/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 The following table is a summary... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 33

The following table is a summary of data taken on the growth of yeast cells in a bioreactor: $$\begin{array}{|c|c|}\hline \text { Time, } t(\mathrm{h}) & \text { Yeast Concentration, } X(\mathrm{g} / \mathrm{L}) \\\\\hline 0 & 0.010 \\\\\hline 4 & 0.048 \\\\\hline 8 & 0.152 \\\\\hline 12 & 0.733 \\\\\hline 16 & 2.457 \\ \hline\end{array}$$ The data can be fit with the function \(X=X_{0} \exp (\mu t)\) where \(X\) is the concentration of cells at any time \(t, X_{0}\) is the starting concentration of cells, and \(\mu\) is the specific growth rate. (a) Based on the data in the table, what are the units of the specific growth rate? (b) Give two ways to plot the data so as to obtain a straight line. Each of your responses should be of the form "plot ______ Versus ________ on _______ axes." (c) Plot the data in one of the ways suggested in Part (b) and determine \(\mu\) from the plot. (d) How much time is required for the yeast population to double?

Short Answer

Expert verified
Units of the growth rate \( \mu \) are 1/hour. Two ways to plot data for a straight line: \(ln(X)\) versus \(t\) on normal axes or \(X\) versus \(t\) on a semi-log axes. \(\mu\) can be determined from the slope of these plots. The time for yeast population to double is \( t = (1/\mu) * \ln (2) \) hours.

Step by step solution

01

Determine The Units of Growth Rate

By analyzing the equation \(X=X_{0} \exp (\mu t)\), where \(X\) is the yeast concentration in \(g/L\), \(t\) is the time in hours, and \(X_{0}\) is the initial concentration. Isolating \(\mu\), the specific growth rate can be done as follows: \( \mu = (1/t) * \ln (X/X_{0}) \). This makes the units of the specific growth rate \(\mu\) to be \(1/hour\).
02

Show Two Ways to Plot The Data

One can plot the data in two ways to get a straight line. First method can be by plotting \(ln(X)\) versus \(t\) on normal axes or the second method can be by plotting \(X\) versus \(t\) on semi-log axes. These two methods provide a straight line because the exponential growth equation is a linear equation in the logarithmic scale.
03

Determine The Growth Rate from the Plot

Taking the first method from step 2 (plotting \(ln(X)\) versus \(t\) on normal axes), the graph will look like a straight line. The slope of this line will be equal to the specific growth rate \(\mu\). One can compute the slope by taking two points from the line and applying the formula \( \mu = \Delta [ln(X)] /\Delta t \). \(\Delta\) denotes the difference between two points.
04

Calculate Time for Population to Double

Starting from the formula \(X=X_{0} \exp (\mu t)\), we can set \(X = 2X_{0}\) because we want to find the time for the population to double. Solving for \(t\), we get \(t = (1/\mu) * \ln (2)\). Therefore, the time it takes for the yeast population to double is \( t = (1/\mu) * \ln (2) \) hours.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Yeast Growth
Yeast growth in a bioreactor is a fascinating example of microbial growth kinetics. This process can be observed by measuring the yeast cell concentration over time. In the context of bioreactors, yeast growth is typically monitored because it is an important factor in processes like fermentation where the production of cell mass is desired. The yeast concentration is expressed in grams per liter (g/L), reflecting how densely packed the yeast cells are in the liquid medium. Fast yeast growth can mean quick product turnaround in industrial applications.

Understanding the growth pattern is crucial. In the lag phase, yeast adjusts to its environment but doesn't grow much. Then comes the exponential (or **logarithmic**) phase where yeast divides rapidly. Eventually, growth slows in the stationary phase due to nutrient depletion or waste accumulation. By monitoring these stages, one can make informed decisions about when to harvest or modify conditions for optimal results.
Exponential Function
The exponential function is a crucial mathematical tool used to describe how yeast grows in a bioreactor. It is represented by the equation: \[ X = X_{0} \exp(\mu t) \] where:
  • \(X\) is the yeast concentration at any given time \(t\).
  • \(X_{0}\) is the initial concentration of yeast.
  • \(\mu\) is the specific growth rate, a constant that shows growth speed.
This equation means yeast growth increases at a rate proportional to its current size. Simply put, the more yeast you already have, the faster it grows. This is typical of exponential growth situations.

