?, sec CA0, mmol/liter CA, mmol/liter
60 35 11 20 11
50 100 100 200 200
20 40 60 80 100
Find the space time needed to treat a feed of CA0= 100 mmol/liter to 80% conversion
(a) in a plug flow reactor. (b) in a mixed flow reactor. Solution: From the data offered, we obtain (a) in a P.F.R: CA0=100 mmol/L, XA=0.8
So CA=20 mmol/L ?p=area of the shaded (b) in a M.F.R.
?m?CA0?CAf?rAf=(100-20)/(-rA20)=2×80=160s.
6.20 Reactant A decomposes with stoichiometry A → R and with rate dependent only on CA. The following data on this aqueous decomposition are obtained in a mixed flow reactor:
?, sec
14 25 29 30 29 27 24 19 15 12 20
CA0 200 190 180 170 160 150 140 130 120 110 101
CA 100 90 80 70 60 50 40 30 20 10 1
Determine which setup, plug flow, mixed flow, or any two-reactor combination gives
minimum ? for 90% conversion of a feed consisting of CA0 = 100. Also find this
? minimum. If a two-reactor scheme is found to be optimum, give CA between stages and ? for each stage.
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Solution :
Draw a
1～ CA curse ?rACA0=100, ZA=0.9, so CA=10
From the
11～ CA curse, we know that when CA∈(10,70), increases as CA ?rA?rA1decrease as CA increase. So when CA∈(10,70), ?rAincreases and that when CA∈(70,100),
we plan to use a P.F.R ?p= area of the shaded region followed with a M.F.R. to treat the feed CA∈(10,70), ?m= area of the shaded region.
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Chapter 10 Choosing the Right Kind of Reactor
10.1 Given the two reactions
1?? A + B ? R -r1 = k1CACB
2 R + B ??? S -r2 =k2CACB
Where R is the desired product and is to be maximized. Rate the four sohemes shown in Fig. P10.1 ---either “good” or “not so good,” please, no complicated calculations, just reason it out.
Solution: Number (d) is good for the formation of intermediate –R which causes no maximum occurring there. rR?(?r1)?(?r2)?(k1?k2)CACB, the same order, so R can’t affect the D.
10.2 Repeat Problem 1 with just one change -r2 = k2CRC2 B Solution:
-r1 = k1CACB n1=2 -r2 = k2CRCB2 n2=3
n1< n2, therefore, low CB favors the reaction of lowest order, mixed flow reactor should be used.
So (a) Not so good. (b) Good.
(c) Not so good. (d) Not so good
10.3 Repeat Problem 1 with just one change
-r2 = k2C2 RCB
Solution:
-r1 = k1CACB n1=2 -r2 = k2CR2CB n2=3
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n1< n2, and the order is 2 for R, low CB and low CR is helpful to get more desired product R. So (a) Not so good. (b) Not so good.
(c) Good. (d) Not so good
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Chapter 11 Basics of Non-ideal Flow
11.1 A pulses input to a vessel gives the results shown in Fig.P11.1.
(a) Check the material balance with the tracer curve to see whether the results are consistent. (b) If the result is consistent, determine t, V and sketch the E curve.
Solution:
(a) Ac?t?0.05?5?0.25,
?Ac?t?MM?1?0.25, 4???0?0, so the results are consistent.
5tCdt?t?0.05dt???2.5min (b) t??Cdt?0.05dt050t?V?, so V?t???2.5min?4L/min?10L
?0.05?4?0.2min?1
E?CpulseM/?
11.2 Repeat problem P11.1 with one change: the tracer curve is now as shown in Fig.P11.2.
Solution: Ac?t?0.05?6?0.30?So the results are not consistent.
M
?
11.3 A pulse input to a vessel gives the results shown in Fig. P11.3.
(a) Are the results consistent? (check the material balance with the experimental tracer curve.)
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