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Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
Numerical methods for engineers 5th e-solution ch 12
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Numerical methods for engineers 5th e-solution ch 12

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  • 1. 1CHAPTER 1212.1 Flow balances can be used to determine Q01 = 6 Q15 = 3 Q12 = 4 Q31 = 1 Q03 = 8 Q25 = 1 Q23 = 1 Q54 = 2 Q55 = 2 Q24 = 2 Q34 = 8 Q44 = 12 Mass balances can be used to determine the following simultaneous equations,  7 0 − 1 0 0   c1   240  − 4 4 0 0 0   c2   0       0 − 1 9 0 0   c3  =  80   0 − 2 − 8 12 − 2  c   0   − 3 − 1 0 0 4   c4   0    5    The solution and the matrix inverse can then be developed. For example, using MATLAB, >> A=[7 0 -1 0 0; -4 4 0 0 0; 0 -1 9 0 0; 0 -2 -8 12 -2; -3 -1 0 0 4]; >> B=[240;0;80;0;0]; >> C=AB C = 36.1290 36.1290 12.9032 20.6452 36.1290 >> inv(A) ans = 0.1452 0.0040 0.0161 0 0 0.1452 0.2540 0.0161 0 0 0.0161 0.0282 0.1129 0 0 0.0591 0.0722 0.0806 0.0833 0.0417 0.1452 0.0665 0.0161 0 0.2500 − −112.2 The relevant coefficients of the matrix inverse are a131 = 0.018868 and a 43 = 0.087479. Therefore a 25% change in the input to reactor 3 will lead to the following concentration changes to reactors 1 and 4: ∆c1 = 0.018868(0.25 ×160) = 0.754717PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 2. 2 ∆c 4 = 0.087479(0.25 ×160) = 3.499142 These can be expressed as percent changes, ∆ c1 0.754717 × 100% = × 100% = 6.56% c1 11.50943 ∆c 4 3.499142 × 100% = × 100% = 20.59% c4 16.9982812.3 Because of conservation of flow: Q01 + Q03 = Q 44 + Q5512.4 Mass balances can be used to determine the following simultaneous equations,  8 0 − 3 0 0   c1   50   − 4 4 0 0 0   c2   0       0 − 2 10 0 0   c3  = 160  0 0 − 7 10 − 3  c 4   0  − 4 − 2 0 0 6  c   0    5    The solution can then be developed. For example, using MATLAB, >> A=[8 0 -3 0 0; -4 4 0 0 0; 0 -2 10 0 0; 0 0 -7 10 -3; -4 -2 0 0 6]; >> B=[50;0;160;0;0]; >> C=AB C = 13.2432 13.2432 18.6486 17.0270 13.243212.5 Flow balances can be used to determine Q01 = 5 Q15 = 3 Q12 = 0 Q31 = −2 Q03 = 8 Q25 = 0 Q23 = −7 Q54 = 0 Q55 = 3 Q24 = 7 Q34 = 3 Q44 = 10 Mass balances can be used to determine the following simultaneous equations,PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 3. 3 5 0 0 0 0  c1   50   0 7 −7 0 0  c 2   0       − 2 0 10 0 0  c3  = 160  0 − 7 − 3 10 0  c 4   0  − 3 0 0 0  3  c   0   5    The solution can then be developed. For example, using MATLAB, >> A=[5 0 0 0 0; 0 7 -7 0 -1; -2 0 10 0 0; 0 -7 -3 10 0; -3 0 0 0 3]; >> B=[50;0;160;0;0]; >> C=AB C = 10.0000 18.0000 18.0000 18.0000 10.000012.6 Mass balances can be written for each of the reactors as 500 − Q13 c1 − Q12 c1 + Q 21 c 2 = 0 Q12 c1 − Q21 c 2 − Q23 c 2 = 0 200 + Q13 c1 + Q23 c 2 − Q33 c3 = 0 Values for the flows can be substituted and the system of equations can be written in matrix form as  130 − 30 0   c1   500       − 90 90 0  c2  =  0   − 40 − 60 120  c3   200      The solution can then be developed. For example, using MATLAB, >> A=[130 -30 0;-90 90 0;-40 -60 120]; >> B=[500;0;200]; >> C=AB C = 5.0000 5.0000PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 4. 4 5.833312.7 Mass balances can be written for each of the lakes as Superior, c1: 180 = 67c1 Michigan, c2: 710 = 36c 2 Huron, c3: 740 + 67c1 + 36c 2 = 161c3 Erie, c4: 3850 + 161c 3 = 182c 4 Ontario, c5: 4720 + 182c 4 = 212c 5 The system of equations can be written in matrix form as  67 0 0 0 0   c1   180   0 36 0 0 0   c 2   710       − 67 − 36 161 0 0   c3  =  740   0 0 − 161 182 0   c 4   3850  0  0 0 − 182 212  c   4720  5    The solution can then be developed. For example, using MATLAB, >> A=[67 0 0 0 0; 0 36 0 0 0; -67 -36 161 0 0; 0 0 -161 182 0; 0 0 0 -182 212]; >> B=[180 710 740 3850 4720]; >> C=AB C = 2.6866 19.7222 10.1242 30.1099 48.113212.8 (a) The solution can be developed using your own software or a package. For example, using MATLAB, >> A=[13.422 0 0 0; -13.422 12.252 0 0; 0 -12.252 12.377 0; 0 0 -12.377 11.797]; >> W=[750.5 300 102 30];PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 5. 