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Solutions Manual c!
to accompany
System Dynamics, First Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Seven
PROPRIETARY AND CONFIDENTIAL
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c!Solutions Manual Copyright 2005 by The McGraw-Hill Companies, Inc.
7-1
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7.1 Note that f1 = pA1 and f2 = pA2. Thus f1 = (A1/A2)f2 = (10/30)60 = 20 lb.Also, A1x1 = A2x2 from conservation of fluid mass. Thus x1 = (A2/A1)x2 = (30/10)6 =
18 in.The work done is f2x2 = 60(6) = 360 lb-in.
7-2
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7.2 The cross-sectional area is A = !(11/2)2 = 95.033 ft2. The net inflow rate is
(1000 " 800)(0.13368) = 26.736 ft3/min
The initial volume is 5A = 475.166 ft3, and the volume after 5 hrs (300 minutes) is
475.166 + 26.736(300) = 8495.97 ft3
Thus the height after 5 hrs is
h =8495.97
A=
8495.9795.033
= 89.4 ft
7-3
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7.3 Summing forces in the horizontal direction and assuming zero acceleration, we obtain
µmg = A(p1 " p2)
orA =
µmg
p1 " p2=
0.6(1000)(3 " 1) # 105
= 0.003 m2
This corresponds to a radius of 30 mm.
7-4
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7.4 Let ft be the tangential force of the surface acting on the cylinder (positive to the left).Summing moments in the clockwise direction about the mass center of the cylinder gives
I" = Rft (1)
From kinematics, if there is no slipping, x = R" and thus x = R".Summing horizontal forces on the cylinder gives
mx = f " ft
Thus ft = f " mx. Substituting this into (1) gives
I" = Rf " mRx = Rf " mR(R")
or(I + mR2)" = Rf = R(p1 " p2)A
With the given values, this becomes!7 + 100(0.4)2
"" = 0.4(3 # 105)0.005 = 600
or" = 26.087
Thus"(t) = 26.087t
7-5
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7.5 Assuming no friction and summing horizontal forces, we obtain
mx = A(p1 " p2)
or60032.2
x =3
144(10)144 = 10
orx = 0.537 ft/sec2
Thusx = 0.537t
andx =
0.5372
t2 = 0.268t2
The volume isV = Ax =
3144
0.067 = 0.001396 ft3
or 2.412 in3.
7-6
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7.6 From conservation of water mass, qo = 10 + 2 = 12 m3/s.From conservation of salt mass,
d
dt(300so) = 2si " qoso
or300
dso
dt= 2si " 12so
7-7
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7.7 From conservation of water mass, qo = 10 + 2 = 12 m3/s.From conservation of salt mass,
d
dt(V so) = 2si " qoso
orV
dso
dt= 2si " 12so
The time constant is # = V/12. Taking the lag time to be four time constants, we have
4# = 4V
12= 20 s
which gives
V =20(12)
4= 60 m3
7-8
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7.8 The capacitance can be computed from (7.2.3).
C =A(h)
g
The liquid surface area isA(h) = 2L
#Dh " h2
ThusC =
2Lg
#Dh " h2
7-9
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7.9 The capacitance can be computed from (7.2.3).
C =A(h)
g
The liquid surface area is A = D1L if h < D2 and
A =2L(h " D2)
tan $+ D1L h $ D2
Thus
C =$ D1L
g h < D22L(h!D2)
g tan ! + D1Lg h $ D2
7-10
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7.10 From (7.2.5),
%A(h)dh
dt= qmi " qmo
where qmo = 0 and qm = qmi here. From Problem 7.8,
A(h) = 2L#
Dh " h2
Thus%2L
#Dh " h2
dh
dt= qm
7-11
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7.11 From (7.2.5),
%A(h)dh
dt= qmi " qmo
where qmo = 0 here. From Problem 7.9,
A(h) =$
D1L h < D22L(h!D2)
tan ! + D1L h $ D2
Thus the model is%D1L
dh
dt= qmi h < D2
%%2L(h " D2)
tan $+ D1L
&dh
dt= qmi h $ D2
7-12
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7.12 For the laminar resistanceR =
128µL
!%D4
where L = 1 m, D = 10!3 m. Thus
R =128(1.58 # 10!5)1!(0.12885)(10!3)4
= 4.996 # 108 m!1s!1
The mass flow rate is
qm =p
R=
1.0133 # 104
4.996 # 108= 2.028 # 10!5 kg/s
The average velocity is found from
v =qm
A%=
2.028 # 10!5
!(10!3/2)2(1.2885)= 20 m/s
The Reynolds number is
Ne =%vD
µ=
1.2885(20)10!3
1.58 # 10!5= 1634
which is less than 2300, so the flow is laminar.The maximum entrance length Le is found from
Ne = 0.06DNe = 0.06(10!3)1634 = 0.098 m
Since this length is much less than the pipe length of 1 m, most of the pipe has laminarflow.
