Numerical Methods for Engineers, 7th edition Solution Manual

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1CHAPTER 11.1We will illustrate two different methods for solving this problem: (1) separation of variables, and (2)Laplace transform.dvcgvdtmSeparation of variables:Separation of variables gives1dvdtcgvmThe integrals can be evaluated asln/cgvmtCcmwhereC= a constant of integration, which can be evaluated by applying the initial condition to yieldln(0)/cgvmCcm which can be substituted back into the solutionlnln(0)//ccgvgvmmtcmcmThis result can be rearranged algebraically to solve forv,( /)( /)(0)1c m tc m tmgvveecwhere the first part is the general solution and the second part is the particular solution for the constantforcing function due to gravity. For the case where,v(0) = 0, the solution reduces to Eq. (1.10)( /)1c m tmgvecLaplace transform solution:An alternative solution is provided by applying Laplace transform to thedifferential equation to give( )(0)( )gcsV svV ssmSolve algebraically for the transformed velocity

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2(0)( )/(/)vgV sscms scm(1)The second term on the right of the equal sign can be expanded with partial fractions(/)(/)/(/)gABA scmBss scmsscms scm(2)By equating like terms in the numerator, the following must hold0cgAAsBsmThe first equation can be solved forA=mg/c. According to the second equation,B= –A, soB= –mg/c.Substituting these back into (2) gives//(/)/gmgcmgcs scmsscmThis can be substituted into Eq. 1 to give(0)//( )//vmgcmgcV sscmsscmTaking inverse Laplace transforms yields( /)( /)( )(0)c m tc m tmgmgv tveeccor collecting terms( /)( /)( )(0)1c m tc m tmgv tveec1.2Att= 8 s, the analytical solution is 41.137 (Example 1.1). The relative error can be calculated withanalyticalnumericalabsolute relative error100%analyticalThe numerical results are:stepv(8)absoluterelative error244.87009.074%142.89314.268%0.541.99012.073%The error versus step size can then be plotted as

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30%2%4%6%8%10%00.511.522.5Thus, 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'dvcgvdtmMultiply both sides bym/c′gives2''m dvm gvcdtcDefine/'amgc22'm dvavcdtIntegrate by separation of variables,22'dvc dtmavA table of integrals can be consulted to find that1221 tanhdxxaaaxTherefore, the integration yields11'tanhvc tCaamIfv= 0 att= 0, then because tanh–1(0) = 0, the constant of integrationC= 0 and the solution is11'tanhvc taamThis result can then be rearranged to yield

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4'tanh'gmgcvtcm(b)Using Euler’s method, the first two steps can be computed as20.22(2)09.81(0)219.6268.1v20.22(4)19.629.81(19.62)236.7528468.1vThe computation can be continued and the results summarized along with the analytical result as:tv-numericaldv/dtv-analytical009.810219.628.5664218.83093436.752845.44627533.72377647.645392.47639843.46492852.598190.87247849.069771054.343140.26963352.059381254.882410.07934953.5897855.105720.02299355.10572A plot of the numerical and analytical results can be developed020406004812v-numericalv-analytical1.4( /)( )(1)c m tgmv tecjumper #1:(12/70) 99.81(70)( )(1)44.9920412v tejumper #2:(15/80)9.81(80)44.99204(1)15te0.187544.9920452.3252.32te0.18750.14006teln 0.140060.1875t ln 0.1400610.4836 s0.1875t1.5Before the chute opens (t< 10), Euler’s method can be implemented as