5 >> AI=inv(A) AI = 0.0745 0 0 0 0.0816 0.0816 0 0 0.0808 0.0808 0.0808 0 0.0848 0.0848 0.0848 0.0848 >> C=AI*W C = 55.9157 85.7411 93.1163 100.2373 (b) The element of the matrix that relates the concentration of Havasu (lake 4) to the loading −1 of Powell (lake 1) is a 41 = 0.084767. This value can be used to compute how much the loading to Lake Powell must be reduced in order for the chloride concentration of Lake Havasu to be 75 as ∆c 4 100.2373 − 75 ∆W1 = −1 = = 297.725 a 41 0.084767 (c) First, normalize the matrix to give 1 0 0 0  [ A] = 1 − 0.91283 0 − 0.9899 0 0   1 0 0  0 1 − 0.95314  The column-sum norm for this matrix is 2. The inverse of the matrix can be computed as  1 0 0 0  −1 1.095495 − 1.09549 [ A] =  0 0  1.084431 − 1.08443 1 0  1.137747 − 1.13775 1.049165 − 1.04917    The column-sum norm for the inverse can be computed as 4.317672. The condition number is, therefore, 2(4.317672) = 8.635345. This means that less than 1 digit is suspect [log10(8.635345) = 0.93628]. Interestingly, if the original matrix is unscaled, the same condition number results.12.9 For the first stage, the mass balance can be written as F1 y in + F2 x 2 = F2 x1 + F1 x1 Substituting x = Ky and rearranging givesPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 6. 6  F  F − 1 + 2 K  y1 + 2 Ky 2 = − y in    F1  F1 Using a similar approach, the equation for the last stage is  F  F y 4 − 1 + 2 K  y 5 = − 2 xin    F1  F1 For interior stages,  F  F y i -1 − 1 + 2 K  y i + 2 Ky i +1 = 0    F1  F1 These equations can be used to develop the following system,  9 − 8 0 0 0   y1   0.1  − 1 9 − 8 0 0   y2   0       0 − 1 9 − 8 0   y3  =  0   0 0 − 1 9 − 8  y 4   0   0 0 0 − 1 9  y   0    5    The solution can be developed in a number of ways. For example, using MATLAB, >> format long >> A=[9 -8 0 0 0; -1 9 -8 0 0; 0 -1 9 -8 0; 0 0 -1 9 -8; 0 0 0 -1 9]; >> B=[0.1;0;0;0;0]; >> Y=AB Y = 0.01249966621272 0.00156212448931 0.00019493177388 0.00002403268445 0.00000267029827 Note that the corresponding values of X can be computed as >> X=4*Y X = 0.04999866485086PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 7. 7 0.00624849795722 0.00077972709552 0.00009613073780 0.00001068119309 Therefore, yout = 0.0000026703 and xout = 0.05. In addition, here is a logarithmic plot of the simulation results versus stage, 0 1 2 3 4 5 1 0.1 0.01 0.001 0.0001 0.00001 0.000001 Y X12.10 Steady-state mass balances for A in each reactor can be written as Qin c A,in − Qin c A,1 − k1V1 c A,1 = 0 Qin c A,1 + Q32 c A,3 − (Qin + Q32 )c A, 2 − k 2V 2 c A, 2 = 0 (Qin + Q32 )c A, 2 + Q43 c A, 4 − (Qin + Q43 )c A,3 − k 3V3 c A,3 = 0 (Qin + Q43 )c A,3 − (Qin + Q43 )c A, 4 − k 4V 4 c A, 4 = 0 Steady-state mass balances for B in each reactor can be written as − Qin c B ,1 + k1V1 c A,1 = 0 Qin c B ,1 + Q32 c B ,3 − (Qin + Q32 )c B , 2 + k 2V2 c A, 2 = 0 (Qin + Q32 )c B , 2 + Q43 c B , 4 − (Qin + Q43 )c B ,3 + k 3V3 c A,3 = 0 (Qin + Q 43 )c B ,3 − (Qin + Q 43 )c B , 4 + k 4V4 c A, 4 = 0 Values for the parameters can be substituted and the system of equations can be written in matrix form as  c A,1   11.875 0 0 0 0 0 0 0   c B,1  10  − 1.875 10 0 0 0 0 0 0   0  − 10 0 26.25 0 −5 0 0 0  c A, 2   0   0 − 10 − 11.25 15 0 −5 0 0  c B,2  =  0   0 0 − 15 0 53 0 − 3 0   c A, 3   0   0 0 0 − 15 − 40 13 0 − 3  c   0   0 0 0 0 − 13 0 15.5 0   B,3   0   0  0 0 0 0 − 13 − 2.5 13   c A, 4   0      c B,4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 8. 8 The solution can then be developed. For example, using MATLAB, >> A=[11.875 0 0 0 0 0 0 0; -1.875 10 0 0 0 0 0 0; -10 0 26.25 0 -5 0 0 0; 0 -10 -11.25 15 0 -5 0 0; 0 0 -15 0 53 0 -3 0; 0 0 0 -15 40 13 0 -3; 0 0 0 0 -13 0 15.5 0; 0 0 0 0 0 -13 -2.5 13]; >> B=[10 0 0 0 0 0 0 0]; >> C=AB C = 0.8421 0.1579 0.3400 0.9933 0.1010 1.8990 0.0847 1.9153 Therefore, to summarize the results reactor A B inflow 1 0 1 0.842105 0.157895 2 0.340047 0.993286 3 0.101036 1.898964 4 0.084740 1.915260 Here is a plot of the results: 1 cA cB 0.8 0.6 0.4 0.2 0 0 1 2 3 412.11 Assuming a unit flow for Q1, the simultaneous equations can be written in matrix form asPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 9. 