A large electrical power facility produces 1600 MW of “waste heat,” which is dis

Question

A large electrical power facility produces 1600 MW of “waste heat,” which is dissipated to the environment in cooling towers by warming air flowing through the towers by 5.00oC . What is the necessary flow rate of air in m3 /s ? (b) Is your result consistent with the large cooling towers used by many large electrical power plants? A large electrical power facility produces 1600 MW of “waste heat,” which is dissipated to the environment in cooling towers by warming air flowing through the towers by 5.00oC . What is the necessary flow rate of air in m3 /s ? (b) Is your result consistent with the large cooling towers used by many large electrical power plants? A large electrical power facility produces 1600 MW of “waste heat,” which is dissipated to the environment in cooling towers by warming air flowing through the towers by 5.00oC . What is the necessary flow rate of air in m3 /s ? (b) Is your result consistent with the large cooling towers used by many large electrical power plants?

Explanation / Answer

(a)

qc = Hc / ( cp (to - tr))        

where

qc = volume of air for cooling (m3/s)

Hc = cooling load (W)

(to - tr) = 5 degrees celsius

Substituting the given variables, we get

The necessry flow rate of air is 3.44×10^5 m3/s

(b) That is very very huge and is too large to be dissipated by heating the air by only 5ºC.

Many of the cooling towers use the mode of circulation of cooler air over warmer water to increase the rate of evaporation.

This would allow much smaller amounts of the air necessary to remove such a huge amount of heat because evaporation removes larger quantities of heat than was considered in the above result.

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