modelamento de sistemas fluidicos

8/18/2019 Modelamento de Sistemas Fluidicos

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7C H A P T E R

Fluid and Thermal Systems

CHAPTER OUTLINE

Part I. Fluid Systems 3977.1 Conservation of Mass 3977.2 Fluid Capacitance 4027.3 Fluid Resistance 4077.4 Dynamic Models of Hydraulic Systems 4117.5 Pneumatic Systems 427Part II. Thermal Systems 4307.6 Thermal Capacitance 4317.7 Thermal Resistance 4327.8 Dynamic Models of Thermal Systems 441

Part III. MATLAB and Simulink Applications 4497.9 MATLAB Applications 4497.10 Simulink Applications 4537.11 Chapter Review 458Reference 458Problems 458

CHAPTER OBJECTIVES

When you have nished this chapter, you should be able to

1. Apply the conservation of mass principle to modelsimple hydraulic and pneumatic systems.

2. Determine the appropriate resistance relation to usefor laminar, turbulent, and orice ow.

3. Develop a dynamic model of hydraulic andpneumatic systems containing one or more uidcontainers.

4. Determine the appropriate thermal resistancerelation to use for conduction, convection, andradiation heat transfer.

5. Develop a model of a thermal process having one ormore thermal storage compartments.

6. Apply MATLAB and Simulink to solve uid andthermal system models.

Auid system uses one or more uids to achieve its purpose. The dampers, shock absorbers, and door closer we saw in Chapter 4 are examples of uid systems

because they depend on the viscous nature of a uid to provide damping. Auid might be either a liquid or a gas. Part I of this chapter concerns the study of uid systems, which can be divided into hydraulics and pneumatics . Hydraulics isthe study of systems in which the uid is incompressible, that is, its density staysapproximately constant over a range of pressures. Pneumatics is the study of systems inwhich the uid is compressible . Hydraulics and pneumatics share a common modeling

396


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7 .1 Conservation of Mass

principle: conservation of mass. It will form the basis of all our models of such systems.Modeling pneumatic systems also requires application of thermodynamics, because thetemperature and density of a gas can change when its pressure changes.

Thus pneumatics provides a bridge to the treatment of thermal systems, which isthe subject of Part II of the chapter. Thermal systems are systems that operate due totemperature differences. They thus involve the ow and storage of thermal energy, or

heat, and conservation of heat energy forms the basis of our thermal models.Part III illustrates applications of MATLAB and Simulink to uid and thermalsystems.

Fluid and thermal systems are more complicated than most electrical and mechan-ical systems. While, for example, there are formulas available to compute the springconstant of typical elastic elements, few formulas are available for the coefcients thatwill appear in our uid and thermal models, and the coefcients’ values often must bedetermined experimentally. For this reason, the methods for developing models fromdata, covered in Chapter 8 and Appendix C, are most important for modeling uid andthermal systems. ■

PART I. FLUID SYSTEMSIn addition to providing damping, other applications of uid systems include actua-tors and processes that involve mixing, heating, and cooling of uids. Active vehiclesuspensions use hydraulic and pneumatic actuators to provide forces that supplementthe passive spring and damping elements. Water supply, waste treatment, and otherchemical processing applications are examples of a general category of uid systemscalled liquid-level systems, because they involve regulating the volumes, and thereforethe levels of liquids in containers such as tanks.

Because all real uids are compressible to some extent, incompressibility is anapproximation. But this approximation is usually sufciently accurate for most liquidsunder typical conditions, and it results in a simpler model of the system. For this reasonwe will begin our study of uid systems with hydraulics.

7.1 CONSERVATION OF MASSWe will avoid complex system models by describing only the gross system behaviorinstead of the details of the uid motion patterns. The study of such motion belongs tothe specialized subject of uid mechanics and will not be treated here.

For incompressible uids, conservation of mass is equivalent to conservation of volume, because the uid density is constant. If we know the mass density ρ and thevolume ow rate, we can compute the mass ow rate. That is, qm = ρ qv , where qmand qv are the mass and volume ow rates. The FPS and SI units for mass ow rateare slug/sec and kg/s, respectively. The units for volume rates are ft 3 /sec and m 3 /s,respectively. Other common units for volume are the U.S. gallon, which is 0.13368 ft 3 ,and the liter, which is 0.001 m 3 .