9 − 2 1 2 0 0 0   Q 2   0 0 0 −2 1 2 0   Q3   0 0 0 0 0 − 2 3   Q 4  =  0   1 1 0 0 0 0   Q5   1 0 1 −1 − 1 0 0   Q   0 0  0 0 1 − 1 − 1  Q6   0  7      These equations can then be solved. For example, using MATLAB, >> A=[-2 1 2 0 0 0; 0 0 -2 1 2 0; 0 0 0 0 -2 3; 1 1 0 0 0 0; 0 1 -1 -1 0 0; 0 0 0 1 -1 -1]; >> B=[0 0 0 1 0 0 ]; >> Q=AB Q = 0.5059 0.4941 0.2588 0.2353 0.1412 0.094112.12 The mass balances can be expressed in matrix form as  c G1   2.8 0 0 0 0 − 0.8 0 0 0 0   c   200  − 2 2 .8 0 0 0 0 − 0 .8 0 0 0   G2   0   0 − 2 2 .8 0 0 0 0 − 0 .8 0 0   cG3   0   0 0 − 2 2 .8 0 0 0 0 − 0 .8 0   c G 4   0   0 0 0 − 2 2.8 0 0 0 0 − 0.8  cG 5  =  0   − 0 .8 0 0 0 0 1 .8 −1 0 0 0   c L1   0   0 − 0.8 0 0 0 0 1 .8 −1 0 0 c   0   0 0 − 0 .8 0 0 0 0 1.8 −1 0   cL2   0   0 0 0 − 0 .8 0 0 0 0 1.8 − 1   L3   0       0 0 0 0 − 0 .8 0 0 0 0 1.8   c L 4   10     c L5  These equations can then be solved. The results are tabulated and plotted below: Reactor Gas Liquid 0 100PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 10. 10 1 95.73328 85.06649 2 90.2475 76.53306 3 83.19436 65.5615 4 74.12603 51.45521 5 62.46675 33.31856 6 10 120 Gas 100 Liquid 80 60 40 20 0 0 1 2 3 4 5 6 712.13 Let xi = the volume taken from pit i. Therefore, the following system of equations must hold 0.55 x1 + 0.25 x 2 + 0.25 x3 = 4800 0.30 x1 + 0.45 x 2 + 0.20 x3 = 5800 0.15 x1 + 0.30 x 2 + 0.55 x3 = 5700 These can then be solved for x1 = 2416.667, x2 = 9193.333, and x3 = 4690.12.14 We can number the nodes as 1 F1 F3 F5 2 F2 F4 3 H2 4 V2 V3 Node 1: ΣFH = 0 = −F1 cos 30° − F5 cos 45° + F3 cos 45° + 1200 ΣFV = 0 = −F1 sin 30° − F5 sin 45° − F3 sin 45° − 600 Node 2: ΣFH = 0 = H 2 + F2 + F1 cos 30° ΣFV = 0 = F1 sin 30° + V2 Node 3:PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 11. 11 ΣFH = 0 = −F4 − F3 cos 45° ΣFV = 0 = V3 + F3 sin 45° Node 4: ΣFH = 0 = −F2 + F4 + F5 cos 45° ΣFV = 0 = F5 sin 45° − 500 These balances can then be expressed in matrix form as F   0.866 0 − 0.707 0 0.707 0 0 0   1   1200   0.5 0 − 0.707 0 0.707 0 0 0   F2   − 600  − 0.866 − 1 0 0 0 −1 0 0   F3   0   − 0.5 0 0 0 0 0 −1 0   F4  =  0   0 0 0.707 1 0 0 0 0   F5   0   0 0 − 0.707 0 0 0 0 − 1  H   0   0 1 0 − 1 − 0.707 0 0 0 V 2   0   0  0 0 0 − 0.707 0 0 0   2   − 500  V     3 This system can be solved for F1 = −292.82 F2 = 1453.59 F3 = −1348.58 F4 = 953.5898 F5 = 707.1068 H2 = −1200 V2 = 146.4102 V3 = 953.5898 Note that the horizontal reactions (H2 = −1200) and the vertical reactions (V2 + V3 = 146.4102 + 953.5898 = 1100) are equal to the negative of the imposed loads. This is a good check that the computation is correct.12.15 We can number the nodes as 500 100 F1 1 2 F7 F2 F5 F4 5 F6 F3 3 H5 4 V5 V3 Node 1: ΣFH = 0 = F1 + F5 cos 45° − F7 cos 45° ΣFV = 0 = −F5 sin 45° − F7 sin 45° − 500 Node 2: ΣFH = 0 = −F1 + F2 cos 30° − F4 cos 60° ΣFV = 0 = −F2 sin 30° − F4 sin 60° − 100PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 12. 12 Node 3: ΣFH = 0 = −F2 cos 30° − F3 ΣFV = 0 = V3 + F2 sin 30° Node 4: ΣFH = 0 = F3 + F4 cos 60° − F5 cos 45° − F6 ΣFV = 0 = F4 sin 60° + F5 sin 45° Node 5: ΣFH = 0 = F6 + F7 cos 45° + H 5 ΣFV = 0 = F7 sin 45° + V5 These balances can then be expressed in matrix form as − 1 0 0 0 − 0.707 0 0.707 0 0 0   F1   0  0  0 0 0 0.707 0 0.707 0 0 0   F2  − 500      1 − 0.866 0 0.5 0 0 0 0 0 0   F3   0       0 0.5 0 0.866 0 0 0 0 0 0   F4   − 100  0 0.866 1 0 0 0 0 0 0 0   F5   0         =   0 − 0.5 0 0 0 0 0 −1 0 0   F6   0  0 0 − 1 − 0.5 0.707 1 0 0 0 0   F7   0       0 0 0 − 0.866 − 0.707 0 0 0 0 0   V3   0    0 0 0 0 0 − 1 − 0.707 0 − 1 0  H 5   0      0  0 0 0 0 0 − 0.707 0 0 − 1  V5   0      This system can be solved for F1 = −348.334 F2 = −351.666 F3 = 304.5517 F4 = 87.56443 F5 = −107.244 F6 = 424.167 F7 = −599.863 V3 = 175.833 H5 = 0 V5 = 424.16712.16 The first two columns of the inverse provide the information to solve this problem F1H F1V F1 0.866025 0.500000 F2 0.250000 −0.433013 F3 −0.500000 0.866025 H2 −1.000000 0.000000 V2 −0.433013 −0.250000 V3 0.433013 −0.750000 F1 = 2000(0.866025) − 2500(0.5) = 482.0508 F2 = 2000(0.25) − 2500(−0.433013) = 1582.532 F3 = 2000(−0.5) − 2500(0.866025) = −3165.06 H2 = 2000(−1)− 2500(0) = −2000PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 13. 