GAS POWER CYCLES 695

dharm

\M-therm\Th13-6.pm5

∴ T 2 ′ =

502.5 288

0.82

−

• 288 = 549.6 K
  Wcompressor = cpa(T 2 ′ – T 1 ) = 1.005 × (549.6 – 288) = 262.9 kJ/kg
  Now, the work output of H.P. turbine = Work input to compressor
  ∴ cpg(T 3 – T 4 ′) = 262.9
  i.e., 1.15(883 – T 4 ′) = 262.9

∴ T 4 ′ = 883 – 262.9

1.15

= 654.4 K

i.e., Temperature of gases entering the power turbine = 654.4 K. (Ans.)
Again, for H.P. turbine :

ηturbine =

TT

TT

34

34

−′

−

i.e., 0.85 =

883 654 4

(^8834)
−
−
.
T
∴ T 4 = 883 –
883 654.4
0.85
F −
HG
I
KJ = 614 K
Now, considering isentropic expansion process 3-4,
T
T
p
p
3
4
3
4
1

F
HG
I
KJ
−γ
γ
or p
p
T
T
3
4
3
4
1
133
883 033
614

F
HG
I
KJ
=F
HG
I
KJ
−
γ
γ
.

. = 4.32

i.e., p 4 =

p 3

432

707

. 432

.

.

= = 1.636 bar

i.e., Pressure of gases entering the power turbine = 1.636 bar. (Ans.)

(ii)Net power developed per kg/s mass flow, P :

To find the power output it is now necessary to calculate T 5 ′.

The pressure ratio,

p

p

4

5

, is given by

p

p

p

p

4

3

3

5

×

i.e.,

p

p

p

p

p

p

4

5

4

3

2

1

=× (Q p 2 = p 3 and p 5 = p 1 )

=^7

432. = 1.62

Then, T

T

p

p

4

5

4

5

(^1033)
′=F 62 33
HG
I
KJ

−γ
γ
()
.

1. 
   
   
   
   1. 1.127
   
   
   
   

∴ T 5 = T^4

127

654 4

127

′ =

1. 
   
   
   
   1. . = 580.6 K.