Sheet2In the combustion of heptane, CO2 is produced. Assume that you want to produce 500 kg ofdry ice per hour and that 50% of the CO2 can be converted into dry ice, as shown in Fig. E1.27How may kilograms of heptane must be burned per hour?500kg/hrDiketahuiC7H16 combustion --> C7H16 direaksikan dng O2menghasilkan CO2pers reaksi --> C7H16+O2-->CO2+H2Ostoikiometri11178DitanyaC7H16 burned per day ?kgJawabBasis1hratau500kg dry iceBM CO244BM C7H16100CO2 total terproduksi=500 kg dry icex1kg CO20.5kg dry ice=1000kg CO2Mol CO2 yang dihasilkan=1000kg CO244kg/kmol CO2=22.7272727273kmol CO2Mol C7H16 yang dibutuhkan=22.7272727273kmol CO2x1kmol C7H167kmol CO2=3.2467532468kmol C7H16Massa C7H16 yang dibutuhkan=3.2467532468kmol C7H16x100kg/kmol C7H16324.6753246753kg C7H16

ReactorC7H16 GasCO2 Solid (50%)CO2 Gas (50%)Other Products

Sheet1A liquified mixture of n-butane, n-pentane and n-hexane has the following compositionin percent.n - C4H1050n - C5H1230n - C6H1420Calculate the weight fraction, mol fraction and mol percent of each component and also theaverage molecular weight of the mixture.DiketahuiKomposisi camp n-butane, n-pentane dan n-hexanen - C4H1050%n - C5H1230%n - C6H1420%Ditanyaa. weight fractionfraksi beratb. mol fractionfraksi molc. average MWBM rata2JawabBASIS
VAIO: VAIO:harus dipilih/ditentukan apabila tidak disebutkan100kgBM C12BM H1Komponenkgweight fractionBMmolmol fractionn - C4H10500.5580.86206896550.5704178642n - C5H12300.3720.41666666670.2757019677n - C6H14200.2860.23255813950.15388016810011.51129377171avg MW =total mass=100kg=66.1684722529total mole1.5112937717kg/kgmol

Sheet3Tips for Solving Problems1. Write the Given and Find statements2. Draw a picture of the system labeling all Inputs and Outputs for each component.3. Assign variables to the unknowns. Use symbols to describe the known parameters and list their values.4. Determine whether the problem is a batch process or rate process. Determine whether components are generated or consumed.5. Write the conservation of mass for each component and for the entire system.6. Use algebra to solve for the unknowns.7. Check your work.1GIVEN:10 gallon fish tanksalt water10gallonInitial concentration: 2 percent saltinitial concentration2%Final concentration: 3.5 percent saltfinal concentration3.5%FIND:Amount of salt to be added2DRAW A PICTURE3ASSIGN VARIABLES4BATCH PROCESS / RATE PROCESS--> BATCH PROCESS5MASS BALANCETotalISW+DS=FSW10galXlbmYlbmr H2O62.3lbm/ft30.1337ft3/gal83.2951lbm+Xlbm=Ylbmeq (1)SaltCsalt.ISW+Csalt.DS=Csalt.FSW0.0283.2951+1X=0.035Y1.665902+1X=0.035Yeq (2)H2OCH2O.ISW+CH2O.DS=CH2O.FSW0.9883.2951+0X=0.965Y81.629198+0X=0.965Yeq (3)6from eq. 381.629198=0.965YY=84.589842487???from eq. 21.665902+1X=0.035YX=2.960644487-1.665902=1.294742487???7CHECKfrom eq.183.2951lbm+Xlbm=Ylbm83.2951lbm+1.294742487lbm=84.589842487lbm

MixerInitial salt waterFinal salt waterDry saltMixerInitial salt water (ISW)Final salt water (FSW)Dry salt (DS)10 gal salt water2% salt98% H2OX lbm dry saltY lbm dry salt3.5% salt96.5% H2O

