FIRST LAW OF THERMODYNAMICS 167

dharm

/M-therm/Th4-4.pm5

(i)Heat rejected, Q :

Using the flow equation,

h 1 + C^1

2

2 + Q = h^2 +

C 22

2 + W ...(i) [Q Z^1 = Z^2 ]

Kinetic energy at inlet = C^1

(^22)
2
50
= 2 m
(^2) /s (^2) = 50 kg m
2s kg
23
2 = 1250 Nm/kg = 1.25 kJ/kg
Kinetic energy at outlet = C^1
(^22)
2
10
=2 1000
1
× = 6.05 kJ/kg
Substituting these values in eqn. (i), we get
1260 + 1.25 + Q = 400 + 6.05 + 800
∴ Q = – 55.2 kJ/kg
i.e., Heat rejected = + 55.2 kJ/kg = 55.2 × 15 kJ/s = 828 kW. (Ans.)
(ii)Inlet area, A :
Using the relation,
m& CA
v

∴ A = vm
C
&.
=045 15×
50
= 0.135 m^2. (Ans.)
+Example 4.37. In an air compressor air flows steadily at the rate of 0.5 kg/s through an
air compressor. It enters the compressor at 6 m/s with a pressure of 1 bar and a specific volume
of 0.85 m^3 /kg and leaves at 5 m/s with a pressure of 7 bar and a specific volume of 0.16 m^3 /kg. The
internal energy of the air leaving is 90 kJ/kg greater than that of the air entering. Cooling water
in a jacket surrounding the cylinder absorbs heat from the air at the rate of 60 kJ/s. Calculate :
(i)The power required to drive the compressor ;
(ii)The inlet and output pipe cross-sectional areas.
Solution. Air flow rate through the compressor, m& = 0.5 kg/s
Velocity of air at the inlet to compressor, C 1 = 6 m/s
Velocity of air at the outlet of compressor, C 2 = 5 m/s
Pressure of air at the inlet to the compressor, p 1 = 1 bar
Qout
Air in
1 2
Air out
Win
Air
compressor
Water out
Water in
Boundary
Fig. 4.46