FIRST LAW OF THERMODYNAMICS 173dharm
/M-therm/Th4-5.pm5Conditions at ‘1’ :
Pressure, p 1 = 7.5 bar = 7.5 × 10^5 N/m^2 , 750°C
Velocity, C 1 = 140 m/s
Enthalpy, h 1 = 950 kJ/kg
Conditions at ‘2’ :
Pressure, p 2 = 2.0 bar = 2 × 10^5 N/m^2 , 550°C
Velocity, C 2 = 280 m/s
Enthalpy, h 2 = 650 kJ/kg
Gas flow, m = 5 kg/s.
Capacity of the turbine :
Considering the flow of gas as 1 kg and neglecting the change in potential energy, we can
write the steady flow energy equation for the turbine as
h 1 + C^12
2
± Q = h 2 +
C 22
2
± W
Q = 0 as the system is adiabatic and W should be taken as +ve since it develops work.∴ h 1 +
C 12
2
= h 2 +C 22
2 + W∴ W = (h 1 – h 2 ) +
CC 12 22
2−
= 10^3 (950 – 650) +
140 280
2(^22) −
= 10^3 × 300 – 29.4 × 10^3
= 270.6 × 10^3 J/kg = 270.6 kJ/kg.
Power capacity of the turbine
= mW& = 5 × 270.6 = 1353 kJ/s
= 1353 kW. (Ans.)
+Example 4.43. 12 kg of air per minute is delivered by a centrifugal air compressor. The
inlet and outlet conditions of air are C 1 = 12 m/s, p 1 = 1 bar, v 1 = 0.5 m^3 /kg and C 2 = 90 m/s,
p 2 = 8 bar, v 2 = 0.14 m^3 /kg. The increase in enthalpy of air passing through the compressor is
150 kJ/kg and heat loss to the surroundings is 700 kJ/min.
Find : (i) Motor power required to drive the compressor ;
(ii)Ratio of inlet to outlet pipe diameter.
Assume that inlet and discharge lines are at the same level.
Solution. Quantity of air delivered by the compressor, m=
12
60
= 0.2 kg/s.
Conditions of air at the inlet 1 :
Velocity, C 1 = 12 m/s
Pressure, p 1 = 1 bar = 1 × 10^5 N/m^2
Specific volume, v 1 = 0.5 m^3 /kg
Conditions of air at the outlet 2 :
Velocity, C 2 = 90 m/s
Pressure, p 2 = 8 bar = 8 × 10^5 N/m^2