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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 constant at 1500°C. Calculate the effectiveness of the heating process when the atmospheric temperature is 15°C. 

Take specific heat of liquid as 6.3 kJ/kg K

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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%.

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