Chapter 16, Problem 16.120P

(a)

Interpretation Introduction

Interpretation:

A Plot of $\mu vs.S$ for $S$ between $0.0and1.0kg/m^3$ has to be drawn.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

$\mu =\frac{{\mu }_{\max }S}{K_s+S}$

Where,

$\begin{array}{l}\mu ={\text{first-order growth rate (s}}^{\text{-1}}\text{)}\\ \\ {\mu }_{\max }=Maximumgrowthrate\\ \\ S=Substrateconcentration\\ \\ K_s=\text{the value of S that gives one-half of the maximum growth rate}\\ {\text{(in kg/m}}^{\text{3}}\text{)}\end{array}$

Explanation of Solution

Given data:

$\begin{array}{l}{\mu }_{\max }=1.5\times {10}^{-4}{s}^{-1}\\ \\ S=0.00to1.0kg/m^3\\ \\ K_s=0.03kg/m^3\end{array}$

Calculation of first-order growth rate:

$\begin{array}{l}\mu =\frac{{\mu }_{\max }S}{K_s+S}=\frac{\left(1.5\times {10}^{-4}{s}^{-1}\right)\left(0.25kg/m^3\right)}{\left(0.03kg/m^3\right)+\left(0.25kg/m^3\right)}=1.34\times {10}^{-4}{s}^{-1}\\ \\ \mu =\frac{{\mu }_{\max }S}{K_s+S}=\frac{\left(1.5\times {10}^{-4}{s}^{-1}\right)\left(0.50kg/m^3\right)}{\left(0.03kg/m^3\right)+\left(0.50kg/m^3\right)}=1.42\times {10}^{-4}{s}^{-1}\\ \\ \mu =\frac{{\mu }_{\max }S}{K_s+S}=\frac{\left(1.5\times {10}^{-4}{s}^{-1}\right)\left(0.75kg/m^3\right)}{\left(0.03kg/m^3\right)+\left(0.75kg/m^3\right)}=1.44\times {10}^{-4}{s}^{-1}\\ \\ \mu =\frac{{\mu }_{\max }S}{K_s+S}=\frac{\left(1.5\times {10}^{-4}{s}^{-1}\right)\left(1.0kg/m^3\right)}{\left(0.03kg/m^3\right)+\left(1.0kg/m^3\right)}=1.46\times {10}^{-4}{s}^{-1}\end{array}$

$\begin{array}{cc}S\left(kg/m^3\right)& \mu \left({10}^{-4}{s}^{-1}\right)\\ 0.25& 1.34\\ 0.50& 1.42\\ 0.75& 1.44\\ 1.00& 1.46\end{array}$

By taking all the above data, a graph between $\mu$ and $S$ can be plotted like below,

Figure.1

(b)

Interpretation Introduction

Interpretation:

The density after one hour if the initial $S$ is $0.30kg/m^3$ with given initial population density of $5.0\times {10}^{3}cells/m^3$  has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

$\mu =\frac{{\mu }_{\max }S}{K_s+S}$

Where,

$\begin{array}{l}\mu ={\text{first-order growth rate (s}}^{\text{-1}}\text{)}\\ \\ {\mu }_{\max }=Maximumgrowthrate\\ \\ S=Substrateconcentration\\ \\ K_s=\text{the value of S that gives one-half of the maximum growth rate}\\ {\text{(in kg/m}}^{\text{3}}\text{)}\end{array}$

The first-order reaction kinetics,

$\ln {\left[A\right]}_t=\ln {\left[A\right]}_0+k_{0}t$

Where,

$\begin{array}{l}{\left[\text{A}\right]}_{\text{t}}\text{=}\text{Concentration}\text{of}\text{reactant}\text{at}\text{time}\text{t}\\ \\ {\left[\text{A}\right]}_{\text{0}}\text{=}\text{Initial}\text{Concentration}\text{of}\text{reactant}\\ \\ {\text{k}}_{\text{0}}\text{=}\text{First-order}\text{rate}\text{constant}\end{array}$

