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
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
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
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
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
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
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
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
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)
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
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
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
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
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
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
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
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
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.
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
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.