Answer to Question #263450 in Mechanical Engineering for Rekteat

Question #263450

Five kg of water vapour are contained at 150 kPa ad 90 per cent quality in a

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

produce a saturated vapor. What will the pressure be at the end of the

heating process?


1
Expert's answer
2021-11-10T00:48:57-0500

Solution;

Given;

"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\u00d71.0436-150\u00d71.0436)=1042.54kJ"

Answers;

"P_2=168.59kPa"

"Q=1042.54kJ"




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