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Numerical Methods for Engineers, 7th edition Solution Manual

Numerical Methods for Engineers, 7th edition Solution Manual provides you with expert textbook solutions that ensure you understand every concept thoroughly.

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    1 CHAPTER 1 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 1 dv dt c g v m     The integrals can be evaluated as ln / c g v m t C c m           where C = a constant of integration, which can be evaluated by applying the initial condition to yield ln (0) / c g v m C c m          which can be substituted back into the solution ln ln (0) / / c c g v g v m m t c m c m                  This result can be rearranged algebraically to solve for v ,   ( / ) ( / ) (0) 1 c m t c m t mg v v e e 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)   ( / ) 1 c m t mg v e c    Laplace transform solution: An alternative solution is provided by applying Laplace transform to the differential equation to give ( ) (0) ( ) g c sV s v V s s m    Solve algebraically for the transformed velocity

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    2 (0) ( ) / ( / ) v g V s 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 0 c g A As Bs m    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 s c m     This can be substituted into Eq. 1 to give (0) / / ( ) / / v mg c mg c V s s c m s s c m      Taking inverse Laplace transforms yields ( / ) ( / ) ( ) (0) c m t c m t mg mg v t v e e c c      or collecting terms   ( / ) ( / ) ( ) (0) 1 c m t c m t mg v t v e e c      1.2 At t = 8 s, the analytical solution is 41.137 (Example 1.1). The relative error can be calculated with analytical numerical absolute relative error 100% analytical    The numerical results are: step v (8) absolute relative error 2 44.8700 9.074% 1 42.8931 4.268% 0.5 41.9901 2.073% The error versus step size can then be plotted as

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    3 0% 2% 4% 6% 8% 10% 0 0.5 1 1.5 2 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, 2 '   dv c g v dt m Multiply both sides by m / c ′ gives 2 ' '   m dv m g v c dt c Define / ' a mg c  2 2 '   m dv a v c dt Integrate by separation of variables, 2 2 '     dv c dt m a v A table of integrals can be consulted to find that 1 2 2 1 tanh dx x a a a x     Therefore, the integration yields 1 1 ' tanh    v c t C a a m If v = 0 at t = 0, then because tanh –1 (0) = 0, the constant of integration C = 0 and the solution is 1 1 ' tanh   v c t a a m This result can then be rearranged to yield

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    4 ' tanh '          g m gc v t c m (b) Using Euler’s method, the first two steps can be computed as 2 0.22 (2) 0 9.81 (0) 2 19.62 68.1 v           2 0.22 (4) 19.62 9.81 (19.62) 2 36.75284 68.1 v           The computation can be continued and the results summarized along with the analytical result as: t v -numerical dv / dt v -analytical 0 0 9.81 0 2 19.62 8.56642 18.83093 4 36.75284 5.446275 33.72377 6 47.64539 2.476398 43.46492 8 52.59819 0.872478 49.06977 10 54.34314 0.269633 52.05938 12 54.88241 0.079349 53.58978  55.10572 0.022993 55.10572 A plot of the numerical and analytical results can be developed 0 20 40 60 0 4 8 12 v-numerical v-analytical 1.4 ( / ) ( ) (1 ) c m t gm v t e c    jumper #1: (12/70) 9 9.81(70) ( ) (1 ) 44.99204 12 v t e     jumper #2: (15/80) 9.81(80) 44.99204 (1 ) 15 t e    0.1875 44.99204 52.32 52.32 t e    0.1875 0.14006 t e   ln 0.14006 0.1875 t   ln 0.14006 10.4836 s 0.1875 t    1.5 Before the chute opens ( t < 10), Euler’s method can be implemented as