13 V2 = 2000(−0.433013) − 2500(−0.25) = −241.025 V3 = 2000(0.433013) − 2500(−0.75) = 2741.02512.17 Σ y =0 F V 2 +V3 =1000 ΣM = 0 1000(cos30°) L1 − V3 L2 Geometry cos30°L1 + cos60°L3 = L2 Since V2 = 250 and V3 = 750, 866 L1 − 750 L2 = 0 0.866 L1 + 0.5 L3 = L2 Therefore, L2 − 0.866 L1 L3 = 0 .512.18 We can number the nodes as 1 500 F1 F3 2 F2 250 3 F6 F5 F4 4 F7 H4 5 V4 V5 Node 1: ΣFH = 0 = −F1 cos 45° − 500 ΣFV = 0 = −F1 sin 45° − F3 Node 2: ΣFH = 0 = F1 cos 45° + F2 + F5 cos 60° − F6 cos 30° ΣFV = 0 = F1 sin 45° − F5 sin 60° − F6 sin 30° Node 3: ΣFH = 0 = −F2 − 250 ΣFV = 0 = F3 − F4 Node 4: ΣFH = 0 = F6 cos 30° + F7 + H 4 ΣFV = 0 = F6 sin 30° + V4 Node 5:PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 14. 14 ΣFH = 0 = −F7 − F5 cos 60° ΣFV = 0 = F4 + F5 sin 60° + V5 These balances can then be expressed in matrix form as  0.707 0 0 0 0 0 0 0 0 0   F1  − 500   0.707  0 1 0 0 0 0 0 0 0   F2   0      − 0.707 − 1 0 0 − 0.5 0.866 0 0 0 0   F3   0       − 0.707 0 0 0 0.866 0.5 0 0 0 0   F4   0   0 1 0 0 0 0 0 0 0 0   F5  − 250        =    0 0 −1 1 0 0 0 0 0 0   F6   0   0 0 0 0 0 − 0.866 − 1 − 1 0 0   F7   0        0 0 0 0 0 − 0.5 0 0 − 1 0  H 4   0     0 0 0 0 0.5 0 1 0 0 0   V4   0       0  0 0 − 1 − 0.866 0 0 0 0 − 1  V5   0      This system can be solved for F1 = −707.107 F2 = −250 F3 = 500 F4 = 500 F5 = −58.0127 F6 = −899.519 F7 = 29.00635 H4 = 750 V4 = 449.7595 V5 = −449.7612.19 We can number the nodes as 1 F2 2 F3 3 F6 F1 F5 F8 F7 H4 4 6 F4 5 F9 V4 V6 5000 Node 1: ΣFH = 0 = −F1 cos 60° + F2 + F5 cos 60° ΣFV = 0 = −F1 sin 60° − F5 sin 60° Node 2: ΣFH = 0 = −F2 + F3 ΣFV = 0 = −F8 Node 3: ΣFH = 0 = −F3 + F6 cos 45° − F7 cos 45° ΣFV = 0 = −F6 sin 45° − F7 sin 45°PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 15. 15 Node 4: ΣFH = 0 = F1 cos 30° + F4 + H 4 ΣFV = 0 = F1 sin 60° + V 4 Node 5: ΣFH = 0 = −F4 − F5 cos 60° + F7 cos 45° + F9 ΣFV = 0 = F5 sin 60° + F8 + F7 sin 45° − 5000 Node 6: ΣFH = 0 = −F6 cos 45° − F9 ΣFV = 0 = F6 sin 45° + V6 Note that F8 = 0. Thus, the middle member is unnecessary unless there is a load with a nonzero vertical component at node 2. These balances can then be expressed in matrix form as  0.5 −1 0 0 − 0.5 0 0 0 0 0 0  F1   0   0.866    0 0 0 0.866 0 0 0 0 0 0  F2   0      0 1 −1 0 0 0 0 0 0 0 0  F3   0        0 0 1 0 0 − 0.707 0.707 0 0 0 0  F4   0   0 0 0 0 0 0.707 0.707 0 0 0 0  F5   0        − 0.5 0 0 −1 0 0 0 0 −1 0 0  F6  =  0  − 0.866 0 0 0 0 0 0 0 0 −1 0  F7   0        0 0 0 1 0.5 0 − 0.707 − 1 0 0 0  F9   0        0 0 0 0 − 0.866 0 − 0.707 0 0 0 0 H 4  − 5000  0 0 0 0 0 0.707 0 1 0 0 0  V 4   0        0 0 0 0 0 − 0.707 0 0 0 0 1   V6   0  This system can be solved for F1 = −3660.25 F2 = −3660.25 F3 = −3660.25 F4 = 1830.127 F5 = −3660.25 F6 = −2588.19 F7 = 2588.19 F9 = 1830.13 H4 = 0 V4 = 3169.87 V6 = 1830.1312.20 (a) Room 1: 0 = Wsmoker + Qa c a − Qa c1 + E13 (c 3 − c1 ) Room 2: 0 = Qb cb + (Qa − Qd )c 4 − Qc c 2 + E 24 (c 4 − c 2 ) Room 3: 0 = Wgrill + Q a c1 + E13 (c1 − c 3 ) + E 34 (c 4 − c 3 ) − Q a c 3 Room 4: 0 = Qa c3 + E 34 (c 3 − c 4 ) + E 24 (c 2 − c 4 ) − Qa c 4PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 16. 