Question

A vessel of volume $1.0m^3$ contains a mixture of liquid water and steam in equilibrium at $1.0$ bar. Given that $90$% of the volume is occupied by the steam, find the fraction of the mixture. Assume at $1.0$ bar, $v_f=0.001m^3/mg$ and $v_g=1.7m^3/kg$

• A. 5.266

Answer 1

The correct option is A 5.266
$\text{Method I:}$
Volume of the mixture,
$V=1m^3$

$\text{Pressure:}p=1\text{bar}$

Volume of steam,
$V_g=90\text{\%}ofV$

$=0.9V=0.9\times 1=0.9m^3$

$\therefore$ Volume of liquid water,

$V_f=V-V_g=1-0.9=0.1m^3$

$V_f=0.001m^3/kg$

$V_g=1.7m^3/kg$

Specific volume of liquid water,
$V_f=\frac{V_f}{m_f}$

$0.001=\frac{0.1}{m_f}$ or $m_f=100kg$

Specific volume of steam,
$v_g=\frac{V_g}{m_g}$

$1.7=\frac{0.9}{m_g}$

$\text{or}{m}_g=0.529kg$

Dryness fraction,
$x=dfrac{m}_g{m}_1+m_g=\frac{0.529}{100+0.529}$
$5.262\times {10}^{-3}$

Method II:
Mass of mixture in the vessel,
$m=m_f+m_g$

$=\frac{V_f}{V_f}+\frac{V_g}{V_g}$

$\frac{0.1}{0.001}+\frac{0.9}{1.7}=100.53kg$

Specific volume of mixture,
$V=\frac{1}{m}=\frac{1}{100.53}$

$=9.947\times {10}^{-3}{m}^3/kg$

Also, $V=V_f+x(V_g-V_f)$

(x = dryness fraction)

$x=\frac{V-V_f}{V_g-V_f}=\frac{0.009947-0.001}{1.7-0.001}$

Dryness fraction,
$x=5.266\times {10}^{-3}$