Question

A 300 MWe (electrical power output) Power Plant having a thermal efficiency of 40% is cooled by sea water. Due to environmental regulations the seawater can only increase temperature by 5 C during the process. How much sea water (minimum) must be used in kg/s for cooling if the plant operates at it's rated capacity?

Answer 1

m = 22,877 kg / s

Step-by-step explanation:

Let's solve this exercise in parts, first look for the amount of heat generated by the plant and then the amount of water to dissipate this heat

The plant generates a power of 300 MW at a rate of 40%, let's use a direct ratio rule to find the heat. If the power is 400 MW it corresponds to 40%, what heat (Q) corresponds to the other 60%

Q = 300 60% / 40%

Q = 450 MW

having the amount of heat generated we can use the calorimeter equation,

Q = m
`c_(e)`
`(T_(f) - T_(o))`

m = Q / c_{e} (T_{f} - T_{o})

let's use the maximum temperature change allowed

(T_{f} - T_{o}) = 5

the specific heat of sea water is 3934 J / kg ºC, note that it is less than that of pure water, due to the salts dissolved in sea water

power and energy are related

W = Q / t

Q = W t

let's calculate

m = 450 10⁶ / (3934 5)

m = 22,877 kg / s