>> CHAPTER 3 :Approximations and Round-Off Errors 55 << Rent $23.49 Due 12/16/2022 Included with your book Feel free to highlight your book Free shipping on rental returns numerical methods for engineers 6th & 7th edition solution manual download pdf. /Resources 37 0 R CHAPTER 7 Roots of Polynomials 176 >> /Resources 62 0 R This download free numerical methods for engineers 6th & 7th edition solution manual ( chapter solutions ) written by Steven Chapra eBook in pdf format is designed to empower them to do that. 0 0 3MB Read more. VBA Function Procedure MATLAB M-File (a) Factorial Function factor (n) Dim x As Long, i As Integer x = 1 For i = 1 To n x = x * i Next i factor = x function fout = factor (n) x = 1; for i = 1:n x = x * i; end fout = x; >> Chapra - Numerical Methods for Engineers, 7th Edition by Steven Chapra and Raymond Canale (9780073397924) Preview the textbook, purchase or get a FREE instructor-only desk copy. Solution Manual of Numerical methods using MATLAB by John. Numerical Methods for Engineers 7th edition pdf Book Hut May 7th, 2018 - Mathematical concepts are explained in Numerical methods for Engineers 7th edition book Read a full eBook review or download free pdf of latest edition Solution manual for Numerical Methods for Engineers 7th May 4th, 2018 - Numerical Methods for Engineers 7th endobj Book contains more than thirty chapters divided into 8 parts. /Count 10 . << ang ti. (xem ton vn), Ti liu hn ch xem trc, xem y mi bn chn Ti xung. download 1 file . CHAPTER 8 Case Studies: Roots of Equations 204 << The seventh edition of Chapra and CanalesNumerical Methods for Engineersretains the instructional techniques that have made the text so successful. 18 0 obj /MediaBox [0 0 612 792] << 0 0 3MB Read more. View the primary ISBN for: Numerical Methods for Engineers 7th Edition Textbook Solutions Solutions by chapter Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Perform the same computation as in Example 1.1 but use Eq. The book covers the standard numerical methods employed by both students and practicing engineers. /Contents 57 0 R << /Parent 9 0 R /CropBox [0 0 612 792] ZIL%d9m{Y6bbz3Q\HGWOD16= ?e)N".^|s9nL ROby~^z!#B Xl5Wa>mC'JZIy78]e:^#A. CHAPTER 5 Bracketing Methods 123 /CropBox [0 0 612 792] << 3 0 obj Authors: Steven Chapra, Raymond Canale Published: McGraw 2009 , McGraw 2014 Edition: 6th , 7th Pages: 517 , 789 Type: pdf Size: 17MB , 13.8MB Gioumeh .com Follow 19 0 obj >> numerical methods for engineers 7th edition chapra solutions this version is then followed by a matlab script and function that does accommodate complex . Carbohydrate digestion and metabolism in Ruminants Carbohydrate Digestion No public clipboards found for this slide. endobj > and click on the required section for solution manuals. /Rotate 0 26 0 obj APPENDIX B: GETTING STARTED WITH MATLAB /Rotate 0 /CropBox [0 0 612 792] /MediaBox [0 0 612 792] endobj 11 0 obj endobj 9789386279521, 9386279525, 9789380250403 . /Rotate 0 /MediaBox [0 0 612 792] solution manual of numerical methods for engineers 6th & 7th edition download pdf https://gioumeh.com/product/numerical-methods-for-engineers-solution/ 1 CHAPTER 1 1.1 We. Chapter 10 Numerical Solution Methods - San Jose State University /Contents 59 0 R /Type /Page Download Free Sample Need Information : Live Chat Add to cart 22 0 obj CHAPTER 16 Case Studies: Optimization 416 Numerical Methods for Engineers 7th Edition. << Numerical Analysis for Scientists and . /Keywords (Numerical Methods for Engineers;Chapra;7th Edition;Solutions Manual) This is the seventh edition of Chapra and Canale's Numerical Methods for Engineers that retains the instructional techniques that have made the text so successful. By accepting, you agree to the updated privacy policy. 1 0 obj The methods included here are of a basic nature and only rely on material which should have been explored prior to the rst endobj endobj This revised edition discusses numerical methods for computing eigenvalues and eigenvectors of large sparse matrices. Solution Manual of Applied Numerical Methods With MATLAB for Engineers & Scie 7 Examples of C Program (Civil Engineering), Beer vector mechanics for engineers statics 10th solutions, mechanics-for-engineers-dynamics-solutions-10th-edition. BIBLIOGRAPHY, If you require any further information, let me know. /Contents 67 0 R 31 0 obj endobj /MediaBox [0 0 612 792] CHAPTER 19 Fourier Approximation 526 Engineering Numerical Methods for Engineers 6th Edition Raymond P Canale, Steven C Chapra ISBN: 9780073401065 Textbook solutions Verified Chapter 1: Mathematical Modeling and Engineering Problem Solving Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5 Exercise 6 Exercise 7 Exercise 8 Exercise 9 Exercise 10 Exercise 11 Exercise 12 Exercise 13 The fifth edition of "Numerical Methods for Engineers with Software and Programming Applications" continues its tradition of excellence. /Resources 52 0 R 9789386279521, 9386279525, 9789380250403 . However it is not from the textbook. CHAPTER 20 Case Studies: Curve Fitting 563 /Pages 4 0 R Format : Digital copy DOC DOCX