640 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-3.pm5

Consider 1 kg of air.

Total heat supplied = Heat supplied during the operation 2-3

+ heat supplied during the operation 3-4

= cv(T 3 – T 2 ) + cp(T 4 – T 3 )

Heat rejected during operation 5-1 = cv(T 5 – T 1 )

Work done = Heat supplied – heat rejected

= cv(T 3 – T 2 ) + cp(T 4 – T 3 ) – cv(T 5 – T 1 )

ηdual = Work done

Heat supplied

=

−+ −− −

−+ −

cT T c T T c T T

cT T c T T

v p v

v p

()()()

()()

32 43 51

32 43

= 1 –

cT T

cT T c T T

v

v p

()

()()

51

32 43

−

−+ −

= 1 –

cT T

TT TT

v()

()()

51

32 43

−

−+ −γ ...(i)

Q γ=

F

HG

I

KJ

c

c

p

v

Compression ratio, r = v

v

1

2

During adiabatic compression process 1-2,

T

T

2

1

=

v

v

(^1) r
2
1
F 1
HG
I
KJ

−
−
γ
()γ ...(ii)
During constant volume heating process,
p
T
3
3

p
T
2
2
or
T
T
3
2

p
p
3
2
= β, where β is known as pressure or explosion ratio.
or T 2 =
T 3
β ...(iii)
During adiabatic expansion process,
T
T
4
5

v
v
5
4
F^1
HG
I
KJ
−γ

r
ρ
F γ
HG
I
KJ
− 1
...(iv)
Q v
v
v
v
v
v
v
v
v
v
v
v
5 r
4
1
4
1
2
2
4
1
2
3
4
==×=×=
F
HG
I
ρ KJ
, beinρ gthe cut off ratio-
During constant pressure heating process,
v
T
v
T
3
3
4
4

T 4 = T 3
v
v
4
3
= ρ T 3 ...(v)
Putting the value of T 4 in the eqn. (iv), we get
ρ
ρ
T γ
T
3 r
5
1
=F
HG
I
KJ
−
or T 5 = ρ. T 3. ρ
γ
r
F
HG
I
KJ
− 1