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    5 10 ( ) ( ) 9.81 ( ) 80 v t t v t v t t             After the chute opens ( t  10), the drag coefficient is changed and the implementation becomes 60 ( ) ( ) 9.81 ( ) 80 v t t v t v t t             Here is a summary of the results along with a plot: Chute closed Chute opened t v dv / dt t v dv / dt 0 -20.0000 12.3100 10 52.5723 -29.6192 1 -7.6900 10.7713 11 22.9531 -7.4048 2 3.0813 9.4248 12 15.5483 -1.8512 3 12.5061 8.2467 13 13.6971 -0.4628 4 20.7528 7.2159 14 13.2343 -0.1157 5 27.9687 6.3139 15 13.1186 -0.0289 6 34.2826 5.5247 16 13.0896 -0.0072 7 39.8073 4.8341 17 13.0824 -0.0018 8 44.6414 4.2298 18 13.0806 -0.0005 9 48.8712 3.7011 19 13.0802 -0.0001 20 13.0800 0.0000 -30 0 30 60 0 5 10 15 20 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 Deposit Withdrawal Interest Balance 1-May $1,522.33 $220.13 $327.26 $15.22 1-Jun $1,430.42 $216.80 $378.51 $14.30 1-Jul $1,283.02 $450.35 $106.80 $12.83 1-Aug $1,639.40 $127.31 $350.61 $16.39 1-Sep $1,432.49 ( b ) ( ) ( ) dB D t W t iB dt   

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    6 ( c ) for t = 0 to 0.5: 220.13 327.26 0.01(1522.33) 91.91 dB dt      (0.5) 1522.33 91.91(0.5) 1476.38 B    for t = 0.5 to 1: 220.13 327.260 0.01(1476.38) 92.37 dB dt      (0.5) 1476.38 92.37(0.5) 1430.19 B    The balances for the remainder of the periods can be computed in a similar fashion as tabulated below: Date Deposit Withdrawal Interest dB / dt Balance 1-May $220.13 $327.26 $15.22 -$91.91 $1,522.33 16-May $220.13 $327.26 $14.76 -$92.37 $1,476.38 1-Jun $216.80 $378.51 $14.30 -$147.41 $1,430.19 16-Jun $216.80 $378.51 $13.56 -$148.15 $1,356.49 1-Jul $450.35 $106.80 $12.82 $356.37 $1,282.42 16-Jul $450.35 $106.80 $14.61 $358.16 $1,460.60 1-Aug $127.31 $350.61 $16.40 -$206.90 $1,639.68 16-Aug $127.31 $350.61 $15.36 -$207.94 $1,536.23 1-Sep $1,432.26 ( d ) As in the plot below, the results of the two approaches are very close. $1,200 $1,300 $1,400 $1,500 $1,600 $1,700 M M J A S Bi-monthly Monthly 1.7 (a) The first two steps are (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 c    The process can be continued to yield t c dc / dt 0 100.0000 -17.5000 0.1 98.2500 -17.1938 0.2 96.5306 -16.8929 0.3 94.8413 -16.5972 0.4 93.1816 -16.3068 0.5 91.5509 -16.0214 0.6 89.9488 -15.7410 0.7 88.3747 -15.4656

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    7 0.8 86.8281 -15.1949 0.9 85.3086 -14.9290 1 83.8157 -14.6678 (b) The results when plotted on a semi-log plot yields a straight line 4.4 4.5 4.6 0 0.2 0.4 0.6 0.8 1 The slope of this line can be estimated as ln(83.8157) ln(100) 0.17655 1    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 students J s kJ 35 ind 80 20 min 60 3,360 kJ ind s min 1000 J       Q 3 3 Mwt (101.325 kPa)(11m 8m 3m 35 0.075 m )(28.97 kg/kmol) 314.796 kg (8.314 kPa m / (kmol K)((20 273.15)K)         PV m RT students 3,360 kJ 14.86571 K (314.796 kg)(0.718 kJ/(kg K)) v Q T mC     Therefore, the final temperature is 20 + 14.86571 = 34.86571 o C . 1.9 The first two steps yield 2 450 450 (0.5) 0 3 sin (0) 0.5 0 ( 0.36) 0.5 0.18 1250 1250 y               2 450 450 (1) 0.18 3 sin (0.5) 0.5 0.18 ( 0.11176) 0.5 0.23588 1250 1250 y                 The process can be continued to give the following table and plot: t y d y / dt t y d y / dt 0 0.00000 -0.36000 5.5 1.10271 0.17761 0.5 -0.18000 -0.11176 6 1.19152 -0.27568 1 -0.23588 0.40472 6.5 1.05368 -0.31002 1.5 -0.03352 0.71460 7 0.89866 0.10616 2 0.32378 0.53297 7.5 0.95175 0.59023 2.5 0.59026 0.02682 8 1.24686 0.69714 3 0.60367 -0.33849 8.5 1.59543 0.32859

