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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

 g  c v dv   dt m

The integrals can be evaluated as c   ln  g  v  m     t C c/m

where C = a constant of integration, which can be evaluated by applying the initial condition to yield c   ln  g  v(0)  m  C  c/m

which can be substituted back into the solution c  c    ln  g  v  ln  g  v(0)  m  m     t c/m c/m

This result can be rearranged algebraically to solve for v, v  v(0)e( c / m )t 



mg 1  e  ( c / m )t 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)

v



mg 1  e  ( c / m )t c



Laplace transform solution: An alternative solution is provided by applying Laplace transform to the differential equation to give sV ( s )  v(0) 

g c  V (s) s m

Solve algebraically for the transformed velocity 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

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 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 gA

c m

0  As  Bs

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 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



1.2 At t = 8 s, the analytical solution is 41.137 (Example 1.1). The relative error can be calculated with absolute relative error 

analytical  numerical  100% analytical

The numerical results are: step

v(8)

2 1 0.5

44.8700 42.8931 41.9901

absolute relative error 9.074% 4.268% 2.073%

The error versus step size can then be plotted 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.

3

10% 8% 6% 4% 2% 0% 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, dv c'  g  v2 dt m

Multiply both sides by m/c′ gives m dv m  g  v2 c ' dt c ' Define a  mg / c ' m dv  a2  v2 c ' dt

Integrate by separation of variables,

a

dv 2

 v2



c'

 mdt

A table of integrals can be consulted to find that

a

dx 2

x

2



1 x tanh 1 a a

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

4

 gc '  gm tanh  t  c'  m 

v

(b) Using Euler’s method, the first two steps can be computed as 0.22 2   v(2)  0  9.81  (0)  2  19.62 68.1   0.22   v(4)  19.62  9.81  (19.62) 2  2  36.75284 68.1  

The computation can be continued and the results summarized along with the analytical result as: t

v-numerical

dv/dt

v-analytical

0 2 4 6 8 10 12 

0 19.62 36.75284 47.64539 52.59819 54.34314 54.88241 55.10572

9.81 8.56642 5.446275 2.476398 0.872478 0.269633 0.079349 0.022993

0 18.83093 33.72377 43.46492 49.06977 52.05938 53.58978 55.10572

A plot of the numerical and analytical results can be developed 60 40 20

v-numerical v-analytical

0 0

4

8

12

gm (1  e ( c / m ) t ) c 9.81(70) (1  e (12/70) 9 )  44.99204 jumper #1: v(t )  12 9.81(80) (1  e (15/80) t ) jumper #2: 44.99204  15

1.4 v(t ) 

44.99204  52.32  52.32e 0.1875 t 0.14006  e 0.1875 t ln 0.14006  0.1875t

t

ln 0.14006  10.4836 s 0.1875

1.5 Before the chute opens (t < 10), Euler’s method can be implemented 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.

5

10   v(t  t )  v(t )  9.81  v(t )  t 80  

After the chute opens (t  10), the drag coefficient is changed and the implementation becomes 60   v(t  t )  v(t )  9.81  v(t )  t 80  

Here is a summary of the results along with a plot: Chute closed dv/dt v -20.0000 12.3100 -7.6900 10.7713 3.0813 9.4248 12.5061 8.2467 20.7528 7.2159 27.9687 6.3139 34.2826 5.5247 39.8073 4.8341 44.6414 4.2298 48.8712 3.7011

t 0 1 2 3 4 5 6 7 8 9

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

Chute opened dv/dt v 52.5723 -29.6192 22.9531 -7.4048 15.5483 -1.8512 13.6971 -0.4628 13.2343 -0.1157 13.1186 -0.0289 13.0896 -0.0072 13.0824 -0.0018 13.0806 -0.0005 13.0802 -0.0001 13.0800 0.0000

60 30 0 0

5

10

15

20

-30

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 1-May

Deposit

Withdrawal

Interest

$220.13

$327.26

$15.22

$216.80

$378.51

$14.30

$450.35

$106.80

$12.83

$127.31

$350.61

$16.39

1-Jun

$1,430.42

1-Jul

$1,283.02

1-Aug

$1,639.40

1-Sep

(b)

Balance $1,522.33

$1,432.49

dB  D (t )  W (t )  iB dt

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.