7.1.1 DENSITY AND PRESSURE

The units for mass density are slug/ft 3 and kg/m 3 . Sometimes one encounters weight density, whose common symbol is γ . Its units are lb/ft 3 or N/m 3 , and it is related tothe mass density as γ =ρ g , where g is the acceleration due to gravity. The massdensity of fresh water near room temperature is 1.94 slug/ft 3 , or 1000 kg/m 3 . The mass


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398 CHAPTER 7 Fluid and Thermal Systems

density of air at sea level and near room temperature is approximately 0.0023 slug/ft 3 or1.185 kg/m 3 .

Pressure is the force per unit area that is exerted by the uid. The FPS and SI unitsof pressure are lb/ft 2 and the Pascal (1 Pa =1 N/m 2), respectively. Another commonunit is psi (lb/in. 2). At sea level near room temperature, atmospheric pressure, usuallyabbreviated pa , is14.7 psi (2117lb/ft 2)or1 .0133 ×10 5 Pa. Gage pressure is thepressuredifference between the absolute pressure and atmospheric pressure, and is often abbre-viated as psig. For example, 3 psig is 17.7 psi absolute (which is abbreviated as psia).

Hydrostatic pressure is the pressure that exists in a uid at rest. It is caused by theweight of the uid. For example, the hydrostatic pressure at the bottom of a column of uid of height h is ρ gh . If the atmospheric pressure above the column of liquid is pa ,then the total pressure at the bottom of the column is ρ gh + pa .

EXAMPLE 7.1.1 A Hydraulic Brake System

■ Problem

Figure 7.1.1 is a representation of a hydraulic brake system. The piston in the master cylindermoves in response to the foot pedal. The resulting motion of the piston in the slave cylindercauses the brake pad to be pressed against the brake drum with a force f 3 . Obtain the expressionfor the force f 3 with the force f 1 as the input. The force f 1 depends on the force f 4 applied by thedriver’s foot. The precise relation between f 1 and f 4 depends on the geometry of the pedal arm.

■ Solution

The forces are related to the pressures and the piston areas as follows: f 1 = p1 A1 and f 2 = p2 A2 .Assuming the system is in static equilibrium after the brake pedal has been pushed, we see that p1 = p2 +ρ gh , where h is the height between points 1 and 2. Thus, if h is small, that is, if the pressure ρ gh is negligible compared to p2 , then p1 = f 1 / A1 = p2 = f 2 / A2 . The forces aretherefore related as f 2 = f 1 A2 / A1 , and if the area A2 of the slave piston is greater than the area A1 of the master piston, the force f 2 will be greater than the force f 1 . So we see that this systemserves to amplify the pedal force.

The force f 3 can be obtained from the lever relation f 3 = f 2 L1 / L2 , assuming staticequilibrium or negligible lever inertia.Figure 7.1.1 A hydraulicbrake system.

Master cylinder

Pivot

Brakepedal

Slave cylinderPivot

Drum

Brakepad

x 1

p1

p2

A2

x 2 L1

L2

f 3

f 4

f 1

f 2

A1


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The tradeoff for force amplication is that the master piston must move a distance greaterthan that of the slave piston. We may see this effect by equating the uid volume displaced byeach piston.

volume = A1 x 1 = A2 x 2Thus x 2 = x 1 A1 / A2 , and so x 2 < x 1 if A1 < A2 .

Conservation of mass can be stated as follows. For a container holding a mass of uid m , the time rate of change ˙m of mass in the container must equal the total massinow rate minus the total mass outow rate. That is,

˙m =qmi −qmo (7.1.1)where qmi is the mass inow rate and qmo is the mass outow rate.

The uid mass m is related to the container volume V by m =ρ V . For an incom-pressible uid, ρ is constant, and thus ˙m =ρ V̇ . Let qv i and qvo be the total volume inowand outow rates. Thus, qmi =ρ qv i , and qmo =ρ qvo . Substituting these relationshipsinto (7.1.1) givesρ V̇ =ρ qv i −ρ qvo

Cancel ρ to obtainV̇ =qv i −qvo (7.1.2)

This is a statement of conservation of volume for the uid, and it is equivalent toconservation of mass, equation (7.1.1), when the uid is incompressible.

