Question

6.86 kg of water vapor are contained at 150 kPa ad 90 percent quality in a

suitable enclosure. Calculate the heat in kJ which must be added in order to just

produce a saturated vapor. NOTE: - INDICATE THE HS and TS DIAGRAM - INDICATE THE INITIAL AND FINAL STATE OF THE SUBSTANCE

Answer 1

$P_1=150kPa$

$x_1=0.9$

$x_2=1$ (At the end,state is saturated vapour)

$V_1=V_2$ (Vapour is in enclosure,so the volume remains constant)

From tables;

At $P_1=150kPa$

$h_f=467.13kJ/kg$

$h_g=2693.1kJ/kg$

$v_f=0.001053m^3/kg$

$v_g=1.1594m^3/kg$

So,at state point 1;

$v_1=v_f+x_1(v_g-v_f)$

$v_1=0.001053+0.9(1.1594-0.001053)=1.0436m^3/kg$

And;

$h_1=h_f+x(h_g-h_f)$

$h_1=467.13+0.9(2693.1-476.13)=2470.503kJ/kg$

But;

$v_1=v_2=1.0436m^3/kg$

Also;

$x_2=1$

Hence;

P₂ can be given by linear interpolation between 150kPa and 175kPa;

From the tables,at 175kPa;

$v_g=1.0037m^3/kg$

$h_g=2700.2kJ/kg$

Hence;

$P_2=150+\frac{1.0436-1.1594}{1.0037-1.1594}(175-150)=168.59kPa$And;

$h_2=2693.1+\frac{1.0436-1.1594}{1.0037-1.1594}(2700.2-2693.1)=2698.38kJ/kg$

Heat addition in the process;

$Q_{12}=m(u_2-u_1)$

$Q_{12}=m(h_2-h_1)-m(p_2v_2-p_1v_1)$

By direct substitution;

$Q_{12}=5(2698.38-2470.503)-5(168.59×1.0436-150×1.0436)=1042.54kJ$

Answers;

$P_2=168.59kPa$

$Q=1042.54kJ$