Question
A tank of volume 0.05 m3 contains a mixture of saturated water and saturated steam at 200°C. The mass of the liquid present is 8 kg. The entropy (in kJ/kg K) of the mixture is ___________ (correct to two decimal places).
Property data for saturated steam and water are:
At 200°C, Psat = 1.5538 MPa
vf = 0.001157 m3/kg, vg = 0.12736 m3/kg
Sfg=4.1014 kJ/kg K, sf =2.3309 kJ/kg K
Answer :
Total volume of tank (V) = 0.05 m3
Masss of liquid m2 = 8 kg
Volume occupied by liquid in tank= m×vf
VL=8×0.001157 m3
VL=0.009256
Vv=V-VL
Vv=0.05-0.009256=4.990744
mv= Vv /vg = 0.04990744/0.12736 = 0.3198 kg
x = mv/(mv+mL) = 0.0384
Specific entropy of mixture (s)
s = sf + xsfg
s = 2.3309 + 0.0384*4.1014
s= 2.4884 kJ/kg-K
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