A Water Supply Tank EXA MPL E

■ Problem

Water is pumped as needed at the mass ow rate qmo ( t ) from the tank shown in Figure 7.1.2a.Replacement water is pumped from a well at the mass ow rate qmi (t ) . Determine the waterheight h ( t ), assuming that the tank is cylindrical with a cross section A.

Figure 7.1.2 A water supply tank.

qmi

qmo

Well

Distribution

(a) (b)

h A

qmoqmi


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■ Solution

We model the system as shown in part (b) of the gure. The volume of water in the tank is Ah,and therefore the mass of water in the tank is ρ Ah , where ρ is the mass density of water. Fromconservation of mass, we have

d dt

(ρ Ah ) =qmi ( t ) −qmo ( t )Since ρ and A are constant, we have

ρ Adhdt =qmi ( t ) −qmo ( t )

which can be integrated as follows:

h ( t ) =h (0) +1

ρ A t

0[qmi (u ) −qmo (u )] du

Once we know the ow rates, we can evaluate the integral.

A common hydraulic actuator is the piston-and-cylinder actuator used on manytypes of heavy equipment, such as thebackhoe shown in Figure7.1.3.When theoperator

moves a handle, hydraulic uid under high pressure is sent through the line to thecylinder. The uid acts on the piston within the cylinder and produces a force thatis equal to the pressure times the piston area. This large force moves the linkage.Example 7.1.3 develops a simple model of such a device.

Figure 7.1.3 A backhoe. Pistonrod

Cylinder

Operatorcontrols

Hydrauliclines

EXAMPLE 7.1.3 A Hydraulic Cylinder

■ Problem

Figure 7.1.4a shows a cylinder and piston connected to a load mass m, which slides on africtionless surface. Part (b) of the gure shows the piston rod connected to a rack-and-piniongear. The pressures p1 and p2 are applied to each side of the piston by two pumps. Assume thepiston rod diameter is small compared to the piston area, so the effective piston area A is thesame on both sides of the piston. Assume also that the piston and rod mass have been lumpedinto m and that any friction is negligible. (a) Develop a model of the motion of the displacement


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5 x R

p1 p2

m A

x

(a)

R I x

Rack

Pinion

m

(b)

A

p1 p2

Figure 7.1.4 A hydraucylinder for (a) translatinmass and for (b) rotatinga pinion gear.

x of the mass in part (a) of the gure, assuming that p1 and p2 are given functions of time.Also, obtain the expression for the mass ow rate that must be delivered or absorbed by the twopumps. (b) Develop a model of the displacement x in part (b) of the gure. The inertia of thepinion and the load connected to the pinion is I .

■ Solution

a. Assuming that p1 > p2 , the net force acting on the piston and mass m is ( p1 − p2) A, andthus from Newton’s law,

m ¨ x =( p1 − p2) ABecause p1 and p2 are given functions of time, we can integrate this equation once toobtain the velocity:

˙ x ( t ) = ˙ x (0) + Am

t

0[ p1(u ) − p2(u )] du

The rate at which uid volume is swept out by the piston is A˙ x , and thus if ˙ x > 0, thepump providing pressure p1 must supply uid at the mass rate ρ A˙ x , and the pumpproviding pressure p2 must absorb uid at the same mass rate.

b. Because we want an expression for the displacement x , we obtain an expression for theequivalent mass of the rack, pinion, and load. The kinetic energy of the system is

KE = 12 m ˙ x 2 + 12 I ˙θ 2 = 12 m + I R2 ˙ x 2

because Rθ̇ = ˙ x .Thus the equivalent mass is

m e =m + I

R2

The required model can now be obtained by replacing m with me in the model developedin part (a).

