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The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr of steam with enthalpy content of 2426 kJ/kg from feed water at temperature 40°C (liquid enthalpy =168 kJ/kg) is (enthalpy of vaporization of water at 100°C = 2258 kJ/kg)
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- 2000
- 2149
- 186
- 1649
- 2000
Correct Option: A
Equivalent evaporation = | × rh | |||
hfgam |
Equivalent evaporation = | = 2000 kg/hr | |
2258 |