696 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-6.pm5

Again, for L.P. turbine :

ηturbine =

TT

TT

45

45

′− ′

′−

i.e., 0.85 =

654 4

654 4 580 6

. 5
..

−′

−

T

∴ T 5 ′ = 654.4 – 0.85 (654.4 – 580.6) = 591.7 K

WL.P. (^) turbine = cpg (T 4 ′ – T 5 ′) = 1.15 (654.4 – 591.7) = 72.1 kJ/kg
Hence net power output (per kg/s mass flow)
= 72.1 kW. (Ans.)
(iii)Work ratio :
Work ratio =
Net work output
Gross work output

• 
  

  72 1
  72 1 262 9
  .
  ..
  = 0.215. (Ans.)
  (iv)Thermal efficiency of the unit, ηthermal =?
  Heat supplied = cpg(T 3 – T 2 ′) = 1.15(883 – 549.6) = 383.4 kJ/kg
  ∴ηthermal =
  Net work output
  Heat supplied

  
  

  721
  383 4
  .

  
  

. = 0.188 or 18.8%. (Ans.)
Example 13.46. In a gas turbine the compressor takes in air at a temperature of 15°C and
compresses it to four times the initial pressure with an isentropic efficiency of 82%. The air is
then passes through a heat exchanger heated by the turbine exhaust before reaching the combustion
chamber. In the heat exchanger 78% of the available heat is given to the air. The maximum
temperature after constant pressure combustion is 600°C, and the efficiency of the turbine is
70%. Neglecting all losses except those mentioned, and assuming the working fluid throughout
the cycle to have the characteristic of air find the efficiency of the cycle.
Assume R = 0.287 kJ/kg K and γ = 1.4 for air and constant specific heats throughout.

Solution. Given : T 1 = 15 + 273 = 288 K, Pressure ratio,

p

p

p

p

2

1

3

4

= = 4, ηcompressor = 82%.

Effectiveness of the heat exchanger, ε = 0.78,

ηturbine = 70%, Maximum temperature, T 3 = 600 + 273 = 873 K.

Efficiency of the cycle, ηcycle :

Considering the isentropic compression 1-2 :

T

T

p

p

2

1

2

1

(^111)
=F 4)^1
HG
I
KJ

γ− −
γ
(
.4
.4 = 1.486
∴ T 2 = 288 × 1.486 = 428 K
Now, ηcompressor =
TT
TT
21
21
−
′−
i.e., 0.82 =
428 288
2 288
−
T′−
∴ T 2 ′ =
428 288
082
−

. + 288 = 459 K