Question

According to steam tables this is impossible. You can have wet steam at 230 degC or wet steam with a density of .025 g/cm^3 but not both. Please help me.

Wet steam at 230°C has a density of 0.025 g-cm-3 Determine x, H, and S

Answer 1

From the saturated steam table at 230 C,

Specific volumes,

Of satuarted vapour Vv = 0.0715102 m³ /kg

Of saturated water Vl = 0.00120901 m³ /kg

Specific Enthalpy,

Hv = 2803.009 kJ/kg

Hl = 990.209 kJ/kg

Specific entropy,

Sv = 6.213 kJ/kg-K

Sl = 2.31 kJ/kg-K

Given: density = 0.025 g/cm³ = 25 kg/m³

1/density = specific volume V = 1 / 25 = 0.04 m³/kg

V = x(Vv) + (1-x)Vl

0.04 = x(0.0715102) + (1-x)*(0.00120901)

.: quality or vapor mass fraction x = 0.55178

Now,

H = xHv + (1-x)Hl

H = 0.55178*2803.009 + ( 1 - 0.55178)*990.209

H = 1990.48 kJ/kg

Similarly,

S = xSv + (1-x)Sl

.: S = 4.598 kJ/kg-K