Numerical Method for engineers-chapter 12

1

CHAPTER 12 12.1 Flow balances can be used to determine Q01 = 6 Q25 = 1 Q34 = 8

Q15 = 3 Q23 = 1 Q44 = 12

Q12 = 4 Q54 = 2

Q31 = 1 Q55 = 2

Q03 = 8 Q24 = 2

Mass balances can be used to determine the following simultaneous equations, 0 −1 0 0   c1  240 7 0 0 0  c 2   0  − 4 4  0 −1 9 0 0  c3  =  80   0 − 2 − 8 12 − 2 c 4   0   − 3 − 1 0 0 4  c   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=A\B C =

36.1290 36.1290 12.9032 20.6452 36.1290

>> inv(A) ans = 0.1452 0.1452 0.0161 0.0591 0.1452

0.0040 0.2540 0.0282 0.0722 0.0665

0.0161 0.0161 0.1129 0.0806 0.0161

0 0 0 0.0833 0

0 0 0 0.0417 0.2500

−1 −1 12.2 The relevant coefficients of the matrix inverse are a13 = 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) = 0.754717 ∆c 4 = 0.087479(0.25 × 160) = 3.499142 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.

2

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 16.99828 12.3 Because of conservation of flow: Q01 + Q03 = Q44 + Q55 12.4 Mass balances can be used to determine the following simultaneous equations, 0 − 3 0 0   c1   50  8 0 0 0  c 2   0  − 4 4  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=A\B C = 13.2432 13.2432 18.6486 17.0270 13.2432

12.5 Flow balances can be used to determine Q01 = 5 Q25 = 0 Q34 = 3

Q15 = 3 Q12 = 0 Q31 = −2Q03 = 8 Q23 = −7Q54 = 0 Q55 = 3 Q24 = 7 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 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.

3 0 0 0 0  c1   50   5 7 − 7 0 0 c 2   0   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=A\B C = 10.0000 18.0000 18.0000 18.0000 10.0000

12.6 Mass balances can be written for each of the reactors as 500 − Q13 c1 − Q12 c1 + Q21c 2 = 0 Q12 c1 − Q21c 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  0  c 2  =  0   − 90 90 − 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=A\B C = 5.0000 5.0000 5.8333

12.7 Mass balances can be written for each of the lakes as


4 Superior, c1: Michigan, c2: Huron, c3: Erie, c4: Ontario, c5:

180 = 67c1 710 = 36c 2 740 + 67c1 + 36c 2 = 161c3 3850 + 161c3 = 182c 4 4720 + 182c 4 = 212c5

The system of equations can be written in matrix form as 0 0 0 0   c1   180   67 36 0 0 0  c 2   710   0 − 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=A\B C = 2.6866 19.7222 10.1242 30.1099 48.1132

12.8 (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]'; >> AI=inv(A) AI =

0.0745 0.0816 0.0808

0 0.0816 0.0808

0 0 0.0808

0 0 0


5 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 ∆W1 =

∆c 4 −1 a 41

=

100.2373 − 75 = 297.725 0.084767

(c) First, normalize the matrix to give 0 1 1 − 0 . 91283  [ A] =  0 − 0.9899 0 0

0 0  0 0   1 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 . 095495 − 1 . 09549 0 0   [ A] =   1.084431 − 1.08443 1 0 1.137747 − 1.13775 1.049165 − 1.04917 −1

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 gives   F F − 1 + 2 K  y1 + 2 Ky 2 = − y in F1  F1  Using a similar approach, the equation for the last stage is


6   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, 0 0   y1  0.1  9 −8 0 0   y 2   0  − 1 9 − 8 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=A\B 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 = 0.04999866485086 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,


7

0

1

2

3

4

5

1 0.1 0.01 0.001 0.0001 0.00001 0.000001 Y

X

12.10 Steady-state mass balances for A in each reactor can be written as Qin c A,in − Qin c A,1 − k1V1c A,1 = 0 Qin c A,1 + Q32 c A,3 − (Qin + Q32 )c A, 2 − k 2V2 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 4V4 c A, 4 = 0 Steady-state mass balances for B in each reactor can be written as − Qin c B ,1 + k1V1c 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 + Q43 )c B ,3 − (Qin + Q43 )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  0 0 0 0 0 0 0   c B ,1  10  11.875  0 0 0 0 0 0 0  − 1.875 10  − 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  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]; 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.

