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Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 500°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 3260 kJ and kinetic and potential energy effects are negligible. Determine the final volume, in m3, and the work for the process, in kJ.

Answer 1

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Given data: Mass of water (m) = 5 kg. P1 = P2 = 5 bar = 0.5 MPa = 500 kPa, T1 = 500 C. Heat transfer for the process (Q) = 3260 kJ. Ignoring (kinetic + potential) energy effects. P-v diagram From, saturated water pressure table A-5 at P1 we can see that given T1 > Tsat hence, water is in superheated state. 1 2. From, superheated water table A-6 at P1, T1: V1 = 0.71095 m3/kg, h1 = 3484.5 kJ/kg For constant pressure heating process: Q = m * (h2 - hu) 3260 = 5 * (h2 - 3484.5) v1 h2 = (3260 / 5) + 3484.5 = 4136.5 kJ/kg. Vn From, superheated water table A-6 at P2 we can see that h2 lies between (ha = 3927.0 kJ/kg, hp = 4158.4 kJ/kg) hence, after using linear interpolation between them and corresponding (VA = 0.89696 m3/kg, VB = 0.98966 m²/kg) we get, v2 = 0.98088 m3/kg. Final volume (V2) = m * V2 = 5 * 0.98088 = 4.9044 m3.... Answer. Work done for constant pressure process: W = p *m* (V2 - V1) = 500 * 5 * (0.98088 -0.71095) = 674.825 kJ.... Answer.