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    8 3.5 0.43443 -0.22711 9 1.75972 -0.17657 4 0.32087 0.25857 9.5 1.67144 -0.35390 4.5 0.45016 0.67201 10 1.49449 -0.04036 5 0.78616 0.63310 -0.5 0.0 0.5 1.0 1.5 2.0 0 2 4 6 8 10 1.10 The first two steps yield 1.5 2 450 150(1 0) (0.5) 0 3 sin (0) 0.5 0 0.12(0.5) 0.06 1250 1250 y               1.5 2 450 150(1 0.06) (1) 0.06 3 sin (0.5) 0.5 0.06 0.13887(0.5) 0.00944 1250 1250 y                The process can be continued to give t y dy / dt t y dy / dt 0 0.00000 -0.12000 5.5 1.61981 0.02876 0.5 -0.06000 0.13887 6 1.63419 -0.42872 1 0.00944 0.64302 6.5 1.41983 -0.40173 1.5 0.33094 0.89034 7 1.21897 0.06951 2 0.77611 0.60892 7.5 1.25372 0.54423 2.5 1.08058 0.02669 8 1.52584 0.57542 3 1.09392 -0.34209 8.5 1.81355 0.12227 3.5 0.92288 -0.18708 9 1.87468 -0.40145 4 0.82934 0.32166 9.5 1.67396 -0.51860 4.5 0.99017 0.69510 10 1.41465 -0.13062 5 1.33772 0.56419 -0.5 0.0 0.5 1.0 1.5 2.0 0 2 4 6 8 10 1.11 When the water level is above the outlet pipe, the volume balance can be written as 2 1.5 out 3sin ( ) 3( ) dV t y y dt   

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    9 In order to solve this equation, we must relate the volume to the level. To do this, we recognize that the volume of a cone is given by V =  r 2 y /3. Defining the side slope as s = y top / r top , the radius can be related to the level ( r = y / s ) and the volume can be reexpressed as 3 2 3 V y s   which can be solved for 2 3 3 s V y   (1) and substituted into the volume balance 1.5 2 2 3 out 3 3sin ( ) 3 dV s V t y dt             (2) For the case where the level is below the outlet pipe, outflow is zero and the volume balance simplifies to 2 3sin ( ) dV 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 3 3 2 (0) 0.8 0.20944 m 3(1.6) V    For the first step, y < y out and Eq. (3) gives 2 (0) 3sin (0) 0 dV dt   and Euler’s method yields (0.5) (0) (0) 0.20944 0(0.5) 0.20944 dV V V t dt       For the second step, Eq. (3) still holds and 2 (0.5) 3sin (0.5) 0.689547 dV dt   (1) (0.5) (0.5) 0.20944 0.689547(0.5) 0.554213 dV V V t dt       Equation (1) can then be used to compute the new level,

