1
CHAPTER 1
1.1 You are given the following differential equation with the initial condition, v(t = 0) = 0,
c
dv
= g − d v2
dt
m
Multiply both sides by m/cd
m dv m
=
g − v2
c d dt c d
Define a = mg / c d
m dv
= a2 − v2
c d dt
Integrate by separation of variables,
dv
cd
∫ a 2 − v 2 = ∫ m dt
A table of integrals can be consulted to find that
∫a
2
1
dx
x
= tanh −1
2
a
a
−x
Therefore, the integration yields
1
v c
tanh −1 = d 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 = d t
a
a m
This result can then be rearranged to yield
v=
⎛ gc d ⎞
gm
t⎟
tanh⎜
⎜ m ⎟
cd
⎝
⎠
1.2 This is a transient computation. For the period from ending June 1:
Balance = Previous Balance + Deposits – Withdrawals
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2
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
Deposit
Withdrawal
1-May
Balance
$ 1512.33
$ 220.13
$ 327.26
$ 216.80
$ 378.61
$ 450.25
$ 106.80
$ 127.31
$ 350.61
1-Jun
$ 1405.20
1-Jul
$ 1243.39
1-Aug
$ 1586.84
1-Sep
$ 1363.54
1.3 At t = 12 s, the analytical solution is 50.6175 (Example 1.1). The numerical results are:
step
1
0.5
absolute
relative error
1.15%
0.61%
v(12)
51.2008
50.9259
where the relative error is calculated with
absolute relative error =
analytical − numerical
× 100%
analytical
The error versus step size can be plotted as
2.0%
1.0%
relative error
0.0%
0
0.5
1
1.5
2
2.5
Thus, halving the step size approximately halves the error.
1.4 (a) The force balance is
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3
dv
c'
=g− v
dt
m
Applying Laplace transforms,
sV − v(0) =
g c'
− V
s m
Solve for
V=
g
v(0)
+
s ( s + c ' / m) s + c ' / m
(1)
The first term to the right of the equal sign can be evaluated by a partial fraction expansion,
g
A
B
= +
s ( s + c ' / m) s s + c ' / m
(2)
g
A( s + c' / m) + Bs
=
s ( s + c ' / m)
s ( s + c ' / m)
Equating like terms in the numerators yields
A+ B=0
g=
c'
A
m
Therefore,
A=
mg
c'
B=−
mg
c'
These results can be substituted into Eq. (2), and the result can be substituted back into Eq. (1) to
give
V=
mg / c'
mg / c'
v(0)
−
+
s
s + c' / m s + c' / m
Applying inverse Laplace transforms yields
v=
mg mg −( c '/ m )t
−
e
+ v(0)e −( c '/ m )t
c'
c'
or
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4
v = v(0)e −( c '/ m )t +
(
mg
1 − e −( c '/ m )t
c'
)
where the first term to the right of the equal sign is the general solution and the second is the
particular solution. For our case, v(0) = 0, so the final solution is
v=
(
mg
1 − e −( c '/ m )t
c'
)
(b) The numerical solution can be implemented as
12.5 ⎤
⎡
v(2) = 0 + ⎢9.81 −
(0) 2 = 19.62
68.1 ⎥⎦
⎣
12.5
⎡
⎤
v(4) = 19.62 + ⎢9.81 −
(19.62)⎥ 2 = 32.0374
68.1
⎣
⎦
The computation can be continued and the results summarized and plotted as:
t
0
2
4
6
8
10
12
v
0
19.6200
32.0374
39.8962
44.8700
48.0179
50.0102
dv/dt
9.81
6.2087
3.9294
2.4869
1.5739
0.9961
0.6304
60
40
20
0
0
4
8
12
Note that the analytical solution is included on the plot for comparison.
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5
1.5 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
1.6 Before the chute opens (t < 10), Euler’s method can be implemented as
10
⎡
⎤
v(t + Δt ) = v(t ) + ⎢9.8 − v(t )⎥ Δt
80
⎣
⎦
After the chute opens (t ≥ 10), the drag coefficient is changed and the implementation becomes
50
⎡
⎤
v(t + Δt ) = v(t ) + ⎢9.8 − v(t )⎥ Δt
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
dv/dt
v
-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
dv/dt
v
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
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6
60
30
0
0
5
10
15
20
-30
1.7 (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
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
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7
ln(8.1707) − ln(10)
= −0.20203
1
Thus, the slope is approximately equal to the negative of the decay rate.