The exponential function creates a curve on a graph that shows rapid increase. To analyze yeast growth, scientists often convert this curve into a straight line using logarithms. This method is essential to simplify the analysis and easily determine important parameters such as the growth rate.
Specific Growth Rate
The specific growth rate, denoted as \(\mu\), is a key concept in understanding yeast growth kinetics. It quantifies the rate at which yeast cells divide and grow in the bioreactor. The units for \(\mu\) are typically \(\text{1/hour}\), demonstrating how much of the yeast population grows per hour.

Determining \(\mu\) involves graphically analyzing the yeast's growth curve. By plotting the logarithm of the yeast concentration against time on normal axes, a straight line is formed. The slope of this line provides the specific growth rate \(\mu\). This makes it easy to calculate \(\mu\) using the formula:\[ \mu = \frac{\Delta [\ln(X)]}{\Delta t} \]where \(\Delta\) denotes the change between two points. Understanding \(\mu\) allows us to anticipate how quickly the yeast population will double, using:\[ t = \frac{1}{\mu} \ln(2) \]This formula indicates that as \(\mu\) increases, the time required for doubling decreases, leading to faster production cycles in applications like food production or pharmaceuticals.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following reactions take place in a batch reactor: \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\) (desired product) \(\mathrm{B}+\mathrm{C} \rightarrow \mathrm{D}\) (hazardous product) As the reaction proceeds, D builds up in the reactor and could cause an explosion if its concentration exceeds 15 mol/L. To ensure the safety of the plant personnel, the reaction is quenched (e.g., by cooling the reactor contents to a low temperature) and the products are extracted when the concentration of \(D\) reaches \(10 \mathrm{mol} / \mathrm{L}\). The concentration of \(C\) is measured in real-time, and samples are periodically taken and analyzed to determine the concentration of D. The data are shown below: $$\begin{array}{|c|c|}\hline C_{\mathrm{C}}(\mathrm{mol} / \mathrm{L}) & C_{\mathrm{D}}(\mathrm{mol} / \mathrm{L}) \\ \hline 2.8 & 1.4 \\\\\hline 10 & 2.27 \\\\\hline 20 & 2.95 \\\\\hline 40 & 3.84 \\\\\hline 70 & 4.74 \\\\\hline 110 & 5.63 \\ \hline 160 & 6.49 \\\\\hline 220 & 7.32 \\\\\hline\end{array}$$ (a) What would be the general form of an expression for \(C_{\mathrm{D}}\) as a function of \(C_{\mathrm{C}} ?\) (b) Derive the expression. (c) At what concentration of \(C\) is the reactor stopped? (d) Someone proposed not stopping the reaction until \(C_{\mathrm{D}}=13 \mathrm{mol} / \mathrm{L},\) and someone else strongly objected. What would be the major arguments for and against that proposal?

The temperature in a process unit is controlled by passing cooling water at a measured rate through a jacket that encloses the unit. The exact relationship between the unit temperature \(T\left(^{\circ} \mathrm{C}\right)\) and the water flow rate \(\phi(\mathrm{L} / \mathrm{s})\) is extremely complex, and it is desired to derive a simple empirical formula to approximate this relationship over a limited range of flow rates and temperatures. Data are taken for \(T\) versus \(\phi\). Plots of \(T\) versus \(\phi\) on rectangular and semilog coordinates are distinctly curved (ruling out \(T=a \phi+b\) and \(T=a e^{b \phi}\) as possible empirical functions), but a log plot appears as follows: A line drawn through the data goes through the points \(\left(\phi_{1}=25 \mathrm{L} / \mathrm{s}, T_{1}=210^{\circ} \mathrm{C}\right)\) and \(\left(\phi_{2}=40 \mathrm{L} / \mathrm{s},\right.\) \(\left.T_{2}=120^{\circ} \mathrm{C}\right)\). (a) What is the empirical relationship between \(\phi\) and \(T ?\) (b) Using your derived equation, estimate the cooling water flow rates needed to maintain the process unit temperature at \(85^{\circ} \mathrm{C}, 175^{\circ} \mathrm{C},\) and \(290^{\circ} \mathrm{C}\). (c) In which of the three estimates in Part (b) would you have the most confidence and in which would you have the least confidence? Explain your reasoning.