6

(c) for t = 0 to 0.5: dB  220.13  327.26  0.01(1522.33)  91.91 dt B (0.5)  1522.33  91.91(0.5)  1476.38 for t = 0.5 to 1: dB  220.13  327.260  0.01(1476.38)  92.37 dt B (0.5)  1476.38  92.37(0.5)  1430.19 The balances for the remainder of the periods can be computed in a similar fashion as tabulated below: Date 1-May 16-May 1-Jun 16-Jun 1-Jul 16-Jul 1-Aug 16-Aug 1-Sep

Deposit Withdrawal $220.13 $327.26 $220.13 $327.26 $216.80 $378.51 $216.80 $378.51 $450.35 $106.80 $450.35 $106.80 $127.31 $350.61 $127.31 $350.61

Interest $15.22 $14.76 $14.30 $13.56 $12.82 $14.61 $16.40 $15.36

dB/dt -$91.91 -$92.37 -$147.41 -$148.15 $356.37 $358.16 -$206.90 -$207.94

Balance $1,522.33 $1,476.38 $1,430.19 $1,356.49 $1,282.42 $1,460.60 $1,639.68 $1,536.23 $1,432.26

(d) As in the plot below, the results of the two approaches are very close. $1,700

Bi-monthly Monthly

$1,600 $1,500 $1,400 $1,300 $1,200 M

M

J

A

S

1.7 (a) The first two steps are c(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

The process can be continued to yield t 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

c 100.0000 98.2500 96.5306 94.8413 93.1816 91.5509 89.9488 88.3747

dc/dt -17.5000 -17.1938 -16.8929 -16.5972 -16.3068 -16.0214 -15.7410 -15.4656

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.

7

0.8 0.9 1

86.8281 85.3086 83.8157

-15.1949 -14.9290 -14.6678

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

4.6

4.5

4.4 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 Qstudents  35 ind  80 m

J s kJ  20 min  60   3,360 kJ ind s min 1000 J

PVMwt (101.325 kPa)(11m  8m  3m  35  0.075 m3 )(28.97 kg/kmol)   314.796 kg RT (8.314 kPa m3 / (kmol K)((20  273.15)K)

T 

Qstudents 3,360 kJ   14.86571 K (314.796 kg)(0.718 kJ/(kg K)) mCv

Therefore, the final temperature is 20 + 14.86571 = 34.86571oC. 1.9 The first two steps yield 450   450 y (0.5)  0  3 sin 2 (0)  0.5  0  (0.36) 0.5  0.18 1250   1250 450   450 y (1)  0.18  3 sin 2 (0.5)  0.5   0.18  ( 0.11176) 0.5  0.23588 1250   1250

The process can be continued to give the following table and plot: t 0 0.5 1 1.5 2 2.5 3

y 0.00000 -0.18000 -0.23588 -0.03352 0.32378 0.59026 0.60367

dy/dt -0.36000 -0.11176 0.40472 0.71460 0.53297 0.02682 -0.33849

t 5.5 6 6.5 7 7.5 8 8.5

y 1.10271 1.19152 1.05368 0.89866 0.95175 1.24686 1.59543

dy/dt 0.17761 -0.27568 -0.31002 0.10616 0.59023 0.69714 0.32859

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

3.5 4 4.5 5

0.43443 0.32087 0.45016 0.78616

-0.22711 0.25857 0.67201 0.63310

9 9.5 10

1.75972 1.67144 1.49449

-0.17657 -0.35390 -0.04036

2.0 1.5 1.0 0.5 0.0 -0.5

0

2

4

6

8

10

1.10 The first two steps yield  450 150(1  0)1.5  y (0.5)  0  3 sin 2 (0)   0.5  0  0.12(0.5)   0.06 1250  1250   450 150(1  0.06)1.5  y (1)  0.06  3 sin 2 (0.5)   0.5  0.06  0.13887(0.5)  0.00944 1250  1250 

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 -0.06000 0.00944 0.33094 0.77611 1.08058 1.09392 0.92288 0.82934 0.99017 1.33772

dy/dt -0.12000 0.13887 0.64302 0.89034 0.60892 0.02669 -0.34209 -0.18708 0.32166 0.69510 0.56419

t 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10

y 1.61981 1.63419 1.41983 1.21897 1.25372 1.52584 1.81355 1.87468 1.67396 1.41465

dy/dt 0.02876 -0.42872 -0.40173 0.06951 0.54423 0.57542 0.12227 -0.40145 -0.51860 -0.13062

