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Access Numerical Methods for Engineers 5th Edition solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality!. solutions manual. Our solution manuals are written by Chegg experts so you can be assured of the highest quality! 5th Edition Why is Chegg Study better than downloaded Numerical Methods for Engineers PDF solution manuals?. numerical methods for engineers 5th edition chapra solution manual . Numerical Methods for Chemical Engineers with MATLAB Applications[PDF] See more.

Chegg Solution Manuals are written by vetted Chegg Math experts, and rated by students - so you know you're getting high quality answers. ClearContents Range "d5". II Using the Roots Function: Thus, there is a negative root. However, as p approaches pmax, the growth rate decreases and the population levels off. Cell I9:

The problem can be solved with Microsoft Excel: Surface Area vs Weight 2. The predicted surface area for a 95 kg human is approximately: Notice that we have assumed that the model consists of a constant plus two bell-shaped curves: The second model can be fit with the Excel Solver: We did this so that the sum of the squares of the residuals would not be miniscule.

Sometimes this will lead the Solver to conclude that it is at the minimum, even though the fit is poor. Although the fit might appear to be OK, it is biased in that it underestimates the low values and overestimates the high ones.

The poorness of the fit is really obvious if we display the results as a log-log plot: Notice that this view illustrates that the model actually overpredicts the very lowest values. The third and fourth models provide a means to rectify this problem.

Because they raise [S] to powers, they have more degrees of freedom to follow the underlying pattern of the data. For example, the third model gives: Finally, the cubic model results in a perfect fit: Thus, the best fit is 2.

However, if we assume that the function has one maximum and no minima within the interval, a check can be included. Here is a VBA program to implement the Golden section search algorithm for maximization and solve Example Add the following line to the beginning of the VBA code: Although it is not really clear from the problem statement, it should be assumed that each unit of product is equivalent to a kg. Then the objective function line can be superimposed for various values of P until it reaches the boundary.

Notice also that material and storage are the binding constraints and that there is some slack in the time constraint.

C2 Subject to constraints: Excel spreadsheet solution: Linear regression gives 0. An alternative approach would to assume that the physically-unrealistic non-zero intercept is an artifact of the measurement method. Therefore, use the exponential solution. The amount of bacteria after 30 days: It is set up to solve Example Select Next i Range "e3". Select Range "d5: ClearContents Range "d5". The following shows the solution for Prob. Offset 2, 0.

Select Next i Range "e4". Value Call Spline x , y , n, xu, yu, dy, d2y ActiveCell.

It is set up to solve Prob. Note that even though we used a slow PC, we had to call the function numerous times to obtain measurable times. These times were then divided by the number of function calls to determine the time per call shown below N time s 32 0. Obtain the R — squared value to determine the goodness of fit. Dye Concentraion vs. Time 4. Row - n 'Input data from sheet Range "a6". The results of a and b are not exact because they include trapezoidal rule evaluations.

Substituting these values into Eqs. Select Range "a1". Select Sheets "Sheet1". Range "a5: However, we can try to fit different order polynomials to see if we can get a decent fit to the data. This yields the surprising result that a 4th-order polynomial results in almost a perfect fit. For example, using the Excel trend line gives: Thus it approximates the quartering of the error that we would expect according to Eq.

One approach is to set up Excel with a series of equally-spaced x values from 0 to Then one of the formulas described in this Part of the book can be used to numerically compute the derivative. For example, I used x values with an interval of 1 and Eq. The resulting plot of the function and its derivative is 1 There might be a slight discrepancy due to roundoff.

The results are approaching a value of The steady state form i. Thus, there is a negative root. However, if we pick a value that is slightly higher a per machine precision , it will gravitate towards the positive root. The simplest is to compute the mass of salt in the tank over time by solving the following differential equation: In addition, a graph of the entire solution is also displayed. Thus, the use of a simple explicit Runge-Kutta scheme would involve using a very small time step in order to maintain a stable solution.

A solver designed for stiff systems was used to generate the solution shown below. Two views of the solution are given. The first is for the entire solution domain. The finite-difference approach is straightforward: The resulting linear system can be solved with an approach like the Gauss-Seidel method. The following table and graph summarize the results. As can be seen, the person died 1. The non-self-starting Heun yielded the following time series of temperature: This result is reinforced when a state-space plot of the calculation is displayed.

This result is also reinforced when a state-space plot of the calculation is displayed. In fact, both the time series and state-space plots are indistinguishable from each other. The 4th-order RK method yields a stable solution. The first few values are shown in the following table.

A plot of the result for x is also shown below. Then we formed a column containing the squared residuals between our predictions and the measured values. Adjusting the diffusion coefficients with the Solver tool minimized the sum of the squares. At first, we assumed that the diffusion coefficients were zero.

For this case the Solver did not converge on a credible answer. This magnitude was based on the fact that the volumes were of this order of magnitude. The resulting simulation did not fit the data very well, but was much better than when we had guessed zero. Some of the Euler calculations are displayed below along with a plot of the fit. Thus, as would be expected from a first-order model, the slope is equal to the growth rate of the population.

This shape occurs because initially the population is increasing exponentially since p is much less than pmax. However, as p approaches pmax, the growth rate decreases and the population levels off.

The resulting estimates are: The fit is: A table showing the first few steps and a graph of the entire solution are shown below. Some selected values are shown, along with a plot of both the analytical and numerical result. A better match would be obtained by using a smaller step or a higher-order method. Some selected values are shown, along with a plot of the numerical result. Note that the table and plot also show the analytical solution for the linear case computed in Prob. The eigenvectors tell us that these correspond to oscillations that coincide 0.

A plot of the result is also shown below. Note, the nonlinear solution is the bolder line. See solution for Prob. For example, for the point at the beginning of the tank, a forward difference can be used to calculate dc Thus, we place nodes as in the following diagram.

The relationships at the oil-water interface can be used to used to eliminate two of the remaining unknowns. Equations i and ii can be used to determine v6o and v6w.

The results are plotted below: The resulting solution is The Ralston method performs much better for this case. The 4th-order RK is much closer to the exact solution. Select Range "a5: ClearContents Range "a5".

Example From the plot, we approximate that it takes about 58 minutes to drain the cylindrical tank. A plot of the entire solution is also displayed x y 0 0 0. The best way to do this is with LU decomposition since we will have to solve the system repeatedly. For the present case, because its easier to display, we will use the matrix inverse to obtain the solution.

If a smaller step size were used, the solution would improve Test the program by duplicating Example Hence, the percentage of intracellular water be of total weight of the body. Chegg Solution Manuals are written by vetted Chegg 1 experts, and rated by students - so you know you're getting high quality answers.

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