7-13
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7.13 From (7.3.1)
R ='
dp
dqm
(
r(1)
The flow rate isqm =
)p
Ro
whereRo =
12%C2
dA2o
Thusp = Roq
2m
and from (1),
R = 2Roqmr = 2Ro
)pr
Ro
But pr = %ghr, so
R = 2Ro
*%ghr
Ro=
1CdAo
#2ghr
7-14
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7.14 a) From (7.3.12),
Cdp
dt= qmi " qmo
whereC =
A
g
andqmo =
)p
R1
ThusA
g
dp
dt= qmi "
)p
R1
b) From conservation of mass,
%Adh
dt= qmi " qmo = qmi "
)p
R1
But p = %gh, so
%Adh
dt= qmi ""
*%gh
R1
7-15
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7.15 a) The model is
Adh
dt= " g
Rh
where A = 20. The time constant is # = RA/g. Taking the time to empty to be 4# , weobtain # = 200/4 = 50 s. Thus
R =#g
A= 24.525 m!1s!1
b) The model is
Adh
dt= 3 " g
Rh
Thushss =
3Rg
=3(24.525)
9.81= 7.5 m
7-16
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7.16 a) The model is
Adh
dt= " g
Rh
The time constant is# =
RA
g=
150(2)32.2
= 93.168 sec
The time to empty is approximately 4# , regardless of the initial height, and is 4# = 372.671sec.
b) The model is
Adh
dt= 0.1 " g
Rh
The time constant is # = 93.168 and the steady-state height is hss = 0.1R/g = 0.466 ft.The response is
h(t) = 0.466+1 " e!t/93.168
,(1)
Setting h(t) = hss/3 = 0.466/3 in (1) gives
13
= 1 " e!t/93.168
ort = "93.168 ln
23
= 37.776 sec
7-17
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7.17 The model isA
dh
dt= qvi " CdAo
#2gh
At steady state,qvi = CdAo
#2gh
so thatCdAo =
qvi%2gh
(1)
Note that 1 cm is 10!2 m and that
1 liter/min =10!3
60= 1.667 # 10!5 m3/s
So (1) in consistent units becomes
CdAo =qvi(1.667 # 10!5)#
2(9.81)h(10!2)= 3.763 # 10!5 qvi%
h(2)
where qvi is in liters/min and h is in cm. We will compute CdAo from (2) for each data pairin the table, and then average the results. The answers are
CdAo =(6.7329 6.7350 6.9899 7.0619 7.1843 7.2870 7.3867 7.9023 8.6029 10.6456) # 104
The mean of the CdAo values is 7.6528 # 104, and the standard deviation is 1.1936 # 104,which is 16% of the mean.
7-18
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7.18 Let p1 " p3 denote the pressure drop from the bottom of the tank to the outlet ofpipe 3. Denote the pressure drop over the length of pipe 1 by !p1, that across pipe 3 by!p3, and that across the component by !p2. The turbulent resistance relation (7.3.3) isRq2
m = p1 " p3. Thus, because the mass flow rate qm is the same through each element,R1q2
m = !p1, R2q2m = !p2, and R3q2
m = !p3. The total pressure drop across all threeelements is the sum of the drops across each element. Thus
p1 " p3 = !p1 + !p2 + !p3 = R1q2m + R2q
2m + R3q
2m
orp1 " p3 = (R1 + R2 + R3) q2
m
and thus the total resistance is R = R1 + R2 + R3, which shows that turbulent resistanceobeys the series law.
b) From conservation of mass,
%Adh
dt= qmi " qmo = qmi "
)p1 " p3
R
Because p1 " p3 = %gh, the model becomes
%Adh
dt= qmi "
*%gh
R
7-19
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7.19 From (7.3.3),
%Adh
dt= qmi "
*%gh
R1
where p = %gh.
7-20
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7.20 If h < D,
%Adh
dt=
1R1
ps + qmi
If h $ D,
%Adh
dt=
1R1
ps + qmi "1
R2%g(h " D)
7-21
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7.21 a) The model is
%Adh
dt= qmi "
%g
Rh
If there is no inflow,
Adh
dt= " g
Rh
The time constant is # = RA/g and the response is
h(t) = h(0)e!t/"
Take the log of both sides:
ln h(t) = ln h(0) " 1#t (1)
Equation (1) has the form of the equation of a straight line:
ln h(t) = mt + b
where m = "1/# and b = ln h(0). Use the least-squares method to find m and b. TheMATLAB code is
t = [0:300:2400];h = [20.2, 17.26, 14.6, 12.4, 10.4, 9, 7.6, 6.4, 5.4];lnh = log(h);coeff = polyfit(t,lnh,1)
where m is given by coeff(1) and b is given by coeff(2). The results are m = "5.48868#10!4 and b = 3.01. Thus, # = 1.822 # 103 and
R =g#
A= 9.778 # 103 ft!1sec!1
b) If h(0) is known to be exactly 20.2 ft, then (1) becomes
ln h(t) " 3.006 = "1#t = mt (2)
Using (1.6.3) from Chapter 1, we have
m9-
i=1
t2i =9-
i=1
ti ln h(ti)
Continue the code above as follows
m = sum(t.*(lnh-log(20.2)))/sum(t.^2)
This gives m = "5.463 # 10!4. Thus, R = 9.824 # 103.
7-22
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7.22 Applying conservation of mass to each tank gives
%A1dh1
dt= qmi "
%g
R1h1
%A2dh2
dt=
%g
R1h1 "
%g
R2h2
Note that % cancels out in the second equation.