16 Substituting the parameters yields  225 0 − 25 0   c1  1400   0  175 0 − 125 c 2   100    =       − 225 0 275 − 50  c3  2000    0 − 25 − 250 275  c 4   0      These can be solved for  c1   8.0996  c  12.3448  2    =  c3  16.8966 c 4  16.4828     (b) The matrix inverse can be determined as 0.004996 0.0000153 0.000552 0.000107  [ A] =  0.003448 0.006207 −1 0.004966 0.000138 0.003448 0.003448 0.004966 0.000966  0.004828 0.00069  0.004828 0.004828 The percent of the carbon monoxide in the kids’ section due to each source can be computed as (i) the smokers −1 c 2,smokers = a 21 Wsmokers = 0.003448(1000) = 3.448 3.448 % smokers = ×100% = 27.93% 12.3448 (ii) the grill −1 c 2,grill = a 31 Wgrill = 0.003448(2000) = 6.897 6.897 % grill = ×100% = 55.87% 12.3448 (iii) the intakes −1 −1 c 2,intakes = a 21 Qa c a + a 22 Qb c b = 0.003448( 200) 2 + 0.006207(50) 2 =1.37931 + 0.62069 = 2 2 % grill = ×100% =16.20% 12.3448 (c) If the smoker and grill loads are increased by 1000 and 3000 mg/hr, respectively, the concentration in the kids’ section will be increased byPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 17. 17 −1 −1 ∆c 2 = a 21 ∆Wsmoker + a 23 ∆Wgrill = 0.003448( 2000 −1000) + 0.003448(5000 − 2000) = 3.448 +10.3448 =13.7931 (d) If the mixing between the kids’ area and zone 4 is decreased to 5, the system of equations is changed to  225 0 − 25 0   c1  1400   0  155 0 − 105 c 2   100    =       − 225 0 275 − 50  c3  2000    0 − 5 − 250 255  c 4   0      which can be solved for  c1   8.1084  c  12.0800  2    =  c 3  16.9760 c 4  16.8800     Therefore, the concentration in the kids’ area would be decreased 0.26483 mg/m 3 or 2.145%.12.21 The coordinates of the connection points are D: (0, 0, 2.4) A: (0.8, –0.6, 0) B: (–0.8, –0.6, 0) C: (0, 1, 0) The lengths of the legs can be computed as DA = 0.8 2 + 0.6 2 + 2.4 2 = 2.6 DB = 0.8 2 + 0.6 2 + 2.4 2 = 2.6 DC = 0 2 +12 + 2.4 2 = 2.6 Assume that each leg is in tension, which mean that each pulls on point D. 20 kN D A B CPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 18. 18 The direction cosines of the vectors A, B and C can be determined as 4 3 12  A:  , − , −   13 13 13   4 3 12  B:  − , − , −   13 13 13   5 12  C:  0, ,−   13 13  Force balances for point D can then be written as 4 4 ∑F x = 13 A− 13 B =0 3 3 5 ∑F y =− 13 A− 13 B + C =0 13 12 12 12 ∑F z =− 13 A− 13 B− 13 C + 20 = 0 Thus, the solution amounts to solving the following system of linear algebraic equations 0.30769 A − 0.30769 B =0 − 0.23077 A − 0.23077 B + 0.38462C = 0 − 0.92308 A − 0.92308 B − 0.92308C = −20 These equations can be solved with Gauss elimination for A = 6.7708, B = 6.7708, and C = 8.125.12.22 The solution can be generated in a number of ways. For example, using MATLAB, >> A=[1 0 0 0 0 0 0 0 1 0; 0 0 1 0 0 0 0 1 0 0; 0 1 0 3/5 0 0 0 0 0 0; -1 0 0 -4/5 0 0 0 0 0 0; 0 -1 0 0 0 0 3/5 0 0 0; 0 0 0 0 -1 0 -4/5 0 0 0; 0 0 -1 -3/5 0 1 0 0 0 0; 0 0 0 4/5 1 0 0 0 0 0; 0 0 0 0 0 -1 -3/5 0 0 0; 0 0 0 0 0 0 4/5 0 0 1]; >> B=[0 0 -74 0 0 24 0 0 0 0]; >> x=AB x = 37.3333 -46.0000 74.0000PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 19. 19 -46.6667 37.3333 46.0000 -76.6667 -74.0000 -37.3333 61.3333 Therefore, in kN AB = 37.3333 BC = −46 AD = 74 BD = −46.6667 CD = 37.3333 DE = 46 CE = −76.6667 Ax = −74 Ay = −37.33333 Ey = 61.333312.23 The simultaneous equations are 1 1 1 0 0 0   i12   0  0 −1 0 1 − 1 0   i52   0  0 0 −1 0 0 1   i32  =  0    0 0 0 0 1 − 1  i65   0   0 5 − 15 0 − 5 − 2  i   0  10 − 5 0 − 25 0 0   54   200    i 43      This system can be solved in a number of ways. For example, using MATLAB, >> A=[1 1 1 0 0 0; 0 -1 0 1 -1 0; 0 0 -1 0 0 1; 0 0 0 0 1 -1; 0 5 -15 0 -5 -2; 10 -5 0 -25 0 0]; >> B=[0 0 0 0 0 200]; >> I=AB I = 5.1185 -4.1706 -0.9479 -5.1185 -0.9479 -0.9479 i21 = 5.1185 i52 = −4.1706 i32 = −0.9479 i65 = −5.1185 i54 = −0.9479 i43 = −0.9479 Here are the resulting currents superimposed on the circuit:PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 20. 20 0.9479 5.118512.24 The current equations can be written as − i 21 − i 23 + i52 = 0 i 23 − i35 + i 43 = 0 − i 43 + i54 = 0 i35 − i52 + i65 − i54 = 0 Voltage equations: V2 − 10 V5 − V 4 i 21 = i54 = 35 15 V 2 − V3 V 3 − V5 i 23 = i35 = 30 7 V 4 − V3 V5 − V 2 i 43 = i52 = 8 10 150 − V5 i65 = 5 − 1 − 1 1 0 0 0 0 0 0  i 21   0  0 0 0    1 0 −1 1 0 0 0 0 0 0  i 23   0     0 0 0 0 −1 1 0 0 0 0 0  i52   0       0 0 −1 1 0 −1 1 0 0 0 0  i35   0   35 0 0 0 0 0 0 −1 0 0 0  i 43  − 10       0 30 0 0 0 0 0 −1 1 0 0  i54  =  0  0 0 0 0 8 0 0 0 1 − 1 0  i65   0       0 0 0 0 0 15 0 0 0 1 − 1 V2   0       0 0 0 7 0 0 0 0 −1 0 1  V3   0  0 0 10 0 0 0 0 1 0 0 − 1 V4   0       0 0 0 0 0 0 5 0 0 0 1  V5  150 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 21. 