PDF RTF in ZIP file. /Type /Page 15 0 obj /Contents 55 0 R endobj PScript5.dll Version 5.2.2 /MediaBox [0 0 612 792] (BME7) Numerical Methods for Engineers Proceedings of the 7th Indian Young Geotechnical Engineers Conference Applied Statistics 3rd Edition Just Ask Edition with Student Workbook Set Computational Methods in Earthquake Engineering 7th Asian-Pacific Conference on Medical and Biological Engineering Numerical Methods for Engineers 7th UK Computer . /Parent 8 0 R /CropBox [0 0 612 792] /Contents 47 0 R endobj /Type /Page 1,237 262 11MB Read more. endobj /Parent 7 0 R >> stream Free access to premium services like Tuneln, Mubi and more. /Parent 7 0 R /CropBox [0 0 612 792] << Solution manual for Numerical Methods for Engineers 7th edition by Steven C Chapra $ 50.00 $ 35.00 Format : Digital copy DOC DOCX PDF RTF in "ZIP file". /Title (Numerical Methods for Engineers 7th Edition Chapra Solutions Manual) endobj endobj Xem thm: Solution manual for numerical methods for engineers 7th edition by chapra , Copyright 2020 123DOC. The text features a broad array of applications that span all engineering disciplines. Edition/Type: 7th Edition /Solution Manual: Author: by Steven Chapra , Raymond Canale : ISBN: ISBN-13: 978-0073397924 ISBN-10: 007339792X : Document Format: /Parent 4 0 R Numerical Methods for Engineers 7th edition View Textbook Solutions ISBN: 007339792X ISBN-13: 9780073397924 Authors: Steven Chapra, Chapra, Raymond Canale Rent From $23.49 Buy From $34.49 Textbook Solutions Only $15.95/mo. We've encountered a problem, please try again. >> Numerical Methods For Engineers [8 ed.] PDF download. Time: Immediately after payment is completed Categories : Solution manuals, TestBank All the chapters are included. This lecture explains the general concepts of how to convert a physical problem into a mathematical and a numerical problem. /MediaBox [0 0 612 792] << Now customize the name of a clipboard to store your clips. Solution Manual for Numerical Methods for Engineers by Steven C. Chapra by Kenneth H Rosen 7th Edition Step by Step Solutions Chapter 1 section 1. fSolution manual Numerical Methods for Partial Differential Equations : Finite Solution manual Advanced Engineering Mathematics (7th Ed., Peter O'Neil) Numerical Methods For Engineers Chapra 7th Edition. >> /Type /Page Numerical Methods for Engineers 7th Edition Chapra Solutions Manual Apr. /Parent 7 0 R /Rotate 0 This is not the textbook. >> IRJET- A Novel Gabor Feed Forward Network for Pose Invariant Face Recogni Decision Making Using The Analytic Hierarchy Process, FEATool Multiphysics Matlab FEM and CFD Toolbox - v1.6 Quickstart Guide. /CropBox [0 0 612 792] /Parent 7 0 R endobj /Resources 76 0 R Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Uploaded by a239711029 Description: Download full file at https://testbanku.eu/Solution-Manual-for-Numerical-Methods-for-Engineers-7th-Edition-by-Chapra Copyright: All Rights Reserved Available Formats Download as PDF, TXT or read online from Scribd Numerical Methods for Engineers . /Count 10 /CropBox [0 0 612 792] CHAPTER 1 :Mathematical Modeling and Engineering Problem Solving Sample of solution manual for_structural edit two. Numerical Methods for Engineers - 7th Edition - Solutions and Answers | Quizlet Expert solutions Numerical Methods for Engineers 7th Edition Raymond P Canale, Steven C Chapra ISBN: 9780073397924 Textbook solutions Verified Chapter 1: Mathematical Modelling and Engineering Problem Solving Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5 https://gioumeh.com/product/numerical-methods-for-engineers-solution/. endobj download 1 file . /PageLabels 3 0 R << /Parent 7 0 R Solution Manuals to Numerical Methods for Engineers - 5th, 6th, 7th and 8th Edition Author (s): Steven Chapra, Raymond Canale Please note that Solution Manuals for 5th, 6th, 7th and 8th Editions are sold separately. endstream endobj Solutions manual of Micro and Nanoscale Fluid Mechanics Transport in Microflu AIOU Code 1423 Solved Assignment 2 Autumn 2022.pptx, AIOU Code 1429 Solved Assignment 1 Autumn 2022.pptx, Module 4 Developmental Stages in Middle and Late Adolescence.pptx, Atomic Absorption Spectroscopy RESEARCH TECHNIQUES IN ANIMAL NUTRITION.pptx, Intorduction To Production MGT UNIT-1.pptx, Module 6 Coping with Stress in Middle Adolescents.pptx. Each chapter was /Producer (Acrobat Distiller 9.4.0 \(Windows\)) CHAPTER 11 Special Matrices and Gauss-Seidel 300 SIMILAR SOLUTION MANUALS : computational fluid dynamics for incompressible flow : fundamentals of finite element analysis solution manual pdf << /MediaBox [0 0 612 792] /Length 2220 6 0 obj Solutions Manual comes in a PDF or Word format and available for download only. Steven Chapra's Applied Numerical Methods with MATLAB, third edition, is written for engineering and science students who need to learn numerical problem solving. /Resources 70 0 R << Solutions Manual - Numerical Methods in Engineering Practice Authors: Amir W. Al-Khafaji Bradley University John R. Tooley Abstract Solutions Manual For Numerical Methods in Engineering. /Resources 78 0 R >> Numerical Methods for Engineers 7th Edition Chapra Solutions Manual Download free sample - get solutions manual, test bank, quizz, answer key. 