7-23
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7.23 a) Applying conservation of mass to each tank gives
%A1dh1
dt= qmi "
%g
R1(h1 " h2)
%A2dh2
dt=
%g
R1(h1 " h2) "
%g
R2h2
Note that % cancels out in the second equation.b) Substituting the given values we obtain
%Adh1
dt= qmi "
%g
R(h1 " h2)
4Adh2
dt=
g
R(h1 " h2) "
g
3Rh2
Applying the Laplace transform with zero initial conditions, we obtain'
As +g
R
(H1(s) "
g
RH2(s) =
Qmi(s)%
" g
RH1(s) +
'4As +
4g3R
(H2(s) = 0
Let b = g/RA. After dividing both equations by A, they can be expressed as
(s + b)H1(s) " bH2(s) =Qmi(s)
%
"bH1(s) +'
4s +43b(
H2(s) = 0
Using Cramer’s method to solve these equations, we obtain
H2(s) =
.....s + b Qmi(s)/%"b 0
..... /D =bQmi(s)
%D
where
D =
.....s + b "b"b 4s + 4b/3
..... = (s + b)(4s +43b) " b2 = 4s2 +
163
bs +13b2
ThusH2(s)Qmi(s)
=b/%
4s2 + (16/3)bs + (1/3)b2
7-24
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7.24 Applying conservation of mass to each tank gives
%Adh1
dt= "%g
R(h1 " h2)
2%Adh2
dt= qmi +
%g
R(h1 " h2) "
%g
3Rh2
If we divide both equations by %A and let b = g/RA, these equations can be expressed as
dh1
dt= "b(h1 " h2)
2dh2
dt=
qmi
%+ b(h1 " h2) "
b
3Rh2
Applying the Laplace transform with zero initial conditions, we obtain
(s + b) H1(s) " bH2(s) = 0
"bH1(s) +'
2s +43b(
H2(s) =Qmi(s)
%
Using Cramer’s method to solve these equations, we obtain
H1(s) =
.....0 "b
Qmi(s)/% 2s + 4b/3
..... /D =bQmi(s)
%D
where
D =
.....s + b "b"b 2s + 4b/3
..... = (s + b)'
2s +43b(" b2 = 2s2 +
103
bs +13b2
ThusH2(s)Qmi(s)
=b/%
2s2 + (10/3)bs + (1/3)b2
The characteristic roots are
s ="10b/3 ±
#100b2/9 " 8b2/34
='"5
6± 1
6%
19(
b = "1.56b, "0.1069b
7-25
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7.25 From Example 7.4.4 the damping constant is given by
c =128µLA2
!D4
Substituting the given and desired values, we obtain
2000 =128(0.9)LA2
!D4
which givesLA2
D4=
2000!0.9(128)
= 54.5415 (1)
So we have three parameters to select: L, A, and D.Let n be the ratio of the piston area A to the area Ao of the hole through the piston.
n =A
Ao(2)
But
Ao = !'
D
2
(2
and thus
A =n!D2
4From (1),
LA2
D4= L
n2!2
16= 54.5416
soL =
16(54.5415)n2!2
=88.419
n2
Now we try various values for n to see if we obtain a reasonable value for the pistonlength L. Using n = 50 gives L = 0.035 m, which is 1.38 in., which is a reasonable length.
Now we pick the piston area A. A cylinder diameter of 0.05 m (1.96 in.) gives
A = !'0.05
2
(2
= 1.963 # 10!3 m2
So from (2),
Ao =A
n=
1.963 # 10!3
50= 3.9 # 10!5 = !
'D
2
(2
Thus the hole diameter isD = 7.1 # 10!3 m
which is about 0.3 in.So one of many possible designs is
piston diameter = 0.05 m
piston length, L = 0.035 mpiston hole diameter, D = 7.1 # 10!3 m
7-26
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7.26 (a) If m1 = 0, a force balance on the spool valve gives
f(t) = k1x + c1x (1)
Considering the masses m2 and m3 to constitute a single rigid body, Newton’s law gives
(m2 + m3)y + c2y + k2y = A(p1 " p2) (2)
where A is the piston area and (p1 " p2) is the pressure di"erence across the piston (referto Figure 7.4.8).
From equation (5) in Example 7.4.9,
p1 " p2 = ps " 2!p (3)
In this problem, unlike Example 7.4.9, we cannot make the assumption that (m2 +m3)(y) = 0 because the load inertia and the forces k2y and c2y are not negligible. Thusp1 &= p2 here.
Assuming that x, y, and !p are small deviations from equilibrium, then the deviationps will be 0 if the supply pressure is constant, and (3) becomes
p1 " p2 = "2!p (4)
and (2) becomes(m2 + m3)y + c2y + k2y = "2A!p (5)
In addition, we can linearize the relation qv = f(x,!p) for the volume flow rate throughthe spool valve, as follows:
qv = B1x + B2!p
From conservation of mass,Ay = qv = B1x + B2!p
This gives
!p =Ay " B1x
B2
Substitute this into (5) to obtain
(m2 + m3)y + c2y + k2y = "2AAy " B1x
B2
Collect terms:
(m2 + m3)y +/
c2 +2A2
B2
0
y + k2y =2AB1
B2x (6)
The system model is given by equations (1) and (6).(b) The total mass of the spool valve (considered to be rigid) is 2m1. Newton’s law
applied to the spool valve gives
2m1x = f(t) " k1x " c1x
7-27
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or2m1x + c1x + k1x = f(t) (7)
The rest of the problem proceeeds as in part (a). The system model is given by equations(6) and (7).
