# A well-mixed fermenter contain

A well-mixed fermenter contains cells initially at concentrationx0. A sterile feed enters the fermenter with volumetric flow rateF; fermentation broth leaves at the same rate. The concentration ofsubstrate in the feed is si. The equation for the rate of cellgrowth is: rx = k1 x and the equation for the rate of substrateconsumption is: rs = k2 x where k1 and k2 are rate constants withdimensions T-1 , rx and rs have dimensions M L -3T -1 , and x isthe concentration of cells in the fermenter.

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

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

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

d) Use the following data to calculate how long it takes for thecell concentration in the fermenter to reach 4.0 g l-1 : F = 2200 lh-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 ofsubstrate. Substitute the result for x from (c) to obtain adifferential equation in which the only variables are substrateconcentration and time. (Do you think you would be able to solvethis equation algebraically?)

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

Given: Rate ofcell growth is rx=k1x and Rate of substrateconsumption is rs=k2x

Assumptions:

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

where F is the Volumetric flowrate at the feed and product stream

xi is the Concentration of cells in the feed

si 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 statemass balance is

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

iis the Mass flowrate of species in inlet stream

0is the Mass flowrate of species in outlet stream

RG is the Rate of species generated

RC is the Rate of species consumed

In this case, the inlet stream has no cells, i=0. Mass flowrate of cells in product stream, 0=Fx. Rate of cell generation,RG=rxV. The rate of cells consumed,RC=0 as cell lysis is negligible.

The unsteady state mass balance for cell is

where rx is the Rate of cell growth

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

Divide by V throughout the equation

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

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

The above differential equation reduces to

This equation relates F, k1 and V at steadystate.

(c) Consider the differential equation

Integrating the above equation for constant V, F andk1

where C is the Integration constant.

At t=0 and x=x0, C=ln x0

Simplifying further

This expression is the for cell concentration as afunction of time.

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

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

(e) For substrate, the rate of mass in isFsi, Rate of substrate generation is zero, Rate of massout is Fs and rate of consumption is rsV.

The unsteady state mass balance for substrate is

where rs is the Rate of substrate consumption

Dividing throughout by V

Substitute for x from the expression obtained in (c)

Here the variables are s and t as F, V, k1 andk2 are constants. Hence it is diffcult to solve theequation algebraically.

Thus, the differential mass balance for substrateis

(f) Consider the equation

The above expression relates s and x at steady stateconditions.

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