2.0 1.5 1.0 0.5 0.0 -0.5

0

2

4

6

8

10

1.11 When the water level is above the outlet pipe, the volume balance can be written as

dV  3sin 2 (t )  3( y  yout )1.5 dt 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

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 = r2y/3. Defining the side slope as s = ytop/rtop, the radius can be related to the level (r = y/s) and the volume can be reexpressed as V

 3s

2

y3

which can be solved for y3

3s 2V

 (1)

and substituted into the volume balance 1.5

 3s 2V  dV  3sin 2 (t )  3  3  yout    dt   

(2)

For the case where the level is below the outlet pipe, outflow is zero and the volume balance simplifies to dV  3sin 2 (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 V (0) 

 3(1.6) 2

0.83  0.20944 m3

For the first step, y < yout and Eq. (3) gives dV (0)  3sin 2 (0)  0 dt

and Euler’s method yields V (0.5)  V (0) 

dV (0)t  0.20944  0(0.5)  0.20944 dt

For the second step, Eq. (3) still holds and dV (0.5)  3sin 2 (0.5)  0.689547 dt dV V (1)  V (0.5)  (0.5) t  0.20944  0.689547(0.5)  0.554213 dt Equation (1) can then be used to compute the new level,

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.

10

y3

3(1.6) 2 (0.554213)



 1.106529 m

Because this level is now higher than the outlet pipe, Eq. (2) holds for the next step dV 1.5 (1)  2.12422  3 1.106529  1  2.019912 dt V (1.5)  0.554213  2.019912(0.5)  1.564169 The remainder of the calculation is summarized in the following table and figure. 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

Qin 0 0.689547 2.12422 2.984989 2.480465 1.074507 0.059745 0.369147 1.71825 2.866695 2.758607 1.493361 0.234219 0.13883 1.294894 2.639532 2.936489 1.912745 0.509525 0.016943 0.887877

V 0.20944 0.20944 0.554213 1.564169 2.421754 2.570439 1.941885 1.12943 0.93041 1.524207 2.345202 2.671715 2.202748 1.340173 0.902598 1.301258 2.136616 2.659563 2.406237 1.577279 0.943467

y 0.8 0.8 1.106529 1.563742 1.809036 1.845325 1.680654 1.40289 1.31511 1.55031 1.78977 1.869249 1.752772 1.48522 1.301873 1.470703 1.735052 1.866411 1.805164 1.568098 1.321233

Qout 0 0 0.104309 1.269817 2.183096 2.331615 1.684654 0.767186 0.530657 1.224706 2.105581 2.431294 1.95937 1.013979 0.497574 0.968817 1.890596 2.419396 2.167442 1.284566 0.5462

dV/dt 0 0.689547 2.019912 1.715171 0.29737 -1.25711 -1.62491 -0.39804 1.187593 1.641989 0.653026 -0.93793 -1.72515 -0.87515 0.79732 1.670715 1.045893 -0.50665 -1.65792 -1.26762 0.341677

3 2.5 2 1.5 1 0.5 0 0

2

4

6 V

8

10

y

1.12 (a) The force balance can be written as: m

dv R2   mg (0)  cd v v dt ( R  x) 2

Dividing by mass gives c dv R2   g (0)  dvv 2 dt m ( R  x) 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

(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 ( R  x)2

dx

Integrating gives v2 R2  g (0) C 2 Rx

Applying the initial condition yields v02 R2  g (0) C 2 R0

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   v02  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  9.81 (6.37  106 ) 2  v(10, 000)  1,500    (10, 000  0)  1,500  (0.00654)10, 000  1434.600 6 2  1,500 (6.37  10  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.

12

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

dv/dx -0.00654 -0.00682 -0.00713 -0.00750 -0.00794 -0.00847 -0.00912 -0.00995 -0.01106 -0.01264 -0.01513

v 1500.000 1434.600 1366.433 1295.089 1220.049 1140.643 1055.973 964.798 865.317 754.742 628.359

v-analytical 1500.000 1433.216 1363.388 1290.023 1212.475 1129.884 1041.049 944.206 836.579 713.299 564.197

For the analytical solution, the value at 10,000 m can be computed as v  1,5002  2(9.81)

(6.37  106 ) 2 (6.37  106  10, 000)

 2(9.81)(6.37  106 )  1433.216

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

1.13 The volume of the droplet is related to the radius as V

4 r 3 3

(1)

This equation can be solved for radius as r

3

3V 4

(2)

The surface area is A  4 r 2

(3)

Equation (2) can be substituted into Eq. (3) to express area as a function of volume  3V  A  4    4 

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.

