GAS POWER CYCLES 689

dharm

\M-therm\Th13-6.pm5

∴ C =

m

m

a

f

+

F

H

G

I

K

(^1) J cp (T 3 – T 2 ′)
∴ 41800 = (90 + 1) × 1.0 × (T 3 – 471)
i.e., T 3 =^41800
91

• 
  

  471 = 930 K
  Again,
  T
  T
  4
  3
  = p
  p
  4
  3
  (^10)
  1 0
  4
  F
  HG
  I
  KJ
  =FHG IKJ
  −γ
  γ
  .4
  .4 = 0.672
  ∴ T 4 = 930 × 0.672 = 624.9 K
  ηturbine =
  TT
  TT
  34
  34
  −′
  −
  0.85 =
  930
  930 624 9
  −′ 4
  −
  T
  .
  ∴ T 4 ′ = 930 – 0.85 (930 – 624.9) = 670.6 K
  Wturbine = mg × cp × (T 3 – T 4 ′)
  (where mg is the mass of hot gases formed per kg of air)
  ∴ Wturbine = 90 1
  90
  F +
  HG
  I
  KJ
  × 1.0 × (930 – 670.6)
  = 262.28 kJ/kg of air.
  Wcompressor = ma × cp × (T 2 ′ – T 1 ) = 1 × 1.0 × (471 – 293)
  = 178 kJ/kg of air
  Wnet = Wturbine – Wcompressor
  = 262.28 – 178 = 84.28 kJ/kg of air.
  Hence power developed, P = 84.28 × 3 = 252.84 kW/kg of air. (Ans.)
  (ii)Thermal efficiency of cycle, ηηηηηthermal :
  Heat supplied per kg of air passing through combustion chamber

  
  

  1
  90 × 41800 = 464.44 kJ/kg of air
  ∴ηthermal =
  Work output
  Heat supplied
  84.28
  464.44
  = = 0.1814 or 18.14%. (Ans.)
  Example 13.41. A gas turbine unit has a pressure ratio of 6 : 1 and maximum cycle
  temperature of 610°C. The isentropic efficiencies of the compressor and turbine are 0.80 and 0.82
  respectively. Calculate the power output in kilowatts of an electric generator geared to the turbine
  when the air enters the compressor at 15°C at the rate of 16 kg/s.
  Take cp = 1.005 kJ/kg K and γ = 1.4 for the compression process, and take cp = 1.11 kJ/kg
  K and γ = 1.333 for the expansion process.
  Solution. Given : T 1 = 15 + 273 = 288 K ; T 3 = 610 + 273 = 883 K ;
  p
  p
  2
  1
  = 6,
  ηcompressor = 0.80 ; ηturbine = 0.82 ; Air flow rate = 16 kg/s