TITLE.PM5

726 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th14-2.pm5

=

n

n− 1 R[(T^4 – T^3 ) – (T^1 – T^2 )]

The value of R can be calculated as follows

c

c

p

v

= γ

∴ cv =

cp

γ

=1.

1.

003

4

= 0.716

R = (cp – cv) = 1.003 – 0.716 = 0.287 kJ/kg K.

∴ Work done = 1.35

0.35

× 0.287 [(482.2 – 282) – (302 – 176.6)] = 82.8 kJ/kg.

∴ C.O.P. =

Heat abstracted

Work done

105.7

82.4

= = 1.27. (Ans.)

Example 14.10. An air refrigeration open system operating between 1 MPa and 100 kPa is
required to produce a cooling effect of 2000 kJ/min. Temperature of the air leaving the cold
chamber is – 5°C and at leaving the cooler is 30°C. Neglect losses and clearance in the compres-
sor and expander. Determine :
(i)Mass of air circulated per min. ;
(ii)Compressor work, expander work, cycle work ;
(iii)COP and power in kW required. (AMIE)
Solution. Refer Fig. 14.7.
Pressure, p 1 = 1 MPa = 1000 kPa ; p 2 = 100 kPa
Refrigerating effect produced = 2000 kJ/min
Temperature of air leaving the cold chamber, T 3 = – 5 + 273 = 268 K
Temperature of air leaving the cooler, T 1 = 30 + 273 = 303 K

Fig. 14.7