The electric output of a power plant is 669 mw. cooling water is the main way heat from the powerplant is rejected, and it flows at a rate of 1.17x 108 l/hr. the water enters the plant at 23.6°c and exits at 29.8°c. (a) what is the power plant's total thermal power? (b) what is the efficiency of the power plant? (a) mwt (megawatt thermal) o answer part (a) (b) answer part (b)

a) 1511 MW

b) 44%

Explanation:

The thermal power will be the electric power plus the heat taken away by the cooling water.

Qt = P + Qc

The heat taken away by the water will be:

Qc = G * Cp * (t1 - t0)

The Cp of water is 4180 J/(kg K)

The density of water is 1 kg/L

Then

G = 1.17 * 10^8 L/h * 1 kg/L * 1/3600 h/s = 32500 kg/s

Now we calculate Qc

Qc = 32500 * 4180 * (29.8 - 23.6) = 842*10^6 W = 842 MW

The total thermal power then is

Qt = 669 + 842 = 1511 MW

The efficiency is

η = P / Qt

η = 669 / 1511 = 44%

answer: (431,2)

step-by-step explanation:

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