sdsolns07

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

PROPRIETARY MATERIALManual 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.

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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



7.2 The cross-sectional area is A = π(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



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



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

Iω = Rft (1)

From kinematics, if there is no slipping, x = Rω and thus x = Rω.Summing horizontal forces on the cylinder gives

mx = f − ft

Thus ft = f − mx. Substituting this into (1) gives

Iω = Rf − mRx = Rf − mR(Rω)

or(I + mR2)ω = Rf = R(p1 − p2)A

With the given values, this becomes[7 + 100(0.4)2

]ω = 0.4(3 × 105)0.005 = 600

orω = 26.087

Thusω(t) = 26.087t

7-5



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



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



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 τ = V/12. Taking the lag time to be four time constants, we have

4τ = 4V

12= 20 s

which gives

V =20(12)

4= 60 m3

7-8



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



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 φ+ D1L h ≥ D2

Thus

C =

{D1L

g h < D22L(h−D2)

g tan φ + D1Lg h ≥ D2

7-10



7.10 From (7.2.5),

ρA(h)dh

dt= qmi − qmo

where qmo = 0 and qm = qmi here. From Problem 7.8,

A(h) = 2L√

Dh − h2

Thusρ2L

√Dh − h2

dh

dt= qm

7-11



7.11 From (7.2.5),

ρA(h)dh

dt= qmi − qmo

where qmo = 0 here. From Problem 7.9,

A(h) =

{D1L h < D2

2L(h−D2)tan φ + D1L h ≥ D2

Thus the model isρD1L

dh

dt= qmi h < D2

ρ

[2L(h − D2)

tan φ+ D1L

]dh

dt= qmi h ≥ D2

7-12



7.12 For the laminar resistanceR =

128µL

πρD4

where L = 1 m, D = 10−3 m. Thus

R =128(1.58 × 10−5)1π(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

Aρ=

2.028 × 10−5

π(10−3/2)2(1.2885)= 20 m/s

The Reynolds number is

Ne =ρvD

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



7.13 From (7.3.1)

R =(

dp

dqm

)

r

(1)

The flow rate isqm =

√p

Ro

whereRo =

12ρC2

dA2o

Thusp = Roq

2m

and from (1),

R = 2Roqmr = 2Ro

√pr

Ro

But pr = ρghr, so

R = 2Ro

√ρghr

Ro=

1CdAo

√2ghr

7-14



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,

ρAdh

dt= qmi − qmo = qmi −

√p

R1

But p = ρgh, so

ρAdh

dt= qmi −−

√ρgh

R1

7-15



7.15 a) The model is

Adh

dt= − g

Rh

where A = 20. The time constant is τ = RA/g. Taking the time to empty to be 4τ , weobtain τ = 200/4 = 50 s. Thus

R =τg

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




Adh

dt= − g

Rh

The time constant isτ =

RA

g=

150(2)32.2

= 93.168 sec

The time to empty is approximately 4τ , regardless of the initial height, and is 4τ = 372.671sec.

b) The model is

Adh

dt= 0.1 − g

Rh

The time constant is τ = 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



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



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,R1q

2m = ∆p1, R2q

2m = ∆p2, and R3q

2m = ∆p3. The total pressure drop across all three

elements 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,

ρAdh

dt= qmi − qmo = qmi −

√p1 − p3

R

Because p1 − p3 = ρgh, the model becomes

ρAdh

dt= qmi −

√ρgh

R

7-19



7.19 From (7.3.3),

ρAdh

dt= qmi −

√ρgh

R1

where p = ρgh.

7-20



7.20 If h < D,

ρAdh

dt=

1R1

ps + qmi

If h ≥ D,

ρAdh

dt=

1R1

ps + qmi −1

R2ρg(h − D)

7-21




ρAdh

dt= qmi −

ρg

Rh

If there is no inflow,

Adh

dt= − g

Rh

The time constant is τ = RA/g and the response is

h(t) = h(0)e−t/τ

Take the log of both sides:

ln h(t) = ln h(0) − 1τt (1)

Equation (1) has the form of the equation of a straight line:

ln h(t) = mt + b

where m = −1/τ 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, τ = 1.822 × 103 and

R =gτ

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 = −1τt = 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



7.22 Applying conservation of mass to each tank gives

ρA1dh1

dt= qmi −

ρg

R1h1

ρA2dh2

dt=

ρg

R1h1 −

ρg

R2h2

Note that ρ cancels out in the second equation.

