midterm2 2006 solutions

midterm2 2006 solutions - Chemical Engineering 140 Midterm...

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Unformatted text preview: Chemical Engineering 140 Midterm Examination #2 Solution 1) Recycle F3 Recycle to feed ratio = 0.2 1.0 MEOH Methanol F4 MEOH/H2O A F1 F4 MEOH/H2O 220 oC 3.5 atm O2 N2 F5 DMM H2O H2 MEOH HCHO DME 0.5 MEOH 0.5 H2O 0.5 MEOH 0.5 H2O F2 Dry air 0.21 O2 0.79 N2 Water MEOH PMM 2a) Reaction (1): CH3OH + 0.5 O2 HCHO + H2O Basis: F1= 100 mol HCHO = 10 mol Dry air introduced into the MEOH converter = F2 = 2*F1=200 mol. O2 in = 200 mol (0.21)= 42 mol 10 mol of HCHO formed, 0.5 (10) = 5 mol of O2 reacted. 2b) F3/F1 = 0.2 F3 = 20 mol F4 = F1+F3 = 100+20 = 120 mol, F4 Composition: 70 mol MEOH, 50 mol H2O. nMEOH nMEOHin 1 2 2 70 1 2 2 nO2 nO2 in 0.51 42 0.51 nHCHO nHCHOin 1 1 nDME nDMEin 2 2 nN 2 nN 2in 158 nH 2O nH 2Oin 1 2 50 1 2 nTotal 320 0.51 kJ [( 98.2 228.6) (2 * 161.9)] G o mol 0.732 LnK (220o C ) J 1 kJ RT 8.314 ( )(220 273) K mol K 1000 J K 2.08 [ DME ][ H 2O ] 2 (50 1 2 ) 2 (50 101 2 ) K 2.08 [CH 3OH ]2 (70 1 2 2 ) 2 (70 10 2 2 ) 2 2c) Solve for ξ2 2 17.3 nMEOH 70 10 2(17.3) 25.4 nO2 42 0.5(10) 37 nHCHO 10 nDME 17.3 nN 2 158 nH 2O 50 10 17.3 77.3 y MEOH 7.82 yO2 11.38 y HCHO 3.08 y DME 5.32 y N 2 48.62 y H 2O 23.78 2d) mol DME/mol MEOH = 17.3/(50+20) = 0.25 2e) ORSAT analysis did not include water. yMEOH 10.25 yO2 14.93 y HCHO 4.04 y DME 6.98 y N 2 63.79 2f) Single pass conversion = (70-25.4)/70*100% = 63.7 3) Reactive distillation column 2 HCHO + 3 CH3OH CH3O(CH2O)3CH3 +H2O +H2 CH3OCH3 + CH3OH CH3OCH2OCH3 + H2 10 mol of HCHO reacts with 15 mol of MEOH 17.3 mol of DME reacts with 17.3 mol MEOH MEOH in F5 = MEOH in F4 – MEOH in F3 = 25.4-20 = 5.4 mol Overall MEOH required to completely convert al HCHO and DME in the reactive distillation column = 32.3 mol Additional mol of MEOH required = 32.3-5.4 mol = 26.9 mol 4) For 100 mol/h for MEOH/H2O feed = 17.3 mol of DMM produced 5000 mol/h of MEOH/H2O feed = 865 mol of DMM produced 5) i) Double the amount of air fed does not change ξ2, since Reaction (2) has the same number of mol of reactant and products. ii) For Reaction (1), rate is first order in MEOH partial pressure and zero order in O2. iii) Rate (reaction1) = kPMEOH = k yMEOH P = k [nMEOH/nT] As a result of diluting the MEOH inside the methanol converter, the rate for Reaction (1) actually decreases. HCHO formation rate decrease as a result!! 6) i) Reaction (2) is exothermic. Decreasing the temperature favors the formation of DME. ii) In the mean time, rate for Reaction (1) decreased, so the selectivity to DEM increase. ...
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