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=ll11 =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
8183 35 n5; n15001bfh therefore, W7 g%10001bfh W18500W24Wg,
W3IW4+W5
WsW6+W7W6+1000 We can combine the water material balances along with knowledge that the water removal rate is equal in each evaporator into one equation;
8500113 +W4+W6 +10003W2 +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.
 Spring '08
 ali

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