Respuesta :
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 = [tex]P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W[/tex]
Power input = 83475 MW
(b) The efficiency of the plant is defined as the ratio of output power to the input power.
[tex]\eta =\frac{Power output}{Power input}[/tex]
[tex]\eta =\frac{716}{83475}=0.858[/tex]
Thus, the efficiency is 85.8 %.
