This set of Chemical Reaction Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Design of Fluid Particle Reactors”.

1. State true or false.

The fluidized bed reactor is used for mixed flow of solids.

a) True

b) False

View Answer

Explanation: Fluidised bed provides high agitation of solids by high fluid velocity. The gas flow is more complicated than mixed flow.

2. Which of the following does not provide plug flow of Solid – fluids?

a) Fluidised bed reactor

b) Countercurrent flow in blast furnaces

c) Crossflow in moving belt

d) Concurrent flow in driers

View Answer

Explanation: Blast furnaces and cement kilns provide plug flow in countercurrent operation. Moving belt feeders are used in furnaces. Cocurrent flow is employed in polymer driers.

3. Which of the following does not control the design of fluid – solid reactor?

a) Reaction kinetics for single particles

b) Density of fluid being treated

c) Size distribution of solids

d) Flow patterns of solids and fluid

View Answer

Explanation: The complexity of the reaction determines the use of flow reactor or plug flow reactor. The variation of temperature conditions determines the flow patterns of fluid by affecting the viscosity. The range of sizes determines the operational factors.

4. If \(\overline{X_{Ri}}\) is the mean conversion of a reactant of particle size R_{i}, R_{m} is the particle of maximum size in the feed and F(R_{i}) is the fraction of R_{i} fed to the reactor, then the mean conversion of solids of a particular size ‘i’ leaving a plug flow reactor converting a mixture of particles of varying sizes is ____

a) \(\overline{X_{(B)Ri}} = ∑_{R(τ)}^{Rm}\)[1- X_{(B)Ri}]\(\frac{F(Ri)}{F} \)

b) 1-\(\overline{X_{(B)Ri}} = ∑_{R(τ)}^{Rm}\)[ X_{Ri}]\(\frac{F(Ri)}{F} \)

c) 1-\(\overline{X_{(B)Ri}} = ∑_{R(τ)}^{Rm}\)[1-X_{(B)Ri}] F (Ri)

d) 1-\(\overline{X_{(B)Ri}} = ∑_{R(τ)}^{Rm}\)[1-X_{(B)Ri}] \(\frac{F(Ri)}{F} \)

View Answer

Explanation: Mean value for fraction of ‘i’ unconverted is the summation of the product of fraction of the reactant unconverted in the particle size Ri and the fraction of feed in the size Ri.

1-\(\overline{X_{(B)Ri}}\) = ∑(fraction of the reactant unconverted in the particle size Ri × the fraction of feed in the size Ri).

5. For mixed flow of particles containing a single unchanging size and uniform gas composition, the fraction unconverted for film resistance controlling is ____

a) \(\int_0^τ\)(-\(\frac{t}{τ}\))\(\frac{e}{t}^\frac{-t}{t}\) dt

b) \(\int_0^τ\)(1 – \(\frac{t}{τ}\))\(\frac{e}{t}^\frac{-t}{t}\) dt

c) \(\int_0^τ\)(1 –\(\frac{t}{τ}\))dt

d) \(\int_0^τ\)(1 – \(\frac{t}{τ}\))\(\frac{e}{t}^\frac{-t}{t}\) dt

View Answer

Explanation: For a mixed flow reactor, the mean residence time, t in the reactor is, E = \(\frac{e}{t}^\frac{-t}{t}\). For a single size of particles converted in time τ, 1-\(\overline{X_{(B)}}\) = ∫

_{0}

^{τ}(1 – X

_{B})\(\frac{e}{t}^\frac{-t}{t}\)dt for an individual particle. For film resistance controlling, X

_{B}= \(\frac{t}{τ}.\)

6. For mixed flow of particles containing a single unchanging size and uniform gas composition, the fraction unconverted for chemical reaction controlling is ____

a) \(\int_0^τ\)(1-\(\frac{t}{τ}\))^{3}\(\frac{e}{t}^\frac{-t}{t}\) dt

b) \(\int_0^τ\)(1-\(\frac{t}{τ}\))^{2}\(\frac{e}{t}^\frac{-t}{t}\) dt

c) \(\int_0^τ\)(1-\(\frac{t}{τ}\))\(\frac{e}{t}^\frac{-t}{t}\) dt

d) \(\int_0^τ\)(1-\(\frac{t}{τ}\))^{0.5}\(\frac{e}{t}^\frac{-t}{t}\) dt

View Answer

Explanation: For chemical reaction controlling, \(\frac{t}{τ}\) = 1-(1- X

_{B})\(^\frac{1}{3}\). Hence, 1-\(\overline{X_{(B)}}\) = ∫

_{0}

^{τ}(1-\(\frac{t}{τ}\))

^{3}\(\frac{e}{t}^\frac{-t}{t}\) dt.

7. If \(\frac{τ}{t}\) = 0.5, average conversion for a particle B of constant size in a mixed flow reactor for film resistance controlling is ____

a) 0.45

b) 0.38

c) 0.536

d) 0.743

View Answer

Explanation: 1-\(\overline{X_{(B)}}\) = 0.5(\(\frac{τ}{\overline{t}}) – \frac{1}{3!}(\frac{τ}{\overline{t}})^2+\frac{1}{4!} (\frac{τ}{\overline{t}})\)

^{3}-…, as obtained by expansion.

1-\(\overline{X_{(B)}}\) = 0.4635 and \(\overline{X_{(B)}}\) = 0.536

8. If \(\frac{τ}{t}=\frac{1}{3},\) average conversion for a particle B of constant size in a mixed flow reactor for chemical reaction controlling is ____

a) 0.92

b) 0.98

c) 0.75

d) 0.76

View Answer

Explanation: \(\overline{X_{(B)}} = 3(\frac{\overline{t}}{τ}) -6 (\frac{\overline{t}}{τ})^2+ 6(\frac{\overline{t}}{τ})^3(1-e^\frac{τ}{\overline{t}}, 1-\overline{X_{(B)}}\) = 0.078 and \(\overline{X_{(B)}}\) = 0.92

9. For mixed flow of a mixture of sizes of unchanging size particles, the mean residence time of a material of a given size is ____

a) t = \(\frac{Weight \, of \, all \, solids \, in \, the \, reactor}{Feed \, rate \, of \, solids \, to \, the \, reactor} \)

b) t = \(\frac{Feed \, rate \, of \, solids \, to \, the \, reactor}{Weight of all solids in the reactor} \)

c) t = \(\frac{Weight \, of \, all \, solids \, in \, the \, reactor}{Reactant \, concentration} \)

d) t = \(\frac{Reactant \, concentration}{Weight \, of \, all \, solids \, in \, the \, reactor} \)

View Answer

Explanation: The mean residence time of a material of a given size is equal to mean residence time of solids in the bed. It is the fraction of weight of all solids in the reactor to the feed rate of solids in the reactor.

10. If weight of all solids in the reactor is 300kg and the feed rate of solids is \(\frac{50kg}{hr},\) then the mean residence time(in hrs) of a material of a given size for mixed flow of a mixture of sizes of unchanging size particles is ____

a) 7

b) 6

c) 4

d) 3

View Answer

Explanation: t = \(\frac{W}{F} = \frac{300}{50}\) = 6 hrs.

**Sanfoundry Global Education & Learning Series – Chemical Reaction Engineering.**

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