TITLE.PM5

268 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-3.pm5

For a perfect gas at constant volume,

p

T

p

T

1

1

2

2

=

1. 4.05
   293

5

2

=

T

or T 2

45 293

105

=. ×

. = 1255.7 K.
At constant volume,
Q = mcv (T 2 – T 1 ) = 0.0312 × 0.718 (1255.7 – 293)
i.e., Q1–2 = 21.56 kJ.
Also, at constant pressure,
Q = m × cp × (T 3 – T 2 ) = 0.0312 × 1.005 (293 – 1255.7)
i.e., Q2–3 = – 30.18 kJ
∴ Net heat flow = Q1–2 + Q2–3 = 21.56 + (– 30.18) = – 8.62 kJ
i.e., Heat rejected = 8.62 kJ. (Ans.)
(ii)Net entropy change :
Referring to Fig. 5.31.
Net decrease in entropy,
S 1 – S 2 = (S 2 – S 3 ) – (S 2 – S 1 )
At constant pressure, dQ = mcp dT, hence

m(s 2 – s 3 ) =

mc dT

T

p

293

1255 7.

z

= 0.0312 × 1.005 × loge

1255 7

293

.

i.e., S 2 – S 3 = 0.0456 kJ/K

At constant volume, dQ = mcv dT, hence

m(s 2 – s 1 ) = mc dT

T

v

293

1255 7.

z

= 0.0312 × 0.718 × loge

1255 7

293

.
= 0.0326 kJ/K
i.e., S 2 – S 1 = 0.0326 kJ/K
∴ m(s 1 – s 3 ) = S 1 – S 3 = (S 2 – S 3 ) – (S 2 – S 1 )
= 0.0456 – 0.0326 = 0.013 kJ/K
Hence, decrease in entropy = 0.013 kJ/K. (Ans.)
Note that since entropy is a property, the decrease in entropy is given by S 1 – S 3 , is inde-
pendent of the process undergone between states 1 and 3.

Example 5.24. 0.04 m^3 of nitrogen contained in a cylinder behind a piston is initially at
1.05 bar and 15°C. The gas is compressed isothermally and reversibly until the pressure is
4.8 bar. Calculate :
(i)The change of entropy,
(ii)The heat flow, and
(iii)The work done.
Sketch the process on a p-v and T-s diagram.
Assume nitrogen to act as a perfect gas. Molecular weight of nitrogen = 28.