A Mixing Process EXA MPL E

■ Problem

A mixing tank is shown in Figure 7.1.5. Pure water ows into the tank of volume V =600 m 3at the constant volume rate of 5 m 3 /s. A solution with a salt concentration of si kg/m 3 owsinto the tank at a constant volume rate of 2 m 3 /s. Assume that the solution in the tank is wellmixed so that the salt concentration in the tank is uniform. Assume also that the salt dissolvescompletely so that the volume of the mixture remains the same. The salt concentration so kg/m 3

in the outow is the same as the concentration in the tank. The input is the concentration si ( t ) ,


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Figure 7.1.5 A mixingprocess. Solution si

so

Water

Mixer

V 5 600 m 3

5 m 3 /s

qv o

2 m 3 /s

whose value may change during the process, thus changing the value of so . Obtain a dynamicmodel of the concentration so .

■ Solution

Two mass species are conserved here: water mass and salt mass. The tank is always full, so themass of water mw in the tank is constant, and thus conservation of water mass gives

dm wdt =5ρ w +2ρ w −ρ w qvo =0

where ρ w is the mass density of fresh water, and qvo is the volume outow rate of the mixedsolution. This equation gives qvo =5 +2 =7 m 3 /s.

The salt mass in the tank is so V , and conservation of salt mass gives

d dt

(so V ) =0(5) +2si −soqvo =2si −7soor, with V

=600,

600dsodt =2si −7so (1)

This is the model. The time constant for the mixing process is 600 / 7 =85 .7 s. Thus, if si isinitially zero and then becomes a nonzero constant value S , the salt concentration in the outowwill eventually become constant at the value 2 S / 7 after approximately 4 (85 .7) =343 s.

7.2 FLUID CAPACITANCESometimes it is very useful to think of uid systems in terms of electrical circuits.Table 7.2.1 gives the uid quantity, its common nomenclature, its linear relation, and

its analogous electrical property. Fluid resistance is the relation between pressure andmass ow rate. Fluid capacitance is the relation between pressure and stored mass.We will concentrate on uid resistance and capacitance. Fluid resistance relates to

energy dissipation while uid capacitance relates to potential energy. Fluid inertancerelates to uid acceleration and kinetic energy.

Fluid systems obey two laws that are analogous to Kirchhoff’s current and voltagelaws; these laws are the continuity and the compatibility laws. The continuity law issimply a statement of conservation of uid mass. This says that the total mass ow into


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7 .2 Fluid Capacitance

Table 7.2.1 Analogous uid and electrical quantities.

Fluid quantity Electrical quantity

Fluid mass, m Charge, QMass ow rate, qm Current, iPressure, p Voltage, vFluid linear resistance, R Electrical resistance, R

R

= p/ qm R

=v/ i

Fluid capacitance, C Electrical capacitance, C C =m/ p C = Q/vFluid inertance, I Electrical inductance, L I = p/( dq m / dt ) L =v/( di / dt )

a junction must equal the total ow out of the junction. This is analogous to Kirchhoff’scurrent law. Flow through tworigid pipes joined together to make onepipe is an examplewhere this applies. If, however, the ow is through exible tubes that can expand andcontract under pressure, then the outow rate is not the sum of the inow rates. Thisis an example where uid mass can accumulate within the system and is analogous tohaving a capacitor in an electrical circuit.

The compatibility law is analogous to Kirchhoff’s voltage law, which states that thesum of signed voltage differences around a closed loop must be zero. It is an expressionof conservation of energy. The compatibility law states that the sum of signed pressuredifferences around a closed loop must be zero.

7.2.1 FLUID SYMBOLS AND SOURCES

Figure 7.2.1 shows the commonly used symbols for uid system elements. Theresistance symbol is used to represent xed resistances, for example, due to pipe ow,orice ow, or a restriction. A valve that can be manually adjusted, such as a faucet,is a variable resistance and has a slightly different symbol. An actuated valve, driven,for example, by an electric motor or a pneumatic device, has a different symbol. Such

valves are usually operated under computer control.Just as there are ideal voltage and current sources in electrical systems, so we useideal pressure and ow sources in our uid system models. An ideal pressure sourceis capable of supplying the specied pressure at any ow rate. An ideal ow source iscapable of supplying the specied ow.