7-28
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7.27 The equivalent mass of the cylinder is
me = m +I
R2
Use this instead of m in equation (9) of Example 7.4.6 to obtain'
m +I
R2
(x = (R1 + R2)%A2x = A(p1 " p2)
7-29
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7.28 Let qv1 and qv2 be the volume flow flow rates into and out of the center section wherethe pressure is p. If the pressure drop %gh in going up a height h from this point to theplate is small compared to the pressure p, then the pressure at the plate is approximatelythe same as p. From Newton’s law applied to the plate,
mx = pA " kx (1)
where the force pA is that due to the liquid pressure p acting on the plate. Assuming thatmx of the plate is small, then the above equation shows that the pressure force equals thespring force.
pA = kx (2)
Thus x = pA/k. Di"erentiate this with respect to time to obtain.
dx
dt=
A
k
dp
dt(3)
From conservation of volume,
d
dt(Ax) = A
dx
dt= qv1 " qv2 (4)
The model for the dynamic behavior of the pressure as a function of the flow rates isobtained by substituting this expression for dx/dt into (3). The result is
A2
k
dp
dt= qv1 " qv2 (5)
Using the resistances upstream and downstream, the flow rates qv1 and qv2 can beexpressed as functions of the upstream and downstream pressures. The mass flow rates inthe sections are
%qv1 =1R
(p1 " p)
%qv2 =1R
(p " p2)
and (5) becomes
%A2
k
dp
dt=
1R
(p1 " p) " 1R
(p " p2)
orR
2%A2
k
dp
dt+ p =
p1 + p2
2This equation can be solved for p as a functions of time if we are given p1 and p2 as timefunctions.
7-30
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7.29 A force balance on the plate gives kx = Ap. The volume swept out by the plate isV = Ax. With V = 30 in3 and p = 1.5 psi, we have that
x =30A
and k =Ap
x=
1.5A2
30
We were given no indication of any size limits or available spring constants, so we are free tochoose a reasonable value for A. For example, using a plate 6 inches in diameter, A = 9!,and this gives k = 40 lb/in. This gives a plate displacement of x = 3.75 in.
7-31
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7.30 a) For the two resistances in series:
qm2 =1
R1 + R2!p
From the straight line on the graph,
30 =1
R1 + R23 # 104
Since R2 = 400, we obtainR1 = 600 N · s/kg · m2
From equation (4) of Example 7.4.10,
%gh =R2
R1 + R2!p
Thush =
'400100
3 # 104(
/9.81% = 1223/%
where %, in kg/m3, is the mass density of the liquid (which was not specified).b)
&qm1 = "1r&(!p)
where "1/r is the slope of the straight line. Thus r = 700.We have
d
dt&h = "b&h
whereb =
'1r
R1 + R2
R2+
1R2
(g
A=
' 1700
1000400
+1
400
( 9.812
= 0.02978
Thusd
dt&h = "0.02978&h
The time constant is 1/0.02978 = 33.58 s.
7-32
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7.31 From (7.2.5),
%A(h)dh
dt= qmi " qmo
whereA(h) = (2L tan ')h
andqmo =
1R
%gh
Thus% (2L tan ')h
dh
dt= qmi "
1R
%gh
7-33
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7.32 From (7.2.5),
%A(h)dh
dt= qmi " qmo
whereA(h) = (2L tan ')h
and
qmo =
*%gh
Ro
Ro =1
2%C2dA2
o
Thusqmo = %CdAo
#2gh
and% (2L tan ')h
dh
dt= qmi " %CdAo
#2gh
7-34
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7.33 From (7.2.5),
%A(h)dh
dt= qmi " qmo
where, from Problem 7.8,A(h) = 2L
#Dh " h2
and
qmo =)
p
R=
*%gh
R
Thus
2%L#
Dh " h2 dh
dt= qmi "
*%gh
R
7-35
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7.34 (a) From conservation of mass and (7.3.9):
%Ah = %qv " CdAo
#2p% = %qv " %CdAo
#2gh
Thus % cancels out of the equation, and we obtain
100h = qv " 0.5%
64.4h = qv "%
16.1h (1)
(b) At steady state, h = 0 and qv =%
16.1h. With qv = 5, this gives
h =q2v
16.1= 1.55 ft
(c) At h = 1.55,
%16.1h '
116.1(1.55) +
12
(16.1h)!1/2...h=1.55
(h " 1.55) = 5 +110
(h " 1.55)
Then (1) becomes
100h = qv " 5 +110
(h " 1.55)
Let x = h " 1.55 and u = qv " 5 to obtain the linearized model:
100x = u " 0.1x
whose time constant is 100(0.1) = 10 sec.
7-36
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7.35 a) The capacitance can be obtained from
C =A(h)
g
where
A(h) = !'
h
tan '
(2
Thus
C =!
g
'h
tan '
(2
b) From (7.2.5),
%A(h)dh
dt= qmi " qmo
where, from part (a),
A(h) = !'
