Give step-by-step solution with explanation and final answer:’

Consider a layer of bacteria B contained between two semipermeable membranes
that allow the passage of a chemical solute S, but do no

Question

Give step-by-step solution with explanation and final answer:’

Consider a layer of bacteria B contained between two semipermeable membranes
that allow the passage of a chemical solute S, but do not allow the passage of bacteria.
The mass transport of B and S has reached steady state. The concentration of S is low
and it is transported by diffusion with a flux js

. dC
Js=-Ds—>
'S Ss dz
where Cg is the concentration of S and Ds (a constant) is its diffusivity. The
movement of the bacteria B is described by the following flux equation
. dCy L dC
Js =-Dg——+ Cs —
® ® dz dz
where jg is the bacteria flux, C 5 1s the concentration of B., Ds (the bacteria diffusivity)
and y are constant. If Cg is maintained at S; and 0 at the upper and lower surfaces of
the bacterial suspension
(a) determine Cy(z).
(b) determine Cg(z). The original bacterial distribution is uniform and equal to Cg.
7=0 - - \— C $s 0
0000 OOD ) (
| 2 ) OO 0 O\U ( LO
co oo ~—
: ~~ .
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C 0
Semipermeable membranes

VIPSolutions · Accepted Answer

Let's solve the problem step by step:

---

## (a) **Determine $ C_S(z) $**

Given:
- Steady-state diffusion of solute S:  
  $$
  j_S -D_Sfrac{d_S}{dz  $$
 Boundary conditions:  
  $ C_S(z=) = S_ $  
  $ C_S(z=L) =  $

At steady state, $ j_S $ is constant (does not depend on $ z $).  
So, integrate:

$$
j_S = -D_S \frac{dC_S}{dz} \implies \frac{dC_S}{dz} = -\frac{j_S}{D_S}
$$

Integrate with respect to $ z $:

$$
C_S(z) = -\frac{j_S}{D_S} z + A
$$

Find constants using boundary conditions:

- At $ z= $: $ C_S() = S_ \implies A = S_ $
- At $ z=L $: $ C_S(L) =  \implies  = -\frac{j_S}{D_S}L + S_ \implies j_S = \frac{D_S S_}{L} $

So,

$$
C_S(z) = S_ \left(1 - \frac{z}{L}\right)
$$

---

## (b) **Determine $ C_B(z) $**

Bacterial flux equation (at steady state, $ j_B =  $ since bacteria cannot cross boundaries):

$$
j_B = -D_B \frac{dC_B}{dz} + \chi C_B \frac{dC_S}{dz}
$$

Set $ j_B =  $:

$$
 = -D_B \frac{dC_B}{dz} + \chi C_B \frac{dC_S}{dz}
\implies \frac{dC_B}{dz} = \frac{\chi}{D_B} C_B \frac{dC_S}{dz}
$$

This is a first-order linear ODE. Rearranging:

$$
\frac{1}{C_B} \frac{dC_B}{dz} = \frac{\chi}{D_B} \frac{dC_S}{dz}
$$

Integrate both sides with respect to $ z $:

$$
\int \frac{1}{C_B} dC_B = \frac{\chi}{D_B} \int \frac{dC_S}{dz} dz
$$

$$
\ln C_B = \frac{\chi}{D_B} C_S + C_1
$$

Exponentiate both sides:

$$
C_B(z) = A \exp\left(\frac{\chi}{D_B} C_S(z)\right)
$$

Where $ A = e^{C_1} $ is a constant. Use the conservation of total bacteria:

If the original distribution is uniform and equal to $ C_{B} $, then the total number of bacteria is conserved:

$$
\int_^L C_B(z) dz = C_{B} L
$$

Substitute $ C_S(z) $:

$$
C_B(z) = A \exp\left[\frac{\chi}{D_B} S_ \left(1 - \frac{z}{L}\right)\right]
$$

Let $ \alpha = \frac{\chi}{D_B} S_ $:

$$
C_B(z) = A \exp\left[\alpha \left(1 - \frac{z}{L}\right)\right]
$$

To find $ A $:

$$
\int_^L C_B(z) dz = A \int_^L \exp\left[\alpha \left(1 - \frac{z}{L}\right)\right] dz = C_{B} L
$$

Let $ u = 1 - \frac{z}{L} \implies dz = -L du $, as $ z =  \rightarrow u = 1, z = L \rightarrow u =  $:

$$
= A L \int_1^ e^{\alpha u} (-du) = A L \int_^1 e^{\alpha u} du = A L \left[\frac{e^{\alpha u}}{\alpha}\right]_^1
$$
$$
= A L \frac{e^{\alpha} - 1}{\alpha}
$$

Set equal to $ C_{B} L $:

$$
A \frac{e^{\alpha} - 1}{\alpha} = C_{B}
\implies A = \frac{C_{B} \alpha}{e^{\alpha} - 1}
$$

---

## **Final Answers**

### **(a) Solute concentration profile:**
$$
\boxed{C_S(z) = S_ \left(1 - \frac{z}{L}\right)}
$$

---

### **(b) Bacteria concentration profile:**
Let $ \alpha = \frac{\chi}{D_B} S_ $:

$$
\boxed{
C_B(z) = \frac{C_{B} \alpha}{e^{\alpha} - 1} \exp\left[\alpha \left(1 - \frac{z}{L}\right)\right]
}
$$

where $ C_{B} $ is the initial (uniform) bacteria concentration.

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