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510()( )9.81( )80v ttv tv tt After the chute opens (t10), the drag coefficient is changed and the implementation becomes60()( )9.81( )80v ttv tv tt Here is a summary of the results along with a plot:Chute closedChute openedtvdv/dttvdv/dt0-20.000012.31001052.5723-29.61921-7.690010.77131122.9531-7.404823.08139.42481215.5483-1.8512312.50618.24671313.6971-0.4628420.75287.21591413.2343-0.1157527.96876.31391513.1186-0.0289634.28265.52471613.0896-0.0072739.80734.83411713.0824-0.0018844.64144.22981813.0806-0.0005948.87123.70111913.0802-0.00012013.08000.0000-3003060051015201.6(a)This is a transient computation. For the period ending June 1:Balance = Previous Balance + Deposits – Withdrawals + InterestBalance = 1522.33 + 220.13 – 327.26 + 0.01(1522.33) = 1430.42The balances for the remainder of the periods can be computed in a similar fashion as tabulated below:DateDepositWithdrawalInterestBalance1-May$1,522.33$220.13$327.26$15.221-Jun$1,430.42$216.80$378.51$14.301-Jul$1,283.02$450.35$106.80$12.831-Aug$1,639.40$127.31$350.61$16.391-Sep$1,432.49(b)( )( )dBD tW tiBdt

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6(c) fort= 0 to 0.5:220.13327.260.01(1522.33)91.91dBdt (0.5)1522.3391.91(0.5)1476.38Bfort= 0.5 to 1:220.13327.2600.01(1476.38)92.37dBdt (0.5)1476.3892.37(0.5)1430.19BThe balances for the remainder of the periods can be computed in a similar fashion as tabulated below:DateDepositWithdrawalInterestdB/dtBalance1-May$220.13$327.26$15.22-$91.91$1,522.3316-May$220.13$327.26$14.76-$92.37$1,476.381-Jun$216.80$378.51$14.30-$147.41$1,430.1916-Jun$216.80$378.51$13.56-$148.15$1,356.491-Jul$450.35$106.80$12.82$356.37$1,282.4216-Jul$450.35$106.80$14.61$358.16$1,460.601-Aug$127.31$350.61$16.40-$206.90$1,639.6816-Aug$127.31$350.61$15.36-$207.94$1,536.231-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$1,700MMJASBi-monthlyMonthly1.7(a)The first two steps are(0.1)1000.175(100)0.198.25 Bq/Lc(0.2)98.250.175(98.25)0.196.5306 Bq/LcThe process can be continued to yieldtcdc/dt0100.0000-17.50000.198.2500-17.19380.296.5306-16.89290.394.8413-16.59720.493.1816-16.30680.591.5509-16.02140.689.9488-15.74100.788.3747-15.4656

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70.886.8281-15.19490.985.3086-14.9290183.8157-14.6678(b)The results when plotted on a semi-log plot yields a straight line4.44.54.600.20.40.60.81The slope of this line can be estimated asln(83.8157)ln(100)0.176551 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.1.8studentsJskJ35 ind8020 min603,360 kJind smin1000 JQ33Mwt(101.325 kPa)(11m8m3m350.075 m )(28.97 kg/kmol)314.796 kg(8.314 kPa m/ (kmol K)((20273.15)K)PVmRTstudents3,360 kJ14.86571 K(314.796 kg)(0.718 kJ/(kg K))vQTmCTherefore, the final temperature is 20 + 14.86571 = 34.86571oC.1.9The first two steps yield2450450(0.5)03sin (0)0.50( 0.36) 0.50.1812501250y  2450450(1)0.183sin (0.5)0.50.18( 0.11176) 0.50.2358812501250y   The process can be continued to give the following table and plot:tydy/dttydy/dt00.00000-0.360005.51.102710.177610.5-0.18000-0.1117661.19152-0.275681-0.235880.404726.51.05368-0.310021.5-0.033520.7146070.898660.1061620.323780.532977.50.951750.590232.50.590260.0268281.246860.6971430.60367-0.338498.51.595430.32859

