Numerical Method for engineering-chapter 1

1

CHAPTER 1 1.1 For body weight: 4.5 + 4.5 + 12 + 4.5 + 33 + TW = 60 TW = 1.5% For total body water: 7.5 + 7.5 + 20 + 7.5 + 2.5 + IW = 100 IW = 55% 1.2 Qstudents = 30 ind × 80

J s kJ × 15 min × 60 × = 2160 kJ ind s min 1000 J

PVMwt (101.325 kPa )(10m × 8m × 3m − 30 × 0.075 m 3 )(28.97 kg/kmol) = = 286.3424 kg RT (8.314 kPa m 3 /( kmol K)((20 + 273.15)K ) Q 2160 kJ ∆T = students = = 10.50615K mC v ( 286.3424 kg)(0.718 kJ/(kg K)) m=

Therefore, the final temperature is 20 + 10.50615 = 30.50615oC. 1.3 This is a transient computation. For the period from ending June 1: Balance = Previous Balance + Deposits – Withdrawals Balance = 1512.33 + 220.13 – 327.26 = 1405.20 The balances for the remainder of the periods can be computed in a similar fashion as tabulated below: Date 1-May

Deposit $ 220.13

Withdrawal

Balance $ 1512.33

$ 327.26

1-Jun

$ 1405.20 $ 216.80

$ 378.61

$ 450.25

$ 106.80

$ 127.31

$ 350.61

1-Jul

$ 1243.39

1-Aug

$ 1586.84

1-Sep

$ 1363.54

1.4 Q1,in = Q2,out + v3,out A3

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

A3 =

1.5

Q1,in − Q2,out v3,out

∑M -∑M in

=

out

40 m 3 /s − 20 m 3 /s = 3.333 m 2 6 m/s

=0

Food + Drink + Air In + Metabolism = Urine + Skin + Feces + Air Out + Sweat Drink = Urine + Skin + Feces + Air Out + Sweat − Food − Air In − Metabolism Drink = 1.4 + 0.35 + 0.2 + 0.4 + 0.2 − 1 − 0.05 − 0.3 = 1.2 L 1.6 v(t ) =

gm (1 − e −( c / m ) t ) c

jumper #1: v(t ) =

9.8(70) (1 − e −(12 / 70 ) 10 ) = 46.8714 12

jumper #2: 46.8714 =

9.8(75) (1 − e − (15 / 75) t ) 15

46.8714 = 49 − 49e −0.2 t 0.04344 = e −0.2 t ln 0.04344 = −0.2t t=

ln 0.04344 = 15.6818 s − 0.2

1.7 You are given the following differential equation with the initial condition, v(t = 0) = v(0), dv c =g− v dt m The most efficient way to solve this is with Laplace transforms sV ( s ) − v(0) =

g c − V ( s) s m

Solve algebraically for the transformed velocity V ( s) =

v(0) g + s + c / m s ( s + c / m)

(1)

The second term on the right of the equal sign can be expanded with partial fractions 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

g A B = + s ( s + c / m) s s + c / m Combining the right-hand side gives g A( s + c / m) + Bs = s ( s + c / m) s ( s + c / m) By equating like terms in the numerator, the following must hold g=A

c m

0 = As + Bs The first equation can be solved for A = mg/c. According to the second equation, B = –A. Therefore, the partial fraction expansion is g mg / c mg / c = − s ( s + c / m) s s +c/m This can be substituted into Eq. 1 to give V ( s) =

v(0) mg / c mg / c + − s +c/m s s +c/m

Taking inverse Laplace transforms yields v(t ) = v(0)e −( c / m ) t +

mg mg −( c / m )t + e c c

or collecting terms v(t ) = v(0)e −( c / m ) t +

(

mg 1 − e − ( c / m )t c

)

