/*! 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 51 The following \((x, y)\) data ar... [FREE SOLUTION] | 91Ó°ÊÓ

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The following \((x, y)\) data are recorded: $$\begin{array}{|c|c|c|c|}\hline x & 0.5 & 1.4 & 84 \\\\\hline y & 2.20 & 4.30 & 6.15 \\\\\hline \end{array}$$ (a) Plot the data on logarithmic axes. (b) Determine the coefficients of a power law expression \(y=a x^{b}\) using the method of least squares. (Remember what you are really plotting \(-\) there is no way to avoid taking logarithms of the data point coordinates in this case.) (c) Draw your calculated line on the same plot as the data.

Short Answer

Expert verified
The described process involves data plotting on logarithmic axes, conversion to linear form to apply least squares method, calculation and substitution of the coefficients to draw the best fit power law curve on the same plot.

Step by step solution

01

- Data Preparation

Plot the data on logarithmic axes using the coordinates provided. Convert each value into its logarithm, because of the logarithmic nature of the axes being used. Use a logarithmic scale on both \(x\) and \(y\) axes.
02

- Achieve Linear Form

We need a linear form for our data to apply the least squares method. Since our aim is to represent the power law \(y=a x^{b}\) in linear form, it'll become \(\log_{10}y = \log_{10}a + b \log_{10}x\) when taking logarithm on both sides. This is similar to the linear equation \(Y = mX + c\), where \(Y=\log_{10}y\), \(X=\log_{10}x\), \(m=b\), and \(c=\log_{10}a\). Hence, we can now apply the least squares method to derive \(a\) and \(b\).
03

- Application of Least Squares Method

The principle of the method of least squares is all about minimizing the sum of the squares of the vertical distances of points from the regression line (the smallest is the sum, the better is the fit). In linear form, \(b\) (slope) is calculated as \(b = \frac{n(\sum XY) - (\sum X)(\sum Y)}{n(\sum X^2) - (\sum X)^2}\) and \(a\) (intercept) is calculated as \(a = \frac{\sum Y - b(\sum X)}{n}\), where \(n\) is the number of points. Substitutes the logarithms of the given data points into the formulas to obtain \(b\) and \(a\).
04

- Plot the Power Law

Now, convert \(\log_{10}a\) back to \(a\) by taking antilog, and keep \(b\). Substitute these coefficients into the original power law equation \(y=a x^{b}\) to represent the line of best fit calculated through least squares. Draw this line on the same log-log plot. This curve with calculated \(a\) and \(b\) values will be the best fit for the given data points.

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Key Concepts

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

Logarithmic Data Plotting
In the realm of chemical processes and data analysis, logarithmic data plotting is an essential technique that allows us to visualize and analyze data sets that span multiple orders of magnitude. Unlike a linear scale, a logarithmic scale increases exponentially, meaning that each step on the scale represents a multiplication by a constant factor rather than a simple addition.

When plotting data on logarithmic axes, you convert the original data values into their logarithms. This can be especially useful when you're dealing with a power law relationship between two variables, which is common in chemical kinetics and other phenomena. By using logarithmic scales on both the x and y axes, you transform the curve of the power law into a straight line, where the exponent in the power law becomes the slope, and the coefficient becomes the y-intercept. This transformation simplifies analysis and, as is required for the method of least squares, makes the line easier to determine.
For example, given the data points with x-values of 0.5, 1.4, and 84, and corresponding y-values of 2.20, 4.30, and 6.15, you'd take the log of these numbers to correctly place them on log-log axes. Do not neglect this step, as it is crucial for the accurate representation of data in a log-log plot. This preparatory step is unmatched for highlighting patterns that may be undetected in untransformed data.
Power Law Expression
Understanding the power law expression is key to grasping how we interpret chemical process data. A power law is an algebraic equation of the form \(y = a x^{b}\), where \(a\) and \(b\) are constants, and \(x\) and \(y\) are the variables. The value \(b\) is known as the exponent, and \(a\) is the coefficient.

These expressions are particularly useful in science and engineering because they describe a variety of phenomena where one variable varies as a power of another. For instance, the rate of a reaction can be proportional to a power of reactant concentration. In the case of our exercise, to convert the power law relationship to a linear form, we take the logarithm of both sides, which yields \(\text{log} y = \text{log} a + b \text{log} x\). This linear form aligns with the straight line equation \(Y = mX + c\), simplifying the computation of the constants \(a\) and \(b\) from the data. Through this conversion, the interdependent nature of the variables becomes more apparent, and we set the stage for a linear regression approach using the least squares method.
Least Squares Method
The least squares method is a foundational statistical tool used in data analysis and curve fitting. It is particularly adept at finding the best-fitting curve or line to a given set of points by minimizing the sum of the squares of the offsets or residuals of points from the curve.

Our primary goal is to find the values of \(a\) and \(b\) in the power law expression that lead to a line that falls as close as possible to our logarithmic data points. To do this, you calculate the slope \(b\) and the y-intercept \(\text{log} a\) using the formulas derived from setting partial derivatives of the sum of squared residuals to zero.
These calculations involve sums of products and squares of the log-transformed x-values and y-values. Once you've determined the slope and y-intercept, you can convert the intercept back to find \(a\) by raising 10 to the power of the intercept. Consequently, we can express our original power law relationship with the computed variables, \(y=ax^b\), which represents the line of best fit. This line is then plotted on the logarithmic graph, visually demonstrating the relationship and enabling further interpretation and analysis of the dataset's trends and behaviors.

