Question

The electric output of a power plant is 716 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows at a rate of 1.35 x 10^8 L/Hr. The water enters the plant at 25.4°C and exits at 30.7°C. (a) What is the power plant's total thermal power? (MWT (Megawatt thermal)
(b) What is the efficiency of the power plant?

Answer 1

Answer:

(a) 83475 MW

(b) 85.8 %

Explanation:

Output power = 716 MW = 716 x 10^6 W

Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3

mass of water, m = Volume  x density = 1.35 x 10^8 x 10^-3 x 1000

                                                               = 1.35 x 10^8 kg

Time, t = 1 hr = 3600 second

T1 = 25.4° C, T2 = 30.7° C

Specific heat of water, c = 4200 J/kg°C

(a) Total energy, Q = m x c x ΔT

Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J

Power = Energy / time

Power input = $P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W$

Power input = 83475 MW

(b) The efficiency of the plant is defined as the ratio of output power to the input power.

$\eta =\frac{Power output}{Power input}$

$\eta =\frac{716}{83475}=0.858$

Thus, the efficiency is 85.8 %.