Mumbai university > MECH > SEM 3 > THERMO

Marks: 4M

Year: Dec 2014

Given: m= 2kg; P= 1.9 Mpa; X = 0.85

Solution:

Properties of steam are determined from steam table at 19bar pressure from pressure table.

i) Finding enthalpy

H = m(hf + X xhfg)

= 2 (896.8 + 0.85 x 1899.3) = 5022.4 kJ

ii) Finding volume

V = mv = m × X × vg = 2 × 0.85 × 0.1047 = 0.178 $m^3$

iii) Finding entropy

S = m(sf + Xsfg)

= 2(2.423 + 0.85 × 3.931) = 11.53 kJ/K