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Most popular questions from this chapter

The following are measured values of a system temperature versus time: $$\begin{array}{|c|c|c|c|c|c|c|}\hline t(\min ) & 0.0 & 2.0 & 4.0 & 6.0 & 8.0 & 10.0 \\\\\hline T\left(^{\circ} \mathrm{C}\right) & 25.3 & 26.9 & 32.5 & 35.1 & 36.4 & 41.2 \\\\\hline\end{array}$$ (a) Use the method of least squares (Appendix A.1) to fit a straight line to the data, showing your calculations. You may use a spreadsheet to evaluate the formulas in Appendix A.1, but do not use any plotting or statistical functions. Write the derived formula for \(T(t)\), and convert it to a formula for \(t(T)\). (b) Transfer the data into two columns on an Excel spreadsheet, putting the \(t\) data (including the heading) in Cells A1-A7 and the \(T\) data (including the heading) in B1-B7. Following instructions for your version of Excel, insert a plot of \(T\) versus \(t\) into the spreadsheet, showing only the data points and not putting lines or curves between them. Then add a linear trendline to the plot (that is, fit a straight line to the data using the method of least squares) and instruct Excel to show the equation of the line and the \(R^{2}\) value. The closer \(R^{2}\) is to 1 , the better the fit.

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.)

Five hundred \(1 \mathrm{b}_{\mathrm{m}}\) of nitrogen is to be charged into a small metal cylinder at \(25^{\circ} \mathrm{C}\), at a pressure such that the gas density is \(12.5 \mathrm{kg} / \mathrm{m}^{3}\). Without using a calculator, estimate the required cylinder volume in \(\mathrm{ft}^{3}\). Show your work.

Bacteria can serve as catalysts for the conversion of low-cost chemicals, such as glucose, into higher value compounds, including commodity chemicals (with large production rates) and high-value specialty chemicals such as pharmaceuticals, dyes, and cosmetics. Commodity chemicals are produced from bacteria in very large bioreactors. For example, cultures up to 130,000 gallons are used to produce antibiotics and other therapeutics, industrial enzymes, and polymer intermediates. When a healthy bacteria culture is placed in a suitable environment with abundant nutrients, the bacteria experience balanced growth, meaning that they continue to double in number in the same fixed period of time. The doubling time of mesophilic bacteria (bacteria that live comfortably at temperatures between \(35^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\) ) ranges anywhere from 20 minutes to a few hours. During balanced growth, the rate of growth of the bacteria is given by the expression \(\frac{d C}{d t}=\mu C\) where \(C(\mathrm{g} / \mathrm{L})\) is the concentration of bacteria in the culture and \(\mu\) is called the specific growth rate of the bacteria (also described in Problem 2.33 ). The balanced growth phase eventually comes to an end, due either to the presence of a toxic byproduct or the lack of a key nutrient. The following data were measured for the growth of a particular species of mesophilic bacteria at a constant temperature: $$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline t(\mathrm{h}) & 1.0 & 2.0 & 3.0 & 4.0 & 5.0 & 6.0 & 7.0 & 8.0 \\\\\hline C(\mathrm{g} / \mathrm{L}) & 0.008 & 0.021 & 0.030 & 0.068 & 0.150 & 0.240 & 0.560 & 1.10 \\\\\hline\end{array}$$ (a) If bacteria are used in the production of a commodity chemical, would a low or high value of \(\mu\) be desirable? Explain. (b) In the rate expression, separate the variables and integrate to derive an expression of the form \(f\left(C, C_{0}\right)=\mu t,\) where \(C_{0}\) is the bacteria concentration that would be measured at \(t=0\) if balanced growth extended back that far. (It might not.) What would you plot versus what on what kind of coordinates (rectangular, semilog, or log) to get a straight line if growth is balanced, and how would you determine \(\mu\) and \(C_{0}\) from the plot? (Review Section 2.7 if necessary.) (c) From the given data, determine whether balanced growth was maintained between \(t=1 \mathrm{h}\) and \(t=8 \mathrm{h} .\) If it was, calculate the specific growth rate. (Give both its numerical value and its units.) (d) Derive an expression for the doubling time of a bacterial species in balanced growth in terms of \(\mu\) [You may make use of your calculations in Part (b).] Calculate the doubling time of the species for which the data are given.

The relationship between the pressure \(P\) and volume \(V\) of the air in a cylinder during the upstroke of a piston in an air compressor can be expressed as \(P V^{k}=C\) where \(k\) and \(C\) are constants. During a compression test, the following data are taken: $$\begin{array}{|c|c|c|c|c|c|c|}\hline P(\mathrm{mm} \mathrm{Hg}) & 760 & 1140 & 1520 & 2280 & 3040 & 3800 \\ \hline V\left(\mathrm{cm}^{3}\right) & 48.3 & 37.4 & 31.3 & 24.1 & 20.0 & 17.4 \\\\\hline\end{array}$$ Determine the values of \(k\) and \(C\) that best fit the data. (Give both numerical values and units.)

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