FIRST LAW OF THERMODYNAMICS 147

dharm

/M-therm/Th4-3.pm5

5 = a + 0.02b ...(i)

1.5 = a + 0.08b ...(ii)

From (i) and (ii) we get, b = – 58.33 and a = 6.167

W1–2 =

1

2

1

2

zzpdV=+()a bV dV

=

1

2

z (.6 167 58 33−. VdV)

= 10 6 167 58 33

2

5 2

002

008

..

.

.

V

V

−×

=^10 6 167 0 08 0 02 58 33

008 002

2

(^510)
(^223)
.(. .) .−− ×(. −. )× − kJ = 19.5 kJ
This is work done by the system. (Ans.)
Alternatively : W1–2 = Area under the process line 1–2
= Area of trapezium 1–2–l-m

515
2
L + × 105
NM
O
QP
.
× (0.08 – 0.02) = 19.5 kJ
lProcess 2 – 3 (constant pressure) :
p 3 = p 2 = 1.5 bar
The volume V 3 can be worked out from the hyperbolic compression 3–1, as follows :
p 1 V 1 = p 3 V 3 or V 3 =
pV
p
11
3
5002
15
= ×.
.
= 0.0667 m^3
∴ W2–3 = p 2 (V 3 – V 2 ) = 1.5 × 10^5 (0.0667 – 0.08) × 10–3 kJ = – 1.995 kJ
lProcess 3 – 1 (hyperbolic process) :
W3–1 = p 3 V 3 loge
V
V
1
3
F
HG
I
KJ
= (10^5 × 1.5) × 0.0667 loge
002
0 0667
.
.
F
HG
I
KJ × 10
–3 kJ = – 12.05 kJ.
This is the work done on the system. (Ans.)
Net work done, Wnet = W1–2 + W2–3 + W3–1
= 19.5 + (– 1.995) + (– 12.05) = 5.445 kJ. (Ans.)
Heat transferred during the complete cycle, zzδδQW= = 5.455 kJ. (Ans.)
Example 4.34. Fig. 4.27 shows a cylinder of 8 cm inside diameter having a piston loaded
with a spring (stiffness = 150 N/cm of compression). The initial pressure, volume and tempera-
ture of air in the cylinder are 3 × 105 N/m^2 , 0.000045 m^3 and 20°C respectively. Determine the
amount of heat added to the system so that piston moves by 3.5 cm.
Assume cv = 0.71 kJ/kg K and R = 0.287 kJ/kg K.
5 1
p=C
p(bar)
0.02 V(m )^3
Reversible expansion
(p=a+bV)
pV=C
2
0.08
1.5 3
m
V = 0.0667 3
l
Fig. 4.26. p-V diagram.
L
N
M
M
M
M
O
Q
P
P
P
P