7-23



7.23 a) Applying conservation of mass to each tank gives

ρA1dh1

dt= qmi −

ρg

R1(h1 − h2)

ρA2dh2

dt=

ρg

R1(h1 − h2) −

ρg

R2h2

Note that ρ cancels out in the second equation.b) Substituting the given values we obtain

ρAdh1

dt= qmi −

ρg

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)ρ

− 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)

ρ

−bH1(s) +(

4s +43b

)H2(s) = 0

Using Cramer’s method to solve these equations, we obtain

H2(s) =

∣∣∣∣∣s + b Qmi(s)/ρ−b 0

∣∣∣∣∣ /D =bQmi(s)

ρD

where

D =

∣∣∣∣∣s + b −b−b 4s + 4b/3

∣∣∣∣∣ = (s + b)(4s +43b) − b2 = 4s2 +

163

bs +13b2

ThusH2(s)Qmi(s)

=b/ρ

4s2 + (16/3)bs + (1/3)b2

7-24



7.24 Applying conservation of mass to each tank gives

ρAdh1

dt= −ρg

R(h1 − h2)

2ρAdh2

dt= qmi +

ρg

R(h1 − h2) −

ρg

3Rh2

If we divide both equations by ρA and let b = g/RA, these equations can be expressed as

dh1

dt= −b(h1 − h2)

2dh2

dt=

qmi

ρ+ 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)ρ

Using Cramer’s method to solve these equations, we obtain

H1(s) =

∣∣∣∣∣0 −b

Qmi(s)/ρ 2s + 4b/3

∣∣∣∣∣ /D =bQmi(s)

ρD

where

D =

∣∣∣∣∣s + b −b−b 2s + 4b/3

∣∣∣∣∣ = (s + b)(

2s +43b

)− b2 = 2s2 +

103

bs +13b2

ThusH2(s)Qmi(s)

=b/ρ

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



7.25 From Example 7.4.4 the damping constant is given by

c =128µLA2

πD4

Substituting the given and desired values, we obtain

2000 =128(0.9)LA2

πD4

which givesLA2

D4=

2000π0.9(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 = π

(D

2

)2

and thus

A =nπD2

4From (1),

LA2

D4= L

n2π2

16= 54.5416

soL =

16(54.5415)n2π2

=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 = π

(0.052

)2

= 1.963 × 10−3 m2

So from (2),

Ao =A

n=

1.963 × 10−3

50= 3.9 × 10−5 = π

(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 m

piston hole diameter, D = 7.1 × 10−3 m

7-26



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 difference 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 6= 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

)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



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



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)ρA2x = A(p1 − p2)

7-29



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 ρgh 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. Differentiate 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

ρqv1 =1R

(p1 − p)

ρqv2 =1R

(p − p2)

and (5) becomes

ρA2

k

dp

dt=

1R

(p1 − p) − 1R

(p − p2)

orR

2ρA2

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



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 = 9π,and this gives k = 40 lb/in. This gives a plate displacement of x = 3.75 in.

7-31



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,

ρgh =R2

R1 + R2∆p

Thush =

(400100

3 × 104)

/9.81ρ = 1223/ρ

where ρ, in kg/m3, is the mass density of the liquid (which was not specified).b)

δqm1 = −1rδ(∆p)

where −1/r is the slope of the straight line. Thus r = 700.We have

d

dtδh = −bδh

whereb =

(1r

R1 + R2

R2+

1R2

)g

A=(

1700

1000400

+1

400

)9.812

= 0.02978

Thusd

dtδh = −0.02978δh

The time constant is 1/0.02978 = 33.58 s.

