TITLE.PM5

720 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th14-1.pm5

∴ W =

425.96 400

10

×
.19
= 16720.7 kJ/h
= 4.64 kJ/s or 4.64 kW
Hence least power = 4.64 kW. (Ans.)
+Example 14.6. The capacity of the refrigerator (working on reversed Carnot cycle) is
280 tonnes when operating between – 10°C and 25°C. Determine :
(i)Quantity of ice produced within 24 hours when water is supplied at 20°C.
(ii)Minimum power (in kW) required.
Solution. (i) Quantity of ice produced :
Heat to be extracted per kg of water (to form ice at 0°C)
= 4.18 × 20 + 335 = 418.6 kJ/kg
Heat extraction capacity of the refrigerator
= 280 tonnes
= 280 × 14000 = 3920000 kJ/h
∴ Quantity of ice produced in 24 hours,

mice =

3920000 24

418 6 1000

×

. × = 224.75 tonnes. (Ans.)
(ii) Minimum power required :
T 1 = 25 + 273 = 298 K
T 2 = – 10 + 273 = 263 K

C.O.P. =

T

TT

2

12

263

− 298 263

=

− = 7.51

Also, C.O.P. =

Net refrigerating effect

Work done /min

=R

W

n

i.e., 7.51 =

3920000

W

∴ W =
3920000
517.
kJ/h = 145 kJ/s
∴ Power required = 145 kW. (Ans.)
Example 14.7. A cold storage plant is required to store 20 tonnes of fish. The temperature
of the fish when supplied = 25°C ; storage temperature of fish required = – 8°C ; specific heat of
fish above freezing point = 2.93 kJ/kg-°C ; specific heat of fish below freezing point = 1.25 kJ/kg-
°C ; freezing point of fish = – 3°C. Latent heat of fish = 232 kJ/kg.
If the cooling is achieved within 8 hours ; find out :
(i)Capacity of the refrigerating plant.
(ii)Carnot cycle C.O.P. between this temperature range.
(iii)If the actual C.O.P. is 13 rd of the Carnot C.O.P. find out the power required to run the
plant.
Solution. Heat removed in 8 hours from each kg of fish
= 1 × 2.93 × [25 – (– 3)] + 232 + 1 × 1.25 [– 3 – ( – 8)]
= 82.04 + 232 + 6.25 = 320.29 kJ/kg