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Numerical Methods For Engineers 5th Edition Solution Manual Pdf Free Download If you ally dependence such a referred Numerical Methods For Engineers 5th Edition Solution Manual Pdf Free Download ebook that will pay for you worth, get the entirely best seller from us currently from several preferred authors. /Type /Page 23 0 obj /Rotate 0 /Contents 43 0 R /Resources 48 0 R /Parent 8 0 R /Resources 74 0 R >> /MediaBox [0 0 612 792] CHAPTER 32 :Case Studies: Partial Differential Equations /Annots [38 0 R] /Contents 53 0 R /Resources 68 0 R Numerical Methods for Engineers - 5th Edition - Solutions and Answers | Quizlet Science Engineering Numerical Methods for Engineers 5th Edition ISBN: 9780072918731 Alternate ISBNs Raymond P Canale, Steven C Chapra Textbook solutions Verified Chapter 1: Mathematical Modeling and Engineering Problem Solving Exercise 1 Exercise 2 Exercise 3 Exercise 4 CHAPTER 21 Newton-Cotes Integration Formulas 603 /CropBox [0 0 612 792] /MediaBox [0 0 612 792] << /CropBox [0 0 612 792] Chapra and Canales unique approach opens each part of the text with sections called Motivation, Mathematical Background, and Orientation Each part closes with an Epilogue containing Trade-Offs, Important Relationships and Formulas, and Advanced Methods and Additional References. Much more than a summary, the Epilogue deepens understanding of what has been learned and provides a peek into more advanced methods. /Contents 63 0 R << 9 0 obj /MediaBox [0 0 612 792] /Rotate 0 /Parent 4 0 R /CropBox [0 0 612 792] Solution manual Workshop Precalculus : Discovery with Graphing Calculators - Preliminary Edition (Nancy Baxter Hastings) Solution manual Introductory Mathematics for Engineering Applications - Revised Preliminary Edition (Kuldip S. Rattan & Nathan W. Klingbeil) . << /Kids [10 0 R 11 0 R 12 0 R 13 0 R 14 0 R 15 0 R 16 0 R 17 0 R 18 0 R 19 0 R] CHAPTER 27 :Boundary-Value and Eigenvalue Problems Clipping is a handy way to collect important slides you want to go back to later. 5 0 obj In some cases, you likewise realize not discover the notice Numerical Methods For Engineers 5th Edition Solution Manual Pdf that you are looking for. Thanks. CHAPTER 17 Least-Squares Regression 456 %PDF-1.5 /Kids [7 0 R 8 0 R 9 0 R] /Parent 7 0 R 776. >> % /MediaBox [0 0 612 792] >> Applied Numerical Analysis 7Ed - Curtis F. Gerald, Patrick O. Wheatley - Solutions Manual - Free download as PDF File (.pdf) or read online for free. endobj /Rotate 0 /Contents 75 0 R 7 0 obj CHAPTER 12 Case Studies: Linear Algebraic Equations 319 /Parent 4 0 R 840. /CropBox [0 0 612 792] Tap here to review the details. Numerical Methods for Engineers 7th Edition Chapra Chapra Solutions Manual only NO Test Bank included on this purchase . > if the solution manual is not present just leave a message in the. H. Mathews, Kurtis D. Fink Published: Prentice 2004 Designed by 123DOC, Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ CHAPTER 1.1 We will illustrate two different methods for solving this problem: (1) separation of variables, and (2) Laplace transform dv c g v dt m Separation of variables: Separation of variables gives g c v dv dt m The integrals can be evaluated as c ln g v m t C c/m where C = a constant of integration, which can be evaluated by applying the initial condition to yield c ln g v(0) m C c/m which can be substituted back into the solution c c ln g v ln g v(0) m m t c/m c/m This result can be rearranged algebraically to solve for v, v v(0)e( c / m )t mg e ( c / m )t c where the first part is the general solution and the second part is the particular solution for the constant forcing function due to gravity For the case where, v(0) = 0, the solution reduces to Eq (1.10) v mg e ( c / m )t c Laplace transform solution: An alternative solution is provided by applying Laplace transform to the differential equation to give sV ( s ) v(0) g c V (s) s m Solve algebraically for the transformed velocity PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ V (s) v(0) g s c / m s ( s c / m) (1) The second term on the right of the equal sign can be expanded with partial fractions g A B A( s c / m) Bs s ( s c / m) s s c / m s ( s c / m) (2) By equating like terms in the numerator, the following must hold gA c m As Bs The first equation can be solved for A = mg/c According to the second equation, B = A, so B = mg/c Substituting these back into (2) gives g mg / c mg / c s ( s c / m) s sc/m This can be substituted into Eq to give V (s) v(0) mg / c mg / c sc/m s sc/m Taking inverse Laplace transforms yields v(t ) v(0)e( c / m )t mg mg (c / m )t e c c or collecting terms v(t ) v(0)e( c / m )t mg e ( c / m )t c 1.2 At t = s, the analytical solution is 41.137 (Example 1.1) The relative error can be calculated with absolute relative error analytical numerical 100% analytical The numerical results are: step v(8) 0.5 44.8700 42.8931 41.9901 