624 ENGINEERING THERMODYNAMICS

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\M-therm\Th13-2.pm5

p (bar)

3

(^112)
1
4
1
V (m )^3
Adiabatics
V = 0.45 m 1 3
21.48
Fig. 13.12
or
p
p
2
1

V
V
1
2
F
HG
I
KJ
γ
= ()rγ or r =
p
p
2
1
1
F
HG
I
KJ
γ

11
1
1
F^14
HG
I
KJ
.
= (11)0.714 = 5.5
Also
T
T
2
1

V
V
1
2
F^1
HG
I
KJ
−γ
= ()rγ−
1
= (.)^55 14 1. − = 1.977~−1.98
∴ T 2 = T 1 × 1.98 = 303 × 1.98 = 600 K. (Ans.)
Applying gas laws to points 1 and 2
pV
T
11
1

pV
T
22
2
∴ V 2 =
T
T
p
p
2
1
1
2
× × V 1 = 600 1 0 45
303 11
××
×
.
= 0.081 m^3. (Ans.)
The heat supplied during the process 2-3 is given by :
Qs = m cv (T 3 – T 2 )
where m =
pV
RT
11
1

110 045
287 303
××^5
×
.
= 0.517 kg
∴ 210 = 0.517 × 0.71 (T 3 – 600)
or T 3 =
210

0. 517 ×0. 71 + 600 = 1172 K. (Ans.)
   For the constant volume process 2-3
   p
   T

3

3

=

p

T

2

2

∴ p 3 =

T

T

3

2

× p 2 =^1172

600

× 11 = 21.48 bar. (Ans.)

V 3 = V 2 = 0.081 m^3. (Ans.)