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83.50.43443-0.2271191.75972-0.1765740.320870.258579.51.67144-0.353904.50.450160.67201101.49449-0.0403650.786160.63310-0.50.00.51.01.52.002468101.10The first two steps yield1.52450150(10)(0.5)03sin (0)0.500.12(0.5)0.0612501250y1.52450150(10.06)(1)0.063sin (0.5)0.50.060.13887(0.5)0.0094412501250y  The process can be continued to givetydy/dttydy/dt00.00000-0.120005.51.619810.028760.5-0.060000.1388761.63419-0.4287210.009440.643026.51.41983-0.401731.50.330940.8903471.218970.0695120.776110.608927.51.253720.544232.51.080580.0266981.525840.5754231.09392-0.342098.51.813550.122273.50.92288-0.1870891.87468-0.4014540.829340.321669.51.67396-0.518604.50.990170.69510101.41465-0.1306251.337720.56419-0.50.00.51.01.52.002468101.11When the water level is above the outlet pipe, the volume balance can be written as21.5out3sin ( )3()dVtyydt

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9In order to solve this equation, we must relate the volume to the level. To do this, we recognize that thevolume of a cone is given byV=r2y/3. Defining the side slope ass=ytop/rtop, the radius can be related tothe level (r=y/s) and the volume can be reexpressed as323Vyswhich can be solved for233sVy(1)and substituted into the volume balance1.5223out33sin ( )3dVs Vtydt(2)For the case where the level is below the outlet pipe, outflow is zero and the volume balance simplifies to23sin ( )dVtdt(3)These equations can then be used to solve the problem. Using the side slope ofs= 4/2.5 = 1.6, theinitial volume can be computed as332(0)0.80.20944 m3(1.6)VFor the first step,y<youtand Eq. (3) gives2(0)3sin (0)0dVdtand Euler’s method yields(0.5)(0)(0)0.209440(0.5)0.20944dVVVtdt For the second step, Eq. (3) still holds and2(0.5)3sin (0.5)0.689547dVdt(1)(0.5)(0.5)0.209440.689547(0.5)0.554213dVVVtdt Equation (1) can then be used to compute the new level,

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10233(1.6) (0.554213)1.106529 myBecause this level is now higher than the outlet pipe, Eq. (2) holds for the next step1.5(1)2.124223 1.10652912.019912dVdt(1.5)0.5542132.019912(0.5)1.564169VThe remainder of the calculation is summarized in the following table and figure.tQinVyQoutdV/dt000.209440.8000.50.6895470.209440.800.68954712.124220.5542131.1065290.1043092.0199121.52.9849891.5641691.5637421.2698171.71517122.4804652.4217541.8090362.1830960.297372.51.0745072.5704391.8453252.331615-1.2571130.0597451.9418851.6806541.684654-1.624913.50.3691471.129431.402890.767186-0.3980441.718250.930411.315110.5306571.1875934.52.8666951.5242071.550311.2247061.64198952.7586072.3452021.789772.1055810.6530265.51.4933612.6717151.8692492.431294-0.9379360.2342192.2027481.7527721.95937-1.725156.50.138831.3401731.485221.013979-0.8751571.2948940.9025981.3018730.4975740.797327.52.6395321.3012581.4707030.9688171.67071582.9364892.1366161.7350521.8905961.0458938.51.9127452.6595631.8664112.419396-0.5066590.5095252.4062371.8051642.167442-1.657929.50.0169431.5772791.5680981.284566-1.26762100.8878770.9434671.3212330.54620.34167700.511.522.530246810Vy1.12(a)The force balance can be written as:22(0) ()ddvRmmgc v vdtRx Dividing by mass gives22(0) ()dcdvRgv vdtmRx 

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11(b)Recognizing thatdx/dt=v, the chain rule isdvdvvdtdxSetting drag to zero and substituting this relationship into the force balance gives22(0)()dvgRdxvRx (c)Using separation of variables22(0) ()Rv dvgdxRx Integrating gives22(0)2vRgCRxApplying the initial condition yields220(0)20vRgCRwhich can be solved forC=v02/2 –g(0)R, which can be substituted back into the solution to give2220(0)(0)22vvRggRRxor2202(0)2(0)RvvggRRx Note that the plus sign holds when the object is moving upwards and the minus sign holds when it isfalling.(d)Euler’s method can be developed as2112(0)()()()() ()iiiiiigRv xv xxxv xRxThe first step can be computed as62629.81(6.3710 )(10, 000)1,500(10, 0000)1,500( 0.00654)10, 0001434.6001,500 (6.37100)v 