absolute relative error 9.074% 4.268% 2.073% The error versus step size can then be plotted as PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 10% 8% 6% 4% 2% 0% 0.5 1.5 2.5 Thus, halving the step size approximately halves the error 1.3 (a) You are given the following differential equation with the initial condition, v(t = 0) = 0, dv c' g v2 dt m Multiply both sides by m/c gives m dv m g v2 c ' dt c ' Define a mg / c ' m dv a2 v2 c ' dt Integrate by separation of variables, a dv v2 c' mdt A table of integrals can be consulted to find that a dx x x 1 a a Therefore, the integration yields v c' 1 t C a a m If v = at t = 0, then because tanh1(0) = 0, the constant of integration C = and the solution is v c' 1 t a a m This result can then be rearranged to yield PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ gc ' gm t c' m v (b) Using Eulers method, the first two steps can be computed as 0.22 v(2) 9.81 (0) 19.62 68.1 0.22 v(4) 19.62 9.81 (19.62) 36.75284 68.1 The computation can be continued and the results summarized along with the analytical result as: t v-numerical dv/dt v-analytical 10 12 19.62 36.75284 47.64539 52.59819 54.34314 54.88241 55.10572 9.81 8.56642 5.446275 2.476398 0.872478 0.269633 0.079349 0.022993 18.83093 33.72377 43.46492 49.06977 52.05938 53.58978 55.10572 A plot of the numerical and analytical results can be developed 60 40 20 v-numerical v-analytical 0 12 gm (1 e ( c / m ) t ) c 9.81(70) (1 e (12/70) ) 44.99204 jumper #1: v(t ) 12 9.81(80) (1 e (15/80) t ) jumper #2: 44.99204 15 1.4 v(t ) 44.99204 52.32 52.32e 0.1875 t 0.14006 e 0.1875 t ln 0.14006 0.1875t t ln 0.14006 10.4836 s 0.1875 1.5 Before the chute opens (t < 10), Eulers method can be implemented as PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 10 v(t t ) v(t ) 9.81 v(t ) t 80 After the chute opens (t 10), the drag coefficient is changed and the implementation becomes 60 v(t t ) v(t ) 9.81 v(t ) t 80 Here is a summary of the results along with a plot: Chute closed dv/dt v -20.0000 12.3100 -7.6900 10.7713 3.0813 9.4248 12.5061 8.2467 20.7528 7.2159 27.9687 6.3139 34.2826 5.5247 39.8073 4.8341 44.6414 4.2298 48.8712 3.7011 t t 10 11 12 13 14 15 16 17 18 19 20 Chute opened dv/dt v 52.5723 -29.6192 22.9531 -7.4048 15.5483 -1.8512 13.6971 -0.4628 13.2343 -0.1157 13.1186 -0.0289 13.0896 -0.0072 13.0824 -0.0018 13.0806 -0.0005 13.0802 -0.0001 13.0800 0.0000 60 30 0 10 15 20 -30 1.6 (a) This is a transient computation For the period ending June 1: Balance = Previous Balance + Deposits Withdrawals + Interest Balance = 1522.33 + 220.13 327.26 + 0.01(1522.33) = 1430.42 The balances for the remainder of the periods can be computed in a similar fashion as tabulated below: Date 1-May Deposit Withdrawal Interest $220.13 $327.26 $15.22 $216.80 $378.51 $14.30 $450.35 $106.80 $12.83 $127.31 $350.61 $16.39 1-Jun $1,430.42 1-Jul $1,283.02 1-Aug $1,639.40 1-Sep (b) Balance $1,522.33 $1,432.49 dB D (t ) W (t ) iB dt PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ (c) for t = to 0.5: dB 220.13 327.26 0.01(1522.33) 91.91 dt B (0.5) 1522.33 91.91(0.5) 1476.38 for t = 0.5 to 1: dB 220.13 327.260 0.01(1476.38) 92.37 dt B (0.5) 1476.38 92.37(0.5) 1430.19 The balances for the remainder of the periods can be computed in a similar fashion as tabulated below: Date 1-May 16-May 1-Jun 16-Jun 1-Jul 16-Jul 1-Aug 16-Aug 1-Sep Deposit Withdrawal $220.13 $327.26 $220.13 $327.26 $216.80 $378.51 $216.80 $378.51 $450.35 $106.80 $450.35 $106.80 $127.31 $350.61 $127.31 $350.61 Interest $15.22 $14.76 $14.30 $13.56 $12.82 $14.61 $16.40 $15.36 dB/dt -$91.91 -$92.37 -$147.41 -$148.15 $356.37 $358.16 -$206.90 -$207.94 Balance $1,522.33 $1,476.38 $1,430.19 $1,356.49 $1,282.42 $1,460.60 $1,639.68 $1,536.23 $1,432.26 (d) As in the plot below, the results of the two approaches are very close $1,700 Bi-monthly Monthly $1,600 $1,500 $1,400 $1,300 $1,200 M M J A S 1.7 (a) The first two steps are c(0.1) 100 0.175(100)0.1 98.25 Bq/L c(0.2) 98.25 0.175(98.25)0.1 96.5306 Bq/L The process can be continued to yield t 0.1 0.2 0.3 0.4 0.5 0.6 0.7 c 100.0000 98.2500 96.5306 94.8413 93.1816 91.5509 89.9488 88.3747 dc/dt -17.5000 -17.1938 -16.8929 -16.5972 -16.3068 -16.0214 -15.7410 -15.4656 PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 0.8 0.9 86.8281 85.3086 83.8157 -15.1949 -14.9290 -14.6678 (b) The results when plotted on a semi-log plot yields a straight line 4.6 4.5 4.4 0.2 0.4 0.6 0.8 The slope of this line can be estimated as ln(83.8157) ln(100) 0.17655 Thus, the slope is approximately equal to the negative of the decay rate If we had used a smaller step size, the result would be more exact 1.8 Qstudents 35 ind 80 m J s kJ 20 60 3,360 kJ ind s 1000 J PVMwt (101.325 kPa)(11m 8m 3m 35 0.075 m3 )(28.97 kg/kmol) 314.796 kg RT (8.314 kPa m3 / (kmol K)((20 273.15)K) T Qstudents 3,360 kJ 14.86571 K (314.796 kg)(0.718 kJ/(kg K)) mCv Therefore, the final temperature is 20 + 14.86571 = 34.86571oC 1.9 The first two steps yield 450 450 y (0.5) 3 sin (0) 0.5 (0.36) 0.5 0.18 1250 1250 450 450 y (1) 0.18 3 sin (0.5) 0.5 0.18 ( 0.11176) 0.5 0.23588 1250 1250 The