21 This system can be solved for i21 = 2.9291 i23 = −0.6457 i52 = 2.2835 i35 = −0.4950 i43 = 0.1507 i54 = 0.1507 i65 = 2.9291 V2 = 112.5196 V3 = 131.8893 V4 = 133.0945 V5 = 135.354312.25 The current equations can be written as i32 − i 25 + i12 = 0 − i32 − i34 + i63 = 0 i34 − i 47 = 0 i 25 + i 65 − i58 = 0 i76 − i63 − i65 = 0 i 47 − i76 + i97 = 0 i58 − i89 − i80 = 0 i89 − i97 = 0 Voltage equations: − 20i 25 + 10i 65 − 5i 63 − 5i32 = 0 5i63 + 20i76 + 5i47 + 20i34 = 0 − 50i58 − 15i89 − 0i97 − 20i 76 − 10i65 = 0 120 − 20i 25 − 50i58 = 40  i32   1 −1 1 0 0 0 0 0 0 0 0 0   i25   0  − 1 0 0 − 1 1 0 0 0 0 0 0 0 i   0  0 0 0 1 0 −1 0 0 0 0 0 0   12   0  0 1 0 0 0 0 1 −1 0 0 0 0   i34   0   0 0 0 0 −1 0 −1 0 1 0 0 0   i63   0  0 0 0 0 0 1 0 0 − 1 1 0 0   i 47  =  0  0 0 0 0 0 0 0 1 0 0 − 1 − 1  i65   0  0 0 0 0 0 0 0 0 0 − 1 1 0  i   0   5 20 0 0 5 0 − 10 0 0 0 0 0   i58   0   0 0 0 − 20 − 5 − 5 0 0 − 20 0 0 0   76   0  0 0 0 0     0 0 10 50 20 0 15 0   i97   0   0 20 0 0 0 0 0 50 0 0 0 0   i89   80    i80  This system can be solved forPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 22. 22 i32 = −2.5670 i25 = 1.0449 i12 = 3.6119 i34 = 1.2287 i63 = −1.3384 i47 = 1.2287 i65 = 0.1371 i58 = 1.1820 i76 = −1.2012 i97 = −2.4299 i89 = −2.4299 i80 = 3.611912.26 Let ci = component i. Therefore, the following system of equations must hold 15c1 + 17c 2 + 19c 3 = 3890 0.30c1 + 0.40c 2 + 0.55c 3 = 95 1.0c1 + 1.2c 2 + 1.5c 3 = 282 These can then be solved for c1 = 90, c2 = 60, and c3 = 80.12.27 First, we can number the loops and assume that the currents are clockwise. R1 R2 R3 R4 R5 + i3 + V1 − i1 i2 − V2 Kirchhoff’s voltage law can be applied to each loop. − V1 + R1i1 + R 4 (i1 − i 2 ) = 0 R 4 (i 2 − i1 ) + R2 i 2 + R5 (i 2 − i3 ) = 0 R5 (i3 − i 2 ) + R3 i3 + V2 = 0 Collecting terms, the system can be written in matrix form as  20 − 15 0   i1   80       − 15 50 − 25  i 2  =  0   0 − 25 45   i3   − 50      This can be solved with a tool like MATLAB, > A=[20 -15 0;-15 50 -25;0 -25 45]; >> B=[80;0;-50]; >> I=AB I = 4.9721 1.2961 -0.3911PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 23. 23 Therefore, I1 = 4.9721, I2 = 1.2961, and I3 = –0.3911.12.28 This problem can be solved by applying Kirchhoff’s voltage law to each loop. − 20 + 4(i1 − i 2 ) + 2(i1 − i3 ) = 0 4(i 2 − i1 ) + 6i 2 + 8(i 2 − i3 ) = 0 8(i3 − i 2 ) + 5i3 + 2(i3 − i1 ) = 0 Collecting terms, the system can be written in matrix form as  6 − 4 − 2  i1   20      − 4 18 − 8  i 2  =  0   − 2 − 8 15   i3   0       This can be solved with a tool like MATLAB, > A=[6 -4 -2;-4 18 -8;-2 -8 15]; >> B=[20;0;0]; >> I=AB I = 5.1759 1.9095 1.7085 Therefore, I1 = 5.1759, I2 = 1.9095, and I3 = 1.7085.12.29 This problem can be solved directly on a calculator capable of doing matrix operations or on MATLAB. >> b=[-200;-250;100]; >> a=[55 0 -25;0 -37 -4;-25 -4 29]; >> b=[-200;-250;100]; >> x=ab x = -2.7278 6.5407 1.9989 Therefore, I1 = −2.7278 A, I3 = 6.5407 A, and I4 = 1.9989 A12.30 This problem can be solved directly on a calculator capable of doing matrix operations or on MATLAB. >> a=[60 -40 0 -40 150 -100 0 -100 130]; >> b=[200PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 24. 24 0 230]; >> x=ab x = 7.7901 6.6851 6.9116 Therefore, I1 = 7.79 A, I2 = 6.69 A, and I3 = 6.91 A.12.31 At steady state, the force balances can be written as 4kx1 − 3kx 2 = m1 g − 3kx1 + 4kx 2 − kx3 = m 2 g − kx 2 + kx3 = m3 g Substituting the parameter values  120 − 90 0   x1  19.6       − 90 120 − 30  x 2  =  29.4  0 − 30 30   x3   24.5      The solution is x1 = 2.45, x2 = 3.049, and x3 = 3.866.12.32 At steady state, the force balances can be written as  30 − 20 0   x1   98       − 20 30 − 10  x 2  =  34.3  0 − 10 − 10  x3  19.6       The solution is x1 = 15.19, x2 = 17.885, and x3 = 19.845.12.33 The force balances can be written as  k1 + k 2 − k 2 0 0   x1   0   − k 2 k 2 + k3 − k3 0   x2   0  =  0 − k 3 k 3 + k 4 − k 4   x3   0        0 0 − k4 k 4   x4   F  Substituting the parameter valuesPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 25. 