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    10 2 3 3(1.6) (0.554213) 1.106529 m y    Because this level is now higher than the outlet pipe, Eq. (2) holds for the next step   1.5 (1) 2.12422 3 1.106529 1 2.019912 dV dt     (1.5) 0.554213 2.019912(0.5) 1.564169 V    The remainder of the calculation is summarized in the following table and figure. t Q in V y Q out dV / dt 0 0 0.20944 0.8 0 0 0.5 0.689547 0.20944 0.8 0 0.689547 1 2.12422 0.554213 1.106529 0.104309 2.019912 1.5 2.984989 1.564169 1.563742 1.269817 1.715171 2 2.480465 2.421754 1.809036 2.183096 0.29737 2.5 1.074507 2.570439 1.845325 2.331615 -1.25711 3 0.059745 1.941885 1.680654 1.684654 -1.62491 3.5 0.369147 1.12943 1.40289 0.767186 -0.39804 4 1.71825 0.93041 1.31511 0.530657 1.187593 4.5 2.866695 1.524207 1.55031 1.224706 1.641989 5 2.758607 2.345202 1.78977 2.105581 0.653026 5.5 1.493361 2.671715 1.869249 2.431294 -0.93793 6 0.234219 2.202748 1.752772 1.95937 -1.72515 6.5 0.13883 1.340173 1.48522 1.013979 -0.87515 7 1.294894 0.902598 1.301873 0.497574 0.79732 7.5 2.639532 1.301258 1.470703 0.968817 1.670715 8 2.936489 2.136616 1.735052 1.890596 1.045893 8.5 1.912745 2.659563 1.866411 2.419396 -0.50665 9 0.509525 2.406237 1.805164 2.167442 -1.65792 9.5 0.016943 1.577279 1.568098 1.284566 -1.26762 10 0.887877 0.943467 1.321233 0.5462 0.341677 0 0.5 1 1.5 2 2.5 3 0 2 4 6 8 10 V y 1.12 (a) The force balance can be written as : 2 2 (0) ( ) d dv R m mg c v v dt R x     Dividing by mass gives 2 2 (0) ( ) d c dv R g v v dt m R x    

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    11 (b) Recognizing that dx / dt = v , the chain rule is dv dv v dt dx  Setting drag to zero and substituting this relationship into the force balance gives 2 2 (0) ( ) dv g R dx v R x    (c) Using separation of variables 2 2 (0) ( ) R v dv g dx R x    Integrating gives 2 2 (0) 2 v R g C R x    Applying the initial condition yields 2 2 0 (0) 2 0 v R g C R    which can be solved for C = v 0 2 /2 – g (0) R , which can be substituted back into the solution to give 2 2 2 0 (0) (0) 2 2 v v R g g R R x     or 2 2 0 2 (0) 2 (0) R v v g g R R x      Note that the plus sign holds when the object is moving upwards and the minus sign holds when it is falling. (d) Euler’s method can be developed as 2 1 1 2 (0) ( ) ( ) ( ) ( ) ( ) i i i i i i g R v x v x x x v x R x                The first step can be computed as 6 2 6 2 9.81 (6.37 10 ) (10, 000) 1,500 (10, 000 0) 1,500 ( 0.00654)10, 000 1434.600 1,500 (6.37 10 0) v                 

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    12 remainder of the calculations can be implemented in a similar fashion as in the following table x v dv / dx v -analytical 0 1500.000 -0.00654 1500.000 10000 1434.600 -0.00682 1433.216 20000 1366.433 -0.00713 1363.388 30000 1295.089 -0.00750 1290.023 40000 1220.049 -0.00794 1212.475 50000 1140.643 -0.00847 1129.884 60000 1055.973 -0.00912 1041.049 70000 964.798 -0.00995 944.206 80000 865.317 -0.01106 836.579 90000 754.742 -0.01264 713.299 100000 628.359 -0.01513 564.197 For the analytical solution, the value at 10,000 m can be computed as 6 2 2 6 6 (6.37 10 ) 1,500 2(9.81) 2(9.81)(6.37 10 ) 1433.216 (6.37 10 10, 000) v         The remainder of the analytical values can be implemented in a similar fashion as in the last column of the above table. The numerical and analytical solutions can be displayed graphically. 0 400 800 1200 1600 0 20000 40000 60000 80000 100000 v-analytical v-numerical 1.13 The volume of the droplet is related to the radius as 3 4 3 r V   (1) This equation can be solved for radius as 3 3 4 V r   (2) The surface area is 2 4 A r   (3) Equation (2) can be substituted into Eq. (3) to express area as a function of volume 2/3 3 4 4 V A         

