Chemical Reaction Engineering Questions and Answers – Applications of Design Equations for Continuous – Flow Reactors

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This set of Chemical Reaction Engineering Quiz focuses on “Applications of Design Equations for Continuous – Flow Reactors”.

1. Find the volume of reactor needed for 70% conversion of A to product. Aqueous feed of A and B (200lit/min, 20mmolA/lit, 50mmolB/lit) is fed to MFR. Reaction kinetics is as follows
A+B → P
-rA = 200CACB mol/lit-min
a) 12.5
b) 3.6
c) 6.4
d) 1.5
View Answer

Answer: c
Explanation:
CA0*XA = CB0*XB
XB = \(\frac{20*0.7}{50}\) = 0.28
\(\frac{τ}{CA0} = \frac{XA}{-rA}\)
τ = \(\frac{0.7*20*10^{-3}}{200*10^{-6}[20(1-0.7)*50(1-0.28)])}\) = 0.032 = \(\frac{V}{v0} \)
V = 0.032*200 = 6.4 lit.
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2. An existing MFR is to be replaced by the one 4 times larger than it. Find the new conversion if a homogeneous liquid phase reaction with reaction kinetics as follows.
A → R
-rA = kCA2 with 30% conversion.
a) 0.67
b) 0.45
c) 0.53
d) 0.92
View Answer

Answer: c
Explanation:
\(\frac{τ}{CA0} = \frac{XA}{-rA} \)
\(\frac{V}{CA0 V0} = \frac{0.3}{k[ CA0(1-0.3)]^2} \)
\(\frac{V}{CA0 V0}\) = kτCA0 = 0.6122
When replaced by new reactor
\(\frac{4τ}{CA0} = \frac{XA}{-rA} \)

\(\frac{4Vk}{CA0 V0} = \frac{XA}{(1-XA)^2} \)
XA = 0.53.

3. Replace the existing MFR with a PFR and calculate the new conversion if a homogeneous liquid phase reaction with reaction kinetics as follows.
A → R
-rA = kCA2 with 30% conversion
a) 0.65
b) 0.43
c) 0.38
d) 0.92
View Answer

Answer: c
Explanation:
\(\frac{τ}{CA0} = \frac{XA}{-rA} \)
\(\frac{V}{CA0 V0} = \frac{0.3}{k[ CA0(1-0.3)]^2} \)
\(\frac{Vk}{CA0 V0} \) = kτCA0 = 0.6122
When replaced with a new PFR
\(\frac{τ}{CA0} = \int_0^{XA} \frac{dXA}{-rA} \)
For 2nd order on Integration we get,
kτCA0 = \(\frac{XA}{1-XA}\) = 0.6122
XA = 0.38.
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4. Find the conversion of a 25 litre PFR if a gaseous feed of pure A (2 mol/lit, 150 mol/min)
Is decomposed, kinetics are as follows.
A → 2.5R
-rA = (15min-1) CA
a) 0.89
b) 0.92
c) 0.45
d) 0.68
View Answer

Answer: b
Explanation:
\(\frac{τ}{CA0} = \frac{V}{FA0}\)
τ = (25*2)/150 = 0.33 min
For a 1st order reaction with variable volume on integration we get
kτ = (1+εA)ln\((\frac{1}{1-XA})\) – εAXA
15*0.33 = (1+1.5) ln\((\frac{1}{1-XA})\) – 1.5XA
XA = 0.92.

5. For a constant volume system, the size of batch reactor is _______________ PFR.
a) smaller than
b) larger than
c) same as
d) cannot be compared
View Answer

Answer: c
Explanation:
For PFR, \(\frac{τ}{CA0} = \int_0^{XA}\frac{dXA}{-rA} \)
For Batch reactor, \(\frac{t}{CA0} = \int_0^{XA}\frac{dXA}{-rA} \)
From the above equations it is clear that theoretically the element of fluid reacts for same length of time in PFR and Batch reactor. But shutdown time as to be considered for actual design.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter