SSC JE Mechanical Engineering 2018 Set 1 Paper

Question : 160 of 200

Marks: +1, -0

Which one of the following is the steady flow energy equation for a boiler?
(A)

h_1+v_1^2/{2gJ}=h_2+v_2^2/{2gJ}

(B)

Q=(h_2-h_1)

(C)

h_1+v_1^2/{2gJ}+Q=h_2+v_2^2/{2gJ}

(D)

W_s=(h_2-h_1)+Q

Solution:

Steady flow energy equation for Boiler

h_1+v_1^2/{2 }+Q=h_2+v_2^2/{2 }+W

for boiler K. E = 0
workdone = 0

h_1 + Q = h_2

[Q =h_2-h_1]

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