Answer to Problem 16.120P

The density after one hour has been determined as $8.2\times 10^{3}cells/m^3.$

Explanation of Solution

Given data:

$\begin{array}{l}{\mu }_{\max }=1.5\times {10}^{-4}{s}^{-1}\\ \\ S=0.30kg/m^3\\ \\ K_s=0.03kg/m^3\\ \\ t=1h=3600s\end{array}$

Calculation of first-order growth rate:

$\mu =\frac{{\mu }_{\max }S}{K_s+S}=\frac{\left(1.5\times {10}^{-4}{s}^{-1}\right)\left(0.25kg/m^3\right)}{\left(0.03kg/m^3\right)+\left(0.25kg/m^3\right)}=1.363\times {10}^{-4}{s}^{-1}.$

According to the first-order kinetics,

$\begin{array}{l}\ln {\left[A\right]}_t=\ln {\left[A\right]}_0+k_{0}t\\ \\ \ln {\left[A\right]}_t=\ln \left[5.0\times {10}^3\right]+\left(1.363\times {10}^{-4}{s}^{-1}\right)\left(3600s\right)\\ \\ \ln {\left[A\right]}_t=9.00808919\\ \\ {\left[A\right]}_t=8.1689\times {10}^3=8.2\times {10}^{3}cells/m^3.\end{array}$

Therefore, the density after one hour is found out to be $8.2\times 10^{3}cells/m^3.$

(c)

Interpretation Introduction

Interpretation:

The density after one hour if the initial $S$ is $0.70kg/m^3$ has to be determined.

Concept Introduction:

The effect of a substrate concentration on first-order growth rate of a microbial population follows the Monod equation:

$\mu =\frac{{\mu }_{\max }S}{K_s+S}$

Where,

$\begin{array}{l}\mu ={\text{first-order growth rate (s}}^{\text{-1}}\text{)}\\ \\ {\mu }_{\max }=Maximumgrowthrate\\ \\ S=Substrateconcentration\\ \\ K_s=\text{the value of S that gives one-half of the maximum growth rate}\\ {\text{(in kg/m}}^{\text{3}}\text{)}\end{array}$

The first-order reaction kinetics,

$\ln {\left[A\right]}_t=\ln {\left[A\right]}_0+k_{0}t$

Where,

$\begin{array}{l}{\left[A\right]}_t=\text{Concentration}\text{of}\text{reactant}\text{at}\text{time}\text{t}\\ \\ {\left[A\right]}_0=\text{Initial}\text{Concentration}\text{of}\text{reactant}\\ \\ k_0=First-\text{order}\text{rate}\text{constant}\end{array}$

Answer to Problem 16.120P

The density after one hour if the initial $S$ is $0.70kg/m^3$ has been determined as $8.4\times 10^{3}cells/m^3.$

Explanation of Solution

Given data:

$\begin{array}{l}{\mu }_{\max }=1.5\times {10}^{-4}{s}^{-1}\\ \\ S=0.70kg/m^3\\ \\ K_s=0.03kg/m^3\\ \\ t=1h=3600s\end{array}$

Calculation of first-order growth rate:

$\mu =\frac{{\mu }_{\max }S}{K_s+S}=\frac{\left(1.5\times {10}^{-4}{s}^{-1}\right)\left(0.70kg/m^3\right)}{\left(0.03kg/m^3\right)+\left(0.70kg/m^3\right)}=1.438356\times {10}^{-4}{s}^{-1}.$

According to the first-order kinetics,

$\begin{array}{l}\ln {\left[A\right]}_t=\ln {\left[A\right]}_0+k_{0}t\\ \\ \ln {\left[A\right]}_t=\ln \left[5.0\times {10}^3\right]+\left(1.438356\times {10}^{-4}{s}^{-1}\right)\left(3600s\right)\\ \\ \ln {\left[A\right]}_t=9.03500135\\ \\ {\left[A\right]}_t=8.39172\times {10}^3=8.4\times {10}^{3}cells/m^3.\end{array}$

Therefore, the density after one hour is found out to be $8.4\times 10^{3}cells/m^3.$