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    13 This result can then be substituted into the original differential equation, 2/3 3 4 4 dV V k dt           (4) The initial volume can be computed with Eq. (1), 3 3 3 4 4 (2.5) 65.44985 mm 3 3 r V      Euler’s method can be used to integrate Eq. (4). For the first step, the result is 2/3 3(65.44985) (0.25) (0) (0) 65.44985 0.08(4) 0.25 4 65.44985 6.28319(0.25) 63.87905                   dV V V t dt Here are the beginning and ending steps t V dV / dt 0 65.44985 -6.28319 0.25 63.87905 -6.18225 0.5 62.33349 -6.08212 0.75 60.81296 -5.98281 1 59.31726 -5.8843 • • • 9 23.35079 -3.16064 9.25 22.56063 -3.08893 9.5 21.7884 -3.01804 9.75 21.03389 -2.94795 10 20.2969 -2.87868 A plot of the results is shown below. We have included the radius on this plot (dashed line and right scale): 0 20 40 60 80 0 2 4 6 8 10 1.6 2 2.4 V r Eq. (2) can be used to compute the final radius as 3 3(20.2969) 1.692182 4 r    Therefore, the average evaporation rate can be computed as min mm 080782 . 0 min 10 mm ) 692182 . 1 5 . 2 (    k which is approximately equal to the given evaporation rate of 0.08 mm/min.

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    14 1.14 The first two steps can be computed as     (1) 70 0.019(70 20) 2 68 ( 0.95)2 68.1 (2) 68.1 0.019(68.1 20) 2 68.1 ( 0.9139)2 66.2722 T T                 The remaining results are displayed below along with a plot of the results. t T dT / dt t T dT / dt 0 70.00000 -0.95000 12.00000 59.62967 -0.75296 2 68.10000 -0.91390 14.00000 58.12374 -0.72435 4 66.27220 -0.87917 16.00000 56.67504 -0.69683 6 64.51386 -0.84576 18.00000 55.28139 -0.67035 8 62.82233 -0.81362 20.00000 53.94069 -0.64487 10 61.19508 -0.78271 50 60 70 80 0 5 10 15 20 1.15 The pair of differential equations to be solved are 1 di R i q dt L CL    dq i dt  or substituting the parameters 40 2, 000 di i q dt    dq i dt  The first step can be implemented by first using the differential equations to compute the slopes 40(0) 2, 000(1) 2, 000 di dt      0 dq dt  Then, Euler’s method can be applied as (0.01) 0 2, 000(0.01) 20 i     (0.01) 1 0(0.01) 1 q   

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    15 For the second step 40( 20) 2, 000(1) 1, 200 di dt       20 dq dt   (0.02) 20 1, 200(0.01) 32 i      (0.02) 1 20(0.01) 0.8 q    The remaining steps are summarized in the following table and plot: t i q di / dt dq / dt 0 0 1 -2000 0 0.01 -20 1 -1200 -20 0.02 -32 0.8 -320 -32 0.03 -35.2 0.48 448 -35.2 0.04 -30.72 0.128 972.8 -30.72 0.05 -20.992 -0.1792 1198.08 -20.992 0.06 -9.0112 -0.38912 1138.688 -9.0112 0.07 2.37568 -0.47923 863.4368 2.37568 0.08 11.01005 -0.45548 470.5485 11.01005 0.09 15.71553 -0.34537 62.12813 15.71553 0.1 16.33681 -0.18822 -277.034 16.33681 -40 -20 0 20 40 0 0.02 0.04 0.06 0.08 0.1 -1 -0.5 0 0.5 1 i q 1.16 ( a ) The solution of the differential equation is 0 t N N e   The doubling time can be computed as the time when N = 2 N 0 , (20) 0 0 2 N N e   ln 2 0.693 0.034657/hr 20 hrs 20 hrs     ( b ) The volume of an individual spherical cell is 3 cell volume 6 d   (1)

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