These ideal sourcesareapproximations to real devices such as pumps. For example,Figure7.2.2 shows the steady-state ow-pressure relation fora centrifugal pump, whereqm is the mass ow rate produced by the pump when the pressure difference across thepump is p. When the outlet pressure is greater than the inlet pressure, p > 0. Such

qsIdeal ow

sourceIdeal pressure

source

p1 p2

ps 5 p2 2 p1

Actuatedvalve

Manuallyadjusted valve

Resistance

Pump

ps

Figure 7.2.1 Fluid sys

symbols.


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Figure 7.2.2 Steady-stateow-pressure relation for acentrifugal pump.

s1

s2

s3

qm

D p

curves depend on the pump speed, labeled s1 , s2 , and so on in the gure. To determinethe operating condition of the pump for a given speed, we need another relation betweenqm and p. This relation depends on the load connected to the pump outlet. We willsee how such a relation is obtained in Example 7.4.8 in Section 7.4.

7.2.2 CAPACITANCE RELATIONS

Fluid capacitance is the relation between stored uid mass and the resulting pressurecaused by the stored mass. Figure 7.2.3 illustrates this relation, which holds for bothpneumatic and hydraulic systems. At a particular reference point ( pr , m r ) the slopeis C , where

C =dmdp p= pr

(7.2.1)

Thus, uid capacitance C is the ratio of the change in stored mass to the change inpressure.

Figure 7.2.3 General uidcapacitance relation and itslinear approximation.

m

mr

p r p

C 1


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For such a container, conservation of mass gives

dmdt =qmi −qmo (7.2.2)

but

dm

dt =dm

dp

dp

dt =C

dp

dt

Thus

C dpdt =

dmdt =qmi −qmo (7.2.3)

Also

dmdt =

dmdh

dhdt =ρ A

dhdt

So

ρ Adh

dt =qmi

−qmo (7.2.4)

Equations (7.2.2), (7.2.3), and (7.2.4) are alternative, but equivalent, hydraulicmodels of a container of uid. They suggest that either pressure p, mass m, or height hcan be chosen as the model’s variable. These variables are all indicators of the system’spotential energy, and as such any one can be chosen as a state variable. If the container’scross-sectional area is constant, then V = Ah and thus the liquid volume V can alsobe used as the model variable.

EXAMPLE 7.2.2 Capacitance of a V-Shaped Trough

■ Problem

(a)Derive thecapacitance of the V -shaped trough shown in Figure7.2.6a. (b)Use thecapacitanceto derive the dynamic models for the bottom pressure p and the height h . The mass inow rateis qmi ( t ) , and there is no outow.

■ Solution

a. From part (b) of the gure, D =2h tan θ , and the vertical cross-sectional area of the liquidis h D / 2. Thus the uid mass is given by

m =ρ V =ρ12

h D L =(ρ L tan θ)h 2

Figure 7.2.6 A V-shaped

trough.

(a) (b)

D

h 2

A

h 2

L


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7 .3 Fluid Resistance

But p =ρ gh and thus,m =(ρ L tan θ)

pρ g

2

= L tan θ

ρ g2 p2

From the denition of capacitance,

C

=dm

dp =2 L tan θ

ρ g 2 p

b. From (7.2.3) with qmo =0, C ˙ p =qmi , or2 L tan θ

ρ g2 p

dpdt =qmi

which is a nonlinear equation because of the product p ˙ p. We can obtain the model for theheight by substituting h = p/ρ g . The result is

(2ρ L tan θ) hdhdt =qmi

7.3 FLUID RESISTANCEFluid meets resistance when owing through a conduit such as a pipe, through a com-ponent such as a valve, or even through a simple opening or orice, such as a hole. Wenow consider appropriate models for each type of resistance.

