SSC JE Mechanical Engineering 25 Sep 2019 Shift 1 Paper

Question : 154 of 200

Marks: +1, -0

A vessel of volume 0.04 m

^{3}

contains a mixture of saturated water and steam at 200°C. The mass of liquid is 5 kg. Find the mass of vapor in the mixture (given

v_f

= 0.0011 and

v_g

= 0.12 m

^{3}

/kg at 200°C).

Solution:

Given:

{V}=0.04 {m}^{3}

{v}_{{f}}=0.0011 {m}^{3} \/ {kg}, {m}_{{f}}=5 {kg},

{v}_{{g}}=0.12 {m}^{3} \/ {kg}

{V}_{{f}}={v}_{{f}} \×{m}_{{f}}=0.0011 \×5=0.0055 {m}^{3}

{V}={V}_{{f}}+{V}_{{g}}={m}_{{f} {V}} {f}+{m}_{{g} {v}_{{g}}}

0.04=5(0.0011)+{m}_{{g}}(0.12)

{m}_{{g}}=0.29 {kg}

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