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rjkz [21]
3 years ago
7

Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°

C. What are the temperature and quality of the wet steam? Calculate the change of entropy of the steam as a result of the process. It is important to note that a throttling process is where the change of enthalpy is zero (that is, AH = 0). Answer: Temperature ~ 198 °C and the quality of 96.76%. AS = 0.8708 kJ/kg-K
Engineering
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

m\cdot (h_{in} - h_{out}) = 0

h_{in} = h_{out}

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

P = 1500\,kPa

T = 198.29\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 6.3068\,\frac{kJ}{kg\cdot K}

x = 0.967

State 2 - Outlet (Superheated Vapor)

P = 200\,kPa

T = 130\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 7.1776\,\frac{kJ}{kg\cdot K}

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

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