# Tubular Reactor with catalytic walls

• Engineering
Homework Statement
Consider a tubular reactor with a non-porous catalyst coated on the wall. A liquid solution
containing reactant A flows through the tube in laminar flow with an inlet concentration, CA0 . The reaction:
A⇌B
occurs homogeneously via a first order reaction and heterogeneously via a Langmuir-Hinshelwood mechanism:

A+S⇌A-S
A-S⇌B-S
B-S⇌B+S

where S represents a surface site and the surface reaction is rate controlling. The reactor is heated electrically which gives a constant heating flux over the entire reactor. Derive the relevant PDE which describe the reactor (2-D model) and give the necessary boundary conditions. Do not solve the equation.
Relevant Equations
2D Heterogeneous Model
My attempt:
Species fluid and solid phase balances for a tubular reactor in which there is a homogeneous first order reaction and a heterogeneous reaction on the external surface of non-porous catalyst coated along the reactor wall.

Mole balance equation for the fluid phase

The heterogeneous reaction is represent by mass transfer to the solid phase.
Dz- Axial diffusion coefficient [m^2/s]
vz- superficial fluid velocity [m/s]
km- mass transfer coefficient [m/s]
am- particle surface area per unit bed volume [m^-1]
k-rate constant [s^-1]

Solid Mole balance equation for the solid phase

Rs- heterogeneous rate of reaction (LHHW)

1)A+S⇌A-S (chemisorption of A)
2)A-S⇌B-S (Surface reaction)
3) B-S⇌B+S (Desorption of B)
2) is the rate determining reaction and 1) and 3) are assumed to be in equilibrium.

There is a single type of site and there is competitive adsorption between A and B.

The rate of adsorption is given by
At equilibrium the net rate of adsorption and desorption are 0. Therefore the surface coverage by A and B are given by.

where θv is the fraction of vacant sites.
Site balance:

Substitute into the rate determining step (rate equation):

We can simplify the above equation using the equilibrium constant for the heterogeneous reaction. At equilibrium the surface reaction is equal to 0.

Kdes- [m^3/mol]
k2 -[mol/(m^3*s)]

Energy Balance for the fluid phase:

Does the catalyst transfer heat from the heat of homogeneous reaction to the fluid and surrounding. In a similar way the fluid phase undergoes a homogeneous reaction where heat is transfer to the catalyst and surrounding.
Where kr- thermal conductivity in the radial direction [W/(m*C)]
kz- thermal conductivity in the axial direction [W/(m*C)]
ρ- Bulk density [kg/m^3]
Cp- specific heat capacity [J/(Kg*C)]
HRx1- Heat of homogeneous reaction [J/mol]
hfs-solid-fluid heat transfer coefficient

Energy balance for the solid phase:

I think I am account for the heat of heterogeneous reaction twice but I am not sure.
HRx2- Heat of Heterogeneous reaction
q''- Electric heat flux

Boundary Conditions (Danckwert Conditions)
I am not so sure about the energy boundary conditions the heat flux at the wall equals the electric heat flux. Is this heat flux compensating the heat flux from both fluid phase and solid phase.

Last edited:
Homework Statement: Consider a tubular reactor with a non-porous catalyst coated on the wall. A liquid solution
containing reactant A flows through the tube in laminar flow with an inlet concentration, CA0 . The reaction:
A⇌B
occurs homogeneously via a first order reaction and heterogeneously via a Langmuir-Hinshelwood mechanism:

A+S⇌A-S
A-S⇌B-S
B-S⇌B+S

where S represents a surface site and the surface reaction is rate controlling. The reactor is heated electrically which gives a constant heating flux over the entire reactor. Derive the relevant PDE which describe the reactor (2-D model) and give the necessary boundary conditions. Do not solve the equation.
To simplify things, let's first look at the case without surface reaction.
Relevant Equations: 2D Heterogeneous Model

My attempt:
Species fluid and solid phase balances for a tubular reactor in which there is a homogeneous first order reaction and a heterogeneous reaction on the external surface of non-porous catalyst coated along the reactor wall.

Mole balance equation for the fluid phase
View attachment 336934

The heterogeneous reaction is represent by mass transfer to the solid phase.
Dz- Axial diffusion coefficient [m^2/s]
vz- superficial fluid velocity [m/s]
km- mass transfer coefficient [m/s]
am- particle surface area per unit bed volume [m^-1]
k-rate constant [s^-1]
The important thing to remember is that the heat flux at the wall due to the electrical heat is fixed. This effect belongs in the boundary conditions, rather than in the differential equation. The same goes for the heterogeneous reaction; it belongs in the boundary condition at r = R rather than in the differential equation for the flow. One more thing: the axial diffusion of mass and the axial diffusion of mass are typically negligible compared to the radial diffusion, and, except for liquid metals, are virtually always neglected. So the differential equation becomes: $$v_z\frac{\partial C_A}{\partial z}= 2 \bar{v}\left[1-\left(\frac{r}{R}\right)^2\right]\frac{\partial C_A}{\partial z}=\frac{D_A}{r}\left(r\frac{\partial C_A}{\partial r}\right)-k_fC_A+k_rC_B$$
$$2 \bar{v}\left[1-\left(\frac{r}{R}\right)^2\right]\frac{\partial C_B}{\partial z}=\frac{D_B}{r}\left(r\frac{\partial C_B}{\partial r}\right)+k_fC_A-k_rC_B$$

