11

A rigid container has volume of 2m^3, and holds steam at 260^{\circ}C.  1/4 of the volume is in liquid point and the remaining at vapor form. Determine the pressure of the steam, and quality of the saturated mixture, and density of the mixture.

Given:

Volume (V)=2m^3

Temperature (T)=260^{\circ}C

Find:

  1. The pressure of the steam.
  2. The quality of the saturated mixture.
  3. The density of the mixture.

Solution

  1. From Table A-4, the properties of Temperature(T) at 220^{\circ}C  v_f=0.001276m^3/kg\text{ and }v_g=0.042175m^3/kg                           P=T_{sat@220^{\circ} C}=4692.3 kPa
  2. The total mass and the quality are determined as
    m_f=\frac{V_f}{v_f}=\frac{0.25x2m^3}{0.001276 m^3/kg}=391.84kg
    m_g=\frac{V_g}{v_g}=\frac{0.75x2m^3}{0.042175m^3/kg}=35.56609kg
    m_t=m_f+m_g=391.84+35.6=427.44kg
    m_t=\frac{m_g}{m_t}=\frac{35.6}{427.44}=0.083286
  3. The density is determined from

v=v_f+x(v_g-v_f)=0.001276+(0.083286)(0.040899)=0.00468231411m^3/kg
\rho=\frac{1}{v}=\frac{1}{0.00468231411}=213.569 kg/m^3

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Thermodynamics Copyright © by Diana Bairaktarova (Adapted from Engineering Thermodynamics - A Graphical Approach by Israel Urieli and Licensed CC BY NC-SA 3.0) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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