Answer:
3.613 bar
1.0290 m3/kg
0.038378 m3/kg
1.363 m3/kg
0.012166 m3/kg
0.002083 m3/kg
0.001177 m3/kg
Explanation:
part a
Table A-3: vf = 1.0435 x 10-3 m
3/kg
, vg = 1.673
m
3
/kg. Since vf < v < vg, the state is in the two phase liquid-vapor region, as shown.
From Table A-3, the pressure is the saturation is pressure at 140 C: p = 3.613 bar.
Answer: p = 3.613 bar
part b
The pressure is higher than the critical pressure, as shown on the diagram. Hence, the state is in the compressed liquid region.
From Table A-5: v = 1.0290 m3
/kg.
Answer: v = 1.0290 m3
/kg
part c
Since the temperature is higher than Tsat at 100 bar, the state is super-heated vapor. T sat = 311.1 C
Interpolating in Table A-4, we get
v = 0.038378 m3/kg
Answer: v = 0.038378 m3/kg
part d
vx = vf + x(vg – vf)
v = 1.0291 x 10-3
+ (0.4)(3.407 - 1.0291 x 10-3
)
v = 1.363 m3/kg
Answer: v = 1.363 m3/kg
a ) T = 440 C , p = 20 MPa
Tsat(@20 MPa) = 365.75 C
T > Tsat(@20 MPa) hence, super-heated steam
Interpolate the results
v = 0.012166 m3/kg
Answer: v = 0.012166 m3/kg
b) T = 160 C , p= 20 MPa
Tsat(@20 MPa) = 365.75 C
T < Tsat(@20 MPa) hence, compressed liquid region
v = 0.002038 m3/kg
Answer: v = 0.002038 m3/kg
c) T = 40 C , p = 2 MPa
Tsat(@2 MPa) = 212.38 C
T < Tsat(@2 MPa) hence, compressed liquid region
v = 0.001177 m3/kg
Answer: v = 0.001177 m3/kg
