Answer:
a) 0.7377
b) 2.36*10^6 W
c) 839.344 KW
d) 2.9144 KW / K
Step-by-step explanation:
Given:
- Q_h Heat flow in from hot reservoir = 3200 KW
-Temperature of source T_h = 825 C
- Temperature of sink T_l = 15 C
Find:
a) thermal efficiency of this engine
b) power delivered by the engine
c) rate is heat rejected to the cold temperature sink
d) entropy change of the sink
Solution:
The thermal efficiency n_th of a Carnot heat engine is given by:
n_th = 1 - (T_l / T_h)
n_th = 1 - (15+273 / 825+273)
n_th = 0.7377
The amount of heat rejected can for a carnot engine is given by:
Q_l / Q_h = T_1 / T_h
Q_l = Q_h *(T_1 / T_h)
Q_l = 3200 *(288 / 1098)
Q_l = 839.344 KW
The net power delivered is determined by W_out:
W_out = Q_h - Q_l
W_out = 3200 - 839.344
W_out = 2.36*10^6 W
The change in entropy of sink dS:
dS = Q_l / T_l
dS = 839.344 / 288
dS = 2.914 KW / K