TITLE.PM5

REFRIGERATION CYCLES 719

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\M-therm\Th14-1.pm5

Solution. T 2 = – 5 + 273 = 268 K ; T 1 = 35 + 273 = 308 K

Heat leakage from the surroundings into the cold storage = 29 kW

Ideal C.O.P. =

T

TT

2

12

268

− 308 268

=

− = 6.7

Actual C.O.P. =

1

3 × 6.7 = 2.233 =

R

W

n

(where Rn = net refrigerating effect, and W = work done)

or 2.233 =
29
W
or W =
29
2 233.
= 12.98 kJ/s
Hence power required to drive the plant = 12.98 kW. (Ans.)
Example 14.4. Ice is formed at 0°C from water at 20°C. The temperature of the brine is

• 8°C. Find out the kg of ice formed per kWh. Assume that the refrigeration cycle used is perfect
  reversed Carnot cycle. Take latent heat of ice as 335 kJ/kg.
  Solution. Latent heat of ice = 335 kJ/kg
  T 1 = 20 + 273 = 293 K
  T 2 = – 8 + 273 = 265 K

C.O.P. =

T

TT

2

12

265

− 293 265

=

−

= 9.46

Heat to be extracted per kg of water (to form ice at 0°C i.e., 273 K), Rn

= 1 × cpw × (293 – 273) + latent heat of ice

= 1 × 4.18 × 20 + 335 = 418.6 kJ/kg

Also, 1 kWh = 1 × 3600 = 3600 kJ

Also, C.O.P. = R

W

n=Refrigerating effect in kJ/kg

Work done in kJ

∴ 9.46 =

mice×418.6

3600

i.e., mice =

9.46 3600

418.6

×
= 81.35 kg
Hence ice formed per kWh = 81.35 kg. (Ans.)
Example 14.5. Find the least power of a perfect reversed heat engine that makes 400 kg of
ice per hour at – 8°C from feed water at 18°C. Assume specific heat of ice as 2.09 kJ/kg K and
latent heat 334 kJ/kg.
Solution. T 1 = 18 + 273 = 291 K
T 2 = – 8 + 273 = 265 K

C.O.P. =

T

TT

2

12

265

− 291 265

=

− = 10.19

Heat absorbed per kg of water (to form ice at – 8°C i.e., 265 K), Rn

= 1 × 4.18 (291 – 273) + 334 + 1 × 2.09 × (273 – 265) = 425.96 kJ/kg

Also, C.O.P. =

Net refrigerating effect

Work done

=

R

W

n

i.e., 10.19 = 425 96 400. ×

W