TITLE.PM5

238 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-1.pm5

Source

Sink

HE

Q = 1500 kJ/min 1

W = 8.2 kW

Q 2 HE = Heat engine

Solution. Heat received by the heat engine,

Q 1 = 1500 kJ/min

=^1500

60

= 25 kJ/s

Work output, W = 8.2 kW = 8.2 kJ/s.

(i) Thermal efficiency, ηth =

W

Q 1

=^82

25

. = 0.328 = 32.8%

Hence, thermal efficiency = 32.8%. (Ans.)
(ii) Rate of heat rejection,
Q 2 = Q 1 – W = 25 – 8.2
= 16.8 kJ/s
Hence, the rate of heat rejection = 16.8 kJ/s.
(Ans.)
+Example 5.2. During a process a system receives 30 kJ of heat from a reservoir and
does 60 kJ of work. Is it possible to reach initial state by an adiabatic process?
Solution. Heat received by the system = 30 kJ
Work done = 60 kJ

p

V

1

A

B

2

Fig. 5.10
Process 1-2 : By first law of thermodynamics,
Q1–2 = (U 2 – U 1 ) + W1–2
30 = (U 2 – U 1 ) + 60 ∴ (U 2 – U 1 ) = – 30 kJ.
Process 2-1 : By first law of thermodynamics,
Q2–1 = (U 1 – U 2 ) + W2–1
∴ 0 = 30 + W2–1 ∴ W2–1 = – 30 kJ.
Thus 30 kJ work has to be done on the system to restore it to original state, by adiabatic
process.

Fig. 5.9