The mass ow rate q̂m through a resistance is related to the pressure difference ˆ pacross the resistance. This relation, ˆ p = f (q̂m ) , is illustrated in general by Figure 7.3.1.We dene the uid resistance R r as the slope of f (q̂m ) evaluated at a reference equi-librium condition ( pr , qmr ) . That is,

Rr =d ˆ p

d q̂m q̂m=q̂mr = d ˆ pd q̂m r

(7.3.1)

If we need to obtain an approximate linear model of the pressure-ow rate relation,

we can use a Taylor series expansion to linearize the expression ˆ pr = f (q̂m ) neara reference operating point ( pr , qmr ) as follows (after dropping the second order andhigher terms in the expansion):

ˆ p = pr + d ˆ pd q̂m r

(qm −qmr ) = pr + Rr (qm −qmr ) (7.3.2)where Rr is the linearized resistance dened by (7.3.1).

p̂

p r

Rr

p

qm

qmr q̂m

1

Figure 7.3.1 General resistance relation and itslinear approximations.


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Referring to Figure7.3.1,we dene a new setof variables p and qm , called deviationvariables , that represent small but nite changes in ˆ p and q̂m from their reference values pr and qmr . From Figure 7.3.1, we see that

p = ˆ p − ˆ pr (7.3.3)qm = q̂m − q̂mr (7.3.4)

In terms of these deviation variables, we can rewrite (7.3.2) asˆ p − ˆ pr = Rr ( q̂m − q̂mr )

or p = Rr qm , orqm =

p Rr

(7.3.5)

which is a linear relation. Thus the resistance Rr is called the linearized resistance.Thus we may think of the linearized relation (7.3.5) as

small pressure change Rr = resulting small change in mass ow rate

The values of pr and qmr depend on the particular application. So the resistance Rr depends on these values, as well as the functional form of f (q̂

m) , which depends on

the application.In a limited number of cases, such as pipe ow under certain conditions,the relation

of ˆ p versus q̂m is linear so that ˆ p = Rq̂m , orq̂m =

ˆ p R

(7.3.6)

where R is the linear resistance.In some other applications the relation is a square-root relation.

q̂m = ˆ p B (7.3.7)where B is a constant that often must be determined empirically.

The relation (7.3.7) gives ˆ p = Bq̂2m and thus

Rr = d ˆ pd q̂m r =2 Bqmr

and

qm = p

2 Bqmr (7.3.8)

This is the linearized model corresponding to the relation (7.3.7). This relation can beexpressed instead in terms of pr as follows. Because qmr =√ pr / B, we obtain

Rr =2 B pr / B =2 Bpr and thusqm =

12√ Bpr p (7.3.9)

When only the curve of ˆ p versus q̂m is available, we can obtain a linearized model bygraphically computing the slope S of the tangent line that passes through the referencepoint ( ˆ pr , q̂mr ) . The equivalent, linearized resistance Rr is the slope S .

The resistance symbol shown in Figure7.3.2 represents all types of uid resistance,whether linear or not. Although the symbol looks like a valve, it can represent uidresistance due to other causes, such as pipe wall friction and orices.


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7 .3 Fluid Resistance

1 5 1 R

1 R1

1 R2

(a)

R1

p1 p2 p3

R2 qm

R 5 R1 1 R2

p1 p3

Rqmqm

qm 5 qm1 1 qm2

R1

R2

qmqm

qm1

qm2

R

qm

(b)

Figure 7.3.2 Combinat(a) series resistances and(b) parallel resistances.

As with electrical resistances, linear uid resistance elements obey the series andparallel combination rules. These are illustrated in Figure 7.3.2. Series uid resistances

carry the same ow rate; parallel uid resistances have the same pressure differenceacross them.

7.3.1 LAMINAR PIPE RESISTANCE

Fluid motion is generally divided into two types: laminar ow and turbulent ow .Laminar ow can be described as “smooth” in the sense that the average uid particlevelocity is the same as the actual particle velocity. If the ow is “rough,” the averageparticle velocity will be less than the actual particle velocity, because the uid particlesmeander while moving downstream. This is turbulent ow. You can see the differencebetween laminar andturbulentow by slightly opening a faucet; theow will be smooth.As you open the faucet more, eventually the ow becomes rough.

If the pipe ow is laminar, the following linear relation applies.

q̂m =ˆ p

R(7.3.10)

or equivalently

qm = p R

(7.3.11)

The laminar resistance for a level pipe of diameter D and length L is given by the Hagen-Poiseuille formula

R =128 µ Lπρ D4

(7.3.12)

where µ is the uid viscosity. The viscosity is a measure of the “stickiness” of the uid.Thus molasses has a higher value of µ than that of water.