13 This result can then be substituted into the original differential equation, dV  3V   k 4   dt  4 

2/3

(4)

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

4 r 3 4 (2.5)3   65.44985 mm3 3 3

Euler’s method can be used to integrate Eq. (4). For the first step, the result is dV  3(65.44985)  (0)  t  65.44985  0.08(4)   dt 4    65.44985  6.28319(0.25)  63.87905

V (0.25)  V (0) 

2/3

 0.25

Here are the beginning and ending steps t 0 0.25 0.5 0.75 1

V 65.44985 63.87905 62.33349 60.81296 59.31726

dV/dt -6.28319 -6.18225 -6.08212 -5.98281 -5.8843

23.35079 22.56063 21.7884 21.03389 20.2969

-3.16064 -3.08893 -3.01804 -2.94795 -2.87868

• • • 9 9.25 9.5 9.75 10

A plot of the results is shown below. We have included the radius on this plot (dashed line and right scale): 80 60 40 20 0

V

2.4

r

2 1.6 0

2

4

6

8

10

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

3

3(20.2969)  1.692182 4

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

mm (2.5  1.692182) mm  0.080782 10 min min

which is approximately equal to the given evaporation rate of 0.08 mm/min. 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

1.14 The first two steps can be computed as T (1)  70   0.019(70  20)  2  68  (0.95)2  68.1 T (2)  68.1   0.019(68.1  20)  2  68.1  (0.9139)2  66.2722

The remaining results are displayed below along with a plot of the results. t 0 2 4 6 8 10

T 70.00000 68.10000 66.27220 64.51386 62.82233 61.19508

dT/dt -0.95000 -0.91390 -0.87917 -0.84576 -0.81362 -0.78271

t 12.00000 14.00000 16.00000 18.00000 20.00000

T 59.62967 58.12374 56.67504 55.28139 53.94069

dT/dt -0.75296 -0.72435 -0.69683 -0.67035 -0.64487

80 70 60 50 0

5

10

15

20

1.15 The pair of differential equations to be solved are

di R 1  i q dt L CL dq i dt or substituting the parameters di  40i  2, 000q dt dq i dt

The first step can be implemented by first using the differential equations to compute the slopes di  40(0)  2, 000(1)  2, 000 dt dq 0 dt

Then, Euler’s method can be applied as i (0.01)  0  2, 000(0.01)  20 q(0.01)  1  0(0.01)  1

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15 For the second step di  40(20)  2, 000(1)  1, 200 dt dq  20 dt i (0.02)  20  1, 200(0.01)  32 q(0.02)  1  20(0.01)  0.8

The remaining steps are summarized in the following table and plot: t

i

q

di/dt

dq/dt

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

0 -20 -32 -35.2 -30.72 -20.992 -9.0112 2.37568 11.01005 15.71553 16.33681

1 1 0.8 0.48 0.128 -0.1792 -0.38912 -0.47923 -0.45548 -0.34537 -0.18822

-2000 -1200 -320 448 972.8 1198.08 1138.688 863.4368 470.5485 62.12813 -277.034

0 -20 -32 -35.2 -30.72 -20.992 -9.0112 2.37568 11.01005 15.71553 16.33681

40

1

20

0.5

0

0 0

0.02

0.04

0.06

0.08

-20

0.1 -0.5

i

-40

q

-1

1.16 (a) The solution of the differential equation is N  N0 et

The doubling time can be computed as the time when N = 2N0, 2N 0  N 0 e  (20)



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

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

 d3 6

(1)

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16

The total volume is volume 

 d3 6

N

(2)

The rate of change of N is defined as dN  N dt

(3)

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

(4)

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

 d3 6

N 0 e t

(5)

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

6  volume  d 3 N0



(6)

The volume of a 500 m diameter tumor can be computed with Eq. 2 as 65,449,847. Substituting this value along with d = 20 m, N0 = 1 and  = 0.034657/hr gives  6  65,449,847  ln    203 (1)  t   278.63 hr  11.6 d 0.034657

(6)

1.17 Continuity at the nodes can be used to determine the flows as follows: Q1  Q2  Q3  0.6  0.4  1.0 m3 s Q10  Q1  1.0 m3 s Q9  Q10  Q2  1.0  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 Q6  Q5  Q7  0.3  0.2  0.1 m3 s

Therefore, the final results are

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.

