GAS POWER CYCLES 683

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\M-therm\Th13-6.pm5

(ii)Heat supplied to air :

For compression process 1-2, we have

T

T

p

p

2

1

2

1

(^1) 14 1
35 14
1

F
HG
I
KJ
=F
HG
I
KJ
γ− −
γ.
.

. = 1.43

∴ T 2 = T 1 × 1.43 = 293 × 1.43 ~− 419 K

∴ Heat supplied to air, Q 1 = cp (T 3 – T 2 ) = 1.005 (873 419& − ) = 456.27 kJ/kg. (Ans.)

(iii)Work available at the shaft, W :

We know that, ηcycle = Work

Heat

output ( )

input ( )

W

Q 1

or 0.30 =

W

456 27.

or W = 0.3 × 456.27 = 136.88 kJ/kg

(iv)Heat rejected in the cooler, Q 2 :

Work output (W) = Heat supplied (Q 1 ) – heat rejected (Q 2 )

∴ Q 2 = Q 1 – W = 456.27 – 136.88 = 319.39 kJ/kg. (Ans.)

(v)Temperature of air leaving the turbine, T 4 :

For expansion (isentropic) process 3-4, we have

T

T

(^3) rp
4
(^1) 14 1
== 35 14
− −
() (.)
.
.
γ
γ = 1.43
∴ T 4 =
T 3
143
873
.. 143
= = 610.5 K. (Ans.)
[Check : Heat rejected in the air cooler at constant pressure during the process 4-1 can also
be calculated as : Heat rejected = m × cp (T 4 – T 1 ) = 1 × 1.005 × (610.5 – 293) = 319.1 kJ/kg]
Example 13.36. A closed cycle ideal gas turbine plant operates between temperature limits
of 800°C and 30°C and produces a power of 100 kW. The plant is designed such that there is no
need for a regenerator. A fuel of calorific 45000 kJ/kg is used. Calculate the mass flow rate of air
through the plant and rate of fuel consumption.
Assume cp = 1 kJ/kg K and γ = 1.4. (GATE, 2000)
Solution. Given : T 1 = 30 + 273 = 303 K ; T 3 = 800 + 273 = 1073 K ; C = 45000 kJ/kg ; cp
= 1 kJ/kg K ; γ = 1.4 ; Wturbine – Wcompressor = 100 kW.
m,m :&&af
Since no regenerator is used we can assume the
turbine expands the gases upto T 4 in such a way that
the exhaust gas temperature from the turbine is equal
to the temperature of air coming out of the compressor
i.e., T 2 = T 4
p
p
p
p
2
1
=^3
4
, p
p
T
T
2
1
2
1
1
=F
HG
I
KJ
−
γ
γ
and p
p
T
T
3
4
3
4
1

F
HG
I
KJ
−
γ
γ
∴
T
T
T
T
T
T
2
1
3
4
3
2

(Q TT 24 = ......assumed)
T (K)
s
2
1
3
4
Fig. 13.55