Problem set # 3 - Solution

Problem set # 3 - Solution - Q1 Diagram Product 1 MW 98 mol...

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Unformatted text preview: Q1 Diagram: Product 1 MW} 98 mol% 02 N2 Air Separator 79 mol% N2 21 mol% ()2 Product 2 02 N2 System: Separator Components: N2 and O2 DOF analysis: Constraints Stream 6 S tern 0 Specified flows Specified compositions Specified system performance Material balance DOF = 6 - 6 = 0 1 (l tonr‘yr Product 1) 1 (80% recovery) Q2a Diagram: Gas 11.4 mol% propane Propane WW) mixer W} 8" fl— 1 liter Air System: mixer Components: Air, Propane (Note that since the air does not undergo any separation or chemical reaction, we can treat it as a single component! The composition of the air is irrelevant for this problem. It would not be wrong to consider air as a mix of 2 components, but it is not necessary.) DOF anal sis: l l l l Specified 1 l l Specified system 0 erformance I Material balance 2 DOF = 4 — 4 = 0 Basis: 1 liter air fed Stream variables: P for propane, A for air. Streams indicated as “in” or “out”. Units: We need to convert from volume (L) to moles. We’ll use the ideal gas law: PV (1 atm)(100 L) 4464 gml RT (0.08205? atm Ugmol K)(2’?3K) Stream composition specifications: ‘00—!“ -0‘1 14 A03: 4' Pout System performance specifications: none Material balances; Ag” - A0“: - 4.464 gmol Pin- our Combining the material balance with the stream composition, we find: inc—“‘sQlM, Pow c0534 gmol- P3,: 4.464 4- P0,“ Converting to pounds: 0.534 gmolx 44 g x 3” -o.0561b propane (or more) must be mixed with 100 L of gmol 454 g air to exceed the flammabith limit. Q2b Diagram: to w w 2 4 6 Raw juice 1 w w Cone juice 85% w Evap 1 S Evap2 S Evap 3 ——’ 40% W 15% s 3 5 7 60% s Components are water (W) and sugar solids (S). Streams are numbered. Basis: 10,0001bfh raw juice fed to the first separator. Units: all in mass (1b) and wt%, no conversion needed DOF anal sis: Variables Constraints Stream 1 1 _S stem 0 S . ‘ cified flows 1 Specified 2 (85% water in raw compositions juice, 40% water in conc juice, solids content is not independent) Specified system 2 (the flow rates of performance water from the 3 evaporators are equal - this provides 2, not 3, independent pieces of information) Material balance 6 (2 per unit) DOF=ll-11 =0 Basis, stream composition and system performance written in terms of stream variables: 31 - 0.15(10,000) -1500 lbr'h W1 = 0.85(10,000)= 8500 lbfh £21- $-15 W; 40 W2 - W4 - W6 Material balances: S; 1500 81-83 -35 n5; n15001bfh -therefore, W7 -g-%-10001bfh W1-8500-W24-Wg, W3IW4+W5 Ws-W6+W7-W6+1000 We can combine the water material balances along with knowledge that the water removal rate is equal in each evaporator into one equation; 8500-113 +W4+W6 +1000-3W2 +1000, or W2 =25001bz’h To answer the specific questions: (a) the flow rate of concentrated juice is S; + W? - 1500 +1000 - 2500 lbfh (b) the amount of water removed per evaporator is 2500 1th (c) the concentration of water in the juice fed to the second evaporator is W3 8500 — 2500 S3 «1» W3 1500 + (8500 - 2500) 0.8 lb waterflb juice ...
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This note was uploaded on 04/05/2010 for the course CHEBE 241 taught by Professor Ali during the Spring '08 term at UBC.

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