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Determine the mass of 0.15 m3 of wet steam at a pressure of 4 bar and dryness fraction 0.8. Also calculate the heat of 1 m3 of steam. |
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Answer» Volume of wet steam, v = 0.15 m3 Pressure of wet steam, p = 4 bar Dryness fraction, x = 0.8 At 4 bar. From steam tables, vg = 0.462 m3/kg, hf = 604.7 kJ/kg, hfg = 2133 kJ/kg ∴ Density = \(\cfrac{1}{xv_g}\) = \(\cfrac{1}{0.8\times0.462}\) = 2.7056 kg/m3 Mass of 0.15 m3 of steam = 0.15 × 2.7056 = 0.4058 kg. Total heat of 1 m3 of steam which has a mass of 2.7056 kg = 2.7056 h (where h is the total heat of 1 kg of steam) = 2.7056 (hf + xhfg) = 2.7056(604.7 + 0.8 × 2133) = 6252.9 kJ. |
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