1.8 The first two steps yield
500 ⎤
⎡ 500
sin 2 (0) −
y (0.5) = 0 + ⎢3
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
y
dy/dt
t
y
dy/dt
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0.00000
-0.20833
-0.27301
-0.03880
0.37474
0.68317
0.69869
0.50281
0.37138
0.52101
0.90991
-0.41667
-0.12936
0.46843
0.82708
0.61686
0.03104
-0.39177
-0.26286
0.29927
0.77779
0.73275
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
1.27629
1.37907
1.21953
1.04012
1.10156
1.44313
1.84656
2.03672
1.93453
1.72973
0.20557
-0.31908
-0.35882
0.12287
0.68314
0.80687
0.38031
-0.20436
-0.40961
-0.04672
2.0
1.5
1.0
0.5
0.0
-0.5
0
2
4
6
8
10
1.9 The first two steps yield
⎡ 500
300(1 + 0)1.5 ⎤
sin 2 (0) −
y (0.5) = 0 + ⎢3
⎥ 0.5 = 0 + [0 − 0.25]0.5 = −0.125
1200
⎣ 1200
⎦
⎡ 500
300(1 − 0.125)1.5 ⎤
sin 2 (0.5) −
y (1) = −0.125 + ⎢3
⎥ 0.5 = −0.08366
1200
⎣ 1200
⎦
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8
The process can be continued to give
t
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
dy/dt
-0.25000
0.08269
0.66580
0.89468
0.48107
-0.22631
-0.59094
-0.31862
0.31541
0.72277
0.50073
y
0.00000
-0.12500
-0.08366
0.24924
0.69658
0.93711
0.82396
0.52849
0.36918
0.52689
0.88827
t
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
y
1.13864
1.05881
0.73834
0.48077
0.52530
0.83973
1.13958
1.14687
0.85981
0.54630
dy/dt
-0.15966
-0.64093
-0.51514
0.08906
0.62885
0.59970
0.01457
-0.57411
-0.62702
-0.11076
1.5
1.0
0.5
0.0
0
2
4
6
8
10
-0.5
1.10 Q1,in = Q2,out + v3,out A3
A3 =
Q1,in − Q2,out
v3,out
=
40 m 3 /s − 20 m 3 /s
= 3.333 m 2
6 m/s
1.11
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
ΔT = students =
= 10.50615K
mC v
(286.3424 kg)(0.718 kJ/(kg K))
m=
Therefore, the final temperature is 20 + 10.50615 = 30.50615oC.
1.12
∑M - ∑M
in
out
=0
Food + Drink + Air In + Metabolism = Urine + Skin + Feces + Air Out + Sweat
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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
1.13 (a) The force balance can be written as:
m
dv
R2
= − mg (0)
+ c' v
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
g ( 0) R 2
dv
=−
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
v 02
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
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10
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( x i +1 ) = v( x i ) + ⎢−
( x − xi )
2 ⎥ i +1
⎣⎢ v( xi ) ( R + x i ) ⎦⎥
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
(6.37 × 10 + 10,000)
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.
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11
1600
v-analytical
v-numerical
1200
800
400
0
0
20000
40000
60000
80000
100000
1.14
Errata: In the first printing, the rate of evaporation should be changed to 0.1 mm/min.
Subsequent printings should show the correct value.
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),
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12
4πr 3 4π (3) 3
=
= 113.0973 mm 3
3
3
V=
Euler’s method can be used to integrate Eq. (4). Here are the beginning and last steps
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
mm
= 0.10103
10 min
min
which is approximately equal to the given evaporation rate of 0.1 mm/min.
1.15 The first two steps can be computed as
T (1) = 68 + [− 0.017 (68 − 21)]1 = 68 + ( −0.799)1 = 67.201
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13
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
t
0
1
2
3
4
5
dT/dt
-0.79900
-0.78542
-0.77206
-0.75894
-0.74604
-0.73336
T
68.00000
67.20100
66.41558
65.64352
64.88458
64.13854
t
6
7
8
9
10
T
63.40519
62.68430
61.97566
61.27908
60.59433
dT/dt
-0.72089
-0.70863
-0.69659
-0.68474
-0.67310
80
60
40
20
0
0
2
4
6
8
10
1.16 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
Q6 = Q5 − Q7 = 0.3 − 0.2 = 0.1
m3
s
Therefore, the final results are
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14
1
0.4
0.6
1
0.1
0.4
0.3
0.1
0.2
0.3
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