17

1

0.4

0.6

0.3

0.1

1

0.1

0.4

0.2

0.3

1.18 (a) Substituting Eq. (1.10) into Eq. (P1.18) gives dx gm  (1  e  ( c / m )t ) dt c

Separation of variables gives



x

0

dx 

gm c

t

 1 e

 ( c / m )t

0

dt

Integration yields x

gm gm 2 t  2 (1  e ( c / m )t ) c c

(b) Euler’s method can be applied for the first step as dv c 12.5 (0)  g  v  9.81  0  9.81 dt m 68.1 dx (0)  v  0 dt dv v(2)  v(0)  (0)t  0  9.81(2)  19.62 dt dx x(2)  x(0)  (0)t  0  0(2)  0 dt

For the second step: dv 12.5 (2)  9.81  19.62  6.2087 dt 68.1 dx (0)  19.62 dt v(4)  19.62  6.2087(2)  32.0374 x(4)  0  19.62(2)  39.24

The remaining steps can be computed in a similar fashion as tabulated below along with the analytical solution: t 0 2

vnum 0.0000 19.6200

xnum 0.0000 0.0000

dv/dt 9.8100 6.2087

dx/dt 0.0000 19.6200

vanal 0.0000 16.4217

xanal 0.0000 17.4242

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.

18

4 6 8 10

32.0374 39.8962 44.8700 48.0179

39.2400 103.3147 183.1071 272.8472

3.9294 2.4869 1.5739 0.9961

32.0374 39.8962 44.8700 48.0179

27.7976 35.6781 41.1372 44.9189

62.3380 126.2949 203.4435 289.7305

(c) 50

400

vnum vanal xnum xanal

40 30

300 200

20 100

10 0

0 0

2

4

6

8

10

1.19 (a) For the constant temperature case, Newton’s law of cooling is written as

dT  0.12(T  10) dt The first two steps of Euler’s methods are dT (0)  t  37  0.12(10  37)(0.5)  37  3.2400  0.50  35.3800 dt T (1)  35.3800  0.12(10  35.3800)(0.5)  35.3800  3.0456  0.50  33.8572 T (0.5)  T (0) 

The remaining calculations are summarized in the following table: t 0:00 0:30 1:00 1:30 2:00 2:30 3:00 3:30 4:00 4:30 5:00

Ta 10 10 10 10 10 10 10 10 10 10 10

T 37.0000 35.3800 33.8572 32.4258 31.0802 29.8154 28.6265 27.5089 26.4584 25.4709 24.5426

dT/dt -3.2400 -3.0456 -2.8629 -2.6911 -2.5296 -2.3778 -2.2352 -2.1011 -1.9750 -1.8565 -1.7451

(b) For this case, the room temperature can be represented as Ta  20  2t

where t = time (hrs). Newton’s law of cooling is written as dT  0.12(T  20  2t ) dt

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19 The first two steps of Euler’s methods are T (0.5)  37  0.12(20  37)(0.5)  37  2.040  0.50  35.9800 T (1)  35.9800  0.12(19  35.9800)(0.5)  35.9800  2.0376  0.50  34.9612

The remaining calculations are summarized in the following table: t 0:00 0:30 1:00 1:30 2:00 2:30 3:00 3:30 4:00 4:30 5:00

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

dT/dt -2.0400 -2.0376 -2.0353 -2.0332 -2.0312 -2.0294 -2.0276 -2.0259 -2.0244 -2.0229 -2.0215

T 37.0000 35.9800 34.9612 33.9435 32.9269 31.9113 30.8966 29.8828 28.8699 27.8577 26.8462

Comparison with (a) indicates that the effect of the room air temperature has a significant effect on the expected temperature at the end of the 5-hr period (difference = 26.8462 – 24.5426 = 2.3036oC). (c) The solutions for (a) Constant Ta, and (b) Cooling Ta are plotted below: 40 Constant Ta Cooling Ta

