TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 289

dharm

/M-therm/th5-4.pm5

Heat

pump

Source

T 1

Sink

T 2

W

Q 1

Q 2

Solution. The arrangement is shown in Fig. 5.45.
For the minimum work absorbed by the heat pump, the heat pump must be reversed
Carnot cycle engine and the required condition is

dQ

T

F

H

I

z K = 0

For infinitely small changes, we can write

cpdTT^1 cpdTT

1

2

2

• = 0
  If T 1 ′ is the final temperature of the high level reservoir, then the above equation can be
  written as

c dT

T

c dT

pT T

T

p T

1 T

1

2

1 2

1

1

F 2

HG

I

KJ

+

F

HG

I

KJ

′

zz = 0

∴ loge

T

T

1

1

F ′

HG

I

KJ

+ loge

T

T

2

1

F

HG

I

KJ = 0 = loge (1)

∴ loge

TT

T

12

12

F ′

HG

I

KJ

= loge (1)

∴ T 1 ′ =

T

T

1

2

2

Now the work given to the heat pump

= Heat rejected at higher level temperature

• heat picked up at lower level temperature

∴ W = cdTc dTp

T

T

p T

T

() ()

1

1

2

′ 1

zz−

= cp [(T 1 ′ – T 1 ) – (T 1 – T 2 )] = cp (T 1 ′ + T 2 – 2T 1 )

Now substituting the value of T 1 ′ in the above equation in terms of

T 1 and T 2

WcT

T

=+−p TT

L

N

M

M

O

Q

P

P

1

2

2

212. Proved.

+Example 5.44. The connections of a reversible engine to three sources at 500 K, 400 K
and 300 K are shown in Fig. 5.46. It draws 1500 kJ/min of energy from the source at 800 K and
develops 200 kJ/min of work.
(i)Determine the heat interactions with the other two sources of heat.
(ii)Evaluate the entropy change due to each heat interaction with the engine.
(iii)Total entropy change during the cycle.
Solution. Refer Fig. 5.46.
Temperature of source 1 = 500 K
Temperature of source 2 = 400 K
Temperature of source 3 = 300 K
Heat energy drawn from source 1, Q 1 = 1500 kJ/min
Work developed, W = 200 kJ/min.

Fig. 5.45