318 ENGINEERING THERMODYNAMICS
DHARM
M-therm\Th6-1.PM5(^5) s
290 K^4
T
1200 K^2
Infinite
source 1
3
6
290 K
400 K
647 K
T
Gas 2 ′
4′ 3′
5′ 6′
1′
s
Fig. 6.9
600 = 3 × 0.81 (T 2 ′ – 400)
∴ T 2 ′ =^600
381 ×0.
- 400 = 646.9 K say 647 K
Available energy with the source
= area 1-2-3-4-1
= (1200 – 290) × 1200600 = 455 kJ
Change in entropy of the gas
= mgcv loge
T
T
2
1
′
′
F
HG
I
KJ
= 3 × 0.81 × loge FH^647400 IK = 1.168 kJ/K
Unavailability of the gas = area 3′- 4′- 5′- 6′- 3′
= 290 × 1.168 = 338.72 kJ
Available energy with the gas
= 600 – 338.72 = 261.28 kJ
∴ Loss in available energy due to heat transfer
= 455 – 261.28 = 193.72 kJ. (Ans.)
+Example 6.6. Calculate the unavailable energy in 60 kg of water at 60°C with respect to
the surroundings at 6°C, the pressure of water being 1 atmosphere.
Solution. Refer Fig. 6.10.
Mass of water, m = 60 kg
Temperature of water, T 1 = 60 + 273 = 333 K
Temperature of surroundings, T 0 = 6 + 273 = 279 K
Pressure of water, p = 1 atm.
If the water is cooled at a constant pressure of 1 atm. from 60°C to 6°C the heat given up
may be used as a source for a series of Carnot engines each using the surroundings as a sink. It
is assumed that the amount of energy received by any engine is small relative to that in the source
and temperature of the source does not change while heat is being exchanged with the engine.