Not all pipe ow is laminar. A useful criterion for predicting the existence of laminar ow is the Reynolds number N e , the ratio of the uid’s inertial forces to theviscosity forces. For a circular pipe,

N e =ρv D

µ(7.3.13)


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where v = qv /(π D2/ 4) , the average uid velocity. For N e > 2300 the ow is oftenturbulent, while for N e < 2300 laminar ow usually exists. The precise value of N eabove which the ow becomes turbulent depends on, for example, the ow conditionsat the pipe inlet. However, the criterion is useful as a rule of thumb.

The resistance formula (7.3.12) applies only if the so-called “entrance length” L e ,which is the distance from the pipe entrance beyond which the velocity prole no

longer changes with increasing distance, is much less than 0 .06 DN e . Because laminarow can be expected only if N e < 2300, Le might be as long as 138 pipe diameters.Of course, for small Reynolds numbers, L e is shorter. The smaller L e is relative to thepipe length, the more reliable will be our resistance calculations.

7.3.2 TORRICELLI'S PRINCIPLE

An orice can simply be a hole in the side of a tank or it can be a passage in a valve. Wesaw an example of orice ow in Example 1.5.2 in Chapter 1, in which we analyzedthe ow rate of water through a small hole in the side of a plastic milk bottle. We foundthat the tted function is f =9.4h 0.558 , where f is the outow rate in ml/s and the waterheight h is in centimeters. It turns out that the empirically determined exponent 0 .558is close to its theoretical value of 0 .5, as we will now demonstrate.

Around 1640 Torricelli discovered that the ow rate through an orice is propor-tional to the square root of the pressure difference. This observation can be simplyderived by considering a mass m of uid a height h above the orice (see Figure 7.3.3).The potential energy of the mass is mgh . As the mass falls toward the orice itspotential energy is converted to kinetic energy mv2 / 2. If all the potential energy isconverted to kinetic energy at the orice, then mgh =mv2/ 2, and the maximum speedthe uid mass can attain through the orice is v =√ 2gh . Because the pressure dropacross the orice is p =ρ gh , we can express the maximum speed as v =√ 2 p/ρ .Thus the mass ow rate qm through the orice of area Ao can be no greater than Aoρv = Aoρ √ 2 p/ρ = Ao√ 2 pρ . The actual ow rate will be less than this valuebecause of friction effects. To account for these frictional effects, we introduce a factorC d in the ow rate equation as follows:

q̂m =C d Ao 2 pρ (7.3.14)The factor C d is the discharge coefcient, which must lie in the range 0 < C d ≤1. Atypical value for water is 0.6.Because p = ρ gh , (7.3.14) can be expressed in terms of the volume ow rate ˆ qvand the height h as follows:

q̂v = C d Ao 2g h0.5

Figure 7.3.3 Derivation of Torricelli's principle.

m

m

h

PE 5 mgh , KE 5 0

AoPE 5 0, KE 5 mv 2

2


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7 . 4 Dynamic Models of Hydraulic Systems

Thus the theoretical value of the exponent (0.5) is close to the value obtained in thebottle experiment.

Equation (7.3.14) depends on the orice area being small enough so that the pres-sure variation over the orice area is negligible compared to the average pressure atthe orice. For a liquid-level system with a circular orice, this implies that the liquidheight above the orice must be large compared to the orice diameter.

The orice relation (7.3.14) can be rearranged in the form of (7.3.7).

q̂m =C d Ao 2ρ ˆ p = ˆ p Ro

(7.3.15)

where the orice resistance is dened as

Ro =1

2ρ C 2d A2o(7.3.16)

7.3.3 TURBULENT AND COMPONENT RESISTANCE

For us, the practical importance of the difference between laminar and turbulent owlies in the fact that laminar ow can be described by the linear relation (7.3.5), whileturbulent ow is described by the nonlinear relation (7.3.7). Components, such asvalves, elbow bends, couplings, porous plugs, and changes in ow area resist owand usually induce turbulent ow at typical pressures, and (7.3.7) is often used tomodel them. Experimentally determined values of B are available for common typesof components.