process can be continued to give the following table and plot: t 0.5 1.5 2.5 y 0.00000 -0.18000 -0.23588 -0.03352 0.32378 0.59026 0.60367 dy/dt -0.36000 -0.11176 0.40472 0.71460 0.53297 0.02682 -0.33849 t 5.5 6.5 7.5 8.5 y 1.10271 1.19152 1.05368 0.89866 0.95175 1.24686 1.59543 dy/dt 0.17761 -0.27568 -0.31002 0.10616 0.59023 0.69714 0.32859 PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 3.5 4.5 0.43443 0.32087 0.45016 0.78616 -0.22711 0.25857 0.67201 0.63310 9.5 10 1.75972 1.67144 1.49449 -0.17657 -0.35390 -0.04036 2.0 1.5 1.0 0.5 0.0 -0.5 10 1.10 The first two steps yield 450 150(1 0)1.5 y (0.5) 3 sin (0) 0.5 0.12(0.5) 0.06 1250 1250 450 150(1 0.06)1.5 y (1) 0.06 3 sin (0.5) 0.5 0.06 0.13887(0.5) 0.00944 1250 1250 The process can be continued to give t 0.5 1.5 2.5 3.5 4.5 y 0.00000 -0.06000 0.00944 0.33094 0.77611 1.08058 1.09392 0.92288 0.82934 0.99017 1.33772 dy/dt -0.12000 0.13887 0.64302 0.89034 0.60892 0.02669 -0.34209 -0.18708 0.32166 0.69510 0.56419 t 5.5 6.5 7.5 8.5 9.5 10 y 1.61981 1.63419 1.41983 1.21897 1.25372 1.52584 1.81355 1.87468 1.67396 1.41465 dy/dt 0.02876 -0.42872 -0.40173 0.06951 0.54423 0.57542 0.12227 -0.40145 -0.51860 -0.13062 2.0 1.5 1.0 0.5 0.0 -0.5 10 1.11 When the water level is above the outlet pipe, the volume balance can be written as dV 3sin (t ) 3( y yout )1.5 dt PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ In order to solve this equation, we must relate the volume to the level To this, we recognize that the volume of a cone is given by V = r2y/3 Defining the side slope as s = ytop/rtop, the radius can be related to the level (r = y/s) and the volume can be reexpressed as V 3s y3 which can be solved for y3 3s 2V (1) and substituted into the volume balance 1.5 3s 2V dV 3sin (t ) yout dt (2) For the case where the level is below the outlet pipe, outflow is zero and the volume balance simplifies to dV 3sin (t ) dt (3) These equations can then be used to solve the problem Using the side slope of s = 4/2.5 = 1.6, the initial volume can be computed as V (0) 3(1.6) 0.83 0.20944 m3 For the first step, y < yout and Eq (3) gives dV (0) 3sin (0) dt and Eulers method yields V (0.5) V (0) dV (0)t 0.20944 0(0.5) 0.20944 dt For the second step, Eq (3) still holds and dV (0.5) 3sin (0.5) 0.689547 dt dV V (1) V (0.5) (0.5) t 0.20944 0.689547(0.5) 0.554213 dt Equation (1) can then be used to compute the new level, PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ y3 3(1.6) (0.554213) 10 1.106529 m Because this level is now higher than the outlet pipe, Eq (2) holds for the next step dV 1.5 (1) 2.12422 1.106529 1 2.019912 dt V (1.5) 0.554213 2.019912(0.5) 1.564169 The remainder of the calculation is summarized in the following table and figure t 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10 Qin 0.689547 2.12422 2.984989 2.480465 1.074507 0.059745 0.369147 1.71825 2.866695 2.758607 1.493361 0.234219 0.13883 1.294894 2.639532 2.936489 1.912745 0.509525 0.016943 0.887877 V 0.20944 0.20944 0.554213 1.564169 2.421754 2.570439 1.941885 1.12943 0.93041 1.524207 2.345202 2.671715 2.202748 1.340173 0.902598 1.301258 2.136616 2.659563 2.406237 1.577279 0.943467 y 0.8 0.8 1.106529 1.563742 1.809036 1.845325 1.680654 1.40289 1.31511 1.55031 1.78977 1.869249 1.752772 1.48522 1.301873 1.470703 1.735052 1.866411 1.805164 1.568098 1.321233 Qout 0 0.104309 1.269817 2.183096 2.331615 1.684654 0.767186 0.530657 1.224706 2.105581 2.431294 1.95937 1.013979 0.497574 0.968817 1.890596 2.419396 2.167442 1.284566 0.5462 dV/dt 0.689547 2.019912 1.715171 0.29737 -1.25711 -1.62491 -0.39804 1.187593 1.641989 0.653026 -0.93793 -1.72515 -0.87515 0.79732 1.670715 1.045893 -0.50665 -1.65792 -1.26762 0.341677 2.5 1.5 0.5 0 V 10 y 1.12 (a) The force balance can be written as: m dv R2 mg (0) cd v v dt ( R x) Dividing by mass gives c dv R2 g (0) dvv dt m ( R x) PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 13 This result can then be substituted into the original differential equation, dV 3V k 4 dt 4 2/3 (4) The initial volume can be computed with Eq (1), V 4 r 4 (2.5)3 65.44985 mm3 3 Eulers method can be used to integrate Eq (4) For the first step, the result is dV 3(65.44985) (0) t 65.44985 0.08(4) dt 4 65.44985 6.28319(0.25) 63.87905 V (0.25) V (0) 2/3 0.25 Here are the beginning and ending steps t 0.25 0.5 0.75 V 65.44985 63.87905 62.33349 60.81296 59.31726 dV/dt -6.28319 -6.18225 -6.08212 -5.98281 -5.8843 23.35079 22.56063 21.7884 21.03389 20.2969 -3.16064 -3.08893 -3.01804 -2.94795 -2.87868 9.25 9.5 9.75 10 A plot of the results is shown below We have included the radius on this plot (dashed line and right scale): 80 60 40 20 V 2.4 r 1.6 10 Eq (2) can be used to compute the final radius as r 3(20.2969) 1.692182 4 Therefore, the average evaporation rate can be computed as k mm (2.5 1.692182) mm 0.080782 10 min which is approximately equal to the given evaporation rate of 0.08 mm/min PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 14 1.14 The first two steps can be computed as T (1) 70 0.019(70 20) 68 (0.95)2 68.1 T (2) 68.1 0.019(68.1 20) 68.1 (0.9139)2 66.2722 The remaining