# sample - Test Bank College

```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
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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
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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
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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 &lt; 10), Euler’s method can be implemented as
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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
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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
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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
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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
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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 = 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 &lt; 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,
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publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their
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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)
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publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their
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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
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) 
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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
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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.
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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
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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
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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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publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their
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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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individual course preparation. If you are a student using this Manual, you are using it without permission.
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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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
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
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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
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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
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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
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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