Initial temperature of fluid, T1 = 20 + 273 = 293 K
Final temperature of fluid, T2 = 80 + 273 = 353 K
Temperature of the furnace, Tf = 1500 + 273 = 1773 K
Atmospheric temperature, T0 = 15 + 273 = 288 K
Specific heat of liquid, cpl = 6.35 kJ/kg K
Increase of availability of the liquid
= b2 – b1 = (h2 – h1) – T0(s2 – s1)
i.e., b2 – b1 = cpl (T2 – T1) – T0 × cpl loge \(\cfrac{T_2}{T_1}\)
= 6.3 (353 – 293) – 288 × 6.3 × loge \(\left(\cfrac{353}{293}\right)\) = 39.98 kJ/kg
Now, the heat rejected by the furnace = Heat supplied to the liquid, (h2 – h1).
If this quantity of heat were supplied to a heat engine operating on the Carnot cycle its thermal efficiency would be,
Work which could be obtained from a heat engine
= Heat supplied × Thermal efficiency
Possible work of heat engine = (h2 – h1) × 0.837
The possible work from a heat engine is a measure of the loss of availability of the furnace.
∴ Loss of availability of surroundings
= (h2 – h1) × 0.837 = cpl (T2 – T1) × 0.837
= 6.3 ( 353 – 293) × 0.837 = 316.38 kJ/kg
Then, effectiveness of the heating process
∈ = \(\cfrac{Increase \,of \,availability of\, the \,liquid}{ Loss \,of \,availability \,of \,surroundings}\)
= \(\cfrac{39.98}{316.38}\) = 0.1263 or 12.63%.