36 32 28 24 0:00

1:00

1.20 (a) dx dy  vx  vy dt dt

2:00

3:00

4:00

5:00

dvx c   vx dt m

dv y dt

g

c vy m

(b) The first step, dx t  0  180(1)  180 dt dy y (1)  y (0)  t  100  0(1)  100 dt dv 12.5 vx (1)  vx (0)  x t  180  180(1)  147.8571 dt 70 dv y 12.5   (0)  (1)  9.81 t  0  9.81  v y (1)  v y (0)  70 dt   x(1)  x(0) 

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20 The second step x(2)  180  147.8571(1)  327.8571 y (1)  100  9.81(1)  90.19 12.5 vx (1)  147.8571  147.8571(1)  121.4541 70 12.5   v y (1)  9.81  9.81  (9.81)  (1)  17.8682 70  

These along with the remaining results can be tabulated as t

x

y

vx

vy

dx/dt

dy/dt

dvx/dt

dvy/dt

0 1 2 3 4 5 6 7 8 9 10

0.0000 180.0000 327.8571 449.3112 549.0771 631.0276 698.3441 753.6398 799.0613 836.3718 867.0197

-100.0000 -100.0000 -90.1900 -72.3218 -47.8343 -17.9096 16.4814 54.5411 95.6145 139.1633 184.7456

180.0000 147.8571 121.4541 99.7659 81.9505 67.3165 55.2957 45.4215 37.3105 30.6479 25.1751

0.0000 9.8100 17.8682 24.4875 29.9247 34.3910 38.0598 41.0734 43.5488 45.5823 47.2526

180.0000 147.8571 121.4541 99.7659 81.9505 67.3165 55.2957 45.4215 37.3105 30.6479 25.1751

0.0000 9.8100 17.8682 24.4875 29.9247 34.3910 38.0598 41.0734 43.5488 45.5823 47.2526

-32.1429 -26.4031 -21.6882 -17.8153 -14.6340 -12.0208 -9.8742 -8.1110 -6.6626 -5.4728 -4.4955

9.8100 8.0582 6.6192 5.4372 4.4663 3.6687 3.0136 2.4755 2.0334 1.6703 1.3720

(c) The following plot indicates that the jumper will hit the ground in about t = 5.6 s at about x = 670 m. -150

y versus x

-100 -50

0

200

400

600

800

6

8

0 50 100 150 -150

y versus t

-100 -50

0

2

4

0 50 100 150

1.21 (a) The force balance can be written as m

dv 1  mg   v v ACd 2 dt

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21

Dividing by mass gives

 ACd dv g vv dt 2m

(1)

The mass of the sphere is sV where V = volume (m3). The area and volume of a sphere are d2/4 and d3/6, respectively. Substituting these relationships gives 3  Cd dv g vv dt 4d  s

dx v dt

(b) The first step for Euler’s method is dv 3(1.3)0.47  9.81  ( 40) 40  10.0363 4(1.2)2700 dt dx  40 dt v  40  10.0363(2)  19.9274 dx  100  40(2)  20 dt

The remaining steps are shown in the following table: t 0 2 4 6 8 10 12 14

x 100.0000 20.0000 -19.8548 -20.2450 18.6049 96.4813 212.7399 366.3269

v -40.0000 -19.9274 -0.1951 19.4249 38.9382 58.1293 76.7935 94.7453

dx/dt -40.0000 -19.9274 -0.1951 19.4249 38.9382 58.1293 76.7935 94.7453

dv/dt 10.0363 9.8662 9.8100 9.7566 9.5956 9.3321 8.9759 8.5404

(c) The results can be graphed as (notice that we have reversed the axis for the distance, x, so that the negative elevations are upwards. 120

v

80 40 0 0 -40

5

10

15

v

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22

-100 0

5

x

10

15

0 100 200 300 x

400

(d) Inspecting the differential equation for velocity (Eq. 1) indicates that the bulk drag coefficient is c' 

 ACd 2

Therefore, for this case, because A = (1.2)2/4 = 1.131 m2, the bulk drag coefficient is c' 

