SSC JE Mechanical Engineering 2014 Set 2 Paper

Question : 109 of 200

Marks: +1, -0

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 temp. 40°C (liquid enthalpy = 168 kj/kg; enthalpy of vaporisation of water at 100°C = 2258 kj/kg) is :

1649
2000
2149
1682

Validate

Solution:

Given data : mass of steam per hour = 2000 kg/hr
liquid enthalpy

h_\f= 168

KJ/kg
enthalpy of evaporation

h_\fg = 2258

KJ/kg
enthalpy of steam per kg
h = 2426 KJ/kg
Equivalent evaporation

M_e={m_a(h-h_\f)}/h_\fg

={2000(2426-168)}/2258

= 2000 Kg/hr.

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