h
tan '
(2
andqmo =
%gh
R
Thus
!%'
h
tan '
(2 dh
dt= qmi "
%gh
R
7-37
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7.36 For an isothermal process, n = 1, and from (7.5.6),
C =V
nRgT=
201(1715)(70 + 460)
= 2.2 # 10!5 slug " ft2/lb
7-38
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7.37 If pi " p < 0, the flow will be out of the tank (and will be negative). Thus,
Cdp
dt= "f(|pi " p|)
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7.38 For the left-hand tank,
C1d(&p1)
dt=
1R1
(&pi " &p1) "1
R2(&p1 " &p2)
For the right-hand tank,
C2d(&p2)
dt=
1R2
(&p1 " &p2)
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7.39 a)C = %V cp = 1000(250 # 10!6)4.18 # 103 = 1045 J/"C
b)E = C(T " To) = 1045(99 " 20) = 8.256 # 104 J
7-41
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7.40 a)V = 15(10)(8) = 1200 ft3
C = %V cp = 0.0023(6.012 # 103)1200 = 1.659 # 104 ft " lb/"F
b)E = C(T " To) = 1.659 # 104(72 " 68) = 6.636 # 104 ft " lb
7-42
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7.41 See Example 7.6.1, which with T (0) = 20, qv = 0.5, and V = 12 gives
T (t) = 20e!t/24 ++1 " e!t/24
,80
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7.42 a) For conduction, the thermal resistance is given by
R =L
kA
Thus
R1 =10 # 10!3
400!(10!3)2= 7.958 "C/W
R2 =5 # 10!3
400!((1.5/2) # 10!3)2= 7.074 "C/W
The total resistance isR = R1 + R2 = 15.032 "C/W
b) The heat flow rate is
q =30R
=30
15.032= 1.995 W
7-44
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7.43 The total resistance isR =
1h1A
+L
kA+
1h2A
R =' 1
85+
3/[16(12)]47
+115
( 1A
=0.0788
A
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7.44 The resistances are in series and thus they add. Using R = L/kA for conduction andR = 1/hA for convection, and letting x be the required thickness of the middle layer, wehave
R =130
/130
+10 # 10!3
0.2+
x
0.04+
20 # 10!3
0.1
0
=130
'0.047634 +
x
0.04
(
whereq =
1R!T
With q = 400 and !T = 40, we have
R =40400
=130
'0.047634 +
x
0.04
(
This gives x = 3.54 m, which is rather large.
7-46
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7.45 a) The resistance formula is
R =L
kA
For the brick,
Rbrick =4/12
0.086(12)2/144= 3.876
For the concrete,
Rconcrete =4/12
0.02(36)2/144= 1.852
The segments are in parallel because they have the same temperature di"erence, so thetotal resistance is given by
1R
=1
Rbrick+
1Rconcrete
which givesR = 1.253 "F " sec/lb " ft
b) The heat flow rate is given by q = !T/R. For the brick,
qbrick =40
Rbrick= 10.32
For the concrete,
qconcrete =40
Rconcrete= 21.6
The total heat flow rate is
q = qbrick + qconcrete = 31.92 ft " lb/sec
7-47
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7.46 The solution procedure follows that of Example 7.7.4.a) Assuming that the temperature inside the pipe wall does not change with time, then
the same heat flow rate occurs in the inner and outer convection layers and in the pipe wall.Thus the three resistances are in series and we can add them to obtain the total resistance.The inner and outer surface areas are
Ai = 2!riL = 2!'1
2
( ' 112
(10 = 2.618 ft2
Ao = 2!roL = 2!'3
4
( ' 112
(10 = 3.927 ft2
The inner convective resistance is
Ri =1
hiAi=
116(2.618)
= 0.0239sec "Fft lb
Ro =1
hoAo=
11.1(3.927)
= 0.2315sec "Fft lb
The conductive resistance of the pipe wall is
Rc =ln
+rori
,
2!Lk=
ln+
3/41/2
,
2!(10)(10.1)= 6.389 # 10!4 sec "F
ft lb
Thus the total resistance is
R = Ri + Rc + Ro = 0.0239 + 6.389 # 10!4 + 0.2315 = 0.256sec "Fft lb
The heat loss from the pipe, assuming that the water temperature is a constant 120" alongthe length of the pipe, is
qh =1R!T =
10.256
(120 " 70) = 195ft lbsec
To investigate the assumption that the water temperature is constant, compute thethermal energy E of the water in the pipe, using the mass density % = 1.94 slug/ft3 andcp = 25, 000 ft-lb/slug-"F:
E = mcpTi = (!r2i L%)cpTi = 3.1743 # 105 ft lb
Assuming that the water flows at 1 ft/sec, a slug of water will be in the pipe for 20 sec. Dur-ing that time it will lose 195(20) = 3900 ft-lb of heat. Because this amount is approximatelyonly 1% of E, our assumption that the water temperature is constant is confirmed.
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7.47 The thermal energy E of the water in the pipe is
E = mcpTi = (!r2i L%)cpTi = !
' 148
(2
6(1.94)(2.5 # 104)Ti = 396.8Ti
From conservation of heat energy,
dE
dt= " 1
R(Ti " 70)
ormcp
dTi
dt= " 1
R(Ti " 70)
The resistance of the inside surface is
Ri =1
6(0.785)= 0.2123
and the total resistance is
R = Ri + Rv + Ro = 0.2123 + 2.15 # 10!4 + 0.77 = 0.925
Thus396.8
dTi
dt= "1.081 (Ti " 70)
or367
dTi
dt+ Ti = 70
For Ti(0) = 120, the solution is
Ti(t) = 50e!t/367 + 70
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7.48 a)C1 = mcp = %V cp = 1.94(1000)2.5 # 104 = 4.85 # 107 ft " lb/"F
b) From conservation of heat energy,
d
dt(C1T1) = " 1
R1(T1 " To)
or4.85 # 107R1
dT1
dt+ T1 = To
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7.49 The steady-state temperature di"erence is 90 " 70 = 20", and the temperature dif-ference has decayed by 98% (to 70.4") in 4000 sec. So we take 4000 sec to be four timeconstants. Thus # = 4000/4 = 1000 sec. This value is confirmed by noting from the datathat it took 1000 sec for the temperature di"erence to decay by 63% (to 77"). From Problem7.48, the model is
4.85 # 107R1dT1
dt+ T1 = To
so# = 4.85 # 107R1 = 1000
ThusR1 = 2.06 # 10!5 "F " sec/ft " lb
7-51
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7.50 From conservation of heat energy,
d
dt(C1T1) = qi " q1
d
dt(C2T2) = q1 " qo
whereq1 =
1R1
(T1 " T2) qo =1
R2(T2 " To)
ThusC1
dT1
dt= qi "
1R1
(T1 " T2)
C2dT2
dt=
1R1
(T1 " T2) "1
R2(T2 " To)
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7.51 a) From conservation of heat energy,
C1dT1
dt= " 1
R1(T1 " To) +
1R2
(T2 " T1) (1)
C2dT2
dt= " 1
R2(T2 " T1) (2)
b) If C2 ' 0, (2) shows that T1 = T2, and (1) becomes
C1dT1
dt= " 1
R1(T1 " To)
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7.52 a) The sphere model is
cp%VdT
dt= " 1
R(T " To)
where R = 1/hA. The time constant is # = cp%V/hA, and the response is
T (t) = 22 + [T (0) " 22]e!t/"
which has the form!T (t) = !T (0)e!t/"
where !T = T " 22. The following MATLAB session computes the answer.
