Question

A well-mixed fermenter contains cells initially at concentration x0. A sterile feed enters the fermenter with volumetric flow rate F; fermentation broth leaves at the same rate. The concentration of substrate in the feed is si. The equation for the rate of cell growth is: rx = k1 x and the equation for the rate of substrate consumption is: rs = k2 x where k1 and k2 are rate constants with dimensions T-1 , rx and rs have dimensions M L -3T -1 , and x is the concentration of cells in the fermenter.

a) Derive a differential equation for the unsteady-state mass balance of cells.

b) From this equation, what must be the relationship between F, k1, and the volume of liquid in the fermenter V at steady state?

c) Solve the differential equation to obtain an expression for cell concentration in the fermenter as a function of time.

d) Use the following data to calculate how long it takes for the cell concentration in the fermenter to reach 4.0 g l-1 : F = 2200 l h-1 V = 10,000 l x0 = 0.5 g l-1 k1 = 0.33 h-1

e) Set up a differential equation for the mass balance of substrate. Substitute the result for x from (c) to obtain a differential equation in which the only variables are substrate concentration and time. (Do you think you would be able to solve this equation algebraically?)

f) At steady state, what must be the relationship between s and x?

Answer 1

Given: Rate of cell growth is r_x=k₁x and Rate of substrate consumption is r_s=k₂x

Assumptions:

1. It is a well mixed fermenter. For a well mixed fermenter, concentration of cells and substrate at the outlet and inside the fermenter remain same.
2. Density of the broth remains constant at the inlet and outlet.
3. Cell lysis is negligible.

where F is the Volumetric flow rate at the feed and product stream

x_i is the Concentration of cells in the feed

s_i is the Concentration of substrate in the feed

V is the Volume of broth in the fermenter

x is the Concentration of cells

s is the Concentration of substrate

(a) The general equation for unsteady state mass balance is

$\frac{dM}{dt}=\hat{M}_{i}-\hat{M}_{0}+R_{G}-R_{C}$

where dM/dt is the Rate of change of mass with time

$\hat{M}$ _i is the Mass flowrate of species in inlet stream

$\hat{M}$ ₀ is the Mass flowrate of species in outlet stream

R_G is the Rate of species generated

R_C is the Rate of species consumed

In this case, the inlet stream has no cells, $\hat{M}$ _i=0. Mass flowrate of cells in product stream, $\hat{M}$ ₀=Fx. Rate of cell generation, R_G=r_xV. The rate of cells consumed, R_C=0 as cell lysis is negligible.

The unsteady state mass balance for cell is

$\frac{d(Vx)}{dt}=-Fx+r_{x}V$

where r_x is the Rate of cell growth

$\Rightarrow \frac{d(Vx)}{dt}=-Fx+(k_{1}x)V$

As flowrate and density of liquid is constant, the volume of liquid in fermenter remains constant.

$\Rightarrow V\frac{d(x)}{dt}=-Fx+(k_{1}x)V$

Divide by V throughout the equation

$\Rightarrow \frac{dx}{dt}=\frac{-Fx}{V}+k_{1}x$

$\mathbf{\frac{dx}{dt}=x(k_{1}-\frac{F}{V})}$

This equation is the differential form of unsteady state mass balance for cells.

(b) For steady state conditions, the accumulation rate is zero, i.e., dx/dt=0.

The above differential equation reduces to

$0=x(k_{1}-\frac{F}{V})$

$\Rightarrow k_{1}-\frac{F}{V}=0$

$\mathbf{k_{1}=\frac{F}{V}}$

This equation relates F, k₁ and V at steady state.

(c) Consider the differential equation

$\frac{dx}{dt}=x(k_{1}-\frac{F}{V})$

$\Rightarrow \frac{dx}{x}=(k_{1}-\frac{F}{V})dt$

Integrating the above equation for constant V, F and k₁

$\Rightarrow ln\ x=(k_{1}-\frac{F}{V})t+C$

where C is the Integration constant.

At t=0 and x=x₀, C=ln x₀

$\Rightarrow ln\ x=(k_{1}-\frac{F}{V})t+ln\ x_{0}$

Simplifying further

$\Rightarrow ln\ x-ln\ x_{0}=(k_{1}-\frac{F}{V})t$

$\Rightarrow ln\frac{x}{x_{0}}=(k_{1}-\frac{F}{V})t$

$\Rightarrow \frac{x}{x_{0}}=e^{(k_{1}-\frac{F}{V})t}$

$\mathbf{x=x_{0}e^{(k_{1}-\frac{F}{V})t}}$

This expression is the for cell concentration as a function of time.

(d) For F=2200 L/h, V=10000 L, x₀=0.5 g/L and k₁=0.33/h, the time required for the cell concentration to reach 4 g/L is

$\Rightarrow 4\ g/L=0.5\ g/L*e^{(0.33/h-\frac{2200\ L/h}{10000\ L})t}$

$\Rightarrow 8=e^{0.11t}$

$\Rightarrow ln8=0.11t$

$\Rightarrow 2.079=0.11t$

$\mathbf{t=18.9\ h}$

Therefore, the time required for cell concentration to reach is 4 g/L is t=18.9 h.

(e) For substrate, the rate of mass in is Fs_i, Rate of substrate generation is zero, Rate of mass out is Fs and rate of consumption is r_sV.

The unsteady state mass balance for substrate is

$\frac{d(Vs)}{dt}=Fs_{i}-Fs-r_{s}V$

where r_s is the Rate of substrate consumption

$\Rightarrow V\frac{ds}{dt}=Fs_{i}-Fs-k_{2}xV$

Dividing throughout by V

$\Rightarrow \frac{ds}{dt}=\frac{Fs_{i}}{V}-\frac{Fs}{V}-k_{2}x$

$\Rightarrow \frac{ds}{dt}=\frac{F}{V}(s_{i}-s)-k_{2}x$

Substitute for x from the expression obtained in (c)

$\Rightarrow \frac{ds}{dt}=\frac{F}{V}(s_{i}-s)-k_{2}(x_{0}-e^{k_{1}-\frac{F}{V}}t)$

Here the variables are s and t as F, V, k₁ and k₂ are constants. Hence it is diffcult to solve the equation algebraically.

Thus, the differential mass balance for substrate is

$\mathbf{\frac{ds}{dt}=\frac{F}{V}(s_{i}-s)-k_{2}(x_{0}-e^{k_{1}-\frac{F}{V}}t)}$

(f) Consider the equation

$\frac{ds}{dt}=\frac{F}{V}(s_{i}-s)-k_{2}x$

At steady state, ds/dt=0

$\Rightarrow \frac{F}{V}(s_{i}-s)=k_{2}x$

$\Rightarrow s_{i}-s=\frac{Vk_{2}}{F}x$

$\Rightarrow -s=-s_{i}+\frac{Vk_{2}}{F}x$

$\mathbf{s=s_{i}-\frac{V}{F}k_{2}x}$

The above expression relates s and x at steady state conditions.