25  150 − 50 0 0   x1   0   − 50 130 − 80 0   x2  =  0   0 − 80 280 − 200  x3   0   0  0 − 200 200   x  1500(9.8)  4    The solution is x1 = 147, x2 = 441, x3 = 624.75, and x4 = 698.25.12.34 The equations can be solved in a number of ways. For example, using MATLAB, >> A=[100 1 0;50 -1 1;25 0 -1]; >> B=[519.72;216.55;108.27]; >> x=AB x = 4.8259 37.1257 12.3786 Therefore, a = 4.8259, T = 37.1257, and R = 12.3786.12.35 In order to solve this problem, we must assume the direction that the blocks are moving. For example, we can assume that the blocks are moving from left to right as shown Force balances can be written for each block: T 392(sin 30o) = 196 392(cos 30o)0.2 = 67.896 40 × 9.8 = 392 –196 – 67.896 + T = 40aPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 26. 26 R T 98(cos 30o)0.5 = 42.435 98(sin 30o) = 49 10 × 9.8 = 98 –49 – T + R – 42.435 = 10a R 73.5 424.352 50 × 9.8 = 490 424.352 − 73.5 – R = 50a Therefore, the system of equations to be solved can be written in matrix form as  40 − 1 0   a   − 263.8964     10 1 − 1 T  =  − 91.43524  50 0 1   R   350.852       The solution is a = –0.044792, T = 262.1047, and R = 353.092. Note that if we had assumed that the blocks were moving from right to left, the system of equations would have been  40 1 0   a   128.1036      10 − 1 1  T  =  6.564755  50 0 − 1  R   − 497.852      The solution for this case is a = –3.63184, T = 273.3772, and R = 316.2604.12.36 In order to solve this problem, we must assume the direction that the blocks are moving. For example, we can assume that the blocks are moving from right to left as shownPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 27. 27 Force balances can be written for each block: T 83.16 103.94 15 × 9.8 = 147 103.94 – T – 83.16 = 15a R 13.86 69.3 10 × 9.8 = 98 T T + 69.3 – R – 13.86 = 10aPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 28. 28 R S 8 × 9.8 = 78.4 R – 78.4 – S = 8a S 5 × 9.8 = 49 S – 49 = 5a Therefore, the system of equations to be solved can be written in matrix form as 15 1 0 0   a   20.789 10 − 1 1 0  T  = 55.437   8 0 − 1 1   R   − 78.4   5 0 0 − 1  S   − 49       The solution is a = –1.347, T = 40.989, R = 109.893, and S = 42.267. Note that if we had assumed that the blocks were moving from left to right, the system of equations would have been 15 − 1 0 0   a   − 187.1005  10 1 − 1 0  T  =  − 83.15576  8 0 1 − 1  S   78.4   5 0 0 1   R      49   The solution for this case is a = –3.759374, T = 130.7098, R = 176.27186, and S = 67.79687.12.37 This problem can be solved in a number of ways. For example, using MATLAB, %prob1237.mPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 29. 29 k1=10;k2=30;k3=30;k4=10; m1=2;m2=2;m3=2; km=[(1/m1)*(k2+k1),-(k2/m1),0; -(k2/m2),(1/m2)*(k2+k3),-(k3/m2); 0,-(k3/m3),(1/m3)*(k3+k4)] x=[0.05;0.04;0.03] kmx=km*x >> prob1237 km = 20 -15 0 -15 30 -15 0 -15 20 x = 0.0500 0.0400 0.0300 kmx = 0.4000 0 0 Therefore, 1 = −0.4, 2 = 0, and 3 = 0 m/s2. x x x12.38 (a) Substituting the parameters gives d 2T + h (Ta − T ) = 0 dx 2 An analytical solution can be derived in a number of ways. One way is to assume a solution of the form T = A + Be λx + Ce λx λλ λλ Differentiating twice gives 2 x 2 x T"= Be + Ce Substituting these into the original differential equation yields λ2 Be λx + λ2 Ce λx + h (Ta − A − Be λx − Ce λx ) = 0 Equating like terms yieldsPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 30. 