Question

Can you explain how the answer is -0.717 kJ/K?

A rigid, 16-L steam cooker is arranged with a pressure relief valve set to release vapor and maintain the pressure once the pressure inside the cooker reaches 150 kPa. Initially, this cooker is filled with water at 175 kPa with a quality of 10 percent. Heat is now added until the quality inside the cooker is 40 percent. The water is stirred at the same time that it is being heated. Determine the minimum entropy change of the heat-supplying source if 100 kJ of work is done on the water as it is being heated. Use steam tables. The minimum entropy change of the heat-supplying source is -0.717 kJ/K.

Answer 1

Sol Given dato :- 162 steam cooker pressure inside the cookeyCp2]= 150 kpa cooker foued with the wateYCR) = 175 kpa Quality (2) - -10% -01 quality inside the cookeycy} 40% 100kJ Initially , steam propertieg at 175 kpa from the steam tabler -0.001057 m®/kg 19, = 1.00 34 m/kg Sf. 104850 kg/kg-K $$9, 5.6865 kJ/kg-K work : > .: Specific volume cvi) = Vf, +X, Vfg, = 4, +4(Vg1 - "F,] -0.001057 +01 (1•0037 -0.00105+) 20-10132 malkg = 0.10132 m2/kg) .: Specific Entropy C5,7 = Sfit X, sfg, = 1.4850 + 0.1(5.6865) 2.05365 kging.K .: S, = 2.05365 kg/kg-k]

:: properties of steam at 150/pa from steam tably at 150 kpa ~ 1.5 bay Vf2 = 0.001053 m®/kg mºlkg = 1.1590 m3 939 592 Sfa 1.433 Hj/kg-K .5.790 kJ/kg. k 7.223 kJ/kg-k .: Specific volume at final condition V2 = Vf2 +*2 Vf92 = Vf2 + 2₂ CVg2 - Vf2] = 0.001053 7 0.4 [1590 - 0.001053) - 0.46423 m3/ng .:V2 = 0.46423 milkg ... Specific entropy (92) = 42+ R2 5892 = 7.433 + 0.405.79) 3.749 kg/kg-k .. 52 3.749 kJ/kg-k| .. mass of water cm, 3 = ŕ where volume CV) - 166 = 16x103 m3 specific volume cy] = 0.10132 m2/kg mi =

16X10-3 0.10132 - 0.1579 .: mass cm, - 0.1579 kg Final V . mass cm₂3 2 16x103 0.46423 out = 0.03446 kg .. mass cm2) - 0-03446 kg Therefore, conservation of mais drm Eto em dt ćno water entering] dm 0 - Em dt out Entropy balance Equation dams) + Sx mouto dt -- mout SUY dt + LSSUY + - 5 dm dt dt dcms) de integrate the above equation neith dt Assur + m2 52 - m,si - Sout Cm -mozo Sout → out Csaturated vaporer) 592 € 7•223 Rging K at 150kpa

::Asuy + m2 52 - m, 5, - Sout [my-mozo → Asur + 60-3446x 3.749)- (0.1579x 2.05365) - 7.223 [0·03446 - 0:1579) 7,0 → ASUY + 0.12919 0.32427 -0.8986 7,0 → Asur + 0. 7052 20 =) osur > -0.7052 kglk .: The minimum Entropy change required 0.7052 -0.717 kJik