GATE Mechanical Engineering (ME) 2018 Shift 1 Solved Paper

Question : 65 of 65

Marks: +1, -0

A tank of volume

0.05 {m}^{3}

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 of two decimal places)
Property data for saturated steam and water are:
At

200^{∘} {C}

, Psat

=1.5538 {MPa}

v_{f}=0.001157 {m}_{3}{/} {kg}, v_{g}=0.12736 {m}_{3} {/} {kg}

s_{f g}=4.1014 {kJ} {/} {kg} {K}, {sf}=2.3309 {kJ}{/} {kg} {K}

Solution:

Total volume of tank

(v)=0.05 {m}^{3}

Means of liquid

(m_{2})=8 {kg}

Volume occupied by liquid in tank

= m×vf

{V}_{ {L}}=8 × 0.001157 {m}^{3}

{V}_{ {L}}=0.009256

{V}_{ {v}}= {V}- {V}_{ {L}}

{V}_{ {v}}=0.05-0.009256=4.990744

{m}_{ {v}}=\;{V_{v}}/{v_{g}}=\;{0.04990744}/{0.12736}=0.3198 {kg}

x=\;{m_{v}}/{m_{v}+m_{L}}=0.0384

Specific entropy of mixture (s)

s=s_{f}+x s_{f g}

s=2.3309+0.0384 × 4.1014

s=2.4884 {kJ} {/} {kg}- {K}

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