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26 February, 14:56

Determine the enthalpy, volume and density of 1.0 kg of steam at a pressure of 0.5 MN/m2 and with a dryness fraction of 0.96

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  1. 26 February, 16:08
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    Enthalpy, hsteam = 2663.7 kJ/kg

    Volume, Vsteam = 0.3598613 m^3 / kg

    Density = 2.67 kg / m^3

    Explanation:

    Mass of steam, m = 1 kg

    Pressure of the steam, P = 0.5 MN/m^2

    Dryness fraction, x = 0.96

    At P = 0.5 MPa:

    Tsat = 151.831°C

    Vf = 0.00109255 m^3 / kg

    Vg = 0.37481 m^3 / kg

    hf = 640.09 kJ/kg

    hg = 2748.1 kJ/kg

    hfg = 2108 kJ/kg

    The enthalpy can be given by the formula:

    hsteam = hf + x * hfg

    hsteam = 640.09 + (0.96 * 2108)

    hsteam = 2663.7 kJ/kg

    The volume of the steam can be given as:

    Vsteam = Vf + x (Vg - Vf)

    Vsteam = 0.00109255 + 0.96 (0.37481 - 640.09)

    Vsteam = 0.3598613 m^3 / kg

    From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg / m^3
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