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18 January, 08:51

A 9 m3 container is filled with 300 kg of r-134a at 10°c. what is the specific enthalpy (kj/kg) of the r-134a in the container? (4 significant figures)

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  1. 18 January, 09:47
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    Given the mass of R-134a m = 300kg; Volume of the container V = 9 cu. meter; Temperature of R-134a T = 10 degrees Celsius;

    Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg.

    At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m / kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.

    Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg
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