SSC JE Mechanical Engineering 27 Oct 2020 Shift 2 Papers

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Question : 113 of 200

Marks: +1, -0

The specific volume of a wet steam at 1600°C, 80 % quality,

v_ f

= 0.011 m

^3

/kg and

v_g

= 0.3071 m

^3

/kg is:

1
1.09
1.25
0.25

Solution:

{x}=80 \%=0.8, {v}_{{f}}=0.011 {m}^{3} \/ {kg}

and

{v}_{{g}}=0.3071 {m}^{3} \/ {kg} .

{v}={v}_{{f}}+{x}({v}_{{g}}-{v}_{{f}})

v = 0.011 + 0.8 (0.3071 - 0.011)

v = 0.248

v = 0.25 m^3\/kg.

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