The first part is the general solution and the second part is the particular solution for the constant forcing function due to gravity. 1.8 At t = 10 s, the analytical solution is 44.87 (Example 1.1). The relative error can be calculated with

absolute relative error =

analytical − numerical × 100% analytical


4 The numerical results are: step 2 1 0.5

absolute relative error 6.90% 3.32% 1.63%

v(10) 47.9690 46.3639 45.6044

The error versus step size can then be plotted as 8% relative error 6% 4% 2% 0% 0

0.5

1

1.5

2

2.5

Thus, halving the step size approximately halves the error. 1.9 (a) You are given the following differential equation with the initial condition, v(t = 0) = 0, dv c' = g − v2 dt m Multiply both sides by m/c′ m dv m = g − v2 c' dt c' Define a = mg / c' m dv = a2 − v2 c' dt Integrate by separation of variables,

∫a

2

dv c' = dt 2 m −v

∫

A table of integrals can be consulted to find that

∫a

2

dx 1 x = tanh −1 2 a a −x


5

Therefore, the integration yields 1 v c' tanh −1 = 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 v c' tanh −1 = t a a m This result can then be rearranged to yield v=

 gc '  gm tanh t  m  c'  

(b) Using Euler’s method, the first two steps can be computed as 0.225 2   v(2) = 0 + 9.8 − (0)  2 = 19.6 68.1   0.225   v(4) = 19.6 + 9.8 − (19.6) 2  2 = 36.6615 68 . 1   The computation can be continued and the results summarized and plotted as: t 0 2 4 6 8 10 12 ∞

v 0 19.6 36.6615 47.3800 52.1461 53.7777 54.2673 54.4622

dv/dt 9.8 8.53075 5.35926 2.38305 0.81581 0.24479 0.07002


6

60 40

vnum vanal

20 0 0

4

8

12

Note that the analytical solution is included on the plot for comparison. 1.10 Before the chute opens (t < 10), Euler’s method can be implemented as 10   v(t + ∆t ) = v(t ) + 9.8 − v(t ) ∆t 80   After the chute opens (t ≥ 10), the drag coefficient is changed and the implementation becomes 50   v(t + ∆t ) = v(t ) + 9.8 − v(t )  ∆t 80   Here is a summary of the results along with a plot: t 0 1 2 3 4 5 6 7 8 9

Chute closed v dv/dt -20.0000 12.3000 -7.7000 10.7625 3.0625 9.4172 12.4797 8.2400 20.7197 7.2100 27.9298 6.3088 34.2385 5.5202 39.7587 4.8302 44.5889 4.2264 48.8153 3.6981

t 10 11 12 13 14 15 16 17 18 19 20

Chute opened v dv/dt 52.5134 -23.0209 29.4925 -8.6328 20.8597 -3.2373 17.6224 -1.2140 16.4084 -0.4552 15.9531 -0.1707 15.7824 -0.0640 15.7184 -0.0240 15.6944 -0.0090 15.6854 -0.0034 15.6820 -0.0013


7

60 30 0 0

5

10

15

20

-30 1.11 (a) The force balance can be written as: m

dv R2 = −mg (0) + cv dt ( R + x) 2

Dividing by mass gives dv R2 c = − g (0) + v 2 dt m ( R + x) (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 dv g (0) R 2 =− dx v ( R + x) 2 (c) Using separation of variables v dv = − g (0)

R2 dx ( R + x) 2

Integrating gives v2 R2 = g (0) +C 2 R+x Applying the initial condition yields v02 R2 = g (0) +C 2 R+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.

8

which can be solved for C = v02/2 – g(0)R, which can be substituted back into the solution to give v2 v2 R2 = g (0) + 0 − g (0) R 2 R+x 2 or v = ± v 02 + 2 g (0)

