Question

Find the dryness fraction, specific volume and internal energy of steam at 7 bar and enthalpy 2550 kJ/kg.

Answer 1

Pressure of steam, p = 7 bar 

Enthalpy of steam, h = 2550 kJ 

From steam tables corresponding to 7 bar pressure : 

h_f = 697.1 kJ/kg, 

h_fg = 2064.9 kJ/kg, 

v_g = 0.273 m³/kg, 

u_f = 696 kJ/kg, 

u_g = 2573 kJ/kg

(i) Dryness fraction, x : 

At 7 bar, h_g = 2762 kJ/kg, hence since the actual enthalpy is given as 2550 kJ/kg, the steam must be in the wet vapour state. 

Now, using the equation, 

h = h_f + xh_fg 

∴ 2550 = 697.1 + x × 2064.9

x = \(\cfrac{2550-697.1}{2064.9}\) = 0.897

Hence, dryness fraction = 0.897.

(ii) Specific volume of wet steam, 

v = xv_g = 0.897 × 0.273 = 0.2449 m³/kg.

(iii) Specific internal energy of wet steam, 

u = (1 – x)u_f + xu_g 

= (1 – 0.897) × 696 + 0.897 × 2573 

= 2379.67 kJ/kg.