8 >> B=[10 0 0 0 0 0 0 0]'; >> C=A\B C = 0.8421 0.1579 0.3400 0.9933 0.1010 1.8990 0.0847 1.9153

Therefore, to summarize the results reactor inflow 1 2 3 4

A

B

1 0.842105 0.340047 0.101036 0.084740

0 0.157895 0.993286 1.898964 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

4

12.11 Assuming a unit flow for Q1, the simultaneous equations can be written in matrix form as − 2  0  0 1  0  0

1 2 0 0 0  Q2  0 0 −2 1 2 0  Q3  0 0 0 0 − 2 3  Q4  = 0 1 0 0 0 0  Q5  1 1 −1 −1 0 0  Q6  0 0 0 1 − 1 − 1 Q  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; 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.

9 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=A\B Q =

0.5059 0.4941 0.2588 0.2353 0.1412 0.0941

12.12 The mass balances can be expressed in matrix form as  cG1  0 0 0 0 − 0.8 0 0 0 0  c  200  2.8 2.8 0 0 0 0 − 0.8 0 0 0  G 2   0   −2  0 −2 2.8 0 0 0 0 − 0.8 0 0  c G 3   0   0 0 −2 2.8 0 0 0 0 − 0.8 0  cG 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   L 3   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 0 1 2 3 4 5 6

Gas 100 95.73328 90.2475 83.19436 74.12603 62.46675

Liquid 85.06649 76.53306 65.5615 51.45521 33.31856 10

120

Gas Liquid

100 80 60 40 20 0 0

1

2

3

4

5

6

7


10

12.13 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 H2

2

F2

F3 F5 F4

3

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: 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 − 0.707 0 0.707 0 0 0   1   1200   0.866 0 − 0.707 0 0.707 0 0 0   F2  − 600  0.5 − 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   V3  This system can be solved for 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.

11 F1 = −292.82 F5 = 707.1068

F3 = −1348.58 V2 = 146.4102

F2 = 1453.59 H2 = −1200

F4 = 953.5898 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 F7 H5

5

100

F1 1

2 F5

3

F3

F6 V5

F2

F4 4

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° − 100 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


12 0 0 0 − 0.707 0 0.707 0 0 0   F1   0  − 1 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       0.5 0 0.866 0 0 0 0 0 0   F4   − 100  0 0 0.866 1 0 0 0 0 0 0 0   F5   0     =   − 0.5 0 0 0 0 0 −1 0 0   F6   0  0 0 0 − 1 − 0.5 0.707 1 0 0 0 0   F7   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 − 0.707 0 0 − 1  V5   0  This system can be solved for F1 = −348.334 F6 = 424.167

F2 = −351.666 F7 = −599.863

F3 = 304.5517 V3 = 175.833

F4 = 87.56443 H5 = 0

F5 = −107.244 V5 = 424.167

12.16 The first two columns of the inverse provide the information to solve this problem F1 F2 F3 H2 V2 V3

F1H 0.866025 0.250000 −0.500000 −1.000000 −0.433013 0.433013

F1V 0.500000 −0.433013 0.866025 0.000000 −0.250000 −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) = −2000 V2 = 2000(−0.433013) − 2500(−0.25) = −241.025 V3 = 2000(0.433013) − 2500(−0.75) = 2741.025 12.17 ΣF y = 0 ΣM = 0 Geometry

V2 + V3 = 1000 1000(cos30°) L1 − V3 L2 cos30° L1 + cos60° L3 = L2

Since V2 = 250 and V3 = 750, 866 L1 − 750 L2 = 0 0.866 L1 + 0.5L3 = L2 Therefore, 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.

13

L3 =

L2 − 0.866 L1 0.5

12.18 We can number the nodes as 1 F1

500 F3

F2

2

250

3 F6

H4

4

F5 F7

V4

F4

5 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: ΣFH = 0 = − F7 − F5 cos 60° ΣFV = 0 = F4 + F5 sin 60° + V5 These balances can then be expressed in matrix form as


14 0 0 0 0 0 0 0 0 0   F1  − 500   0.707  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 0 0.866 0.5 0 0 0 0   F4   0  − 0.707 0  0 1 0 0 0 0 0 0 0 0   F5  − 250    =   0 −1 1 0 0 0 0 0 0   F6   0   0  0 0 0 0 0 − 0.866 − 1 − 1 0 0   F7   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 0 − 1 − 0.866 0 0 0 0 − 1  V5   0  This system can be solved for F1 = −707.107 F6 = −899.519

F2 = −250 F7 = 29.00635

F3 = 500 H4 = 750

F4 = 500 V4 = 449.7595

F5 = −58.0127 V5 = −449.76

12.19 We can number the nodes as 1

F1 H4

F2

F5

2

F3

F8

3 F6 F7 6

4 V4

F4

5

F9

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° Node 4: ΣFH = 0 = F1 cos 30° + F4 + H 4 ΣFV = 0 = F1 sin 60° + V4 Node 5: ΣFH = 0 = − F4 − F5 cos 60° + F7 cos 45° + F9 ΣFV = 0 = F5 sin 60° + F8 + F7 sin 45° − 5000 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.