Sheet4Diketahuimassa CO2 padat yang dihasilkan500kg/hrprosentase CO2 padaat/CO2 semua50%pers reaksiC7H16+O2-->CO2+H2O11178Ditanyamassa C7H16 yang dibutuhkanJawabBasis1jam1mol CO2 padat yang dihasilkan=500=11.3636363636mol442mol CO2 keseluruhan yang dihasilkan=11.3636363636=22.7272727273mol50%3mol C7H16 yang dibutuhkan --> berdasar persamaan stoikiometriC7H16+O2-->CO2+H2Ostoikiometri11178mol3.246753246835.714285714322.727272727325.9740259744massa C7H16 yang dibutuhkan=3.2467532468x100

Sheet5Tips for Solving Problems1. Write the Given and Find statements2. Draw a picture of the system labeling all Inputs and Outputs for each component.3. Assign variables to the unknowns. Use symbols to describe the known parameters and list their values.4. Determine whether the problem is a batch process or rate process. Determine whether components are generated or consumed.5. Write the conservation of mass for each component and for the entire system.6. Use algebra to solve for the unknowns.7. Check your work.1,2,3GivenFINDHitunglah massa distilat per kg limbah!4BATCH PROCESS5MASS BALANCETOTALF=D+W1=D+WEQ1ETHANOLXf . F=Xd . D+Xw . W0.35=0.85 D+0.05 wEQ2H2O(1-Xf) . F=(1-Xd) . D+(1-Xw ). W0.65=0.15 D+0.95 wEQ36SOLUTIONS(EQ2 & EQ3)A0.850.05B0.350.150.950.65A-11.1875-0.0625-0.18751.0625X0.375D0.625W7CHECKF=D+w0.350.318750.031250.650.056250.5937510.3750.625

1 kg Feed (F):35% Etanol65% H2ODistilat (D):85% Etanol15% H2OWaste(W): 5% Etanol95% H2O

Sheet6A23B80.250.350.550.10.34-3-20.350.20.40.30.30.40.450.050.60.4A-10.16666666670.16666666670.16666666670.16666666670.2222222222-0.11111111110.2222222222-0.1111111111-3.44.60.62.85-4.151.85X = A-1.B11.550.55-1.4520.60.350.05A0.800-1B1.2XNG1.60-100EA00.21-0.5-14.9W0.20.790093.9DI0.60490045940.3024502297-0.60490045940.1607963247-0.1531393568-0.07656967840.15313935681.22511485450.9678407351-0.5160796325-0.96784073510.2572741194-0.51607963250.2419601838-0.48392036750.128637059712.8606431853115.6049004594NGEAWDIFG20.577029096512.8606431853115.604900459420.57702909659.08851454821009.0885145482C10.288514548200-9.08851454821.2H220.57702909650-20.577029096500O2024.2770290965-10.2885145482-9.08851454824.9N22.572128637191.32787136290093.9100CO2 teradsorb88.34%kebutuhan O220.5770290965ekses air3.717.98%

Sheet7A natural gas analyzes CH4, 80.0 percent and N2, 20.0 percent. It is burned under a boilerTips for Solving Problemsand most of the CO2 is scrubbed out of the flue gas for the production of dry ice. The exit1. Write the Given and Find statementsgas from the scrubber analyzes CO2, 1.2 percent; O2, 4.9 percent; and N2, 93.9 percent.2. Draw a picture of the system labeling all Inputs and Outputs for each component.Calculate the3. Assign variables to the unknowns. Use symbols to describe the known parameters and list their values.(a) Percentage of the CO2 absorbed.4. Determine whether the problem is a batch process or rate process. Determine whether components are generated or consumed.(b) Percent excess air used.5. Write the conservation of mass for each component and for the entire system.6. Use algebra to solve for the unknowns.7. Check your work.1,2,3GIVEN

