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.

Expert's answer

Let the initial pressure and temperature of steam be P_{1 } and T_{1} respectively.

Given as P_{1}=12 bar=12x10^{5} N/m^{2}

From steam table : At P_{1}=12 bar=12x10^{5} N/m^{2}

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 P_{2}=1bar=1x10^{5} N/m^{2} and T_{2} =140^{o}C.

During throttling process the enthalpy of steam remains constant.

Therefore, h_{1}=h_{2}

From steam table : At P_{2}=1bar=1x10^{5} N/m^{2} and T_{2} =140^{o}C.(Steam is super heated)

h_{2}= 2756.7 KJ/kg

If X=dryness fraction of steam in main

h_{1}=h_{f}+X. h_{fg}=h_{2}

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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