results are displayed below along with a plot of the results t 10 T 70.00000 68.10000 66.27220 64.51386 62.82233 61.19508 dT/dt -0.95000 -0.91390 -0.87917 -0.84576 -0.81362 -0.78271 t 12.00000 14.00000 16.00000 18.00000 20.00000 T 59.62967 58.12374 56.67504 55.28139 53.94069 dT/dt -0.75296 -0.72435 -0.69683 -0.67035 -0.64487 80 70 60 50 10 15 20 1.15 The pair of differential equations to be solved are di R i q dt L CL dq i dt or substituting the parameters di 40i 2, 000q dt dq i dt The first step can be implemented by first using the differential equations to compute the slopes di 40(0) 2, 000(1) 2, 000 dt dq 0 dt Then, Eulers method can be applied as i (0.01) 2, 000(0.01) 20 q(0.01) 0(0.01) PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 15 For the second step di 40(20) 2, 000(1) 1, 200 dt dq 20 dt i (0.02) 20 1, 200(0.01) 32 q(0.02) 20(0.01) 0.8 The remaining steps are summarized in the following table and plot: t i q di/dt dq/dt 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 -20 -32 -35.2 -30.72 -20.992 -9.0112 2.37568 11.01005 15.71553 16.33681 1 0.8 0.48 0.128 -0.1792 -0.38912 -0.47923 -0.45548 -0.34537 -0.18822 -2000 -1200 -320 448 972.8 1198.08 1138.688 863.4368 470.5485 62.12813 -277.034 -20 -32 -35.2 -30.72 -20.992 -9.0112 2.37568 11.01005 15.71553 16.33681 40 20 0.5 0 0.02 0.04 0.06 0.08 -20 0.1 -0.5 i -40 q -1 1.16 (a) The solution of the differential equation is N N0 et The doubling time can be computed as the time when N = 2N0, 2N N e (20) ln 0.693 0.034657/hr 20 hrs 20 hrs (b) The volume of an individual spherical cell is cell volume d3 (1) PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 16 The total volume is volume d3 N (2) The rate of change of N is defined as dN N dt (3) If N = N0 at t = 0, Eq can be integrated to give N N e t (4) Therefore, substituting (4) into (2) gives an equation for volume volume d3 N e t (5) (c) This equation can be solved for time ln t volume d N0 (6) The volume of a 500 m diameter tumor can be computed with Eq as 65,449,847 Substituting this value along with d = 20 m, N0 = and = 0.034657/hr gives 65,449,847 ln 203 (1) t 278.63 hr 11.6 d 0.034657 (6) 1.17 Continuity at the nodes can be used to determine the flows as follows: Q1 Q2 Q3 0.6 0.4 1.0 m3 s Q10 Q1 1.0 m3 s Q9 Q10 Q2 1.0 0.6 0.4 m3 s Q4 Q9 Q8 0.4 0.3 0.1 m3 s Q5 Q3 Q4 0.4 0.1 0.3 m3 s Q6 Q5 Q7 0.3 0.2 0.1 m3 s Therefore, the final results are PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 0.4 0.6 0.3 0.1 17 0.1 0.4 0.2 0.3 1.18 (a) Substituting Eq (1.10) into Eq (P1.18) gives dx gm (1 e ( c / m )t ) dt c Separation of variables gives x dx gm c t 1 e ( c / m )t dt Integration yields x gm gm t (1 e ( c / m )t ) c c (b) Eulers method can be applied for the first step as dv c 12.5 (0) g v 9.81 9.81 dt m 68.1 dx (0) v dt dv v(2) v(0) (0)t 9.81(2) 19.62 dt dx x(2) x(0) (0)t 0(2) dt For the second step: dv 12.5 (2) 9.81 19.62 6.2087 dt 68.1 dx (0) 19.62 dt v(4) 19.62 6.2087(2) 32.0374 x(4) 19.62(2) 39.24 The remaining steps can be computed in a similar fashion as tabulated below along with the analytical solution: t vnum 0.0000 19.6200 xnum 0.0000 0.0000 dv/dt 9.8100 6.2087 dx/dt 0.0000 19.6200 vanal 0.0000 16.4217 xanal 0.0000 17.4242 PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 10 32.0374 39.8962 44.8700 48.0179 39.2400 103.3147 183.1071 272.8472 18 3.9294 2.4869 1.5739 0.9961 32.0374 39.8962 44.8700 48.0179 27.7976 35.6781 41.1372 44.9189 62.3380 126.2949 203.4435 289.7305 (c) 50 400 vnum vanal xnum xanal 40 30 300 200 20 100 10 0 10 1.19 (a) For the constant temperature case, Newtons law of cooling is written as dT 0.12(T 10) dt The first two steps of Eulers methods are dT (0) t 37 0.12(10 37)(0.5) 37 3.2400 0.50 35.3800 dt T (1) 35.3800 0.12(10 35.3800)(0.5) 35.3800 3.0456 0.50 33.8572 T (0.5) T (0) The remaining calculations are summarized in the following table: t 0:00 0:30 1:00 1:30 2:00 2:30 3:00 3:30 4:00 4:30 5:00 Ta 10 10 10 10 10 10 10 10 10 10 10 T 37.0000 35.3800 33.8572 32.4258 31.0802 29.8154 28.6265 27.5089 26.4584 25.4709 24.5426 dT/dt -3.2400 -3.0456 -2.8629 -2.6911 -2.5296 -2.3778 -2.2352 -2.1011 -1.9750 -1.8565 -1.7451 (b) For this case, the room temperature can be represented as Ta 20 2t where t = time (hrs) Newtons law of cooling is written as dT 0.12(T 20 2t ) dt PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 19 The first two steps of Eulers methods are T (0.5) 37 0.12(20 37)(0.5) 37 2.040 0.50 35.9800 T (1) 35.9800 0.12(19 35.9800)(0.5) 35.9800 2.0376 0.50 34.9612 The remaining calculations are summarized in the following table: t 0:00 0:30 1:00 1:30 2:00 2:30 3:00 3:30 4:00 4:30 5:00 Ta 20 19 18 17 16 15 14 13 12 11 10 dT/dt -2.0400 -2.0376 -2.0353 -2.0332 -2.0312 -2.0294 -2.0276 -2.0259 -2.0244 -2.0229 -2.0215 T 37.0000 35.9800 34.9612 33.9435 32.9269 31.9113 30.8966 29.8828 28.8699 27.8577 26.8462 Comparison with (a) indicates that the effect of the room air temperature has a significant effect on the expected temperature at the end of the 5-hr period (difference = 26.8462 24.5426 = 2.3036oC) (c) The solutions for (a) Constant