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A vessel having a volume of 0.6 m3 contains 3.0 kg of liquid water and water vapour mixture in equilibrium at a pressure of 0.5 MPa. Calculate : (i) Mass and volume of liquid ; (ii) Mass and volume of vapour |
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Answer» Volume of the vessel, V = 0.6 m3 Mass of liquid water and water vapour, m = 3.0 kg Pressure, p = 0.5 MPa = 5 bar Thus, specific volume, v = \(\cfrac Vm\) = \(\cfrac{0.6}{3.0}\) = 0.2 m3/kg At 5 bar : From steam tables, vfg = vg – vf = 0.375 – 0.00109 = 0.3739 m3/kg We know that, v = vg – (1 – x) vfg, where x = quality of the vapour. 0.2 = 0.375 – (1 – x) × 0.3739 (1 – x) = \(\cfrac{(0.375-0.2)}{0.3739}\) = 0.468 x = 0.532 (i) Mass and volume of liquid, mliq. = ? Vliq. = ? mliq. = m(1 – x) = 3.0 × 0.468 = 1.404 kg. Vliq. = mliq. vf = 1.404 × 0.00109 = 0.0015 m3. (ii) Mass and volume of vapour, mvap. = ? Vvap. = ? mvap. = m.x = 3.0 × 0.532 = 1.596 kg Vvap. = mvap. vg = 1.596 × 0.375 = 0.5985 m3. |
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