Batch Growth Kinetics from Two Semilog Data Points: 04-BS-13, December 2019, Q9

Question 9: Batch Growth Kinetics from Two Semilog Data Points (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Point	$t$ (h)	$x$ (g/l)
1	0.5	3.5
2	15	10.6

Find. (a) the equation $x(t)$; (b) the specific growth rate $\mu$.

Approach. A straight line on semilog ($\ln x$ vs. $t$) paper means exponential growth, $x=x_0 e^{\mu t}$; two points on that line give two equations for the two unknowns $\mu$ and $x_0$.

(a) Specific growth rate from the two points. Since $\ln x_2-\ln x_1=\mu(t_2-t_1)$: $$\mu=\frac{\ln(x_2/x_1)}{t_2-t_1}=\frac{\ln(10.6/3.5)}{15-0.5}=\frac{\ln(3.0286)}{14.5}=\frac{1.1079}{14.5}=\boxed{0.0764\ \text{h}^{-1}}.$$
Back out $x_0$ (concentration at $t=0$). $$x_0=\frac{x_1}{e^{\mu t_1}}=\frac{3.5}{e^{0.0764(0.5)}}=\frac{3.5}{1.0390}=\boxed{3.369\ \text{g/l}}.$$ Check: $x_0e^{\mu t_2}=3.369\,e^{0.0764(15)}=3.369(3.146)=10.60$ g/l, matching $x_2$ exactly.
(a) Equation relating $x$ and $t$. $$\boxed{x(t)=3.369\,e^{0.0764\,t}}\quad(x\text{ in g/l},\ t\text{ in h}).$$
(b) Specific growth rate. $\mu=0.0764\ \text{h}^{-1}$ (the exponent found in part (a), by definition of exponential growth—the same value applies at any point along the straight-line segment).

Quantity	Result
Specific growth rate $\mu$	0.0764 h^-1
$x_0$ (extrapolated to $t=0$)	3.369 g/l
Equation relating $x$ and $t$	$x(t)=3.369\,e^{0.0764t}$

Topic: Extracting exponential-growth-phase kinetics ($\mu$, $x_0$) from two points on a semilog plot.

Key relations: $x=x_0e^{\mu t}\Leftrightarrow\ln x=\ln x_0+\mu t$ (a straight line on semilog paper); $\mu=\dfrac{\ln(x_2/x_1)}{t_2-t_1}$ from any two points on that line.

Why this works: a straight line on a $\ln x$ vs. $t$ (semilog) plot is the defining signature of balanced exponential growth with a constant specific growth rate; the slope of that line is $\mu$ by construction, so any two points on it are sufficient to solve for both $\mu$ and the (extrapolated) intercept $x_0$.

Common pitfall: using $\log_{10}$ (common log, matching most commercial semilog graph paper) instead of $\ln$ without converting — the slope read directly off semilog graph paper is in $\log_{10}$ units and must be multiplied by $\ln 10=2.303$ to obtain $\mu$ in the natural-log convention used here.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 6 (batch growth kinetics, exponential phase, specific growth rate determination).

Question 9 of 9: Batch Growth Kinetics from Two Semilog Data Points

Question 9: Batch Growth Kinetics from Two Semilog Data Points (20 marks)