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

(a) 83475 MW

(b) 85.8 %

Step-by-step 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 = (3 * 10^(12))/(3600)=8.35 * 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 =(716)/(83475)=0.858`

Thus, the efficiency is 85.8 %.