1.3(1.131)0.47 kg  0.3455 2 m

1.22 (a) A force balance on a sphere can be written as: m

dv  Fgravity  Fbuoyancy  Fdrag dt

where

Fgravity  mg

Fbuoyancy  Vg

Fdrag  3 dv

Substituting the individual terms into the force balance yields m

dv  mg  Vg  3 dv dt

Divide by m dv Vg 3 dv g  dt m m

Note that m = sV, so dv  g 3 dv g  dt s  sV

The volume can be represented in terms of more fundamental quantities as V = d3/6. Substituting this relationship into the differential equation gives the final differential equation

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

 dv   g 1  dt  s 

 18 v  2  s d

(b) At steady-state, the equation is   0  g 1   s 

 18 v  2  s d

which can be solved for the terminal velocity v 

g s   2 d 18 

This equation is sometimes called Stokes Settling Law. (c) Before computing the result, it is important to convert all the parameters into consistent units. For the present problem, the necessary conversions are d  10 μm 

 s  2.65

m  105 m 6 10 μm

g 106 cm3 g kg   3  2650 3 cm3 m3 10 kg m

 1

g 106 cm3 g kg   3  1000 3 3 3 cm m 10 kg m

  0.014

g 100 cm kg kg    0.0014 cm s m 1000 g ms

The terminal velocity can then computed as v 

9.81 2650  1000 m (1 105 ) 2  6.42321 105 18 0.0014 s

(d) The Reynolds number can be computed as Re 

 dv 1000(105 )6.42321 105   0.0004588  0.0014

This is far below 1, so the flow is very laminar. (e) Before implementing Euler’s method, the parameters can be substituted into the differential equation to give dv 18(0.0014)  1000   9.81 1  v  6.108113  95,094v  2 2650 2650(0.00001) dt  

The first two steps for Euler’s method are v(3.8147  106 )  0  (6.108113  95,094(0))  3.8147  106  2.33006  105

v(7.6294  106 )  2.33006  105  (6.108113  95,094(2.33006 105 ))  3.8147  10 6  3.81488  105

The remaining steps can be computed in a similar fashion as tabulated and plotted below:

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24

t

v

dv/dt

t

v

dv/dt

0 3.8110–6 –6 7.6310 –5 1.1410 1.5310–5 1.9110–5

0 2.33E-05 3.81E-05 4.76E-05 5.36E-05 5.75E-05

6.108113 3.892358 2.480381 1.580608 1.007233 0.641853

2.2910–5 2.6710–5 3.0510–5 3.4310–5 3.8110–5

5.99E-05 6.15E-05 6.25E-05 6.31E-05 6.35E-05

0.409017 0.260643 0.166093 0.105842 0.067447

1.23 (a) A force balance on a sphere can be written as:

m

dv 1  mg  Vg   v v ACd dt 2

(b) Dividing by mass gives  Vg  ACd dv vv g  dt m 2m

The mass of the sphere is sV where V = volume (m3). The area and volume of a sphere are d2/4 and d3/6, respectively. Substituting these relationships gives  dv   3 Cd  g 1    vv dt  s  4 s d

(c) At steady state, for a sphere falling downward    3  Cd 2 0  g 1    v  s  4s d

which can be solved for v

4 g s d    1   3  Cd   s 

Substituting the parameters gives v

4(9.81)2700(0.01)  1000  m 1    0.68783 3(1000)0.47  2700  s

(d) Before implementing Euler’s method, the parameters can be substituted into the differential equation to give dv  1000  3(1000)0.47 2  9.811  v  6.176667  13.055556v 2  dt  2700  4(2700)(0.01)

The first two steps for Euler’s method are

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.

25

v(0.03125)  0  (6.176667  13.055556(0) 2 )0.03125  0.193021 v(0.0625)  0.193021  (6.176667  13.055556(0.193021) 2 )0.03125  0.370841

The remaining steps can be computed in a similar fashion as tabulated and plotted below: t

v

dv/dt

t

v

dv/dt

0 0.03125 0.0625 0.09375 0.125

0.000000 0.193021 0.370841 0.507755 0.595591

6.176667 5.690255 4.381224 2.810753 1.545494

0.15625 0.1875 0.21875 0.25

0.643887 0.667761 0.678859 0.683860

0.763953 0.355136 0.160023 0.071055

0.0625

0.125

0.8 0.6 0.4 0.2 0.0 0

0.1875

0.25

1.24 Substituting the parameters into the differential equation gives dy 10000   4 x3  12(4) x 2  12(4)2 x  dx 24(2  1011 )0.000325  2.5641 105  x3  12 x 2  48 x 

