654 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-3.pm5

p 5 = p 4 ×

V

V

n

4

5

F

HG

I

KJ

= p 4 ×

1357

() 5125

.

ren ().

= = 4.77 bar. (Ans.)

Also T

T

V

V r

n

e

n

5

4

4

5

1

11251

11

5

=

F

HG

I

KJ

==

−

()−−(). = 0.668

∴ T 5 = T 4 × 0.668 = 2163.4 × 0.668 = 1445 K or 1172°C. (Ans.)

(ii)Mean effective pressure, pm :

Mean effective pressure is given by

pm =^1

34 3 11

44 55 22 11

V

pV V pV pV

n

pV pV

s n

()−+ −

−

− −

−

L

N

M

O

Q

P

=

1

1

1

(^311)
45 21
()
()
r
p ppr
n
ppr
c n
cc
−
−+ −
−
− −
−
L
N
M
O
Q
ρ P
ρ
Now, rc = ρ, ρ = 1.8, n = 1.25, p 1 = 1 bar, p 2 = 15.59 bar, p 3 = 35.7 bar,
p 4 = 35.7 bar, p 5 = 4.77 bar
∴ pm =
1
91
35 7 1 8 1 35 7 1 8 4 77 9
125 1
15 59 1 9
() 125 1
.(. ) ...
.
.
−.
−+ ×− ×
−
− −×
−
L
N
M
O
Q
P

1
8 [28.56 + 85.32 – 26.36] = 10.94 bar
Hence mean effective pressure = 10.94 bar. (Ans.)
(iii)Efficiency of the cycle :
Work done per cycle is given by W = pmVs
Here, Vs = π/4D^2 L = π/4 × 0.25^2 × 0.4 = 0.0196 m^3
∴ W =
10 94 10 0 0196
1000
..××^5
kJ/cycle = 21.44 kJ/cycle
Heat supplied per cycle = mQs,
where m is the mass of air per cycle which is given by
m = pV
RT
11
1
where V 1 = Vs + Vc = r
r
c
c−^1
Vs
r VV
V
V
V
V V
r
VV V
r
V
r
r
r
V
c sc
c
s
c
c s
c
s s
c
s
c
c
c
s
= + =+ =
−
∴=+
−
=+
−
F
HG
I
KJ

−
L
N
M
M
M
M
M
O
Q
P
P
P
P
P
1
1
1
1 1
(^111)
or

9
91 − × 0.0196 = 0.02205 m
3
∴ m =
1 10 0 02205
287 303
××^5
×
.
= 0.02535 kg/cycle
∴ Heat supplied per cycle
= mQs = 0.02535[cv(T 3 – T 2 ) + cp(T 4 – T 3 )]
= 0.02535[0.71(1201.9 – 524.8) + 1.0(2163.4 – 1201.9)]
= 36.56 kJ/cycle
Efficiency =
Work done per cycle
Heat supplied per cycle

21 44
36 56
.
.
= 0.5864 or 58.64%. (Ans.)