# 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 k

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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

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Concept:

The entropy of a mixture:

s = sf + x sfg

where x is the dryness fraction.

$x = \frac{{{m_s}}}{{{m_f} + {m_s}}}$

where ms is the mass of the steam and mf is the mass of liquid

V = Vf + Vg

V = mfvf + mgvg

Calculation:

Given, m= 8 kg,

vf = 0.001157 m3/kg, vg = 0.12736 m3/kg

sfg = 4.1014 kJ/kg, K, sf = 2.3309 kJ/kg K

V = Vf + Vg = 0.05 m3

mfvf + mgvg = 0.05

(8) (0.001157) + mg(0.12736) = 0.05

mg = 0.32 kg

$x = \frac{{{m_s}}}{{{m_f} + {m_s}}} = \frac{{0.32}}{{8 + 0.32}} = 0.0385$

s = sf + x sfg = 2.3309 + 0.0385 × 4.1014 = 2.488 kJ/kg

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