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15 September, 15:27

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.

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Answers (1)
  1. 15 September, 16:41
    0
    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
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