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12remainder of the calculations can be implemented in a similar fashion as in the following tablexvdv/dxv-analytical01500.000-0.006541500.000100001434.600-0.006821433.216200001366.433-0.007131363.388300001295.089-0.007501290.023400001220.049-0.007941212.475500001140.643-0.008471129.884600001055.973-0.009121041.04970000964.798-0.00995944.20680000865.317-0.01106836.57990000754.742-0.01264713.299100000628.359-0.01513564.197For the analytical solution, the value at 10,000 m can be computed as62266(6.3710 )1,5002(9.81)2(9.81)(6.3710 )1433.216(6.371010, 000)vThe remainder of the analytical values can be implemented in a similar fashion as in the last column of theabove table. The numerical and analytical solutions can be displayed graphically.040080012001600020000400006000080000100000v-analyticalv-numerical1.13The volume of the droplet is related to the radius as343rV(1)This equation can be solved for radius as334Vr(2)The surface area is24Ar(3)Equation (2) can be substituted into Eq. (3) to express area as a function of volume2/3344VA

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13This result can then be substituted into the original differential equation,2/3344dVVkdt (4)The initial volume can be computed with Eq. (1),33344(2.5)65.44985 mm33rVEuler’s method can be used to integrate Eq. (4). For the first step, the result is2/33(65.44985)(0.25)(0)(0)65.449850.08(4)0.25465.449856.28319(0.25)63.87905  dVVVtdtHere are the beginning and ending stepstVdV/dt065.44985-6.283190.2563.87905-6.182250.562.33349-6.082120.7560.81296-5.98281159.31726-5.8843•••923.35079-3.160649.2522.56063-3.088939.521.7884-3.018049.7521.03389-2.947951020.2969-2.87868A plot of the results is shown below. We have included the radius on this plot (dashed line and right scale):02040608002468101.622.4VrEq. (2) can be used to compute the final radius as33(20.2969)1.6921824rTherefore, the average evaporation rate can be computed asminmm080782.0min10mm)692182.15.2(kwhich is approximately equal to the given evaporation rate of 0.08 mm/min.

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141.14The first two steps can be computed as(1)700.019(7020)268( 0.95)268.1(2)68.10.019(68.120)268.1( 0.9139)266.2722TT    The remaining results are displayed below along with a plot of the results.tTdT/dttTdT/dt070.00000-0.9500012.0000059.62967-0.75296268.10000-0.9139014.0000058.12374-0.72435466.27220-0.8791716.0000056.67504-0.69683664.51386-0.8457618.0000055.28139-0.67035862.82233-0.8136220.0000053.94069-0.644871061.19508-0.7827150607080051015201.15The pair of differential equations to be solved are1diR iqdtLCL dqidtor substituting the parameters402, 000diiqdt dqidtThe first step can be implemented by first using the differential equations to compute the slopes40(0)2, 000(1)2, 000didt  0dqdtThen, Euler’s method can be applied as(0.01)02, 000(0.01)20i (0.01)10(0.01)1q

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15For the second step40( 20)2, 000(1)1, 200didt  20dqdt (0.02)201, 200(0.01)32i  (0.02)120(0.01)0.8qThe remaining steps are summarized in the following table and plot:tiqdi/dtdq/dt001-200000.01-201-1200-200.02-320.8-320-320.03-35.20.48448-35.20.04-30.720.128972.8-30.720.05-20.992-0.17921198.08-20.9920.06-9.0112-0.389121138.688-9.01120.072.37568-0.47923863.43682.375680.0811.01005-0.45548470.548511.010050.0915.71553-0.3453762.1281315.715530.116.33681-0.18822-277.03416.33681-40-200204000.020.040.060.080.1-1-0.500.51iq1.16(a) The solution of the differential equation is0tNN eThe doubling time can be computed as the time whenN= 2N0,(20)002NN eln 20.6930.034657/hr20 hrs20 hrs(b) The volume of an individual spherical cell is3cell volume6d(1)

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