7-32



7.31 From (7.2.5),

ρA(h)dh

dt= qmi − qmo

whereA(h) = (2L tan θ)h

andqmo =

1R

ρgh

Thusρ (2L tan θ)h

dh

dt= qmi −

1R

ρgh

7-33



7.32 From (7.2.5),

ρA(h)dh

dt= qmi − qmo

whereA(h) = (2L tan θ)h

and

qmo =

√ρgh

Ro

Ro =1

2ρC2dA2

o

Thusqmo = ρCdAo

√2gh

andρ (2L tan θ)h

dh

dt= qmi − ρCdAo

√2gh

7-34



7.33 From (7.2.5),

ρA(h)dh

dt= qmi − qmo

where, from Problem 7.8,A(h) = 2L

√Dh − h2

and

qmo =√

p

R=

√ρgh

R

Thus

2ρL√

Dh − h2dh

dt= qmi −

√ρgh

R

7-35



7.34 (a) From conservation of mass and (7.3.9):

ρAh = ρqv − CdAo

√2pρ = ρqv − ρCdAo

√2gh

Thus ρ 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 ≈

√16.1(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



7.35 a) The capacitance can be obtained from

C =A(h)

g

where

A(h) = π

(h

tan θ

)2

Thus

C =π

g

(h

tan θ

)2

b) From (7.2.5),

ρA(h)dh

dt= qmi − qmo

where, from part (a),

A(h) = π

(h

tan θ

)2

andqmo =

ρgh

R

Thus

πρ

(h

tan θ

)2 dh

dt= qmi −

ρgh

R

7-37



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



7.37 If pi − p < 0, the flow will be out of the tank (and will be negative). Thus,

Cdp

dt= −f(|pi − p|)

7-39



7.38 For the left-hand tank,

C1d(δp1)

dt=

1R1

(δpi − δp1) −1

R2(δp1 − δp2)

For the right-hand tank,

C2d(δp2)

dt=

1R2

(δp1 − δp2)

7-40



7.39 a)C = ρV 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



7.40 a)V = 15(10)(8) = 1200 ft3

C = ρV 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



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

7-43



7.42 a) For conduction, the thermal resistance is given by

R =L

kA

Thus

R1 =10 × 10−3

400π(10−3)2= 7.958 ◦C/W

R2 =5 × 10−3

400π((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



7.43 The total resistance isR =

1h1A

+L

kA+

1h2A

R =(

185

+3/[16(12)]

47+

115

)1A

=0.0788

A

7-45



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

)=

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



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 difference, 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



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 = 2πriL = 2π(

12

)(112

)10 = 2.618 ft2

Ao = 2πroL = 2π(

34

)(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

)

2πLk=

ln(

3/41/2

)

2π(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 =1

0.256(120 − 70) = 195

ft 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 ρ = 1.94 slug/ft3 andcp = 25, 000 ft-lb/slug-◦F:

E = mcpTi = (πr2i Lρ)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.

7-48



7.47 The thermal energy E of the water in the pipe is

E = mcpTi = (πr2i Lρ)cpTi = π

(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

7-49



7.48 a)C1 = mcp = ρV 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

7-50



7.49 The steady-state temperature difference 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 τ = 4000/4 = 1000 sec. This value is confirmed by noting from the datathat it took 1000 sec for the temperature difference to decay by 63% (to 77◦). From Problem7.48, the model is

4.85 × 107R1dT1

dt+ T1 = To

soτ = 4.85 × 107R1 = 1000

ThusR1 = 2.06 × 10−5 ◦F − sec/ft − lb

7-51



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)

7-52



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)

7-53



7.52 a) The sphere model is

cpρVdT

dt= − 1

R(T − To)

where R = 1/hA. The time constant is τ = cpρV/hA, and the response is

T (t) = 22 + [T (0) − 22]e−t/τ

which has the form∆T (t) = ∆T (0)e−t/τ

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 coefficients 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 affecting the process, butonly for the first 30 seconds or so.

