A → 3 R, -rA = (0.6min-1)CA
Find the conversion of A in a 50% A – 50% inert feed (?0 = 180 liter/min, CA0 =300 mmol/liter) to a 1 m3 mixed flow reactor.
4?2Solution: V?1m3, M.F.R. ?A??1
2According to page 91 eq.11, ??V?CAoXA?CAoXA
So we obtain XA?0.667
?o?rA0.6C1?XAAo1?XA?XA(1?XA)100L0.6(1?X?0
A)180L/min25
Chapter 6 Design for Single Reactions
6.1 A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in a
series. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the concentration in the exit stream of the second reactor. The reaction is second-order with respect to A and V2/V1 =2.
Solution:
V2/V1 = 2, ?1 =
C?CA1V1C?CA2V = A0 , ?2 = 2 = A1 ?01?rA?rA2?02CA0=1mol/l , CA1=0.5mol/l , ?01??02, -rA1=kC2 A1 ,
CA0?CA1CA1?CA2-rA2=kC2 = A2 (2nd-order) , 2×22kCA1kCA2 So we obtain 2×(1-0.5)/(k0.52)=(0.5-CA2)/(kCA22)
CA2= 0.25 mol/l
6.2 Water containing a short-lived radioactive species flows continuously through a
well-mixed holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed stream. This is not bad, but we’d like to lower it still more.
One of our office secretaries suggests that we insert a baffle down the middle of the
tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this would help? If not, tell why; if so calculate the expected activity of the exit stream compared to the entering stream.
Solution: 1st-order reaction, constant volume system. From the information offered about
the first reaction,
we obtain
?1=
If a baffle is added,
V1?01?CA0?CA1kCA11CA0?CA07? 1k?CA07?2??21??22?V21?021?V22?022=
V1CA0?CA21CA2?CA22?=
?01kCA21kCA226CA0=7=6/k …… ① 1kCA0726
CA0?CA21kCA211V1C?CA222?=3/k=A21 …… ② kCA22?021Combining equation ① and ② we obtain:
CA21= 0.25CA0 ;
1CA0 16So it will help, and the expected activity of the exit stream is 1/16 of the feed.
CA22=0.25CA21=
6.3 An aqueous reactant stream (4 mol A/liter) passes through a mixed flow reactor
followed by a plug flow reactor. Find the concentration at the exit of the plug flow reactor if in the mixed flow reactor CA = 1 mol/liter. The reaction is second-order with respect to A, and the volume of the plug flow unit is three times that of the mixed flow unit.
Solution: Constant volume system and 2nd-order reaction:
?m?Vm?0Vp=
CA0?CA1CA0?CA14?1=>=3/k …… ① 2?rA1kkCA1?p??0?3Vm?0=9/k= -?CAfCA1CAfCdCAA= -?1?rAk?211?1) …… ② dCA=(kCAfCombining equation. ① and ② we obtain:
CAf= 0.1 mol/liter
6.4 Reactant A (A → R,CA0=26 mol/m3) passes in steady flow through four equal-size
mixed flow reactors in series (?what must be ?Solution:
total
=2 min). When steady state is achieved the
3
concentration of A is found to be 11, 5, 2, 1 mol/m3 in the four units. For this reaction,
plug so as to reduce CA from CA0 = 26 to CAf = 1 mol/m
?
?m??m1??m2??m3??m4=
CA0?CA1CA1?CA2CA2?CA3CA3?CA4===
?rA1?rA2?rA3?rA4CA0=26mol/liter, CA1=11 mol/liter, CA2=5 mol/liter, CA3= 2mol/liter, CA4=1mol/liter So we abtain: 15/(-rA1) = 6/(-rA2) = 3/(-rA3) = 1/(-rA4) We postalate the reaction rate is 1 unit when CA4=1 mol/liter So we obtain
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CA, mol -rA 1/(-rA)
11 30 1/30
5 12 1/12
CAfCA0dCdCAA=? ?rACAf?rA2 6 1/6
1 2 1/2
?p???So we obtain ?p?2.63 min.
CA06.6 At 100℃ pure gaseous A reacts away with stoichiometry 2A → R + S in a constant
volume batch reactor as follows:
t, sec pA, atm
0 1.00
20 0.96
40 0.80
60 0.56
80 0.32
100 0.18
120 0.08
140 0.04
160 0.02
What size of plug flow reactor operating at 100℃ and 1 atm can treat 100 moles A/hr in a feed consisting of 20% inserts to obtain 95% conversion of A? Solution:
FA0= 100 mol/hr = 0.0278 mol A/s, CA0 = (1 atm×101.3)/(8.314×37.3)=0.0327mol/l
t, sec pA, atm
0 1.00
20 0.96
40 0.80
60 0.56
80 0.32
100 0.18
120 0.08
140 0.04
160 0.02
CA,mol/l 0.0327 0.03136 0.0261 0.0285 0.01045 0.00588 0.00261 0.001307 0.0006533
??Now , pA0=0.8 atm, CA0=0.0261 mol/l
??When the conversion of A is 90%, CA=0.001305mol/l , pA=0.04 atm So ?=?0.04??0.8=140-40=100 s So V =??0=?
6.7 We wish to treat 10 liters/min of liquid feed containing 1mol A/liter to 99% conversion.
The stoichiometry and kinetics of the reaction are given by
A → R, -rA =
CAmol
0.2?CAliter?minFA0=100×0.027/0.0261=106.4 L CA0 Suggest a good arrangement for doing this using two mixed flow reactors, and find
the size of the two units needed. Sketch the final design chosen.
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Solution:
?0=10l/min , CA0=1 mol/l, when ZA=0.99 CAf =0.91mol/l (1) two for equal-she M.F.R. ?=
Vi?0=
CA0?CA1CA1?CAf=
?rAf?rA1 -rA1=CA1/(0.2+CA1) , -rAf = CAf/(0.2+CAf) = 0.0476, So we obtain CA1=0.121mol/l, ?i=2.333min
Vt=2Vi=2?i?0=2×2.333×10=46.65 L
(2)
-/rA L K N M 0 Z2 Z1 ZA When the area of rectangle KLMN is maximum, the volume of reactors needed its minimum.
?=
V1?0?V2?0=
CA0?CA1CA1?CA2?=2.7+1.89=4.59
?rA1?rA2So
?V=4.59×10=45.9
d(?Slope of curse =
1)rA =0.2/(1-ZA)2 dx?(?Slope of LN =
1)rA?ZA10.221?(?1)1?ZAi=
ZAiSo when slope of curse = slope of LN, ZA1=0.9,
CA1=0.1 mol/L, -rA1=1/3 , 1/(-rA) = 3
6.8 From steady-state kinetics runs in a mixed flow reactor, we obtain the following data
on the reaction A → R.
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