AVAILABILITY AND IRREVERSIBILITY 327DHARM
M-therm\Th6-2.PM5
the water evaporates at 200°C. The unavailable energy becomes the area under PW. Therefore, the
increase in unavailable energy due to irreversible heat transfer is represented by the area under
NW.
Now, Q 1 = T 1 ∆s = T 1 ′∆s′
∆
∆s
sT
T′=
′1
1
W = Work done in cycle LMNP
= (T 1 – T 0 ) ∆s ...per unit mass
W′ = Work done in cycle RTWP
= (T 1 – T 0 ) ∆s′ ...per unit mass
The fraction of energy that becomes unavailable due to irreversible heat transfer,
WW
WTTsTTs
TTsTs s
TTs−′= −−−′
− =′−
−()()
()()
()10 10
100
10∆∆
∆∆∆
∆=F ′−
HI
K
− =′−F
HGI
KJ
−T ss
TTT TT
TT0
100 11
10∆∆ 1 1()()=TT TTT T01 1′−−′
11 0()
() =303 673 473
473 673 303()
()−
− = 0.346
Hence the fraction of energy that becomes unavailable = 0.346 or 34.6%. (Ans.)
Example 6.15. A liquid is heated at approximately constant pressure from 20°C to 80°C
by passing it through tubes which are immersed in a furnace. The furnace temperature is con-
stant at 1500°C. Calculate the effectiveness of the heating process when the atmospheric tempera-
ture is 15°C.
Take specific heat of liquid as 6.3 kJ/kg K.
Solution. Initial temperature of fluid, T 1 = 20 + 273 = 293 K
Final temperature of fluid, T 2 = 80 + 273 = 353 K
Temperature of the furnace, Tf = 1500 + 273 = 1773 K
Atmospheric temperature, T 0 = 15 + 273 = 288 K
Specific heat of liquid, cpl = 6.35 kJ/kg K
Increase of availability of the liquid
= b 2 – b 1 = (h 2 – h 1 ) – T 0 (s 2 – s 1 )
i.e., b 2 – b 1 = cpl (T 2 – T 1 ) – T 0 × cpl logeT
T2
1= 6.3 (353 – 293) – 288 × 6.3 × loge353
293F
HGI
KJ = 39.98 kJ/kg
Now, the heat rejected by the furnace = Heat supplied to the liquid, (h 2 – h 1 ).
If this quantity of heat were supplied to a heat engine operating on the Carnot cycle its
thermal efficiency would be,ηth =^1F − 0
H
GI
K
JT
Tf =
1 288
1773F −
HGI
KJ = 0.837 (or 83.7%)
∴ Work which could be obtained from a heat engine
= Heat supplied × Thermal efficiency