Question

A rigid tank of 0.5 m^3 volume contains wet steam at 80 C and 1% quality. The tank is then heated until the water inside becomes saturated vapor. Determine the final temperature and pressure within the tank.

Answer: 271 C, 5.588MPa.

Answer 1

Initial State before heating: Total volume of tank (V) = Temperature (T) = Quality (xi) = 0.5 m3 80 C 1% = 0.01 Since quality is mentioned, that means the initial state is a mixture of saturated liquid water and saturated water vapor. We first need to find mass of total mixture for which we need to find initial specific volume. From saturated steam table, Specific Volume of saturated liq water at 80 C (VLI) Specific Volume of saturated water vapor at 80 C (Vv1) = = 0.001029 m/Kg 3.407 m²/Kg Sp. Volume of mixture using quality is given as: Initial specific volume (vi) = (1-x) * (VLI) +x* (VI) (1 -0.01) * 0.001029 +0.01 * 3.407 0.03508871 m²/Kg Mass of mixture (m) Total volume of tank Initial specific volume of mixture Total volume of tankisture - 0.5 0.03508871 Kg 14.2496 Kg Final State after heating: Tank is rigid. Thus its volume will not change. Total volume of tank after heating (V2) = V = 0.5 m Mass will also not change since no mass is escaping or added. m = 14.2496 Kg Final volume of tank Final specific volume (v2) = 0.5 14.2496 0.03508871 m/Kg (Same as vi)

We know that the final state is saturated vapor. Thus, we need to find T and P from saturated steam table which corresponds to specific vapor volume of 0.03508871 m/Kg This exact value of specific volume is not available. Thus, we need to find 2 values of specific volume of saturated vapor in which this value lies. From saturated steam table, (Pa, Ta, va) = (5.4987, 270, 0.03564) (Po, Tb, vb) = (5.9418, 275, 0.03279) 0.03508871 lies between 0.03564 and 0.03279. We need to use Linear Interpolation to find corresponding T and P Using Linear Interpolation, T = T + T5-T2 * (V2-v.) (vb-va) 270+ - (275-270) (0.03279-0.03564) * (0.03508871 -0.03564) 270.97 °C (ANSWER) P2 = P. + (Pb-Pa) . (V2-va) (vb-va) 5.4987 + *(0.03508871 – 0.03564) (0.03279-0.03564) (5.9418-5.4987) 5.584 Mpa (ANSWER)