# Assessment of Cooling Tower Performance

Yesterday, I just got an opportunity to inspect some of cooling tower equipment, such as motors, fans, fan stack, low oil level switch, and so on. It was great. I got more understanding about the system, which is very valuable for me as young chemical engineer.

Before I do the inspection, I tried to learn something about cooling tower. I found from this link about how to assess a cooling tower. In this post I want to give you an example of my own cooling tower.

There are several parameters to assess the performance of cooling towers. Note: CT = cooling tower; CW = cooling water

Range

This is the difference between cooling tower inlet and outlet temperature. A high cooling tower range means that cooling tower has been able to reduce temperature effectively. The formula is:

CT range (oC) = CW inlet temperature (oC) – CW outlet temperature (oC)

Example:

Inlet and outlet cooling water temperature of my cooling tower is 31 and 41oC, respectively. Then CT range is 41-31 = 10oC

Approach

This is the difference between the cooling tower outlet cold water temperature and ambient wet bulb temperature. The lower the approach the better the cooling tower performance. The ‘approach’ is a better indicator of cooling tower performance.

CT approach (oC) = CW outlet temperature (oC) – CW wet bulb temperature (oC)

Example:

Wet bulb temperature = 27.4 oC, therefore CT approach is 31-27.4 = 3.6oC

Effectiveness

This is the ratio between the range and the ideal range (in percentage), i.e. difference between cooling water inlet temperature and ambien wet bulb temperature. The higher this ratio, the higher the cooling tower effectiveness.

CT effectiveness (%) = 100 x (CW in temp – CW out temp) / (CW in temp – WB temp)

Example:

CT effectiveness of my cooling tower = 100 x (41-31) / (41-27.4) = 73.53%

Cooling Capacity

This is the heat rejected in kCal/hr, given as product of mass flow rate of water, specific heat and temperature difference.

Example:

Flow rate of water = 1225 m3/hr

Density = 1000 kg/m3

Specific heat =4.2 kJ/kgoC

Temperature difference = 10 oC

Heat rejected = 1225 x 1000 x 4.2 x 10 =  51,450,000 kJ/hr = 12,348,000 kCal/hr

Evaporation Loss

This is the water quantity evaporated for cooling duty. Theoretically the evaporation quantity works out to 1.8 m3 for every 1,000,000 kCal heat rejected. The following formula can be used:

Evaporation loss (m3/hr)= 0.00085 x 1.8 x circulation rate (m3/hr) x (CW inlet temperature – CW outlet temperature)

Example:

Evaporation loss =0.00085 x 1.8 x 1225 x (41-31) = 18.74 m3/hr

Cycles of Concentration

This is the ratio of dissolved solids in circulating water to the dissolved solids in make up water. Based on the vendor of my cooling tower, cycles of concentration is usually 5-7.

Blow down Losses

It depends upon cycles of concentration and the evaporation losses and is given by formula:

Blow down = Evaporation loss /(Cycles of concentration -1)

Example:

By using cycles of concentration = 7,

Blow down = 18.74/(7-1) = 3.12 m3/hr