15 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 −1 0 0 − 0.5 0 0 0 0 0  0.5  0.866 0 0 0 0.866 0 0 0 0 0   0 1 −1 0 0 0 0 0 0 0  0 1 0 0 − 0.707 0.707 0 0 0  0  0 0 0 0 0 0.707 0.707 0 0 0  0 0 −1 0 0 0 0 −1 0  − 0.5 − 0.866 0 0 0 0 0 0 0 0 −1  0 0 1 0.5 0 − 0.707 − 1 0 0  0  0 0 0 − 0.866 0 − 0.707 0 0 0  0  0 0 0 0 0 0.707 0 1 0 0  0 0 0 0 − 0.707 0 0 0 0  0

0  F1   0  0  F2   0  0  F3   0      0  F4   0  0  F5   0      0  F6  =  0  0  F7   0      0  F9   0      0  H 4  − 5000 0  V 4   0      1   V6   0 

This system can be solved for F1 = −3660.25 F6 = −2588.19 V6 = 1830.13

F2 = −3660.25 F7 = 2588.19

F3 = −3660.25 F9 = 1830.13

F4 = 1830.127 H4 = 0

F5 = −3660.25 V4 = 3169.87

12.20 (a) Room 1: 0 = Wsmoker + Qa c a − Qa c1 + E13 (c3 − c1 ) Room 2: 0 = Qb cb + (Qa − Qd )c 4 − Qc c 2 + E 24 (c 4 − c 2 ) Room 3: 0 = Wgrill + Qa c1 + E13 (c1 − c3 ) + E 34 (c 4 − c3 ) − Qa c3 Room 4: 0 = Qa c3 + E 34 (c3 − c 4 ) + E 24 (c 2 − c 4 ) − Qa c 4 Substituting the parameters yields 0 − 25 0   c1  1400   225  0 175 0 − 125 c 2   100    =  − 225 0 275 − 50  c3  2000   − 25 − 250 275  c 4   0   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.

16 These can be solved for  c1   8.0996  c  12.3448  2    =  c 16 . 8966  3   c 4  16.4828 (b) The matrix inverse can be determined as 0.004996 0.0000153 [ A] = 0.003448 0.006207 0.004966 0.000138 0.004828 0.00069 −1

0.000552 0.003448 0.004966 0.004828

0.000107  0.003448 0.000966 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 % smokers =

3.448 × 100% = 27.93% 12.3448

(ii) the grill −1 c 2,grill = a 31 Wgrill = 0.003448( 2000) = 6.897 % grill =

6.897 × 100% = 55.87% 12.3448

(iii) the intakes −1 −1 c 2,intakes = a 21 Qa c a + a 22 Qb cb = 0.003448(200) 2 + 0.006207(50) 2 = 1.37931 + 0.62069 = 2 % grill =

2 × 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 by −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


17 0 − 25 0   c1  1400   225  0 155 0 − 105 c 2   100    =  − 225 0 275 − 50  c3  2000   − 5 − 250 255  c 4   0   0 which can be solved for  c1   8.1084  c  12.0800  2    =  c 16 . 9760  3   c 4  16.8800 Therefore, the concentration in the kids’ area would be decreased 0.26483 mg/m3 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

C

The direction cosines of the vectors A, B and C can be determined as 3 12  4 A:  , − , −   13 13 13 


18 3 12   4 B:  − , − , −  13 13 13   12   5 C:  0, , −  13 13   Force balances for point D can then be written as

∑F ∑F ∑F

4 4 A− B =0 13 13

x

=

y

=−

3 3 5 A− B + C =0 13 13 13

z

=−

12 12 12 A − B − C + 20 = 0 13 13 13

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 − 0.92308B − 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=A\B x = 37.3333 -46.0000 74.0000 -46.6667 37.3333 46.0000 -76.6667 -74.0000 -37.3333 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.

