TITLE.PM5

130 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-2.pm5

Solution. Heat received by the system,

Q = 50 kJ

Change in volume ∆ V = 0.14 m^3

Pressure = 1.2 × 10^5 N/m^2

Mass lifted in the surroundings = 90 kg

Distance through which lifted = 5.5 m

Work done during adiabatic process = – 110 kJ.

(i) Q = ∆ E + W ...(i)

Now, W = p.∆ V + Wnet

=

1 2 10 14

1000

90 8

1000

F. ××^5

HG

I

KJ

+F ××

HG

I

KJ

0. 5.5 9. kJ

= 16.8 + 4.85 = 21.65 kJ
But [from (i)], ∆ E = Q – W
= 50 – 21.65 = 28.35 kJ. (Ans.)
(ii) Since the process is adiabatic,
Q = 0
and ∆ E = – W
= – (– 110) = 110 kJ. (Ans.)
(iii) Change in internal energy,
∆ E = Q – W
= 50 – [(– 110) + 21.65] = 138.35 kJ. (Ans.)
+Example 4.16. A fluid system undergoes a non-flow frictionless process following the

pressure-volume relation as p =
5
V + 1.5 where p is in bar and V is in m

(^3). During the process the
volume changes from 0.15 m^3 to 0.05 m^3 and the system rejects 45 kJ of heat. Determine :
(i)Change in internal energy ;
(ii)Change in enthalpy.
Solution. Pressure-volume relation : p = V^5 + 1.5
Initial volume, V 1 = 0.15 m^3
Final volume, V 2 = 0.05 m^3
Heat rejected by the system, Q = – 45 kJ
Work done is given by,
WpdVV V dV
V
==+F
HG
I
zz KJ

.^55
1

2

1

2

1.

=+F

HG

I

KJ

=+−

L

N

M

O

Q

z P ×

(^5552510)
1
015 21
005
015
005
5
V
dV V
V

1. loge 1. (VV) N-m
   .

.

.

.

=+−L

NM

O

QP

10 5 005

015

(^5) log. 505 15
.
e 1. 0.()0. = 10^5 (– 5.49 – 0.15) N-m
= – 5.64 × 10^5 N-m = – 5.64 × 10^5 J[Q 1 Nm = 1 J]
= – 564 kJ.