Question

5kg of steam is condensed in a condenser following reversible constant pressure process from 0.75 bar and 150ºC state. At the end of process steam gets completely condensed. Determine the heat to be removed from steam and change in entropy. Also sketch the process on T-s diagram and shade the area representing heat removed.

Answer 1

At state 1:P₁ = 0.75 bar and T₁ = 150°C Applying SF’EE to the control volume Q – W_s = m_f [(h₂ – h₁)+ 1/2(V²₂  – V¹₂) + g(z₂ – z₁)] neglecting the changes in kinetic and potential energies.i.e.; 

W_s = 1/2(V₂² – V₁²) = g(z₂ – z₁) = 0 i.e.; Q = m [(h₂ – h₁)] 

From super heat steam table at P₁ = 0.75 bar and T₁ = 150°C 

We have(75 – 50)/(100 – 50) = (h₁ – 2780.08)/(2776.38 – 2780.08) h₁ = 2778.23 KJ/kg

Also, entropy at state (1) (75 – 50)/(100 – 50) = (s₁ – 7.94)/(7.6133 – 7.94)

s₁ = 7.77665 KJ/kgK at state (2), the condition is saturated liquid.

From saturated steam table h₂ = h_f = 384.36 kJ/kg s₂ = s_f = 1.2129 kJ/kgK

Q = 384.36 – 2778.23 = –2393.87kJ/kg-ve sign shows that heat is rejected by system Total heat rejected 

= 5 × 2393.87 = 11.9693 MJ similarly, total change in entropy 

= m(s₂ – s₁) = 5(l.2129 – 7.7767) = – 32.819 kJ/K.