A heat exchanger (boiler) produces dry steam at 100 degrees C from feed water at 30 degrees C at a rate of 1.5kg/s. The specific heat capacity of water is 4187J/Kg and its specific latent heat of vapourisation is 2,257,000 J/kg- The heat exchanger receives heat energy at a rate of 4MW from the fuel used. Determine the thermal eficency from the boiler.Formulas or guidance would be appreciated thankyou.
the water must be raised 70 degrees C (1.5 kg/s)(4,187 J/kg C)(70 C) = 439,635 Joules per second = 0.440 Megawatts add to that the heat of vaporisation (1.5 kg/s)(2,257,000 J/kg)= 3,385,500 Joules per second = 3.386 Megawatts Total power required 3.825 Megawatt efficiency = 3.825/4.00 = 95.63%
Thermal efficiency = Useful output/total input X 100% Thermal energy required to boil and evaporate 1kg of water = (Rise in temperature)(Specific heat capacity of water) + (Specific latent heat of vapourisation) = (100-70)(4187) + (2257000) = 2550090J Therefore, Since 1.5Kg of water is turned to dry stream per second, Useful Output = (1.5)(2550090) = 3825135J Total Input = 4000000 Hence, Thermal efficiency of boiler = 3825135/4000000 X 100%= 95.628375%= 95.6% (3 significant figures) Hope this helps.