7-54



7.53 From conservation of heat energy,

CdT

dt=

1R

(To − T )

whereR =

1hA

C = ρV cp

The surface area isA = 4πr2 = 4π(30 × 10−3)2 = 1.131 × 10−2

and the volume isV =

43πr3 =

43π(30 × 10−3)3 = 1.13 × 10−4

The resistance isR =

1hA

= 0.2947

The capacitance isC = ρV 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 τ = 114.2. The response is

T (t) = T (0)e−t/τ + To(1 − e−t/τ ) = 400e−t/τ + 50(1 − e−t/τ ) = 50 + 350e−t/τ

The time to reach 130◦ is found from

50 + 350e−t/τ = 130

which givest = −τ ln(0.2286) = 168.5 s

7-55



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 = 4πr2 = 4π(30 × 10−3)2 = 1.131 × 10−2

and the volume isV1 =

43πr3 =

43π(30 × 10−3)3 = 1.13 × 10−4

R =1

hA=

1300(1.131 × 10−2)

= 0.2947

C1 = ρV1cp = 8900(1.13 × 10−4)385 = 387.5

C2 = ρV2cp = 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 τ1 = 114.2 and τ2 = 9.3125× 104. Applying the Laplace transform to each equationgives

τ1sT1(s) − τ1T1(0) = T2(s) − T1(s)

τ2sT2(s) − τ2T2(0) = T1(s) − T2(s)

These have the solutionT1(s) =

bs + c

s(s + a)

wherea =

τ1 + τ2

τ2τ20.008767

b = T1(0) = 400

c =T1(0)

τ2+

T2(0)τ1

= 0.4421

7-56



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.

7-57



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 are

(R1 + R2) −R1 0R3 −(R2 + R3) R2

0 −R4 (R3 + R4)

T1

T2

T3

=

R2Ti

0R3To)

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



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 =

R1 0 0 10 R2 0 −10 0 R3 −11 −1 −1 0

b =

pa

−pb

−pc

0

x =

q1

q2

q3

p1

(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.

7-59



7.57 For the given values

CdA = 0.5π(2 × 10−2)2 = 2π × 10−4

and the differential equation is

π(6h − h2)dh

dt= −2π × 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 differential 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)’)

7-60



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 different, 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 difficulties 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 3

x 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.

7-61



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

Sal

t Mas

s y

Figure : for Problem 7.58.

7-62



(b) The variable coefficient 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.

7-63



7.59 The relation between height h and the volume inflow rate r is

100dh

dt= r

Thush(t) =

1100

∫ 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



7.60 This problem requires both analytical and numerical methods.(a) Let

b =13π

(R

H

)2

=13π

(1.54

)2

=3π64

Then V = bh3. When the cup is full, h = 4 in, and the water volume is V = b(4)3 = 64b =3π.

Let q be the flow rate (q = 2 cubic in/sec). From mass conservation,

dV

dt= q

andV (t) =

∫ t

0q(t) dt =

∫ t

02 dt = 2t

Equating the two expressions for the water volume gives 2t = 64b, or t = 32b = 3π/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) =∫ t

0q(t) dt =

∫ t

02(1 − e−2t) dt =

[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 = 3π

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.

7-65



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.

7-66



7.62 Using the values given, the equation can be reduced to

dh

dt= − CdA

√2gh

π(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.


7-67



7.63 From conservation of fluid mass, 100h = q, where q is the flow rate. Thus

h(t) =1

100

∫ 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.


7-68



7.64 From conservation of mass, the rate of change of water mass in the cup is

d(ρV )dt

= ρq

orρdV )dt

= ρq

where ρ is the mass density of the water. We see that ρ cancels out of the equation, whichcan then be expressed as

dV

dt= π

(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

π(

RH

)2h2

=q

π(

964

)h2

a) With q = 2 the equation becomes

dh

dt=

2

π(

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


By experimenting with the Stop time, we find that the height is essentially equal to 4 in.after 4.7 sec.

7-69



Figure : for Problem 7.64a

7-70



b) With q a function of time, we must modify the model in part (a). The equationbecomes

dh

dt=

q(t)

π(

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 π

(964

)h2. In the Divide block enter the number of inputs as */ (to multiply by q(t) and

to divide by π(

964

)h2). You can plot the results by typing


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

7-71



7.65 Use the model developed in Example 7.10.1 and change the Relay settings.

7-72



7.66 The model is shown in the following figure. In the Relay block, set the Switch onpoint to 5.5 and the Switch off point to 4.5. Set the Output when on to 0 and the Outputwhen off to 50. Set the Initial condition of the Integrator to 1. Set the gain of the Gainblock to 1/ρA = 1/2000. Set the gain of the other block to ρg/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



7-73



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