TITLE.PM5

620 ENGINEERING THERMODYNAMICS

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\M-therm\Th13-2.pm5

Fig. 13.9. Atkinson cycle.

Now, T

T

p

p

5

3

5

3

1

=

F

HG

I

KJ

−γ

γ

or T 5 = 1842 ×

10

36 9

14 1

. 14
.

.

F.

HG

I

KJ

−

= 657 K

∴ηAtkinson = 1 –

1 4 657 300

1842 614 4

.( )

(.)

−
− = 0.5929 or 59.29%.
∴ Improvement in efficiency = 59.29 – 51.16 = 8.13%. (Ans.)
Example 13.11. A certain quantity of air at a pressure of 1 bar and temperature of 70°C is
compressed adiabatically until the pressure is 7 bar in Otto cycle engine. 465 kJ of heat per kg of
air is now added at constant volume. Determine :
(i)Compression ratio of the engine.
(ii)Temperature at the end of compression.
(iii)Temperature at the end of heat addition.
Take for air cp = 1.0 kJ/kg K, cv = 0.706 kJ/kg K.
Show each operation on p-V and T-s diagrams.
Solution. Refer Fig. 13.10.
Initial pressure, p 1 = 1 bar
Initial temperature, T 1 = 70 + 273 = 343 K
Pressure after adiabatic compression, p 2 = 7 bar
Heat addition at constant volume, Qs = 465 kJ/kg of air
Specific heat at constant pressure, cp = 1.0 kJ/kg K
Specific heat at constant volume, cv = 0.706 kJ/kg K

∴γ =

c

c

p

v

=

1.

0.

0

706 = 1.41

(i)Compression ratio of engine, r :

According to adiabatic compression 1-2

p 1 V 1 γ = p 2 V 2 γ