TITLE.PM5

GAS POWER CYCLES 705

dharm
\M-therm\Th13-6.pm5

As per given conditions :T 1 = T 3 , T 2 ′ = T 4 ′

T

T

p

p

2

1

2

1

1

=

F

HG

I

KJ

−γ

γ

and px = pp 12 =×1. 56 = 3 bar

Now, T 2 = T 1 × p

T

2

1

1

F

HG

I

KJ

−γ

γ

= 293 ×^3

5

14 1

14

1.

F

HG

I

KJ

. −
. = 357 K

ηcompressor (L.P.) =

TT

TT

21

21

−

′−

0.82 =

357 293

2 293

−

T′−

∴ T 2 ′ = 357 293

082

−

.

+ 293 = 371 K

i.e., T 2 ′ = T 4 ′ = 371 K

Now, T

T

p

p

p

px

5

6

5

6

1

2

11

1

=

F

HG

I

KJ

=

F

HG

I

KJ

γ−−

γ

.4

.4 Q pp

ppx

52

6

=

=

L

N

M

O

Q

P

1023 6

3

219

6

0286

T

=F

HG

I

KJ

=

.

1.

∴ T 6 =

1023

1 219. = 839 K

ηturbine (H.P.) =

TT

TT

56

56

−′

−

0.82 =

1023

1023 839

−′ 6

−

T

∴ T 6 ′ = 1023 – 0.82 (1023 – 839) = 872 K
T 8 ′ = T 6 ′ = 872 K as ηturbine (H.P.) = ηturbine (L.P.)
and T 7 = T 5 = 1023 K

Effectiveness of regenerator, ε =

TT

TT

′− ′

′− ′

4
84
where T′ is the temperature of air coming out of regenerator

∴ 0.70 =

T′−

−

371
872 371 i.e., T′ = 0.70 (872 – 371) + 371 = 722 K
Net work available, Wnet = [WT(L.P.) + WT(L.P.)] – [WC(H.P.) + WC(L.P.)]
= 2 [WT(L.P.) – WC(L.P.)] as the work developed by each turbine is
same and work absorbed by each compressor is same.
∴ Wnet = 2cp [(T 5 – T 6 ′) – (T 2 ′ – T 1 )]
= 2 × 1.005 [(1023 – 872) – (371 – 293)] = 146.73 kJ/kg of air
Heat supplied per kg of air without regenerator
= cp(T 5 – T 4 ′) + cp(T 7 – T 6 ′)
= 1.005 [(1023 – 371) + (1023 – 872)] = 807 kJ/kg of air