t = [0:15:135,180:60:960];DeltaT = [95,93,92,90,89,88,87,86,85,84,82,79,...
76,73,71,69,67,65,62,61,59,57,56,54]-22;p = polyfit(t,log(DeltaT),1)tau = -1/p(1);cp = 500; rho = 7920;d = 0.025;r = d/2;V = (4/3)*pi*r^3;A = 4*pi*r^2;h = cp*rho*V/(A*tau)
The regression coe#cients are p = [-0.0008 4.2533]. The answer is h = 13.69 J/(m2 sK).
b) The Biot number is
NB =hL
k=
hr/3k
=13.69(0.025/2)/3
400= 1.426 # 10!4
Because NB is much less than 0.1, the lumped parameter model can be considered accurate.c) Radiation heat transfer, which is dependent on T 4, is thus more significant at higher
temperatures, and does not give an exponential response. A plot of the data and theregression curve shows that the curve is a good fit. The greatest error occurs when T > 90"
and is less than 3". This indicates that radiation heat transfer is a"ecting the process, butonly for the first 30 seconds or so.
7-54
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7.53 From conservation of heat energy,
CdT
dt=
1R
(To " T )
whereR =
1hA
C = %V cp
The surface area isA = 4!r2 = 4!(30 # 10!3)2 = 1.131 # 10!2
and the volume isV =
43!r3 =
43!(30 # 10!3)3 = 1.13 # 10!4
The resistance isR =
1hA
= 0.2947
The capacitance isC = %V cp = 8900(1.13 # 10!4)385 = 387.5
Thus the model is387.5
dT
dt=
10.2947
(50 " T )
or114.2
dT
dt= 50 " T
The time constant is # = 114.2. The response is
T (t) = T (0)e!t/" + To(1 " e!t/" ) = 400e!t/" + 50(1 " e!t/" ) = 50 + 350e!t/"
The time to reach 130" is found from
50 + 350e!t/" = 130
which givest = "# ln(0.2286) = 168.5 s
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7.54 Let T1 be the sphere tempewrature and T2 be the bath temperature. From conservationof heat energy
C1dT1
dt=
1R
(T2 " T1)
C2dT2
dt=
1R
(T1 " T2)
where R is the surface convective resistance.The sphere’s surface area is
A = 4!r2 = 4!(30 # 10!3)2 = 1.131 # 10!2
and the volume isV1 =
43!r3 =
43!(30 # 10!3)3 = 1.13 # 10!4
R =1
hA=
1300(1.131 # 10!2)
= 0.2947
C1 = %V1cp = 8900(1.13 # 10!4)385 = 387.5
C2 = %V2cp = 7900(0.1)400 = 3.16 # 105
SoRC1 = 114.2 RC2 = 9.3125 # 104
Thus114.2
dT1
dt= T2 " T1
9.3125 # 104 dT2
dt= T1 " T2
Let #1 = 114.2 and #2 = 9.3125# 104. Applying the Laplace transform to each equationgives
#1sT1(s) " #1T1(0) = T2(s) " T1(s)
#2sT2(s) " #2T2(0) = T1(s) " T2(s)
These have the solutionT1(s) =
bs + c
s(s + a)
wherea =
#1 + #2
#2#20.008767
b = T1(0) = 400
c =T1(0)#2
+T2(0)#1
= 0.4421
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The response is
T1(t) =c
a+
ab " c
ae!at = 50.43 + 349.6e!0.008767t
The temperature T1 will reach 130" at
t =1
0.008767ln 0.2276 = 168.8 s
Contrast this result with the result of Problem 7.53, which is t = 168.5 s. Thus theassumption of a constant bath temperature in Problem 7.53 is very justified.
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7.55 The last three equations in Example 7.7-1 are:
(R1 + R2)T1 " R1T2 = R2Ti
R3T1 " (R2 + R3)T2 + R2T3 = 0
"R4T2 + (R3 + R4)T3 = R3To
In matrix form these are2
34(R1 + R2) "R1 0
R3 "(R2 + R3) R2
0 "R4 (R3 + R4)
5
67
2
34T1
T2
T3
5
67 =
2
34R2Ti
0R3To)
5
67
Once T1 is computed, qh can be computed from q = (T1 " Ti)/R1.The script file is:
R = [0.036,4.01,0.408,0.038];Ti = 20;To = -10;A = [R(1)+R(2),-R(1),0;R(3),-(R(2)+R(3)),R(2);0,-R(4),R(3)+R(4)];b = [R(2)*Ti;0;R(3)*To];T = A\bq = (1/R(1))*(Ti - T(1))
The results are T = [19.7596,"7.0214,"9.7462] "C and q = 6.6785 watts/m2. Thus thetotal heat loss is 10(6.6785) = 66.785 W.