19 61.3333

Therefore, in kN BC = −46 CE = −76.6667

AB = 37.3333 DE = 46

BD = −46.6667 Ay = −37.33333

AD = 74 Ax = −74

CD = 37.3333 Ey = 61.3333

12.23 The simultaneous equations are 1 1 0 0 0  i12   0  1 1 − 1 0  i52   0   0 −1 0 0 0 −1 0 0 1  i32  =  0  0 0 0 0 1 − 1  i 65   0  0 5 − 15 0 − 5 − 2 i54   0  10 − 5 0 − 25 0 0  i  200  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=A\B 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:

0.9479

5.1185

12.24 The current equations can be written as


20 − i 21 − i 23 + i52 = 0 i 23 − i35 + i 43 = 0 − i 43 + i54 = 0 i35 − i52 + i65 − i54 = 0 Voltage equations: i 21 =

V2 − 10 35

i54 =

V5 − V 4 15

i 23 =

V 2 − V3 30

i35 =

V3 − V5 7

i 43 =

V 4 − V3 8

i52 =

V5 − V 2 10

i65 =

150 − V5 5

0 0 0 − 1 − 1 1 0 1 0 −1 1 0  0 0 0 0 −1 1  0 −1 1 0 −1 0  35 0 0 0 0 0  0 0 0  0 30 0 0 0 0 0 8 0  0 0 0 0 15 0  0 0 7 0 0 0 0 0 10 0 0 0  0 0 0 0 0 0

0 0 0 0 0  i 21   0  0 0 0 0 0  i 23   0  0 0 0 0 0  i52   0      1 0 0 0 0  i35   0  0 −1 0 0 0  i 43  − 10     0 −1 1 0 0  i54  =  0  0 0 1 − 1 0  i 65   0      0 0 0 1 − 1 V2   0      0 0 −1 0 1  V3   0  0 1 0 0 − 1 V4   0      5 0 0 0 1  V5   150 

This system can be solved for i21 = 2.9291 i54 = 0.1507 V5 = 135.3543

i23 = −0.6457 i65 = 2.9291

i52 = 2.2835 V2 = 112.5196

i35 = −0.4950 V3 = 131.8893

i43 = 0.1507 V4 = 133.0945

12.25 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 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.

21 i76 i 47 i58 i89

− i63 − i76 − i89 − i97

− i65 = 0 + i97 = 0 − i80 = 0 =0

Voltage equations: − 20i 25 + 10i 65 − 5i 63 − 5i32 = 0 5i63 + 20i76 + 5i47 + 20i34 = 0 − 50i58 − 15i89 − 0i97 − 20i76 − 10i 65 = 0 120 − 20i 25 − 50i58 = 40

 1 −1 − 1 0 0 0 0 1 0 0 0 0 0 0 0 0  5 20 0 0 0 0  0 20 

i32  1 0 0 0 0 0 0 0 0 0  i 25   0  0 −1 1 0 0 0 0 0 0 0  i   0  0 1 0 −1 0 0 0 0 0 0   12   0  0 0 0 0 1 −1 0 0 0 0  i34   0  0 0 −1 0 −1 0 1 0 0 0  i 63   0  0 0 0 1 0 0 −1 1 0 0  i 47  =  0  0 0 0 0 0 1 0 0 − 1 − 1 i65   0  0 0 0 0 0 0 0 −1 1 0  i58   0  0 0 5 0 − 10 0 0 0 0 0  i   0  0 − 20 − 5 − 5 0 0 − 20 0 0 0   76   0  0 0 0 0 10 50 20 0 15 0  i97   0  0 0 0 0 0 50 0 0 0 0  i89  80 i80 

This system can be solved for i32 = −2.5670 i47 = 1.2287 i89 = −2.4299

i25 = 1.0449 i65 = 0.1371 i80 = 3.6119

i12 = 3.6119 i58 = 1.1820

i34 = 1.2287 i76 = −1.2012

i63 = −1.3384 i97 = −2.4299

12.26 Let ci = component i. Therefore, the following system of equations must hold 15c1 + 17c 2 + 19c3 = 3890 0.30c1 + 0.40c 2 + 0.55c3 = 95 1.0c1 + 1.2c 2 + 1.5c3 = 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.


22 R1 R4

+ −

V1

i1

R3

R2 R5 i2

i3

+ − V2

Kirchhoff’s voltage law can be applied to each loop. − V1 + R1i1 + R4 (i1 − i 2 ) = 0 R4 (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 0   i1   80   20 − 15 − 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=A\B I =

4.9721 1.2961 -0.3911

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, 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.

