Question

Water at 0.5 quality is heated from 0.2 MPa to 200°C at a fixed volume. What is the final pressure? Choose the closest value

a. 285 kPa

b. 172 kPa

c. 505 kPa

d. 1156 kPa

e. 795 kPa

Answer 1

State 1 : The properties of steam at P=0.2 MPa is

T= 120.2 C , Vf = 0.001061 m^3/kg and Vg = 0.88540 m^3/kg

So the specific volume at 0.5 quality is given by ,

V1 = Vf + x(Vg - Vf) = 0.001061 + 0.5(0.88540 - 0.001061)

V1 = 0.443205 m^3/kg

Now for the fixed volume , V1 = V2 = 0.443205 m^3/kg

State 2 : T= 200 C

As we know that as we move up in the vapour dome the dome starts to narrow up and the specific volume keeps on decreasing.

So at T = 200 C , the saturation conditions are ,

P = 15.549 bar , Vf = 0.001156 m^3/kg and Vg = 0.12716 m^3/kg

So, V2 = Vf + x(Vg - Vf)

0.4432305 = 0.001156 + x(0.12716-0.001156)

x = 3.5

Since x>1 so we need to take the conditions from the superheated vapour .

So superheated property at T = 200 C and V = 0.4432305 m^3/kg

The pressure P = 4.81347 Bar

P2 = 481.347 KPa

So it is closest to option (c)