TITLE.PM5

652 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th13-3.pm5

or T 5 = T^4
227

3286

.. 227

= = 1447.5 K or 1174.5°C

Heat rejected during constant volume process 5-1,

= m cv (T 5 – T 1 )

= (0.00854 + 0.0004) × 0.71 (1447.5 – 373) = 6.713 kJ

Work done = Heat supplied – Heat rejected

= (2.898 + 14.66) – 6.713 = 10.845 kJ

∴ Thermal efficiency,

ηth. =

Work done

Heat supplied

=

+

10 845

2 898 14 66

.
(.. ) = 0.6176 or 61.76%. (Ans.)
+Example 13.29. The compression ratio and expansion ratio of an oil engine working on
the dual cycle are 9 and 5 respectively. The initial pressure and temperature of the air are 1 bar
and 30°C. The heat liberated at constant pressure is twice the heat liberated at constant volume.
The expansion and compression follow the law pV1.25 = constant. Determine :
(i)Pressures and temperatures at all salient points.
(ii)Mean effective pressure of the cycle.
(iii)Efficiency of the cycle.
(iv)Power of the engine if working cycles per second are 8.
Assume : Cylinder bore = 250 mm and stroke length = 400 mm.
Solution. Refer Fig. 13.25.

VC VS

V(m )^3

p (bar)

1 (30°C)

5

pV1.25= Constant

1

2

3 4

Fig. 13.25

Initial temperature, T 1 = 30 + 273 = 303 K

Initial pressure, p 1 = 1 bar

Compression and expansion law,

pV1.25 = Constant