30 λ2 Be λx = h Be λT λ2 Ce λx = h Ce λT h Ta = h A The first two equations give λ =± h. The equation third gives A = Ta. Therefore, the solution is T = Ta + Be h x + Ce − h x The unknown constants can be evaluated from the boundary conditions 40 = 20 + B + C 200 = 20 + Be 0.02 (10 ) + Ce − 0.02 (10 ) These simultaneous equations can be solved for B = 45.25365 and C = −25.25365. Therefore, the analytical solution is T = 20 + 45.25365e 0.141421x − 25.25365e −0.141421 x (b) Substituting the parameters into the finite-difference equation gives − Ti −1 + 2.08Ti − Ti +1 =1.6 An analytical solution can be derived in a number of ways. A nice approach is to employ L This equation can be written for each node to yield the following system of equations,  2.08 − 1 0 0   T1   41.6   − 1 2.08 − 1 0  T2  =  1.6   0 − 1 2.08 − 1  T3   1.6   0  0 − 1 2.08 T   201.6  4    These can be solved for T1 = 61.0739, T2 = 85.4338, T3 = 115.0283, and T4 = 152.2252. A plot of the results is shown below (circles). In addition, the plot also shows the analytical solution (line) that was developed in (a):PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 31. 31 200 150 100 numerical 50 analytical 0 0 2 4 6 8 1012.39 Substituting centered difference finite differences, the Laplace equation can be written for the node (1, 1) as T21 − 2T11 + T01 T12 − 2T11 + T10 0= 2 + ∆x ∆y 2 Because the grid is square (∆x = ∆y), this equation can be expressed as 0 = T21 − 4T11 + T01 + T12 + T10 The boundary node values (T01 = 100 and T10 = 75) can be substituted to give 4T11 − T12 − T21 = 175 The same approach can be written for the other interior nodes. When this is done, the following system of equations results  4 − 1 − 1 0   T11  175  − 1 4 0 − 1  T12  = 125  − 1 0 4 − 1  T21   75   0 − 1 − 1 4  T   25     22    These equations can be solved using the Gauss-Seidel method. For example, the first iteration would be 175 + T12 + T21 175 + 0 + 0 T11 = = = 43.75 4 4 125 + T11 + T22 125 + 43.75 + 0 T12 = = = 42.1875 4 4 75 + T11 + T22 75 + 43.75 + 0 T21 = = = 29.6875 4 4PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 32. 32 25 + T12 + T21 25 + 42.1875 + 29.6875 T22 = = = 24.21875 4 4 The computation can be continued as follows: iteration unknown value εa maximum ε a 1 x1 43.75 100.00% x2 42.1875 100.00% x3 29.6875 100.00% x4 24.21875 100.00% 100.00% 2 x1 61.71875 29.11% x2 52.73438 20.00% x3 40.23438 26.21% x4 29.49219 17.88% 29.11% 3 x1 66.99219 7.87% x2 55.37109 4.76% x3 42.87109 6.15% x4 30.81055 4.28% 7.87% 4 x1 68.31055 1.93% x2 56.03027 1.18% x3 43.53027 1.51% x4 31.14014 1.06% 1.93% 5 x1 68.64014 0.48% x2 56.19507 0.29% x3 43.69507 0.38% x4 31.22253 0.26% 0.48% Thus, after 5 iterations, the maximum error is 0.48% and we are converging on the final result: T11 = 68.64, T12 = 56.195, T21 = 43.695, and T22 = 31.22.12.40 Find the unit vectors:  2 ˆ − 1ˆ − i j 4kˆ  A  = − ˆ 0.218i 0.436 ˆ j − 0. 8   12  2 2 + +4 2  ˆ + 2i 1 ˆ − j 4kˆ  B  = + ˆ 0.436i 0.218 ˆ j − 0. 8   12  +2 2 +4 2  Sum moments about the origin: ∑M ox = 50(2) − 0.436 B (4) − 0.218 A(4) = 0 ∑M oy = 0.436 A(4) − 0.218 B (4) = 0 Solve for A and B using equations 9.10 and 9.11: a11 x1 + a12 x 2 = b1 In the form a 21 x 2 + a 22 x 2 = b2PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 33. 33 − 0.872 A − 1.744 B = −100 1.744 A − 0.872 B = 0 Plug into equations 9.10 and 9.11: a 22 b1 − a12 b2 87.2 A= = = 22.94 N a11 a 22 − a12 a 21 3.80192 a11b2 − a 21b1 174.4 B= = = 45.87 N a11 a 22 − a12 a 21 3.80192PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 34. 33 − 0.872 A − 1.744 B = −100 1.744 A − 0.872 B = 0 Plug into equations 9.10 and 9.11: a 22 b1 − a12 b2 87.2 A= = = 22.94 N a11 a 22 − a12 a 21 3.80192 a11b2 − a 21b1 174.4 B= = = 45.87 N a11 a 22 − a12 a 21 3.80192PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 35. 33 − 0.872 A − 1.744 B = −100 1.744 A − 0.872 B = 0 Plug into equations 9.10 and 9.11: a 22 b1 − a12 b2 87.2 A= = = 22.94 N a11 a 22 − a12 a 21 3.80192 a11b2 − a 21b1 174.4 B= = = 45.87 N a11 a 22 − a12 a 21 3.80192PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 36. 33 − 0.872 A − 1.744 B = −100 1.744 A − 0.872 B = 0 Plug into equations 9.10 and 9.11: a 22 b1 − a12 b2 87.2 A= = = 22.94 N a11 a 22 − a12 a 21 3.80192 a11b2 − a 21b1 174.4 B= = = 45.87 N a11 a 22 − a12 a 21 3.80192PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.

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