Question

Two boilers one with superheater and other without superheater are delivering equal quantities of steam into a common main. The pressure in the boilers and main is 20 bar. The temperature of steam from a boiler with a superheater is 350°C and temperature of the steam in the main is 250°C. 

Determine the quality of steam supplied by the other boiler. Take c_ps = 2.25 kJ/kg.

Answer 1

Boiler B₁. 20 bar, 350°C : 

Enthalpy, h₁ = h_g1 = + c_ps (T_sup – T_s ) 

= 2797.2 + 2.25(350 – 212.4) 

= 3106.8 kJ/kg ...(i) 

Boiler B₂. 20 bar (temperature not known) :

 h₂ = h_f2 + x₂h_fg2

= (908.6 + x₂ × 1888.6) kJ/kg

Main. 20 bar, 250°C. 

Total heat of 2 kg of steam in the steam main

= 2[h_g + c_ps (T_sup – T_s)]

= 2[2797.2 + 2.25 (250 – 212.4)] = 5763.6 kJ

Equating (i) and (ii) with (iii), we get 

3106.8 + 908.6 + x₂ × 1888.6 = 5763.6 

4015.4 + 1888.6x₂ = 5763.6

x₂ = \(\cfrac{5763.6-40154}{1888.6}\) = 0.925 

Hence, quality of steam supplied by the other boiler = 0.925.