<p> Problem 1: Let K denote K2CrO4 and W denote water. The process flow chart is shown below: </p><p>M 1 100% W (vapor)</p><p>4500 kg/h M Crystallizer 2 M 33.3% K Evaporator & Filter 5 49.4% K 66.7% W 36.4% K 50.6% W 63.6% W</p><p>Two Streams Combined M3 (crystals) 100% K M4 (solution M5) 36.4% K 63.6% W</p><p>The filter cake is a mixture of solid K crystals and a liquid solution with known K and W mass fractions. The crystals constitute 95% by mass of the filter cake stream (M3+M4). Please determine a) Flow of M2. b) Flow of M3. c) Flow of M5.</p><p>Solution Evaporator Balance In-Out=Acc K balance 4500[kg/h]*0.333-M2*0.494=0 M2=4500*0.333/0.494= 3033.4 kg/h</p><p>Crystallizer Balance In-Out=Acc Overall Balance M2-M3-M4-M5=0 3033.4 kg/h-M3-M4-M5=0 K Balance 3033.4 kg/h *0.494-M3*1.0-M4*0.364-M5*0.364=0 W Balance 3033.4 kg/h *0.506-M3*0 -M4*0.636-M5*0.636=0 Define a quantity M6=M4+M5 since it has the same composition as both stream 3 and 4. Use this substitution on Evaporator W Balance. 3033.4 kg/h *0.506-M3*0 -M6*0.636 =0 M6=3033.4 kg/h *0.506/0.636=2413.4 kg/h</p><p>Use overall balance to determine M3 3033.4 kg/h-M3-M6=0 M3=3033.4 kg/h-M6=3033.4 kg/h-2413.4 kg/h = 620 kg/h crystals M3/(M3+M4)=0.95 gives 0.05*M3=0.95*M4 0.05*620 kg/h/0.95=M4=32.6 kg/h</p><p>Evaporator Overall Balance M2-M3-M4-M5=0 3033.4 kg/h-620 kg/h-32.6 kg/h-M5=0 M5= 2380.8 kg/h</p><p>Problem 2: Perform a degrees of freedom analysis for the splitter in the following process flow sheet. N 5 0.983 DA 0.017 W(v) Air N (mole) N N Split N =100 mole 1 Mix 2 Conditioner 4 6 0.96 DA 0.977 DA 0.983 DA 0.983 DA 0.04 W(v) 0.023 W(v) 0.017 W(v) 0.017 W(v)</p><p>DA = dry air N W(v) = water vapor 3 W(l) = liquid water 1.0 W(l)</p><p>Solution: DoF = 1</p><p>Problem 3: Calculate the theoretical air for the combustion of 100 kg of dried coal with ultimate analysis.</p><p>Element Ultimate Analysis Mole Weight (gm/mole) C 71.2% 12 H 4.8% 1 S 4.3% 32 O 9.5% 16 Ash 10.2% -</p><p>Solution: Basis: 100 kg coal</p><p>Element Ultimate Analysis Mole Weight (kg/kmol) kmol C 71.2% 12 5.933 H 4.8% 1 4.8 S 4.3% 32 0.134 O 9.5% 16 Ash 10.2% - Reactions C + 02  CO2 H + ¼ O2  ½ H2O S + O2  SO2</p><p>Total O2 required for complete combustion= 5.933+4.8/4+0.134=7.267 kmol Minus the amount of O in the coal = ½ * (9.5/16)=0.297*kmol Equals 7.267-0.297=6.970 kmol O2/100 kg coal. Theoretical Air = Theoretical O2/0.21= 6.970 kmol O2/0.21=33.19 kmol Air</p>