TITLE.PM5

REFRIGERATION CYCLES 721

dharm

\M-therm\Th14-1.pm5

Heat removed by the plant /min

=

320 20 1000

8

.29××

= 800725 kJ/h

(i) Capacity of the refrigerating plant =

800725

14000

= 57.19 tonnes. (Ans.)

(ii) T 1 = 25 + 273 = 298 K

T 2 = – 8 + 273 = 265 K

∴ C.O.P. of reversed Carnot cycle

=

T

TT

2

12

265

− 298 265

=

− = 8.03. (Ans.)

(iii) Power required :

Actual C.O.P. =

1

3 × Carnot C.O.P. =

1

3 × 8.03 = 2.67

But actual C.O.P. = Net refrigerating effect/min

Work done /min

=R

W

n

2.67 =

800725

W

kJ/h

∴ W =
800725
672.
= 299897 kJ/h = 83.3 kJ/s
∴ Power required to run the plant = 83.3 kW. (Ans.)
Example 14.8. A heat pump is used for heating the interior of a house in cold climate. The
ambient temperature is – 5°C and the desired interior temperature is 25°C. The compressor of
heat pump is to be driven by a heat engine working between 1000°C and 25°C. Treating both
cycles as reversible, calculate the ratio in which the heat pump and heat engine share the heating
load. (P.U.)
Solution. Refer Fig. 14.3. Given : T 1 = 1000 + 273 = 1273 K ; T 2 = 25 + 273 = 298 K ;
T 3 = – 5 + 273 = 268 K ; T 4 = 25 + 273 = 298 K

The ratio in which the heat pump and heat engine share the heating load,

Q

Q

4

1

:

Since both the cycles are reversible, therefore,

Q

Q

3

4

=

T

T

3

4

and

Q

Q

T

T

2

1

2

1

=

or

Q

Q

3

4

=^268

298

or Q 3 =^268

298

Q 4 and

Q

Q

2

1

298

1273

=

Heat engine drives the heat pump,

∴ W = (Q 1 – Q 2 ) = Q 4 – Q 3

Dividing both sides by Q 1 , we have

1 –

Q

Q

2

1

=

QQ

Q

43

1

−

1 –^298

1273

=

QQ

Q

44

1

268

− 298