638 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-3.pm5

Mean effective pressure, pm :

pm = Work done by the cycle

Swept volume

Work done = Heat added – heat rejected

Heat added = mcp (T 3 – T 2 ), and

Heat rejected = mcv (T 4 – T 1 )

Now assume air as a perfect gas and mass of oil in the air-fuel mixture is negligible and is

not taken into account.

Process 1-2 is an adiabatic compression process, thus

T

T

V

V

2

1

1

2

1

=

F

HG

I

KJ

−γ

or T 2 = T 1 ×

V

V

1

2

F 14 1

HG

I

KJ

. −
(since γ = 1.4)

or T 2 = 300 × (15.3)0.4 = 893.3 K

Also, pV 11 γγ=p V2 2 ⇒ p 2 = p 1 × V

V

1

2

F

HG

I

KJ

γ

= 1 × (15.3)1.4 = 45.56 bar

Process 2-3 is a constant pressure process, hence

V

T

V

T

2

2

3

3

= ⇒ T 3 = VT

V

32

2

= 2.04 × 893.3 = 1822.3 K

Assume that the volume at point 2 (V 2 ) is 1 m^3. Thus the mass of air involved in the process,

m pV

RT

==××

×

(^22) =
2
45.56 10^51
287 893.3
17 77.kg
Q V
V
V
V
V
V
V
V
V
V
V
V
V
V
4
3
1
3
1
2
2
3
3
2
1
2
3
4
15 3
75
204
==×
=×= =
L
N
M
M
M
M
O
Q
P
P
P
or P
.
.
.
Process 3-4 is an adiabatic expansion process, thus
T
T
V
V
4
3
3
4
(^1) 14 1
1
75

F
HG
I
KJ
=F
HG
I
KJ
γ− −
.
.
= 0.4466
or T 4 = 1822.3 × 0.4466 = 813.8 K
∴ Work done = mcp (T 3 – T 2 ) – mcv (T 4 – T 1 )
= 17.77 [1.005 (1822.3 – 893.3) – 0.718 (813.8 – 300)] = 10035 kJ
∴ pm =
Work done
Swept volume

10035 10035
15 3
10035
()(. ).VV 12 − VV 22 14 3

−

= 701.7 kN/m^2 = 7.017 bar. (Ans.)
(massumed)Q V 2 = 1 3
Ratio of maximum pressure to mean effective pressure

p
pm
2 45 56
7 017
=.
.
= 6.49. (Ans.)
Cycle efficiency, ηcycle :
ηcycle =
Work done
Heat supplied