Ta, and (b) Cooling Ta are plotted below: 40 Constant Ta Cooling Ta 36 32 28 24 0:00 1:00 1.20 (a) dx dy vx vy dt dt 2:00 3:00 4:00 5:00 dvx c vx dt m dv y dt g c vy m (b) The first step, dx t 180(1) 180 dt dy y (1) y (0) t 100 0(1) 100 dt dv 12.5 vx (1) vx (0) x t 180 180(1) 147.8571 dt 70 dv y 12.5 (0) (1) 9.81 t 9.81 v y (1) v y (0) 70 dt x(1) x(0) PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 20 The second step x(2) 180 147.8571(1) 327.8571 y (1) 100 9.81(1) 90.19 12.5 vx (1) 147.8571 147.8571(1) 121.4541 70 12.5 v y (1) 9.81 9.81 (9.81) (1) 17.8682 70 These along with the remaining results can be tabulated as t x y vx vy dx/dt dy/dt dvx/dt dvy/dt 10 0.0000 180.0000 327.8571 449.3112 549.0771 631.0276 698.3441 753.6398 799.0613 836.3718 867.0197 -100.0000 -100.0000 -90.1900 -72.3218 -47.8343 -17.9096 16.4814 54.5411 95.6145 139.1633 184.7456 180.0000 147.8571 121.4541 99.7659 81.9505 67.3165 55.2957 45.4215 37.3105 30.6479 25.1751 0.0000 9.8100 17.8682 24.4875 29.9247 34.3910 38.0598 41.0734 43.5488 45.5823 47.2526 180.0000 147.8571 121.4541 99.7659 81.9505 67.3165 55.2957 45.4215 37.3105 30.6479 25.1751 0.0000 9.8100 17.8682 24.4875 29.9247 34.3910 38.0598 41.0734 43.5488 45.5823 47.2526 -32.1429 -26.4031 -21.6882 -17.8153 -14.6340 -12.0208 -9.8742 -8.1110 -6.6626 -5.4728 -4.4955 9.8100 8.0582 6.6192 5.4372 4.4663 3.6687 3.0136 2.4755 2.0334 1.6703 1.3720 (c) The following plot indicates that the jumper will hit the ground in about t = 5.6 s at about x = 670 m -150 y versus x -100 -50 200 400 600 800 50 100 150 -150 y versus t -100 -50 50 100 150 1.21 (a) The force balance can be written as m dv mg v v ACd dt PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 21 Dividing by mass gives ACd dv g vv dt 2m (1) The mass of the sphere is sV where V = volume (m3) The area and volume of a sphere are d2/4 and d3/6, respectively Substituting these relationships gives Cd dv g vv dt 4d s dx v dt (b) The first step for Eulers method is dv 3(1.3)0.47 9.81 ( 40) 40 10.0363 4(1.2)2700 dt dx 40 dt v 40 10.0363(2) 19.9274 dx 100 40(2) 20 dt The remaining steps are shown in the following table: t 10 12 14 x 100.0000 20.0000 -19.8548 -20.2450 18.6049 96.4813 212.7399 366.3269 v -40.0000 -19.9274 -0.1951 19.4249 38.9382 58.1293 76.7935 94.7453 dx/dt -40.0000 -19.9274 -0.1951 19.4249 38.9382 58.1293 76.7935 94.7453 dv/dt 10.0363 9.8662 9.8100 9.7566 9.5956 9.3321 8.9759 8.5404 (c) The results can be graphed as (notice that we have reversed the axis for the distance, x, so that the negative elevations are upwards 120 v 80 40 0 -40 10 15 v PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ -100 22 x 10 15 100 200 300 x 400 (d) Inspecting the differential equation for velocity (Eq 1) indicates that the bulk drag coefficient is c' ACd Therefore, for this case, because A = (1.2)2/4 = 1.131 m2, the bulk drag coefficient is c' 1.3(1.131)0.47 kg 0.3455 m 1.22 (a) A force balance on a sphere can be written as: m dv Fgravity Fbuoyancy Fdrag dt where Fgravity mg Fbuoyancy Vg Fdrag 3 dv Substituting the individual terms into the force balance yields m dv mg Vg 3 dv dt Divide by m dv Vg 3 dv g dt m m Note that m = sV, so dv g 3 dv g dt s sV The volume can be represented in terms of more fundamental quantities as V = d3/6 Substituting this relationship into the differential equation gives the final differential equation PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ dv g 1 dt s 23 18 v s d (b) At steady-state, the equation is g 1 s 18 v s d which can be solved for the terminal velocity v g s d 18 This equation is sometimes called Stokes Settling Law (c) Before computing the result, it is important to convert all the parameters into consistent units For the present problem, the necessary conversions are d 10 m s 2.65 m 105 m 10 m g 106 cm3 g kg 2650 cm3 m3 10 kg m 1 g 106 cm3 g kg 1000 3 cm m 10 kg m 0.014 g 100 cm kg kg 0.0014 cm s m 1000 g ms The terminal velocity can then computed as v 9.81 2650 1000 m (1 105 ) 6.42321 105 18 0.0014 s (d) The Reynolds number can be computed as Re dv 1000(105 )6.42321 105 0.0004588 0.0014 This is far below 1, so the flow is very laminar (e) Before implementing Eulers method, the parameters can be substituted into the differential equation to give dv 18(0.0014) 1000 9.81 v 6.108113 95,094v 2650 2650(0.00001) dt The first two steps for Eulers method are v(3.8147 106 ) (6.108113 95,094(0)) 3.8147 106 2.33006 105 v(7.6294 106 ) 2.33006 105 (6.108113 95,094(2.33006 105 )) 3.8147 10 6 3.81488 105 The remaining steps can be computed in a similar fashion as tabulated and plotted below: PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 24 t v dv/dt t v dv/dt 3.81106 6 7.6310 5 1.1410 1.53105 1.91105 2.33E-05 3.81E-05 4.76E-05 5.36E-05 5.75E-05 6.108113 3.892358 2.480381 1.580608 1.007233 0.641853 2.29105 2.67105 3.05105 3.43105 3.81105 5.99E-05 6.15E-05 6.25E-05 6.31E-05 6.35E-05 0.409017 0.260643 0.166093 0.105842 0.067447 1.23 (a) A force balance on a sphere can be written as: m dv mg Vg v v ACd dt (b) Dividing by mass gives Vg ACd dv vv g dt m 2m The mass of the sphere is sV