TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 249

dharm

/M-therm/th5-2.pm5

(ii) Efficiency of Carnot engine,

ηcarnot engine =

TT

T

13

1

−

=

600 400

600

−

= 0.3333 = 33.33%. (Ans.)

C.O.P.refrigerator =

T

TT

2

32 −

=
300
400 300− = 3. (Ans.)
Example 5.17. A heat pump working on a reversed carnot cycle takes in energy from a
reservoir maintained at 5°C and delivers it to another reservoir where temperature is 77°C. The
heat pump derives power for its operation from a reversible engine operating within the higher
and lower temperatures of 1077°C and 77°C. For 100 kJ/kg of energy supplied to reservoir at
77°C, estimate the energy taken from the reservoir at 1077°C. (U.P.S.C., 1994)
Solution. Given : T 3 = 5 + 273 = 278 K ; T 2 = T 4 = 77 + 273 = 350 K ;
T 1 = 273 + 1077 = 1350 K ;
Energy taken from the revervoir at 1077°C, Q 1 :

T= T

350 K

24

Pump Engine T = 1350 K 1

T = 278 K 3

Q 4

Q 3

Q 1

Q 2

Fig. 5.19

For reversible engine, η =

QQ

Q

12

1

−

=

TT

T

12

1

−

...(i)

or^12

1

−Q

Q =

1 2

1

−T

T

∴

Q

Q

T

T

1

2

1

2

=

For reversible heat pump, C.O.P. = Q

QQ

T

TT

4

43

4

− 43

=

−

...(ii)

Since work for running the pump is being supplied by the engine

∴ Q 1 – Q 2 = Q 4 – Q 3

or

Q

T

TTQ

T

(^1) TT
1
12 4
4
()( )−= 43 − [From (i) and (ii)]