R2 − 2 g (0) 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  g (0)  R2 v( xi +1 ) = v( xi ) + − ( x − xi ) 2  i +1  v( xi ) ( R + xi )  The first step can be computed as  9.8 (6.37 × 10 6 ) 2  v(10,000) = 1,400 + − (10,000 − 0) = 1,400 + ( −0.007)10,000 = 1,330 6 2   1,400 (6.37 × 10 + 0)  The remainder of the calculations can be implemented in a similar fashion as in the following table x 0 10000 20000 30000 40000 50000 60000 70000 80000 90000 100000

v 1400.000 1330.000 1256.547 1179.042 1096.701 1008.454 912.783 807.413 688.661 549.864 376.568

dv/dx -0.00700 -0.00735 -0.00775 -0.00823 -0.00882 -0.00957 -0.01054 -0.01188 -0.01388 -0.01733 -0.02523

v-analytical 1400.000 1328.272 1252.688 1172.500 1086.688 993.796 891.612 776.473 641.439 469.650 174.033

For the analytical solution, the value at 10,000 m can be computed as v = 1,400 2 + 2(9.8)

(6.37 × 10 6 ) 2 − 2(9.8)(6.37 × 10 6 ) = 1,328.272 6 2 (6.37 × 10 + 10,000)


9 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. 1600 v-analytical v-numerical

1200

800

400

0 0

20000

40000

60000

80000

100000

1.12 (a) The first two steps are c(0.1) = 10 − 0.2(10)0.1 = 9.8 Bq/L c(0.2) = 9.8 − 0.2(9.8)0.1 = 9.604 Bq/L The process can be continued to yield t 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

c 10.0000 9.8000 9.6040 9.4119 9.2237 9.0392 8.8584 8.6813 8.5076 8.3375 8.1707

dc/dt -2.0000 -1.9600 -1.9208 -1.8824 -1.8447 -1.8078 -1.7717 -1.7363 -1.7015 -1.6675 -1.6341

(b) The results when plotted on a semi-log plot yields a straight line


10

2.4 2.3 2.2 2.1 2 0

0.2

0.4

0.6

0.8

1

The slope of this line can be estimated as ln(8.1707) − ln(10) = −0.20203 1 Thus, the slope is approximately equal to the negative of the decay rate. 1.13 The first two steps yield 500   500 y (0.5) = 0 + 3 sin 2 (0) − 0.5 = 0 + [ 0 − 0.41667] 0.5 = −0.20833 1200   1200

[

]

y (1) = −0.20833 + sin 2 (0.5) − 0.41667 0.5 = −0.27301 The process can be continued to give t 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

y 0.00000 -0.20833 -0.27301 -0.03880 0.37474 0.68317 0.69869 0.50281 0.37138 0.52101 0.90991

dy/dt -0.41667 -0.12936 0.46843 0.82708 0.61686 0.03104 -0.39177 -0.26286 0.29927 0.77779 0.73275

t 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10

y 1.27629 1.37907 1.21953 1.04012 1.10156 1.44313 1.84656 2.03672 1.93453 1.72973

dy/dt 0.20557 -0.31908 -0.35882 0.12287 0.68314 0.80687 0.38031 -0.20436 -0.40961 -0.04672


11

2.0 1.5 1.0 0.5 0.0 -0.5

0

2

4

6

8

10

1.14 The first two steps yield  500 300(1 + 0)1.5  y (0.5) = 0 + 3 sin 2 (0) −  0.5 = 0 + [ 0 − 0.25] 0.5 = −0.125 1200  1200   300(1 − 0.125)1.5  y (1) = −0.125 + sin 2 (0.5) −  0.5 = −0.08366 1200   The process can be continued to give t 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10

y 0.00000 -0.12500 -0.08366 0.24924 0.69658 0.93711 0.82396 0.52849 0.36918 0.52689 0.88827 1.13864 1.05881 0.73834 0.48077 0.52530 0.83973 1.13958 1.14687 0.85981 0.54630

dy/dt -0.25000 0.08269 0.66580 0.89468 0.48107 -0.22631 -0.59094 -0.31862 0.31541 0.72277 0.50073 -0.15966 -0.64093 -0.51514 0.08906 0.62885 0.59970 0.01457 -0.57411 -0.62702 -0.11076


12

1.5 1.0 0.5 0.0 -0.5

0

2

4

6

8

10

1.15 The volume of the droplet is related to the radius as V=

4πr 3 3

(1)

