636 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th13-3.pm5

For the adiabatic (or isentropic) process 1-2

p 1 V 1 γ = p 2 V 2 γ or

p

p

V

V

(^2) r
1
1
2

F
HG
I
KJ

γ
()γ
∴ p 2 = p 1. ()rγ= 1 × (15)1.4 = 44.31 bar. (Ans.)
Also,
T
T
2
1

V
V
1
2
F^1
HG
I
KJ
−γ
= ()rγ−^1141 =( )^15. − = 2.954
∴ T 2 = T 1 × 2.954 = 300 × 2.954 = 886.2 K. (Ans.)
V 2 = Vc =
V
r
s
−

1 −
0 00942
15 1
.
= 0.0006728 m^3. (Ans.)
p 2 = p 3 = 44.31 bar. (Ans.)
% cut-off ratio =
ρ−
−
1
r 1
8
100

ρ−
−
1
15 1
i.e., ρ = 0.08 × 14 + 1 = 2.12
∴ V 3 = ρ V 2 = 2.12 × 0.0006728 = 0.001426 m^3. (Ans.)
V
VVVsc
3
3
0 08 0 08 0 00942 0 0006728 0 001426^3
can also be calculated as follows :
=+=× +.....m=
L
N
M
M
O
Q
P
P
For the constant pressure process 2-3,
V
T
V
T
3
3
2
2

∴ T 3 = T 2 × V
V
3
2
= 886.2 ×
0 001426
0 0006728
.
.
= 1878.3 K. (Ans.)
For the isentropic process 3-4,
pV pV
ppV
V
p
33 44
433
4
(^31)
1
1
707
44.31
07
γγ
γ

=×
F
HG
I
KJ
=×

(. )
()
.4

7. .4 2.866 bar. (Ans. )

Q

Q

V

V

V

V

V

V

V

V

V

V

r VV

4

3

4

2

2

3

1

2

2

3

41

15

212

707

=×=×

====

L

N

M

M

M

M

O

Q

P

P

P

ρ,.. P

Also,

T

T

4

3

=

V

V

3

4

(^1) 14 1
1
707
F
HG
I
KJ
=F
HG
I
KJ
γ− −
.
.
= 0.457
∴ T 4 = T 3 × 0.457 = 1878.3 × 0.457 = 858.38 K. (Ans.)
V 4 = V 1 = 0.0101 m^3. (Ans.)
(ii)Theoretical air standard efficiency :
ηdiesel = 1 –
11
1
1
1
14 15
212 1
(^1141) 212 1
14
γ
ρ
γ ρ
γ
(). ( )
(. )

..

.

r −−

−

−

L

N

M

M

O

Q

P

P

=−

−

−

L

N

M

M

O

Q

P

P

= 1 – 0.2418 × 1.663 = 0.598 or 59.8%. (Ans.)