23

> A=[6 -4 -2;-4 18 -8;-2 -8 15]; >> B=[20;0;0]; >> I=A\B 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=a\b

x =

-2.7278 6.5407 1.9989

Therefore, I1 = −2.7278 A, I3 = 6.5407 A, and I4 = 1.9989 A 12.30 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=[200 0 230]; >> x=a\b 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 − kx 3 = m 2 g − kx 2 + kx 3 = m3 g


24 Substituting the parameter values 0   x1  19.6   120 − 90 − 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 0   x1   98   30 − 20 − 20 30 − 10  x 2  = 34.3 − 10 − 10  x3  19.6   0 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  − k2  0   0

− k2 k 2 + k3 − k3 0

0 − k3 k3 + k4 − k4

0   x1   0  0 x2   0  = − k 4   x3   0    F  k 4 x4   

Substituting the parameter values 0 0   x1   0  150 − 50   0 x2  =  0 − 50 130 − 80   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=A\B 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


25

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 = 40a 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


26 424.352 − 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 0   a   128.1036  40 1 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 shown

Force balances can be written for each block: T

83.16 103.94 15 × 9.8 = 147


27 103.94 – T – 83.16 = 15a R

13.86 69.3 10 × 9.8 = 98

T

T + 69.3 – R – 13.86 = 10a R

8 × 9.8 = 78.4

S

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 0 0   a  20.789 15 1 0  T  55.437  10 − 1 1  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.


28 Note that if we had assumed that the blocks were moving from left to right, the system of equations would have been 0   a   − 187.1005  15 − 1 0 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.m 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

x =

-15 30 -15

0 -15 20

0.0500 0.0400 0.0300

kmx = 0.4000 0 0

Therefore, x1 = −0.4, x2 = 0, and x3 = 0 m/s2. 12.38 (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


29 T = A + Be λx + Ce λx Differentiating twice gives T " = λ 2 Be λx + λ 2 Ce λx Substituting these into the original differential equation yields

λ2 Be λx + λ 2 Ce λx + h' (Ta − A − Be λx − Ce λx ) = 0 Equating like terms yields

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

+ Ce −

h'x

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 0.141421 x − 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, 0 0  T1   41.6  2.08 − 1 0  T2  =  1.6   − 1 2.08 − 1  0 − 1 2.08 − 1  T3   1.6  0 − 1 2.08 T4  201.6  0  


30 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): 200 150 100 numerical analytical

50 0 0

2

4

6

8

10

12.39 Substituting centered difference finite differences, the Laplace equation can be written for the node (1, 1) as 0=

T21 − 2T11 + T01 ∆x

2

+

T12 − 2T11 + T10 ∆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 0 − 1 T12  = 125 − 1 4 − 1 0 4 − 1 T21   75       0 − 1 − 1 4  T22   25  These equations can be solved using the Gauss-Seidel method. For example, the first iteration would be T11 =

175 + T12 + T21 175 + 0 + 0 = = 43.75 4 4

T12 =

125 + T11 + T22 125 + 43.75 + 0 = = 42.1875 4 4


31

T21 =

75 + T11 + T22 75 + 43.75 + 0 = = 29.6875 4 4

T22 =

25 + T12 + T21 25 + 42.1875 + 29.6875 = = 24.21875 4 4

The computation can be continued as follows: iteration 1

2

3

4

5

unknown x1 x2 x3 x4 x1 x2 x3 x4 x1 x2 x3 x4 x1 x2 x3 x4 x1 x2 x3 x4

value 43.75 42.1875 29.6875 24.21875 61.71875 52.73438 40.23438 29.49219 66.99219 55.37109 42.87109 30.81055 68.31055 56.03027 43.53027 31.14014 68.64014 56.19507 43.69507 31.22253

εa 100.00% 100.00% 100.00% 100.00% 29.11% 20.00% 26.21% 17.88% 7.87% 4.76% 6.15% 4.28% 1.93% 1.18% 1.51% 1.06% 0.48% 0.29% 0.38% 0.26%

maximum ε a

100.00%

29.11%

7.87%

1.93%

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:  1iˆ − 2 ˆj − 4kˆ A  2 2 2  1 +2 +4

  = 0.218iˆ − 0.436 ˆj − 0.873kˆ  

 2iˆ + 1 ˆj − 4kˆ B  2 2 2  1 +2 +4

  = 0.436iˆ + 0.218 ˆj − 0.873kˆ  

Sum moments about the origin:

∑M ∑M

ox

= 50( 2) − 0.436 B ( 4) − 0.218 A( 4) = 0

oy

= 0.436 A(4) − 0.218B (4) = 0


32

Solve for A and B using equations 9.10 and 9.11: a x +a x =b In the form a11 x1 + a12 x2 = b1 21 2 22 2 2 − 0.872 A − 1.744 B = −100 1.744 A − 0.872 B = 0 Plug into equations 9.10 and 9.11: A=

a 22 b1 − a12 b2 87.2 = = 22.94 N a11 a 22 − a12 a 21 3.80192

B=

a11b2 − a 21b1 174.4 = = 45.87 N a11 a 22 − a12 a 21 3.80192


Numerical Method for engineers-chapter 12

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