Assignment 6 Solutions - CHEE 2.41 A4, ISOLUf’lO/Ug \A...

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Unformatted text preview: CHEE 2.41 A4, ISOLUf’lO/Ug \A ‘ Q1 ’ The saturation pressures at 80°C are tabulated and plotted. C Psax (mm H ) M ompound g w Alkanes: n-C5 2754 72 n~C6 1068 86 n-C7 428 100 n-C8 175 114 Alcohols: n-Cl 1339 32 n-C2 ’ 811 46 n-C3 381 60 n-C4 163 74 Water , 355 18 Saturation pressure, mm Hg 0 20 40 so so Molar mass. g/gmol 100 120 Saturation pressure drops rapidly as molar mass increases. The alkanes have the highest saturation pressure at a given molar mass. This is because attractive hydrogen bonding interactions for the alcohols keep them in the liquid phase compared to the alkanes, which have only weaker van der Waals and similar forces to hold together the liquid phase. Water is a very anomalous compound due to its strong hvdrogen bonding. Q2) (a).The dewpoint temperature corresponds to the saturated vapour at ya = 0.35 and 905°C. (b) Reading across from ya = 0.35 at 905°C, we find x5 = 0.06. (c) The bubblepoint temperature corresponds to the saturated liquid at x5 = 0.35 and 815°C. (d) Reading across from x5 = 0.35 and 815°C, we find ya = 0.60. Q3 )The flow diagram is shown, using as a basis 100 kg/h feed. Compositions are given as wt%. MIBK S 25 kgA 75 kg w solvent extraction unit 05 35* > A W M We no longer assume that the MIBK and water are totally immiscible. The raffinate R and extract E stream compositions are related by the phase equilibrium information in Figure 5.15 (and Table B.l3). Reading from the phase envelope in Figure 5.15, we see that a saturated water-rich solution containing 6 wt% acetic acid would also have ~2% MIBK and 92 wt% water. Following along an imaginary tie-line that falls in-between the two dashed tie lines in Figure 5.15, we estimate that this solution would be in equilibrium with an organic phase of 5 wt% acetic acid, 92 wt% MIBK and 3 wt% water. Thus, we find Raffinate: 6 wt% acetic acid, 2 wt% MIBK, 92 wt% water Extract: 5 wt% acetic acid, 92 wt% MIBK, 3 wt% water We write material balances on all 3 components: 25 = 0.06n’1R + 0.052%}; ms = 0.02mR + 0.92mE We solve the first two equations simultaneously to find: m5 = 387 kg/h mR = 69 km (at our chosen basis of 100 kg/h feed) Then we find the solvent flow rate from the 3rd equation: n'zS = 357 kg/h Now we do a final check by using the total mass balance equation: Ii’lS‘i-Ihp =rhE+n'zR -357 +100 = 387 + 69 (I) 457 = 456 (close enough)! Thus, we need 3.57 kg solvent per kg feed. ‘ Q4) Since 95% of the MAb should be adsorbed, then 5% or 5 nmol/L MAb, are left in the liquid phase. There are a total of 100Xl or 100 nmol MAb fed, so by material balance there must be 95 nmol adsorbed to the beads. This is shown on the flow diagram. 5nM MAb t 95ml MAb beads Separator beads 100M! MAb 1L From the phase equilibrium diagram, a fluid at 5 nmol/L MAB is in equilibrium with 70 nmng adsorbed MAb. Therefore, the total grams of beads required is 95/70 or 1.357 g beads. as“) n “I. q,_v (a) From steam table, the enthalpy at 320 °C is found by linear interpolation A 2 H320 = 3074.6 + g3(317513 - 3074.6) = 3115.1 kJ/kg The total enthalpy change due to increasing the temperature from 100 to 320°C is All = mm) = (1155(3115-4' 25%;.‘e)w/tg =' 4 36. a L.) (b) Using the apprfioximate heat capacity of 33.6 J/grnol °C from Table B. 17 AH = mAH =( 1 kg)(33.6 kJ/kgmol °C)(320-100°C)(1 kgmol/IS kg) = 53:13 3?} 4 ’0' 1' or about 6% too low. 4W" .1 _—.—.W:,; _ (c) Using the polynomial expression from Table B.l7 “(8 593 -5 2 I « ‘ AH: AH=——————— 32.24+0.001924T+1.055-10 T ~3.596-109T3 T=430.8kJ, m 18kg/kmol35’;( )d which is within 2% of the true answer. 616) The balanced reaction is C2H6(g) + 3.502(g) —> 2CO2(g) + 3H20(£) A131: = 21,in = (-1)(—83.82) + (—3.5)(0) + (2)(—393.5) + 3(—285.84) = —156o.7 kJ/gmol In Table B.3 AH: = —1428.6 kJ/gmol for combustion of ethane, but this is with water vapor as the product. With liquid water as the product, _ AH: = -—1428.6 + 3(—44.0) = 1560.6 kJ/gmol which is identical to the value obtained above. ...
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This note was uploaded on 04/05/2010 for the course CHEBE 241 taught by Professor Ali during the Winter '08 term at The University of British Columbia.

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Assignment 6 Solutions - CHEE 2.41 A4, ISOLUf’lO/Ug \A...

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