7-58
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7.56 (a)
q2 =1
R2(p1 " pb)
q3 =1
R3(p1 " pc)
b) Rearrange the equations by bringing all the unknowns to the left side.
R1q1 + p1 = pa
R2q2 " p1 = "pb
R3q3 " p1 = "pc
q1 " q2 " q3 = 0
These equations have the form Ax = b where
A =
2
3334
R1 0 0 10 R2 0 "10 0 R3 "11 "1 "1 0
5
6667 b =
2
3334
pa
"pb
"pc
0
5
6667 x =
2
3334
q1
q2
q3
p1
5
6667
(c) The script file is
pa = 30*144;pb = 25*144; pc = 20*144;R = [10000,14000,14000];A = [R(1),0,0,1;0,R(2),0,-1;0,0,R(3),-1;1,-1,-1,0];b = [pa;-pb;-pc;0];format longx = A\b
When this file is run it gives the following output.
x =1.0e+003 *0.000063529411760.000006050420170.000057478991603.68470588235294
Thus q1 = 6.35 # 10!2, q2 = 6.05 # 10!3, q3 = 5.75 # 10!2 ft3/sec, and p1 = 3685 lb/ft2.
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7.57 For the given values
CdA = 0.5!(2 # 10!2)2 = 2! # 10!4
and the di"erential equation is
!(6h " h2)dh
dt= "2! # 10!4
%19.62h
ordh
dt= "2 # 10!4
%19.62h
6h " h2
(a) The greatest outflow rate occurs when the water level is the highest (h = 5). Thus usingh = 5 in the di"erential equation, we can obtain an lower bound on the time required todrain the tank. When h = 5,
dh
dt= "2 # 10!4
%19.62h
6h " h2= "3.9618 # 10!4
This implies that h(t) = "3.9618# 10!4t+5, and h = 0 at t = 5/(3.9618 # 10!4) = 12, 620s, or 210 minutes. Thus the tank will empty in no less than 210.34 minutes.
Instead of a lower bound on the estimate, we can obtain a higher estimate by using themid-point value for h; namely, h = 5/2 = 2.5. This gives
dh
dt= "2 # 10!4
%19.62h
6h " h2= "1.6008 # 10!4
This implies that h(t) = "1.6008# 10!4t+5, and h = 0 at t = 5/(1.6008 # 10!4) = 31, 234s, or 521 minutes.
(b) To use the ode45 solver, solve for the derivative:
dh
dt= "2 # 10!4
%19.62h
6h " h2
and create the following function file:
function hdot = tank(t,h)hdot = -(0.0002*sqrt(19.62*h))/(6*h-h^2);
Then use the ode45 solver in the following script file.
[t, h] = ode45(’tank’, [0, 25200], 5);plot(t,h),xlabel(’t (seconds)’),ylabel(’ h (feet)’)
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Start with a final time of something more than 12,620 s, and run the file until the plotshows the height approaching zero. The time to empty, which is 25,200 s or 420 minutes,was found this way. The estimate of 521 minutes obtained with the mid-point height is notmuch di"erent, and establishes confidence in the numerical result.
Note that if you choose a final time somewhat larger than 25,200 s, the denominator inthe expression for dh/dt becomes zero because h = 0, and the expression for dh/dt becomesundefined. This causes di#culties for the numerical algorithm. Thus it is best to start witha small value for the final time, and increase it. The plot is shown in the figure.
0 0.5 1 1.5 2 2.5 3x 104
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
t (seconds)
h (f
eet)
Figure : for Problem 7.57.
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7.58 (a) Write the equation asdy
dt= 4 " 2y
10 + 2tThen create the following function file.
function ydot = salt(t,y)ydot = 4-2*y/(10+2*t);
The following file solves the problem using the ode45 solver.
[t, h] = ode45(’salt’, [0, 10], 0);plot(t,h),xlabel(’Time t’),ylabel(’Salt Mass y’)
The plot is shown in the figure.
0 1 2 3 4 5 6 7 8 9 100
5
10
15
20
25
30
Time t
Salt
Mas
s y
Figure : for Problem 7.58.
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(b) The variable coe#cient 2/(10 + 2t) varies from 2/10 to 2/30 as t varies from 0 to10. Its mid-point value is 4/30. Using this value the model becomes
dy
dt+
430
y = 4
The step response is
y(t) =4
4/30
+1 " e!30t/4
,= 30
+1 " e!30t/4
,
This equation predicts that y(10) = 30. The plot shows that the numerical solution givesy(10) = 27 approximately. Thus we can have confidence that the numerical solution iscorrect.
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7.59 The relation between height h and the volume inflow rate r is
100dh
dt= r
Thush(t) =
1100
8 t
0r dt
The MATLAB m-file is
t = [0:10];r = [0,80,130,150,150,160,165,170,160,140,120];for k=2:11
h(k) = (1/100)*trapz(t(1:k),r(1:k));endplot(t,h)
The answer for the final height is given by h(11) and is 13.65 ft. The plot is shown below.
0 1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
12
14
t (sec)
h (ft
)
Figure : for Problem 7.59
7-64
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7.60 This problem requires both analytical and numerical methods.(a) Let
b =13!
