Question

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)

Answer 1

a) 1511 MW

b) 44%

Step-by-step 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%