TITLE.PM5

(Ann) #1

610 ENGINEERING THERMODYNAMICS


dharm
\M-therm\Th13-1.pm5


p 3 = p 2 × V
V

2
3

F
HG

I
KJ

γ
= 12 × V
V

1
4

F
HG

I
KJ

γ
Q V
V

V
V

4
1

3
2

=
L
N

M


O
Q

P


= 12 ×^1
6

F^14
HG

I
KJ

.
= 0.97 bar. (Ans.)

Hence

pTT
p
pTT
p

112
2
334
4

===
=
===
=

U
V

||


W


||


18 bar
12 bar
0.97 bar
1.46 bar

(Ans. )

683 K

333.2 K

(ii)Change in entropy :
Change in entropy during isothermal expansion,

S 2 – S 1 = mR loge
V
V

2
1

F
HG

I
KJ
=
pV
T

11
1

loge
V
V

2
1

F
HG

I
KJ

Q pV mRT
mR pV
T

=
=

L


N


M
M

O


Q


P
or P

=
18 10 018
10 683

5
3

××
×

.
loge (1.5) = 0.192 kJ/K. (Ans.)

(iii)Mean thermal efficiency of the cycle :

Heat supplied, Qs = p 1 V 1 loge
V
V

2
1

F
HG

I
KJ
= T 1 (S 2 – S 1 )
= 683 × 0.192 = 131.1 kJ

Heat rejected, Qr = p 4 V 4 loge
V
V

3
4

F
HG

I
KJ
= T 4 (S 3 – S 4 ) because increase in entropy during heat addition
is equal to decrease in entropy during heat rejection.
∴ Qr = 333.2 × 0.192 = 63.97 kJ

∴ Efficiency, η =
QQ
Q

sr
s


= 1 –
Q
Q

r
s
= 1 –
63.97
131.1
= 0.512 or 51.2%. (Ans.)
(iv)Mean effective pressure of the cycle, pm :
The mean effective pressure of the cycle is given by
pm = Work done per cycle
Stroke volume
V
V

3
1
= 6 × 1.5 = 9
Stroke volume, Vs = V 3 – V 1 = 9V 1 – V 1 = 8V 1 = 8 × 0.18 = 1.44 m^3

∴ pm = ()QQ J
V

sr
s

−× = ()QQ
V

sr
s

−× (^1) ()Q J= 1
= ()131.1 63.97^10
44 10
3
5
−×



  1. ×


= 0.466 bar. (Ans.)
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