Question

A turbine is supplied with steam at a pressure of 32 bar and a temperature of 410°C. The steam then expands isentropically to a pressure of 0.08 bar. Find the dryness fraction at the end of expansion and thermal efficiency of the cycle. 

If the steam is reheated at 5.5 bar to a temperature of 395°C and then expanded isentropically to a pressure of 0.08 bar, what will be the dryness fraction and thermal efficiency of the cycle ?

Answer 1

From Mollier chart : 

h₁ = 3250 kJ/kg 

h₂ = 2170 kJ/kg Heat drop (or work done) = h₁ – h₂ 

= 3250 – 2170 = 1080 kJ/kg

Heat supplied = h₁ – h_f2 

= 3250 – 173.9 [h_f2 = 173.9 kJ/kg at 0.08 bar] 

= 3076.1 kJ/kg

Thermal efficiency = \(\cfrac{Work \,done}{Heat \,supplied}\) = \(\cfrac{1080}{3076.1}\) = 0.351 or 35.1%.

Exhaust steam condition, x₂ 

= 0.83 (From Mollier chart).

Second case. Refer

From Mollier chart : 

h₁ = 3250 kJ/kg ; 

h₂ = 2807 kJ/kg ; 

h₃ = 3263 kJ/kg ; 

h₄ = 2426 kJ/kg. 

Work done = (h₁ – h₂) + (h₃ – h₄) = (3250 – 2807) + (3263 – 2426) = 1280 kJ/kg 

Heat supplied = (h₁ – h_f4 ) + (h₃ – h₂) 

= (3250 – 173.9) + (3263 – 2807) = 3532 kJ/kg

Thermal efficiency

Condition of steam at the exhaust, 

x₄ = 0.935 [From Mollier chart].