Question

liquid ammonia at a temperature of 26 celcius available at the expansion valve, undergoes isenthalpic process until its temperature becomes 2 degree celcius. find the percentage of liquid vaporized while lowing through the expansion valve

Answer 1

Solution;

$T_1=26°c=299K$

$T_2=2°c=275K$

For isenthalpic process;

$\Delta h=0$

From the table for ammonia;

$\def\arraystretch{1.5} \begin{array}{c:c:c} Temperature (°c) & h_f(kJ/kg)& h_g(kJ/kg)\\ \hline 26& 322.471&1483. 131\\ \hdashline 2 & 209.255& 1463.80 \end{array}$

Sine the process is isenthalpic;

$h_1=h_2$

$h_{f_1}+x(h_{g_1}-h_{f_1})=h_{f_2}+x(h_{g_2}-h_{f_2})$

But $x_1=0$

$322.471+0=209.255+x_2(1463.80-209.255)$

$x_2(1254.545)=113.216$

$x_2=0.0902$

Percentage of liquid vapourized;

$x_2×100=0.0902×100$

=9.02%