Answer to Question #133333 in Mechanical Engineering for lucky mojeremane

Question

B. The pressure in a steam main is 12 bar. A sample of steam is drawn off and passed through a throttling calorimeter, the pressure and temperature at exit from the calometer being 1 bar and 140 degree C respectively. Calculate the dryness fraction of the steam in the main, stating any assumptions made in the throttling process.

Answer 1

Let the initial pressure and temperature of steam be P₁ and T₁ respectively.

Given as P₁=12 bar=12x10⁵ N/m²

From steam table : At P₁=12 bar=12x10⁵ N/m²

h_f=798.33 kJ/kg, h_g=2783.7kJ/kg, h_fg=1985.4 kJ/kg

Steam is throttled to final state having pressure and temperature as P₂=1bar=1x10⁵ N/m² and T₂ =140^oC.

During throttling process the enthalpy of steam remains constant.

Therefore, h₁=h₂

From steam table : At P₂=1bar=1x10⁵ N/m² and T₂ =140^oC.(Steam is super heated)

h₂= 2756.7 KJ/kg

If X=dryness fraction of steam in main

h₁=h_f+X. h_fg=h₂

798.33+X(1985.4)=2756.7

Solving for X => X=0.986

Ans: Dryness fraction of steam in main(X)=0.986