TITLE.PM5

132 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-2.pm5

∴ ∆U = 3.64 (100 × 10^3 × 0.906 – 500 × 10^3 × 0.25) J [Q 1 Pa = 1 N/m^2 ]
= 3.64 × 10^5 (0.906 – 5 × 0.25) J
= – 3.64 × 10^5 × 0.344 J = – 125.2 kJ
i.e., ∆U = – 125.2 kJ. (Ans.)
For a quasi-static process
WpdV==z pV^11 n−−pV 1 2 2

= ×× −××

−

(.)= −

()

.

.

500 10 25 100 10 0 906

25 1

125 90 6

025

(^33) 0.

1. kJ= 137.6 kJ
   ∴ Q = ∆U + W
   = – 125.2 + 137.6 = 12.4 kJ
   i.e., Q = 12.4 kJ. (Ans.)
   (ii) Here Q = 32 kJ
   Since the end states are the same, ∆U would remain the same as in (i)
   ∴ W = Q – ∆U = 32 – (– 125.2)
   = 157.2 kJ. (Ans.)
   (iii) The work in (ii) is not equal to ∫ p dV since the process is not quasi-static.
   +Example 4.18. The properties of a system, during a reversible constant pressure non-
   flow process at p = 1.6 bar, changed from v 1 = 0.3 m^3 /kg, T 1 = 20°C to v 2 = 0.55 m^3 /kg, T 2 = 260°C.
   The specific heat of the fluid is given by
   cp = 1.5
   75
   T45

• 
  


• 
  

  F
  HG
  I
  KJ
  kJ/kg°C, where T is in °C.
  Determine : (i) Heat added/kg ; (ii) Work done/kg ;
  (iii)Change in internal energy/kg ; (iv)Change in enthalpy/kg.
  Solution. Initial volume, v 1 = 0.3 m^3 /kg
  Initial temperature, T 1 = 20°C
  Final volume, v 2 = 0.55 m^3 /kg
  Final temperature, T 2 = 260°C
  Constant pressure, p = 1.6 bar
  Specific heat at constant pressure, cp =^15
  75
  45

  
  

. +

F

HG

I

T KJ kJ/kg°C

(i) The heat added per kg of fluid is given by

QcdT

T

p dT

T

T

==+

+

F

HG

I

zz 1 KJ

2

15 75

20 45

260

.

=+1 5 75 + 45

20

260

.log()TTe

= 1.5 (260 – 20) + 75 × loge

260 45

20 45

+

+

F

H

I

K = 475.94 kJ

∴ Heat added = 475.94 kJ/kg. (Ans.)