Question

An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 15 kg of water vapor with steam quality 20% at 20 bar and the other part is evacuated. The partition is removed, and the water vapor expands into the entire tank.a) Determine the final temperature and pressure using higher accuracy calculation. b) Recalculate (a) by assuming the gas behave as ideal gas

Answer 1

Any LLL 15 Hy evacuated 2=0. p=20 bar luated From saturated water pressure table at 20 bar (7-5) T59+ = Ti = '212.38'C Ve = 01001177 m2, ky Vg -0,099 587 m3 Ing. 0 - 906 12 831 Uff = 1693 MOIRE U = Up + x Upg = 906.19+ 02 X1693 = 1244.72 KJING Vj = Vftor ( vg-VA) = 0,001177 +0.210.091567-01001177) = 0.020959 m3 1 Kg N = muy = 0, 312885 m3 Vfinal = 2V, = 0.62577 m3 Veinu = 2 V; = 0.62577 = 0.041718 milky Vfinul - 15 "From tatble at U = Ufinal = 1244-72 kg 1 kg & Vpinul a 00041718 m3 1 kg PE 1260 KPU St=1880 TO

.. at UF Uft a vez = 5045+012 (2024.6) gog.42 e V=0.356 m3kg water table. From Saturated T = 102°C Pa = 111 - PQI as gdeal gay. of it behave ther' Du=h cvet OU = CV (Ta - To) L'OU=0 :. To - Ti = bude ou a 212-38 Pova = Pivo To Ti Pa = () CentePu = 20 [ 12 = 10 bar