TITLE.PM5

282 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-4.pm5

Mean absolute temperature, Tmean = TT^12

2

290 666 2

2

+ = +. = 478.1 K

Approximate change of entropy = Q

Tmean

=−90 04

478 1

.

.

= – 0.188 kJ/kg K

∴ % age error = 0 199 0 188

0 199

.. 100

.

− × = 5.53%. (Ans.)

l The approximate value of entropy change is lower, because in the relation Q = Tav × dS
actual value of heat transferred is substituted instead of approximate value (i.e., Area
under the straight line) which is higher.
Example 5.36. 1.2 m^3 of air is heated reversibly at constant pressure from 300 K to 600 K,
and is then cooled reversibly at constant volume back to initial temperature. If the initial pres-
sure is 1 bar, calculate :
(i)The net heat flow.
(ii)The overall change in entropy.
Represent the processes on T-S plot.
Take cp = 1.005 kJ/kg K and R = 0.287 kJ/kg K
Solution. Given : V 1 = 1.2 m^3 ; p 1 = p 2 ; T 1 = 300 K ; T 2 = 600 K ; p 1 = 1 bar ;
cp = 1.005 kJ/kg K ; R = 0.287 kJ/kg K
Fig. 5.42 shows the T-S plot of the processes.
(i)The net heat flow, Q :

Mass of air, m = pV

RT

11

1

110 12^5

0 287 1000 300

= ××

××

.

(. )

= 1.394 kg

Q = Q1-2 + Q2-3

= mcp(T 2 – T 1 ) + mcv(T 3 – T 2 )

= mcp(T 2 – T 1 ) + mcv(T 1 – T 2 ) ...(Q T 1 =T 3 )

= m(T 2 – T 1 )(cp – cv) = m(T 2 – T 1 ) × R

= 1.394(600 – 300) × 0.287 = 120 kJ. (Ans.)

(ii)The overall change in entropy :

Entropy change during constant pressure process 1-2,

S 2 – S 1 = mcp loge

T

T

2

1

F

HG

I

KJ

= 1.394 × 1.005 loge 600

300

F

HG

I

KJ

= 0.9711 kJ/K

Entropy change during constant volume process 2-3 ;

S 3 – S 2 = mcv loge

T

T

mc R T

p e T

3

2

1

2

F

HG

I

KJ

=−

F

HG

I

KJ

()log

= 1.394 × (1.005 – 0.287) loge

300

600

F

HG

I

KJ = – 0.6938 kJ/K

2

T

S

600 K

300 K

V=C

p=C

1

3

Fig. 5.42