HEAT BALANCE
$$2 \bar{v}\rho C_p\left[1-\left(\frac{r}{R}\right)^2\right]\frac{\partial T}{\partial z}=\frac{k}{r}\left(r\frac{\partial T}{\partial r}\right)-(k_fC_A-k_rC_B)\Delta H_{AB}$$

OK so far?

Last edited:
Yes. If we now consider a heterogeneous reaction, the mole and energy balance equations remain the same.
$$C_{A}(r,0)=C_{A_{0}}; T_{f}(r,0)=T_{0}=T_{s}(r,0)\hspace{0.2cm} \text{at z=0}$$
$$\frac{\partial C_{A} }{\partial z}=\frac{\partial T_{f} }{\partial z}=\frac{\partial T_{s} }{\partial z}=0 \hspace{0.2cm} \text{at z=L}$$
At the centerline, the symmetry condition, the symmetry condition gives the zero flux condition. Therefore: $$\frac{\partial C_{A} }{\partial r}=\frac{\partial T_{f} }{\partial r}=\frac{\partial T_{s} }{\partial z}=0 \hspace{0.2cm} \text{at r=0}$$
The heterogeneous reaction appears in the boundary condition
$$-D_{AB}\frac{\partial C_{A} }{\partial r}=R_{s} \hspace{0.2cm} \text{at r=r_0}$$
At the wall, I am unsure how to write the heat flux boundary condition. There is a catalyst coated on the wall, onto which reactants adsorb and are converted into products. Depending on whether the reaction is endothermic or exothermic, heat is released or absorbed by the surrounding environment and the fluid filling the void space and the remaining bulk fluid. There is a heat flux at the wall due to electrical heating, which is transferred through the reactor wall and absorbed by the catalyst and the fluid.

Yes. If we now consider a heterogeneous reaction, the mole and energy balance equations remain the same.
$$C_{A}(r,0)=C_{A_{0}}; T_{f}(r,0)=T_{0}=T_{s}(r,0)\hspace{0.2cm} \text{at z=0}$$
$$\frac{\partial C_{A} }{\partial z}=\frac{\partial T_{f} }{\partial z}=\frac{\partial T_{s} }{\partial z}=0 \hspace{0.2cm} \text{at z=L}$$
This condition is not needed. Plus, the concentration derivative and temperature derivatives are certainly not zero at z = L
At the centerline, the symmetry condition, the symmetry condition gives the zero flux condition. Therefore: $$\frac{\partial C_{A} }{\partial r}=\frac{\partial T_{f} }{\partial r}=\frac{\partial T_{s} }{\partial z}=0 \hspace{0.2cm} \text{at r=0}$$
The Ts doesn't exist at r = 0
The heterogeneous reaction appears in the boundary condition
$$-D_{AB}\frac{\partial C_{A} }{\partial r}=R_{s} \hspace{0.2cm} \text{at r=r_0}$$
This assumes that Rs is the rate of production of B at the surface and minus the rate of consumption of A at the surface.
At the wall, I am unsure how to write the heat flux boundary condition. There is a catalyst coated on the wall, onto which reactants adsorb and are converted into products. Depending on whether the reaction is endothermic or exothermic, heat is released or absorbed by the surrounding environment and the fluid filling the void space and the remaining bulk fluid. There is a heat flux at the wall due to electrical heating, which is transferred through the reactor wall and absorbed by the catalyst and the fluid.
At the wall , $$k\frac{\partial T}{\partial r}=q_e+R_s\Delta H_R$$
where ##q_e## is the rate of electrical heating per unit area of tube surface and ##R_s## is the heterogeneous reaction rate per unit area of surface.

Chestermiller said:
This condition is not needed. Plus, the concentration derivative and temperature derivatives are certainly not zero at z = L
Are the concentration and temperature axial derivatives not equal to zero because the composition of A is changing due to the homogeneous reaction?
Chestermiller said:
This assumes that Rs is the rate of production of B at the surface and minus the rate of consumption of A at the surface.
Will Rs be given by the following formula?View attachment 336940

Are the concentration and temperature axial derivatives not equal to zero because the composition of A is changing due to the homogeneous reaction?
And the heterogeneous reaction, which is captured by the radial concentration gradient.

Tangential, unless you have very good reason for using such a bare tube-column, such as induction heating or 'pigging', a 'packed' column, be it with coated pellets or turbulence-inducing baffles, would seem a better option...

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