat of vaporisation in kJ/kg
Î”g = moisture content differential between supply air and room air, ie (gr â€“ gs)

From psychrometric data the space moisture content gr is 0.00738 kg/kg. The supply air moisture content gs is given by:Â

gs = 0.00738 â€“ 0.000753 = 0.00663 kg/kg

The summer supply condition is therefore 1.626 kg/s at 14ÂºC and ag of 0.00663 kg/kg.

The humid volume at this condition is 0.8218 m3/kg, so the required volume flow rate is:

V= m x v = 1.626 Ã— 0.8218 = 1.336 m3/s

Step2:

Considering the minimum outdoor air requirement of 325 l/s this gives a value of 25% outdoor air and 75% recirculated.

Coil capacity - A mix condition can now be found and the cooling coil sized.

m x tm =Â  (mr Ã— tr) + (mo Ã— to)

1.626 Ã— tm = (0.75 Ã— 1.626 Ã— 20) + (0.25 Ã— 1.626 Ã— 28)

tm = 22ÂºC

m x gm =Â  (mr Ã— gr) + (mo Ã— go)

1.626 Ã— gm =(0.75 Ã— 1.626 Ã— 0.00738) + (0.25 Ã— 1.626 Ã—0.01210).

gm = 0.00856 kg/kg

Step3:

Q = m Î”h = (ha â€“ hm)

From psychrometric data, using t and g
• Enthalpy at mix condition is 43.88 kJ/kg
• Enthalpy at supply condition is 30.83 kJ/kg
Qc = 1.626 Ã— (43.88 â€“ 30.83)

= 21.22 kWÂ

Cooling Coil TR Calculated: 6TR

Cooling medium mass flow rate - Calculate the required mass flow rates of chilled water for water to air heating coil.

Assume flow and return chilled water temperatures of 6 and 12ÂºC

Qc = w cp Î”t

Where Î”t = (tf â€“ tr)

21.22 = wÂ  x 4.2 Ã— 6

w = 0.84 kg/s = Cooling water flow rate through the coil

Assumptions:

â€¢ Fan and duct gains have been ignored
â€¢ Coil contact factors have not been included in this simple example
â€¢ The apparatus dew point) of the coil has also not been considered.Â

Ref: CIBSE Guide