TITLE.PM5

146 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-3.pm5

= (15 × 10^5 ) × 0.15 × loge (4)

= 311916 J = 311.9 kJ

Considering constant volume process 2–3,

we get

V 2 = V 3 = 4 × 0.15 = 0.6 m^3

p

T

p

T

2

2

3

3

=

or, p 3 = p 2 × T

T

3

2

375 290

550

=×. = 1.98 bar

W2–3 = 0

... since volume remains constant

Consider polytropic process 3–1 :

p 3 V 3 n = p 1 V 1 n or

p

p

V

V

n

1

3

3

1

=

F

HG

I

KJ

Taking log on both sides, we get

loge (p 1 /p 3 ) = n loge (V 3 /V 1 )

or, n = log ( / )
log ( / )

log ( /. )

log (

e

e

e

e

pp

VV

13

31

15 1 98

4)

= = 1.46

W3–1 = pV pV

n

33 11

55

1

19810 061510 015

146 1

−

−

= ××−××

−

...
(. )
= – 230869 J or – 230.87 kJ
∴Net work done= W1–2 + W2–3 + W3–1
= 311.9 + 0 + (– 230.87) = 81.03 kJ. (Ans.)
For a cyclic process, zzδδQW=
∴ Heat transferred during the cycle = 81.03 kJ. (Ans.)
Example 4.33. A system consisting of 1 kg of an ideal gas at 5 bar pressure and 0.02 m^3
volume executes a cyclic process comprising the following three distinct operations : (i) Reversible
expansion to 0.08 m^3 volume, 1.5 bar pressure, presuming pressure to be a linear function of
volume (p = a + bV), (ii) Reversible cooling at constant pressure and (iii) Reversible hyperbolic
compression according to law pV = constant. This brings the gas back to initial conditions.
(i)Sketch the cycle on p-V diagram.
(ii)Calculate the work done in each process starting whether it is done on or by the system
and evaluate the net cyclic work and heat transfer.
Solution. Given : m = 1 kg ; p 1 = 5 bar ; V 1 = 0.02 m^3 ; V 2 = 0.08 m^3 ; p 2 = 1.5 bar.
(i)p-V diagram : p-V diagram of the cycle is shown in Fig. 4.26.
(ii)Work done and heat transfer :
l Process 1-2 (Linear law) :
p = a + bV ...(Given)
The values of constants a and b can be determined from the values of pressure and volume
at the state points 1 and 2.

15 1

p 2

V=C

p (bar)

0.15 V(m )^3

pV=C

pV = Cn

p 3 3 (290 K)

V=V 23

2

Fig. 4.25