Mass of steam in the cooker = 1.5 kg 

Pressure of steam, p = 5 bar 

Initial dryness fraction of steam, x₁ = 1 

Final dryness fraction of steam, x₂ = 0.6 

Heat to be rejected : 

Pressure and temperature of the steam at the new state : 

At 5 bar. From steam tables,

t_s = 151.8°C ; hf = 640.1 kJ/kg ; 

h_fg = 2107.4 kJ/kg ; v_g = 0.375 m³/kg 

Thus, the volume of pressure cooker

= 1.5 × 0.375 = 0.5625 m³ 

Internal energy of steam per kg at initial point 1,

u₁ = h₁ – p₁v₁ 

= (h_f + h_fg) – p₁v_g1 (\(\because\)v₁ = vg₁) 

= (640.1 + 2107.4) – 5 × 10⁵ × 0.375 × 10^–3 

= 2747.5 – 187.5 = 2560 kJ/kg

Also, V₁ = V₂ (V₂ = volume at final condition)

0.5625 = 1.5[(1 – x₂) v_f2 + x₂v_g2]

= 1.5 x₂v_g2 (\(\because\)vf₂ negligible) 

= 1.5 × 0.6 × v_g2

v_g2 = \(\cfrac{0.5625}{1.5\times0.6}\) = 0.625 m³/kg. 

From steam tables corresponding to 0.625 m³/kg,

p₂ ~ 2.9 bar, t_s = 132.4°C, h_f = 556.5 kJ/kg, h_fg = 2166.6 kJ/kg 

Internal energy of steam per kg, at final point 2, 

u₂ = h₂ – p₂v₂ 

= (h_f2 + x₂h_fg2) - p₂xv_g2   (\(\because\) v₂ = xv_g2)

= (556.5 + 0.6 × 2166.6) – 2.9 × 10⁵ × 0.6 × 0.625 × 10^–3 

= 1856.46 – 108.75 = 1747.71 kJ/kg. 

Heat transferred at constant volume per kg 

= u₂ – u₁ = 1747.71 – 2560 = – 812.29 kJ/kg 

Thus, total heat transferred 

= – 812.29 × 1.5 = – 1218.43 kJ.  

Negative sign indicates that heat has been rejected.