694 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th13-6.pm5

power shaft. The isentropic efficiencies of compressor, and the H.P. and L.P. turbines are 0.82,
0.85 and 0.85 respectively. If the maximum cycle temperature is 610°C, calculate :
(i)The pressure and temperature of the gases entering the power turbine.
(ii)The net power developed by the unit per kg/s mass flow.
(iii)The work ratio.
(iv)The thermal efficiency of the unit.
Neglect the mass of fuel and assume the following :
For compression process cpa = 1.005 kJ/kg K and γ = 1.4
For combustion and expansion processes : cpg = 1.15 kJ/kg K and γ = 1.333.
Solution. Given : T 1 = 15 + 273 = 288 K, p 1 = 1.01 bar, Pressure ratio =

p

p

2

1

= 7,

ηcompressor = 0.82, ηturbine (H.P.) = 0.85, ηturbine (L.P.) = 0.85,

Maximum cycle temperature, T 3 = 610 + 273 = 883 K

2 ′^3

1 4 ′

C.C.

Air

inlet

Generator

Power

turbine

Exhaust

L.P.

T

H.P.

C T

T

2 ′

2

4 ′

4

3

s

7.07 bar

5 ′

5

1

1.636 bar

1.01 bar

5 ′

(a) (b)

Fig. 13.63

(i) Pressure and temperature of the gases entering the power turbine, p 4 ′′′′′ and T 4 ′′′′′:

Considering isentropic compression 1-2,

T

T

p

p

2

1

2

1

(^111)
=F 7 1
HG
I
KJ

γ− −
γ
()
.4
.4 = 1.745
∴ T 2 = 288 × 1.745 = 502.5 K
Also ηcompressor =
TT
TT
21
21
−
′−
0.82 =
502.5 288
2 288
−
T′−