For H2O, determine the specified property at the indicated state.

(a) T = 140°C, v = 0.5 m3/kg. Find p, in bar.

(b) p = 30 MPa, T = 80°C. Find v, in m3/kg.

(c) p = 10 MPa, T = 600°C. Find v, in m3/kg.

(d) T = 80°C, x = 0.4. Find p, in bar, and v, in m3/kg.

For H2O, determine the specific volume at the indicated state, in m3/kg.

(a) T = 440°C, p = 20 MPa.

(b) T = 160°C, p = 20 MPa.

(c) T = 40°C, p = 2 MPa.

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