## Cooling Tower Calculation

**Experimental Data**

Cooling water flow rate – 4134 M3/hr

Cooling water inlet Temp – 44.0 °C

Cooling water exit Temp – 35.0 °C

Inlet air-wet bulb – 30.0 °C

Inlet air-dry bulb – 38.8 °C

Exit air-wet bulb – 40.7 °C

Exit air-dry bulb – 42.0 °C

Now follow step by step procedure for the calculation.

**Step-1**

Calculate waterside actual heat load, which is as below

Qw = 4134 x 1000 x (44 – 35) / 1000000

= 37.21 Gcal/Hr

**Step-2**

Calculate absolute humidity at wet bulb of inlet air, which is at 30°C in this case. This is a function of wet bulb temperature only.

The equation for the same is

1.4478310678E-10*(Tw^7)-2.6920*10e-8*(Tw^6)+1.99053*10e-6*(Tw^5)-6.65614*10e-5* (Tw^4)+0.00131879344*(Tw^3)+0.00125483272*(Tw^2)+0.291649083*Tw+3.802441

Where Tw is wet bulb temperature in °C. So,

H1 = 27.29 Kg/ ‘000Kg of dry air

**Step-3**

Calculate absolute humidity at dry bulb of inlet air, which is at 38.8°C in this case. It will give you saturation level of humidity, say H2.

**Step-4**

Find out &% Saturation. Of course it can be done from Psychometric charts but then you wont be able to use powerful Excel Tool for simulation of your cooling tower that’s why these equations are generated.

You can also use any good Excel Add-IN for Psycho properties if available.

Here, it will be %Sat = H1/H2

**Step-5**

Based on % Saturation find out the enthalpy content of moist air at inlet condition. Again I did it using self-developed equations ~10 years back.

I found it to be Hin = 26.196 Kcal/Kg of wet air.

**Step-6**

Similarly find out the moist air enthalpy at exit condition, which is

Hex = 41.630 Kcal/Kg of wet air

**Step-7**

Similarly, find out the absolute humidity at wet bulb for exit condition, which is 50.74 Kg/ ‘000 kg of dry air in this case.

**Step-8**

Calculate airflow based on heat load and enthalpy difference, which shall be as below

A = 4134000 x (44-35)/(41.630 – 26.196)

= 2410652 Kg/hr

Now based on Absolute Humidity difference, calculate amount of water evaporated as below

W = 2756000 X (50.74 – 27.29)/1000

= 64654 Kg/hr

**Step-9**

Now heat required for evaporation of this water can be calculated based on average latent heat of water evaporation at the inlet & exit temperature.

Average water temperature = 39.5 °C

Latent heat = 575.33 Kcal/Kg

Hev = 64654 x 575.33

= 37.20 Gcal/Hr

This is matching with the heat load of waterside hence, calculation is correct due to accurate temperature measurements.

So L/G comes out to be = 1.715 in this case.

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