Derive the partial differential equation for unsteady-state unidirectional diffusion accompanied by an nth-order chemical reaction (rate constant k): ∂Ca/∂t = D∂²CA/∂y² - kCA^n where CA is the molar concentration of reactant at position y at time t. Explain why, when applying the equation to

Question

Derive the partial differential equation for unsteady-state unidirectional diffusion accompanied by an nth-order chemical reaction (rate constant k):

[latex]\frac{∂C_{A}}{∂t} =D\frac{∂^{2}C_{A}}{∂y^{2}} -kC_{A}^{n}[/latex]

where [latex]C_{A}[/latex] is the molar concentration of reactant at position y at time t. Explain why, when applying the equation to reaction in a porous catalyst particle, it is necessary to replace the molecular diffusivity D by an effective diffusivity [latex]D_{e}[/latex]. Solve the above equation for a first-order reaction under steady-state conditions, and obtain an expression for the mass transfer rate per unit area at the surface of a catalyst particle which is in the form of a thin platelet of thickness 2L.
Explain what is meant by the effectiveness factor ; for a catalyst particle and show that it is equal to [latex](1/\phi)[/latex] tanh [latex]\phi[/latex] for the platelet referred to previously where [latex]\phi[/latex] is the Thiele modulus [latex]L\sqrt{(k/D_{e})} [/latex].

For the case where there is a mass transfer resistance in the fluid external to the particle (mass transfer coefficient [latex]h_{D}[/latex]), express the mass transfer rate in terms of the bulk concentration [latex]C_{Ao}[/latex] rather than the concentration [latex]C_{AS}[/latex] at the surface of the particle. For a bed of catalyst particles in the form of flat platelets it is found that the mass transfer rate is increased by a factor of 1.2 if the velocity of the external fluid is doubled.

The mass transfer coefficient [latex]h_{D}[/latex] is proportional to the velocity raised to the power of 0.6. What is the value of [latex]h_{D}[/latex] at the original velocity?

k = 1.6 × [latex]10^{-3}[/latex] [latex]s^{-1}[/latex] , [latex]D_{e} = 10^{-8}[/latex] m²/s

catalyst pellet thickness (2L) = 10 mm.

Accepted Answer

(i) The partial differential equation for unsteady-state diffusion accompanied by chemical reaction is derived in Volume 1 as equation 10.170

(ii) The molecular diffusivity D must be replaced by an effective diffusivity [latex]D_{e}[/latex] because of the complex internal structure of the catalyst particle which consists of a multiplicity of interconnected pores, and the molecules must take a tortuous path. The effective distance the molecules must travel is consequently increases. Furthermore, because the pores are very small, their dimensions may be less than the mean free path of the molecules and Knudsen diffusion effects may arise
(iii) Equation 10.170 is solved in Volume 1 to give equation 10.199 for a catalyst particle in the form of a flat platelet
(iv) The effectiveness factor is the ratio of the actual rate of reaction to
that which would be achieved in the absence of a mass-transfer resistance. For a platelet, it is evaluated in terms of the Thiele modulus as equation 10.202
(v) For the case, where there is an external mass transfer resistance, the reaction rate is expressed in terms of the bulk concentration as equation 10.222

[latex]\frac{∂C_A}{∂t}=D\frac{∂^{2}C_{A}}{∂y^{2}} -kC_{A}^{n}[/latex] (10.170)

[latex](N_{A})_{y=L}=-C_{Ai}\sqrt{kD_{e}} anh \lambda L[/latex] (10.199)

[latex]\eta =\frac{1}{\phi } anh \phi[/latex] (10.202)

[latex]\Re _{v }=\frac{kC_{Ao}}{\frac{1}{\eta } +\frac{k}{h_{D}} } L[/latex] (10.222)

[latex]R_{v}=\frac{kC_{Ao}}{(1/\eta )+(kL/h_{D})}[/latex]

For k = 1.6 × [latex]10^{-3}[/latex] [latex]s^{-1}[/latex], [latex]D_{e} = 10^{-8}[/latex] m²/s, L = 5 × [latex]10^{-3} [/latex] m:

[latex]\phi =L\sqrt{\frac{k}{D_{e}} } =5 imes 10^{-3}\sqrt{\frac{1.6 imes 10^{-3}}{10^{-3}} } =2[/latex]

[latex]\eta =\frac{1}{\phi } anh \phi=\frac{1}{2} anh 2=\frac{0.96} {2} =0.48[/latex]

∴ [latex]\frac{1}{\eta kL} =\frac{1}{0.48 imes 1.6 imes 10^{-3} imes 5 imes 10^{-3}} =\frac{1}{3.84 imes 10^{-6}} [/latex]= 0.260 × 10⁶.

If the original value of mass transfer coefficient is [latex]h_{D}[/latex], the new value at twice original velocity = [latex]h_{D}(2)^{0.6} = 1.516 h_{D}[/latex]. Given that the overall rate is increased by a factor of 1.2:

[latex]{\left\lgroup\frac{1}{0.260 imes 10^{6}+\frac{1}{1.516h_{D}} } ight group}/{\left\lgroup\frac{1}{0.260 imes 10^{6}+\frac{1}{h_{D}} } ight group }[/latex] = 1.2

[latex]\frac{\left(0.260 imes 10^{6}+\frac{1}{h_{D}} ight) }{\left(0.260 imes 10^{6}+0.66\frac{1}{h_{D}} ight) }[/latex] = 1.2

[latex]0.260 imes 10^{6}+\frac{1}{h_{D}} = 0.312 imes 10^{6}+0.792\frac{1}{h_{D}}[/latex]

[latex]\frac{1}{h_{D}}= 0.25 imes 10^{6}[/latex] and: [latex]h_{D}=4.0 imes 10^{-6}[/latex] m/s.

Question 10.46: Derive the partial differential equation for unsteady-state ...

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