where V = volume (m3) The area and volume of a sphere are d2/4 and d3/6, respectively Substituting these relationships gives dv 3 Cd g 1 vv dt s s d (c) At steady state, for a sphere falling downward Cd g 1 v s 4s d which can be solved for v g s d 1 Cd s Substituting the parameters gives v 4(9.81)2700(0.01) 1000 m 1 0.68783 3(1000)0.47 2700 s (d) Before implementing Eulers method, the parameters can be substituted into the differential equation to give dv 1000 3(1000)0.47 9.811 v 6.176667 13.055556v dt 2700 4(2700)(0.01) The first two steps for Eulers method are PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 25 v(0.03125) (6.176667 13.055556(0) )0.03125 0.193021 v(0.0625) 0.193021 (6.176667 13.055556(0.193021) )0.03125 0.370841 The remaining steps can be computed in a similar fashion as tabulated and plotted below: t v dv/dt t v dv/dt 0.03125 0.0625 0.09375 0.125 0.000000 0.193021 0.370841 0.507755 0.595591 6.176667 5.690255 4.381224 2.810753 1.545494 0.15625 0.1875 0.21875 0.25 0.643887 0.667761 0.678859 0.683860 0.763953 0.355136 0.160023 0.071055 0.0625 0.125 0.8 0.6 0.4 0.2 0.0 0.1875 0.25 1.24 Substituting the parameters into the differential equation gives dy 10000 x3 12(4) x 12(4)2 x dx 24(2 1011 )0.000325 2.5641 105 x3 12 x 48 x The first step of Eulers method is dy 2.5641105 (0)3 12(0) 48(0) dx y (0.125) 0(0.125) The second step is dy 2.5641 10 5 (0.125)3 12(0.125) 48(0.125) 0.000149 dx y (0.25) 0.000149(0.125) 1.86361 105 The remainder of the calculations along with the analytical solution are summarized in the following table and plot Note that the results of the numerical and analytical solutions are close x 0.125 0.25 0.375 y-Euler 0 1.86E-05 5.47E-05 dy/dx 0.000149 0.000289 0.00042 y-analytical x y-Euler dy/dx y-analytical 2.125 0.001832 0.001472 0.001925 9.42E-06 2.25 0.002016 0.001504 0.002111 3.69E-05 2.375 0.002204 0.001531 0.002301 8.13E-05 2.5 0.002395 0.001554 0.002494 PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ 0.5 0.625 0.75 0.875 1.125 1.25 1.375 1.5 1.625 1.75 1.875 0.000107 0.000175 0.000257 0.000352 0.000459 0.000578 0.000707 0.000845 0.000992 0.001147 0.00131 0.001478 0.001653 0.000542 0.000655 0.000761 0.000859 0.000949 0.001032 0.001108 0.001177 0.00124 0.001298 0.001349 0.001395 0.001436 26 0.000141 0.000216 0.000305 0.000406 0.000519 0.000643 0.000777 0.00092 0.001071 0.00123 0.001395 0.001567 0.001744 2.625 0.00259 0.001574 2.75 0.002787 0.001591 2.875 0.002985 0.001605 0.003186 0.001615 3.125 0.003388 0.001624 3.25 0.003591 0.00163 3.375 0.003795 0.001635 3.5 0.003999 0.001638 3.625 0.004204 0.00164 3.75 0.004409 0.001641 3.875 0.004614 0.001641 0.004819 0.001641 0.00269 0.002887 0.003087 0.003288 0.003491 0.003694 0.003898 0.004103 0.004308 0.004513 0.004718 0.004923 0.001 0.002 0.003 y-Euler y-analytical 0.004 0.005 0.006 1.25 [Note that students can easily get the underlying equations for this problem off the web] The volume of a sphere can be calculated as Vs r 3 The portion of the sphere above water (the cap) can be computed as Va h2 3r h Therefore, the volume below water is h2 Vs r 3r h 3 Thus, the steady-state force balance can be written as 4 h2 3 s g r f g r 3r h Cancelling common terms gives 4 s r f r 3 h2 3r h Collecting terms yields PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at https://TestbankDirect.eu/ f h3 r f h s f 27 43 r 1.26 [Note that students can easily get the underlying equations for this problem off the web] The total volume of a right circular cone can be calculated as Vt r22 H The volume of the frustum below the earths surface can be computed as Vb H h1 r r22 r1r2 Archimedes principle says that, at steady state, the downward force of the whole cone must be balanced by the upward buoyancy force of the below ground frustum, H h1 2 r2 Hg g r1 r2 r1r2 g b 3 (1) Before proceeding we have too many unknowns: r1 and h1 So before solving, we must eliminate r1 by recognizing that using similar triangles (r1/h1 = r2/H) r2 h1 H r1 which can be substituted into Eq (1) (and cancelling the gs) H h1 r2 2 r22 h1 r2 h1 b r2 H g H 3 H Therefore, the equation now has only unknown: h1, and the steady-state force balance can be written as 4 h2 3 s g r f g r 3r h Cancelling common terms gives 4 s r f r 3 h2 3r h and collecting terms yields f h3 r f h s f 43 r PROPRIETARY MATERIAL The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation If you are a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at a student using this Manual, you are using it without permission Full file at https://TestbankDirect.eu/ Solution Manual for Numerical Methods for Engineers 7th Edition by Chapra Full file at, B su tp, Thnh vin, ng k, link full download solution manual for negotiation 7th edition by roy lewicki and david saunders and bruce barry, Solution manual for compensation 5th edition by milkovich newman gerhart yap, link full download solution manual for accounting 9th edition by hoggett, Solution manual for compensation 12th edition by newman gerhart milkovich, Download solution manual for 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