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# Answer to Question #133333 in Mechanical Engineering for lucky mojeremane

Question #133333
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.
1
2020-09-17T06:38:54-0400

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.

Therefore, h1=h2

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

h1=hf+X. hfg=h2

798.33+X(1985.4)=2756.7

Solving for X => X=0.986

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

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