The first step of Euler’s method is dy  2.5641105  (0)3  12(0) 2  48(0)   0 dx y (0.125)  0  0(0.125)  0

The second step is dy  2.5641 10 5  (0.125)3  12(0.125) 2  48(0.125)   0.000149 dx y (0.25)  0  0.000149(0.125)  1.86361 105

The remainder of the calculations along with the analytical solution are summarized in the following table and plot. Note that the results of the numerical and analytical solutions are close. x 0 0.125 0.25 0.375

y-Euler 0 0 1.86E-05 5.47E-05

dy/dx 0 0.000149 0.000289 0.00042

y-analytical x y-Euler dy/dx y-analytical 0 2.125 0.001832 0.001472 0.001925 9.42E-06 2.25 0.002016 0.001504 0.002111 3.69E-05 2.375 0.002204 0.001531 0.002301 8.13E-05 2.5 0.002395 0.001554 0.002494

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.

26

0.5 0.625 0.75 0.875 1 1.125 1.25 1.375 1.5 1.625 1.75 1.875 2

0.000107 0.000175 0.000257 0.000352 0.000459 0.000578 0.000707 0.000845 0.000992 0.001147 0.00131 0.001478 0.001653

0.000542 0.000655 0.000761 0.000859 0.000949 0.001032 0.001108 0.001177 0.00124 0.001298 0.001349 0.001395 0.001436

0

0.000141 0.000216 0.000305 0.000406 0.000519 0.000643 0.000777 0.00092 0.001071 0.00123 0.001395 0.001567 0.001744

1

2.625 0.00259 0.001574 2.75 0.002787 0.001591 2.875 0.002985 0.001605 3 0.003186 0.001615 3.125 0.003388 0.001624 3.25 0.003591 0.00163 3.375 0.003795 0.001635 3.5 0.003999 0.001638 3.625 0.004204 0.00164 3.75 0.004409 0.001641 3.875 0.004614 0.001641 4 0.004819 0.001641

2

3

0.00269 0.002887 0.003087 0.003288 0.003491 0.003694 0.003898 0.004103 0.004308 0.004513 0.004718 0.004923

4

0 0.001 0.002 0.003 y-Euler y-analytical

0.004 0.005 0.006

1.25 [Note that students can easily get the underlying equations for this problem off the web]. The volume of a sphere can be calculated as 4 Vs   r 3 3 The portion of the sphere above water (the “cap”) can be computed as

Va 

 h2 3

 3r  h 

Therefore, the volume below water is

4  h2 Vs   r 3   3r  h  3 3 Thus, the steady-state force balance can be written as

4 3

4

 h2

3

3

s g  r 3   f g   r 3 



 3r  h   0 

Cancelling common terms gives

4 3

4

s r 3   f  r 3  3

 h2  3r  h   0 3 

Collecting terms yields

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.

27

f



h3  r  f h 2   s   f

3

 43 r 3  0

1.26 [Note that students can easily get the underlying equations for this problem off the web]. The total volume of a right circular cone can be calculated as

1 Vt   r22 H 3 The volume of the frustum below the earth’s surface can be computed as

Vb 

  H  h1  3

r

2 1

 r22  r1r2



Archimedes’ principle says that, at steady state, the downward force of the whole cone must be balanced by the upward buoyancy force of the below ground frustum,

  H  h1  2 2 1 2  r2 Hg  g  r1  r2  r1r2 g b 3 3





(1)

Before proceeding we have too many unknowns: r1 and h1. So before solving, we must eliminate r1 by recognizing that using similar triangles (r1/h1 = r2/H)

r2 h1 H

r1 

which can be substituted into Eq. (1) (and cancelling the g’s)

  H  h1    r2 2 2 r22  1 2   h1   r2  h1  b  r2 H  g   H  3 3 H   Therefore, the equation now has only 1 unknown: h1, and the steady-state force balance can be written as

4 3

4

 h2

3

3

s g  r 3   f g   r 3 



 3r  h   0 

Cancelling common terms gives

4 3

4

s r 3   f  r 3  3

 h2  3r  h   0 3 

and collecting terms yields

f 3



h3  r  f h 2   s   f

 43 r 3  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.