'R
H
(2
=13!
'1.54
(2
=3!64
Then V = bh3. When the cup is full, h = 4 in, and the water volume is V = b(4)3 = 64b =3!.
Let q be the flow rate (q = 2 cubic in/sec). From mass conservation,
dV
dt= q
andV (t) =
8 t
0q(t) dt =
8 t
02 dt = 2t
Equating the two expressions for the water volume gives 2t = 64b, or t = 32b = 3!/2 = 4.7s. So we do not need MATLAB for this part of the problem.
(b) If q(t) = 2(1 " e!2t), then
V (t) =8 t
0q(t) dt =
8 t
02(1 " e!2t) dt =
9
2t " 2e!2t
"2
:.....
t
0
= 2t + e!2t " 1
The time to fill is found by equating the two expressions for the volume:
2t + e!2t " 1 = 64b = 3!
We can solve this for t by using the fzero function. First define the function cup:
function f = cup(t)f = 2*t+exp(-2*t)-1 - 3*pi;
Use the fzero function with the answer from part (a) as the starting guess:
(fzero(’cup’,4.7)ans =5.2124
Thus it will take about 5.2 sec to fill the cup.
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7.61 a) The model is just like that shown in Figure 7.10.2 except that the two SSR blocksare not present. Create a subsystem block and save it.
b) Create a Simulink model like that shown in Figure 7.10.3, using the subsystem blockcreated in part (a).
c) In the MATLAB Command window, type the following parameter values.
(A_1 = 2;A_2 = 5;R_1 = 400;R_2 = 600;(rho = 1000; g = 9.81; q_1 = 50;(h10 = 1.5;h20 = 0.5;
Then run the simulation. A Stop time of 2000 s shows the complete response.
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7.62 Using the values given, the equation can be reduced to
dh
dt= " CdA
%2gh
!(2rh " h2)= "8.8589 # 10!4
%h
(6h " h2)
The model is shown in the following figure. Set the Initial condition of Integrator to 5. Inthe Fcn block type -8.8589*10^(-4)*sqrt(u(1))/(6*u(1)-u(1)^2) for the expression.You can plot the results by typing
(plot(tout,simout),xlabel(#t#),ylabel(#x#)
By experimenting with the Stop time, we find that the height is essentially 0 after 25,230 s.
Figure : for Problem 7.62
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7.63 From conservation of fluid mass, 100h = q, where q is the flow rate. Thus
h(t) =1
100
8 t
0q(t) dt
The model is shown in the following figure. We assume that the flow rate remains constantat the previous value for 1 min. Thus in the Lookup Table block, we select the Look-upmethod to be Use Input Below. In this block, the Vector of input values is [0:10] and theVector of output values is [0,80,130,150,50,160,165,170,160,140,120]. The height isapproximately 13 ft after 10 min.
Figure : for Problem 7.63
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7.64 From conservation of mass, the rate of change of water mass in the cup is
d(%V )dt
= %q
or%dV )dt
= %q
where % is the mass density of the water. We see that % cancels out of the equation, whichcan then be expressed as
dV
dt= !
'R
H
(2
h2 dh
dt= q
Using the values given for R and H and solving for dh/dt we obtain
dh
dt=
q
!+
RH
,2h2
=q
!+
964
,h2
a) With q = 2 the equation becomes
dh
dt=
2
!+
964
,h2
The model is shown in the following figure. Set the Initial condition of the Integratorto a small positive number, say 0.01, to avoid a singularity (because h appears in thedenominator of dh/dt. In the Fcn block type 2/pi*(9/64)*u(1)^2) for the expression.You can plot the results by typing
(plot(tout,simout),xlabel(#t#),ylabel(#x#)
By experimenting with the Stop time, we find that the height is essentially equal to 4 in.after 4.7 sec.
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Figure : for Problem 7.64a
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b) With q a function of time, we must modify the model in part (a). The equationbecomes
dh
dt=
q(t)
!+
964
,h2
The model is shown in the following figure. Set the Initial condition of the Integratorto a small positive number, say 0.01, to avoid a singularity (because h appears in thedenominator of dh/dt. In the Fcn block type pi*(9/64)*u(1)^2 for the expression. In theFcn1 block type 2*(1-exp(-2*u(1))) for the expression. The Clock block provides theinput time t to compute the expression 2(1 " e!2t). We use the Divide block to divide q(t)by !
+964
,h2. In the Divide block enter the number of inputs as */ (to multiply by q(t) and
to divide by !+
964
,h2). You can plot the results by typing
(plot(tout,simout),xlabel(#t#),ylabel(#x#)
By experimenting with the Stop time, we find that the height is essentially equal to H = 4after 5.21 sec.
Figure : for Problem 7.64b
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7.65 Use the model developed in Example 7.10.1 and change the Relay settings.
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7.66 The model is shown in the following figure. In the Relay block, set the Switch onpoint to 5.5 and the Switch o" point to 4.5. Set the Output when on to 0 and the Outputwhen o" to 50. Set the Initial condition of the Integrator to 1. Set the gain of the Gainblock to 1/%A = 1/2000. Set the gain of the other block to %g/R. Set the Stop time to1000. Using a value of R = 400, the height does not reach the desired band of 4.5 to 5.5ft because the inflow cannot compensate for the loss due to the small resistance. However,for a larger value, say R = 4000, the height oscillates between the desired values. You canplot the results by typing
(plot(tout,simout),xlabel(#t#),ylabel(#x#)
Figure : for Problem 7.66
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