TITLE.PM5

122 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-2.pm5

(i)For isothermal process :

W1–2 = pdv.

1

2

z = p^1 v^1 loge^

p

p

1

2

F

HG

I

KJ

Fig. 4.14

as p 1 v 1 = p 2 v 2 for isothermal process

∴ W1–2 = – 10^5 × 1.8 loge

110

510

5

5

×

×

F

HG

I
KJ
= – 2.897 × 10^5 = – 289.7 kJ/kg.
(– ve sign indicates that the work is supplied to the air)
∴ Work done on the air = 289.7 kJ/kg. (Ans.)
(ii) Since temperature is constant,
∴ u 2 – u 1 = 0
∴ Change in internal energy = zero. (Ans.)
(iii) Again, Q1–2 = (u 2 – u 1 ) + W
= 0 + (– 289.7) = – 289.7 kJ
(– ve sign indicates that heat is lost from the system to the surroundings)
∴ Heat rejected = 289.7 kJ/kg. (Ans.)
Example 4.7. A cylinder containing the air comprises the system. Cycle is completed as
follows :
(i) 82000 N-m of work is done by the piston on the air during compression stroke and
45 kJ of heat are rejected to the surroundings.
(ii) During expansion stroke 100000 N-m of work is done by the air on the piston.
Calculate the quantity of heat added to the system.
Solution. Refer Fig. 4.15.
Compression stroke. Process 1-2 :
Work done by the piston on the air, W1–2 = – 82000 N-m (= – 82 kJ)
Heat rejected to the system, Q1–2 = – 45 kJ
Now, Q1–2 = (U 2 – U 1 ) + W

• 45 = (U 2 – U 1 ) + (– 82)
  ∴ (U 2 – U 1 ) = 37 kJ ...(i)