A?B?R, CA?50.find XA, XB, CB
XA?1?CA50?1??0.75, CAo200XB?bCAoXA?1.5?1, which is impossible. CBoSo CB?CBo?100
4.4 Given a gaseous feed, CA0 = CB0 =100, A +2B→ R, CB = 20. Find XA, XB, CA. Solution: Given a gaseous feed, CAo?CBo?100，A?2B?R, CBo?20, Find XA, XB, CA
XB?0, V?100A?100B?200 XB?1, V?50A?100R?150
?B?150?200?0.25?100??0.25, ??A???0.5
1200?10021?100?0.842100?202XB??0.842, XA??0.421 100?0.25?201001?XA1?0.421?100??73.34
1??AXA1?0.5?0.421CA?CAo
4.6 Given a gaseous feed, T0 =1000 K, π0＝5atm, CA0=100, CB0=200, A +B→5R,T =400 K,
π=4atm, CA =20. Find XA, XB, CB.
Solution: Given a gaseous feed, To?1000K, ?0?5atm, CAo?100, CBo?200
A?B?5R, T?400K, ??4atm, CA?20, find XA, XB, CB.
?A?a?CT?0400?5600?300??0.5 ?1, ?B?ABo?2,
bCAoT0?1000?4300According to eq page 87,
20
CAT?0201??0.5CAoT0?100XA???0.818
20CAT?01?1??0.51??A100CAoT0?1?XB?bCAoXA100?0.818??0.409
aCBo200CBobT??XA)0CAo(200?0.818)?200CaT?0CB?Ao?100?130
1??AXA1?1?0.818(4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-liter popcorn
to be operated
in steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream produces 28 liter/min
of mixed exit stream. Independent tests show that when raw corn pops its volume goes
from 1 to 31.
With this information determine what fraction of raw corn is popped in the unit.
31?111Solution: ?A??30, CAo?1a.u., CA?CAo?a.u.
1282811?CAo?CA28?46.5% ?XA??1CAo??ACA1?30?28
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Chapter 5 Ideal Reactor for a single Reactor
5.1 Consider a gas-phase reaction 2A → R + 2S with unknown kinetics. If a space velocity
of 1/min is needed for 90% conversion of A in a plug flow reactor, find the corresponding space-time and mean residence time or holding time of fluid in the plug flow reactor.
1Solution: ???1min,
sVarying volume system, so t can’t be found.
5.2 In an isothermal batch reactor 70% of a liquid reactant is converted in 13 min. What
space-time and space-velocity are needed to effect this conversion in a plug flow reactor and in a mixed flow reactor? Solution: Liquid reaction system, so ?A?0 According to eq.4 on page 92, t?CAo? Eq.13, ?M.F.R?XA0dCA?13min ?rACAo?CACAoXA?, ?M.F.R can’t be certain. ?rA?rAXA Eq.17, ?P.F.R?CAo?
0dXA, so ?P.F.R?tB.R?13min ?rA5.4 We plan to replace our present mixed flow reactor with one having double the bolume.
For the same aqueous feed (10 mol A/liter) and the same feed rate find the new conversion. The reaction are represented by
A → R, -rA = kC1.5 A
Solution: Liquid reaction system, so ?A?0
C?CAXXAV???A, Ao? FAoCAo?rACAo(?rA)k[CAo(1?XA)]1.5??Now we know: V??2V, FAo?FAo, CAo?CAo, XA?0.7 So we obtain
22
?XA2XA2V ???.51.5?FAo?1.5kC11.51?XA)Ao()FAokCAo(1?XA)V??XA???(1?XA)1.52?0.7?8.52
(1?0.7)1.5?XA?0.794
5.5 An aqueous feed of A and B (400liter/min, 100 mmol A/liter, 200 mmol B/liter) is to be
converted to product in a plug flow reactor. The kinetics of the reaction is represented by
mol liter?min Find the volume of reactor needed for 99.9% conversion of A to product.
A +B→ R, -rA = 200CACB
Solution: Aqueous reaction system, so ?A?0
Vt1??According to page 102 eq.19,
FAoCAoCAo?XAf0XAfdXdCAA?? 0?rA?rA??V?o?CAo?XAfXAf0dXA, ?o?400liter/min, ?rA?V?CAo?o?00.999dXdXAA?0.1?400??124.3L
0?rA?rA
5.9 A specific enzyme acts as catalyst in the fermentation of reactant A. At a given enzyme
concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (CA0 =2 mol/liter ). The kinetics of the fermentation at this enzyme concentration is given by
??? R , -rA = A ?enzyme0.1CAmol
1?0.5CAmin?literSolution: P.F.R, according to page 102 eq.18, aqueous reaction, ??0
XAdXVA?? 0FAo?rA?V?FAo?XA01?0.5CA1dXA?125?2(Ln?XA)\\
0.1CA1?XA 23
?125(Ln1?0.95)?986.4L 0.05
5.11 Enzyme E catalyses the fermentation of substrate A (the reactant) to product R. Find
the size of mixed flow reactor needed for 95% conversion of reactant in a feed stream (25 liter/min ) of reactant (2 mol/liter) and enzyme. The kinetics of the fermentation at this enzyme concentration are given by
A ?enzyme??? R , -rA =
0.1CAmol
1?0.5CAmin?literSolution: ?o?25L/min, CAo?2mol/L, FAo?50mol/min, XA?0.95 Constant volume system, M.F.R., so we obtain
??V?o?CAoXA??rA2?0.95?199.5min,
0.1?0.05?21?0.5?0.05?2V???o?199.5min?25L/min?4.9875m3
5.14 A stream of pure gaseous reactant A (CA0 = 660 mmol/liter) enters a plug flow reactor at
a flow rate of FA0 = 540 mmol/min and polymerizes the as follows
mmol3A → R, -rA = 54
liter?minHow large a reactor is needed to lower the concentration of A in the exit stream to CAf = 330 mmol/liter?
CA13301??11?CAo2660?0.75 Solution: ?A?3??, XA??C2330131??AA1??3660CAo0-order homogeneous reaction, according to page 103 eq.20
k??kV?o?kVCAo?CAoXA FAoSo we obtain
V?FAo10.75CAoXA?540??7.5L CAok54
5.16 Gaseous reactant A decomposes as follows:
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