A process instrument reading, \(Z\) (volts), is thought to be related to a process stream flow rate \(\dot{V}(\mathrm{L} / \mathrm{s})\) and pressure \(P(\mathrm{kPa})\) by the following expression: $$Z=a \dot{V}^{b} P^{c}$$ Process data have been obtained in two sets of runs- -one with \(\dot{V}\) held constant, the other with \(P\) held constant. The data are as follows: $$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \text { Point } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\\hline \dot{V}(\mathrm{L} / \mathrm{s}) & 0.65 & 1.02 & 1.75 & 3.43 & 1.02 & 1.02 & 1.02 \\\\\hline P(\mathrm{kPa}) & 11.2 & 11.2 & 11.2 & 11.2 & 9.1 & 7.6 & 5.4 \\\\\hline Z(\text { volts }) & 2.27 & 2.58 & 3.72 & 5.21 & 3.50 & 4.19 & 5.89 \\\\\hline\end{array}$$ (a) Suppose you had only performed runs \(2,3,\) and \(5 .\) Calculate \(a, b,\) and \(c\) algebraically from the data for these three runs. (b) Now use a graphical method and all the data to calculate \(a, b,\) and \(c .\) Comment on why you would have more confidence in this result than in that of Part (a). (Hint: You will need at least two plots.)

The Reynolds number is a dimensionless group defined for a fluid flowing in a pipe as \(R e=D u \rho / \mu\) where \(D\) is pipe diameter, \(u\) is fluid velocity, \(\rho\) is fluid density, and \(\mu\) is fluid viscosity. When the value of the Reynolds number is less than about \(2100,\) the flow is laminar- -that is, the fluid flows in smooth streamlines. For Reynolds numbers above \(2100,\) the flow is turbulent, characterized by a great deal of agitation. Liquid methyl ethyl ketone (MEK) flows through a pipe with an inner diameter of 2.067 inches at an average velocity of \(0.48 \mathrm{ft} / \mathrm{s}\). At the fluid temperature of \(20^{\circ} \mathrm{C}\) the density of liquid \(\mathrm{MEK}\) is \(0.805 \mathrm{g} / \mathrm{cm}^{3}\) and the viscosity is 0.43 centipoise \(\left[1 \mathrm{cP}=1.00 \times 10^{-3} \mathrm{kg} /(\mathrm{m} \cdot \mathrm{s})\right]\). Without using a calculator, determine whether the flow is laminar or turbulent. Show your calculations.

A hygrometer, which measures the amount of moisture in a gas stream, is to be calibrated using the apparatus shown here: Steam and dry air are fed at known flow rates and mixed to form a gas stream with a known water content, and the hygrometer reading is recorded; the flow rate of either the water or the air is changed to produce a stream with a different water content and the new reading is recorded, and so on. The following data are taken: $$\begin{array}{cc}\hline \begin{array}{c}\text { Mass Fraction } \\\\\text { of Water, } y\end{array} & \begin{array}{c}\text { Hygrometer } \\\\\text { Reading, } R\end{array} \\\\\hline 0.011 & 5 \\\0.044 & 20 \\\0.083 & 40 \\\0.126 & 60 \\\0.170 & 80 \\ \hline\end{array}$$ (a) Draw a calibration curve and determine an equation for \(y(R)\). (b) Suppose a sample of a stack gas is inserted in the sample chamber of the hygrometer and a reading of \(R=43\) is obtained. If the mass flow rate of the stack gas is \(1200 \mathrm{kg} / \mathrm{h}\), what is the mass flow rate of water vapor in the gas?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.