How to calculate the cooling capacity of a chiller. Chillers provide chilled water which is then used to provide air conditioning within buildings. The amount of cooling they produce varies and it’s important to know how much cooling a chiller is producing or is able to produce. There is a video tutorial at the bottom of the page also.

Firstly to perform this calculation we need to know a few things.

- The volume flow rate of water into the evaporator
- The inlet and outlet chilled water temperature

We then need to lookup the properties of water for the following

- The density of the water at the average temperature (inlet temp + outlet temp)/2
- The Specific Heat Capacity of the chilled water at the average temperature (inlet temp + outlet temp)/2

A recommended website to look these properties up is: PeaceSoftware.de

Let look at how to calculate the cooling capacity. We’ll first look at how to calculate in metric units and then imperial.

**Metric units:**

The water flow rate of chilled water into the evaporator is 0.0995m3/s, the inlet temperature is 12*c and the outlet temperature is 6*c. This means the average temperature is 9*c so we lookup the water properties at this temperature to find the density of 999.78kg/m3 and a specific heat capacity of 4.19kJ/kg/K.

Using the energy equation of Q = ṁ x Cp x ΔT we can calculate the cooling capacity.

Q = (999.78kg/m3 x 0.0995m3/s) x 4.19kJ/kg/K x ((12*c+273.15K) – (6*c+273.15K))

*We add 273.15K to the celcius to convert it to units of Kelvin. The Specific heat capacity (Cp) is measured in units of kJ per kg per Kelvin.*

This gives us a final answer of Q = 2,500kW of cooling. Full calculations are shown below.

Now lets look at how to calculate the cooling capacity of a chiller in imperial units

**Imperial units:**

The flow rate of chilled water into the evaporator is measured as 12,649ft3/h and the chilled water inlet temperature is 53.6*F the outlet temperature is 42.8*F. The average temperature is 48.2*F so we need to calculate the water properties at this temperature.

A good website for this is peacesoftware.de the although we will need to convert the units to imperial so for that we will use Specific heat capacity and density of water

This will give us a specifi heat capacity of 1.0007643BTU/lb.F and density of 62.414lb/Ft3

Using the energy equation of Q = ṁ x Cp x ΔT we can calculate the cooling capacity.

Q = (16,649FT3/h x 62.414lb/ft3) x 1.0007643BTU/lb.F x (53.6F – 42.8F)

Giving us a cooling capacity of 8,533,364BTU/h. see full calculations below.

Hi Paul,

Good day. Need your advice on heat recovery system from water chiller to heat water in building for hot water supply. Requirement parameters are:

1) heat circulating water from ambient inlet 27 degC to supply outlet 60 degC;

2) current chiller design 35 kW;

3) chilled water supply inlet 14.8 degC and return outlet 8 degC.

Please advise with justification/calculations if this system chiller is able to perform to its requirements.

(Hot water supply to roof storage tanks usage 12 cu.metres per day)

Also, the heat loss on a 50mm diameter with 25mm insulation over a distance from chiller to roof storage tanks is est. 100m in length. What’s the possible heat loss?

Appreciate for prompt reply Paul. Thank you & cheers.

Best regards,

Tony