3 years ago

Help please really fast!!

Engineering

1 answer:

ra1l [238]3 years ago

6 0

Answer:368 hdhtygtÿ

901 vuiøöńč

Explanation:

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

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

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

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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:

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

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