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
hf=798.33 kJ/kg, hg=2783.7kJ/kg, hfg=1985.4 kJ/kg
Steam is throttled to final state having pressure and temperature as P2=1bar=1x105 N/m2 and T2 =140oC.
During throttling process the enthalpy of steam remains constant.
From steam table : At P2=1bar=1x105 N/m2 and T2 =140oC.(Steam is super heated)
h2= 2756.7 KJ/kg
If X=dryness fraction of steam in main
Solving for X => X=0.986
Ans: Dryness fraction of steam in main(X)=0.986