This equation can be solved for radius as r =3

3V 4π

(2)

The surface area is A = 4πr 2

(3)

Equation (2) can be substituted into Eq. (3) to express area as a function of volume  3V  A = 4π    4π 

2/3

This result can then be substituted into the original differential equation, dV  3V  = − k 4π   dt  4π 

2/3

(4)

The initial volume can be computed with Eq. (1), V=

4πr 3 4π (3) 3 = = 113.0973 mm 3 3 3

Euler’s method can be used to integrate Eq. (4). Here are the beginning and last steps 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

t 0 0.25 0.5 0.75 1

V 113.0973 110.2699 107.4898 104.7566 102.07

dV/dt -11.3097 -11.1204 -10.9327 -10.7466 -10.5621

38.29357 36.92003 35.57954 34.27169 32.99609

-5.49416 -5.36198 -5.2314 -5.1024 -4.97499

• • • 9 9.25 9.5 9.75 10

A plot of the results is shown below:

100 75 50 25 0 0

2

4

6

8

10

Eq. (2) can be used to compute the final radius as r =3

3(32.99609) = 1.9897 4π

Therefore, the average evaporation rate can be computed as k=

(3 − 1.9897) mm 60 min mm = 0.10103 10 min hr min

which is approximately equal to the given evaporation rate. 1.16 The first two steps can be computed as T (1) = 68 + [ − 0.017(68 − 21)] 1 = 68 + ( −0.799)1 = 67.201 T ( 2) = 67.201 + [ − 0.017(67.201 − 21)] 1 = 68 + (−0.78542)1 = 66.41558 The remaining results are displayed below along with a plot 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.

14

t 0 1 2 3 4 5 6 7 8 9 10

T 68.00000 67.20100 66.41558 65.64352 64.88458 64.13854 63.40519 62.68430 61.97566 61.27908 60.59433

dT/dt -0.79900 -0.78542 -0.77206 -0.75894 -0.74604 -0.73336 -0.72089 -0.70863 -0.69659 -0.68474 -0.67310

80 60 40 20 0 0

2

4

6

8

10

1.17 (a) The solution of the differential equation is N = N 0 e µt The doubling time can be computed as the time when N = 2N0, 2 N 0 = N 0 e µ ( 20 )

µ=

ln 2 0.693 = = 0.034657/hr 20 hrs 20 hrs

(b) The volume of an individual spherical cell is cell volume =

πd 3 6

(1)

The total volume is volume =

πd 3 N 6

(2)

The rate of change of N is defined as PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

15

dN = µN dt

(3)

If N = N0 at t = 0, Eq. 3 can be integrated to give N = N 0 e µt

(4)

Therefore, substituting (4) into (2) gives an equation for volume volume =

πd 3 N 0 e µt 6

(5)

(c) This equation can be solved for time ln t=

6 × volume πd 3 N 0 µ

(6)

The volume of a 500 µm diameter tumor can be computed with Eq. 2 as 65,449,847. Substituting this value along with d = 20 µm, N0 = 1 and µ = 0.034657/hr gives  6 × 65,449,847   ln  π 20 3 (1)   t= = 278.63 hr = 11.6 d 0.034657

(6)

1.18 Continuity at the nodes can be used to determine the flows as follows: Q1 = Q2 + Q3 = 0.6 + 0.4 = 1

Q10 = Q1 = 1

m3 s

m3 s

Q9 = Q10 − Q2 = 1 − 0.6 = 0.4

m3 s

Q4 = Q9 − Q8 = 0.4 − 0.3 = 0.1

m3 s

Q5 = Q3 − Q4 = 0.4 − 0.1 = 0.3

m3 s


16

Q6 = Q5 − Q7 = 0.3 − 0.2 = 0.1

m3 s

Therefore, the final results are 1

0.4 0.6

1

0.1 0.4

0.3 0.1

0.2

0.3


Numerical Method for engineering-chapter 1

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