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Find the dryness fraction, specific volume and internal energy of steam at 7 bar and enthalpy 2550 kJ/kg. |
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Answer» Pressure of steam, p = 7 bar Enthalpy of steam, h = 2550 kJ From steam tables corresponding to 7 bar pressure : hf = 697.1 kJ/kg, hfg = 2064.9 kJ/kg, vg = 0.273 m3/kg, uf = 696 kJ/kg, ug = 2573 kJ/kg (i) Dryness fraction, x : At 7 bar, hg = 2762 kJ/kg, hence since the actual enthalpy is given as 2550 kJ/kg, the steam must be in the wet vapour state. Now, using the equation, h = hf + xhfg ∴ 2550 = 697.1 + x × 2064.9 x = \(\cfrac{2550-697.1}{2064.9}\) = 0.897 Hence, dryness fraction = 0.897. (ii) Specific volume of wet steam, v = xvg = 0.897 × 0.273 = 0.2449 m3/kg. (iii) Specific internal energy of wet steam, u = (1 – x)uf + xug = (1 – 0.897) × 696 + 0.897 × 2573 = 2379.67 kJ/kg. |
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