CO2

CH480%CO21.2%N220%O24.9%N293.9%O2N2FIND% CO2 ABSORBED% EXCES AIR4BATCH PROCESS5BASIS100kgmol FGNM TOTALNGA- W- DI=FGNM.KOMPC0.8000-11.2H3.200-200O00.42-1-212.2N0.401.5800187.86INVERSE A0.60490045940.1512251149-0.30245022970.0803981623A^-1-0.1531393568-0.03828483920.07656967840.61255742730.9678407351-0.2580398162-0.48392036750.1286370597-0.51607963250.1209800919-0.24196018380.0643185299MULT12.134762634NGCEK115.7886676876ANM TOTAL19.4156202144WNG+A=W+DI+FG8.5078101072DI127.9234303216127.92343032167CH4 IN NATURAL GASCO2 IN DRY ICE9.71kgmol8.5078101072kgmolCO2 ABSORBED8.51kgmol87.6388187784%REACTIONCH4 +2O2-->CO2 +2H2OO2 REQUIRED=2*9.71kgmol=19.42kgmolO2 IN AIR (A)=48.6312404288kgmolEXCESS O2=29.22kgmol=150.4748233216%

BURNERNatural Gas (NG)Flue Gas (FG)Dry Ice (DI)Air (A)

Sheet8A synthesis gas analyzing CO2, 4.5 percent; CO, 26 percent; H2, 13 percent; CH4, 0.5percent; and N2, 56 percent, is burned in a furnace with 10 percent excess air. Calculatethe Orsat analysis of the flue gas1,2,3GIVEN

H2O

CO24.5%n CO2?CO26%n O2?H213%n N2?CH40.5%10% EXCESS AIRN256%O221%N279%FINDORSAT ANALYSIS OF FLUE GAS4BATCH PROCESS / STEADY STATE5BASIS100kgmol SG6O2 REQUIREDCO +0.5O2 -->1CO2261326CH4 +2O2 -->1CO2 +2H2O0.510.51H2 +0.5O2 -->1H2O136.513

TOTAL20.5kgmol

O2 EXCESS2.05kgmol

O2 ENTER22.55kgmolN2 ENTER22.55*79%=84.830952381kgmol21%EXCESS AIR (A) =22.55+84.830952381=107.380952381kgmolNM TOTALINOUTGENCONSACCOUTORSAT ANALYSISNM. KOMPCO24.5-n CO2+26.5-0=03117.83%CO26-n CO+0-26=00-H213-n H2+0-13=00-CH40.5-n CH4+0-0.5=00-N2140.830952381-n N2+0-0=0140.83095238180.99%H2O0-n H2O+14-0=014-O222.55-n O2+0-20.5=02.051.18%187.880952381100.00%kgmol7CHECKINOUTNM ELEMENC4.5+26+0.5=3131H26+2=2828O9+26+45.1=80.180.1N281.6619047619=281.6619047619281.6619047619

BURNERSyn Gas (SG)Flue Gas (FG)Water (W)Air (A)

Sheet91,2,3GIVEN

C3H615%C3H3N?NH37%H2O?O278%N2?N2O2?FIND1LR2EXCESS REACTANT3RATIO C3H3N/NH3, JIKA KONVERSI = 30%4COMPOSITION PRODUCT4BASISFEED100kgmol4BASISFEED100kg5FEEDMOL5FEEDMASSAmolC3H64215C3H642150.3571428571NH3177NH31770.4117647059O23278O232782.4375N2140N21400C3H61NH31.5O21C3H3N3H2OC3H61NH31.5O21C3H3N3H2OM15778M0.35714285710.41176470592.4375R7710.5721R0.35714285710.35714285710.53571428570.35714285711.0714285714SR8067.5-7-21SR00.05462184871.9017857143-0.3571428571-1.0714285714LRNH3LRC3H6EXCESSEXCESSC3H6114.2857142857NH315.2941176471O2642.8571428571O2355KONVERSI30%C3H61NH31.5O21C3H3N3H2OM15778R2.12.13.152.16.3SR12.94.974.852.16.3RATIO MOL C3H3N/NH3=2.1X 100 =30%7

ReactorFeedProduct

Sheet10