7.4 DYNAMIC MODELS OF HYDRAULIC SYSTEMSIn this section we consider a number of hydraulic system examples dealing with liquid-level systems, dampers, actuators, pumps, and nonlinear systems.

7.4.1 LIQUID LEVEL SYSTEMS

In liquid-level systems energy is stored in two ways: as potential energy in the massof liquid in the tank, and as kinetic energy in the mass of liquid owing in the pipe.In many systems, the mass of the liquid in the pipes is small compared to the liquidmass in the tanks. If the mass of liquid in a pipe is small enough or is owing at a smallenough velocity, the kinetic energy contained in it will be negligible compared to thepotential energy stored in the liquid in the tank. If the kinetic energy of the liquid issignicant, more advanced uid-ow theory is required. This is usually not the casefor the scope of applications considered here.

Liquid-Level System with an Orice EXA MPL E

■ Problem

The cylindrical tank shown in Figure 7.4.1 has a circular bottom area A. The volume inow ratefrom the ow source is q̂vi ( t ) , a given function of time. The orice in the side wall has an area Ao and discharges to atmospheric pressure pa . Develop a model of h , the deviation of the liquidheight from a reference equilibrium value h r , assuming that h1 > L.


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Figure 7.4.1 A liquid-levelsystem with an orice.

h1

hĥ

hr

L A

pa

q̂ v i

■ Solution

If the inow rate ˆqvi is held constant at the rate qvir , the liquid level eventually becomes constantat the height h r . Using the orice ow relation (7.3.14), this height can be found from

ρ qvir =C d Ao 2ρ(ρ gh r )or

h r =1

2g qvir C d Ao

2

So if we are given qvir , we can determine h r , or vice versa.Noting that h 1 =h +h r + L, the rate of change of liquid mass in the tank is

d (ρ Ah1)dt =ρ A

d (h +h r + L)dt =ρ A

dhdt

Conservation of mass implies that

ρ Adhdt =ρ q̂vi −C d Ao

2 ˆ pρ (1)

where the pressure drop across the orice is

ˆ p = pa +ρ g(h +h r ) − pa =ρ g(h +h r )Therefore, equation (1) becomes

ρ Adhdt =ρ q̂vi −C d Ao 2gρ 2(h +h r )

Canceling ρ gives the desired model.

Adhdt = q̂vi −C d Ao

2g(h +h r )

or, because ĥ =h +h r ,

Ad ĥdt = q̂vi −C d Ao 2gĥ (2)

Note that the height L does not appear in the model because the liquid below the orice doesnot affect the pressure at the orice.


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Linearizing a Model EXA MPL E

■ Problem

Consider the liquid-level system with an orice, treated in Example 7.4.1. The model is givenby equation (2) of that example.

Ad ĥdt = q̂vi − q̂vo = q̂vi −C d Ao 2gĥConsider the case where A =2 ft 2 and C d Ao√ 2g =6. Estimate the system’s time constant for

two cases: (i) the inow rate is held constant at q̂vi = 12 ft 3 /sec and (ii) the inow rate is heldconstant at q̂vi =24 ft 3 /sec.■ Solution

Substituting the given values, we obtain

2d ĥdt = q̂vi − q̂vo = q̂vi −6 ĥ (1)

When the inow rate is held constant at the value q̂vir , the liquid height ĥ reaches an equilibrium

value h r that can be found from the preceding equation by setting ˆh = hr and d

ˆh / dt equal tozero. This gives 36 h r =q 2vir .

The two cases of interest to us are (i) h r =(12)2 / 36 =4 ft and (ii) h r =(24)2 / 36 =16 ft.Figure 7.4.2 is a plot of the outow ow rate ˆ qvo =6 ĥ through the orice as a function of theheight ĥ . The two points corresponding to ĥ =4 and ĥ =16 are indicated on the plot.

In Figure 7.4.2 two straight lines are shown, each passing through one of the points of interest ( ĥ =4 and ĥ =16) and having a slope equal to the slope of the curve at that point. Thegeneral equation for these lines is

q̂vo =6 ĥ =6 ĥ r +d q̂vod ĥ r

( ĥ − ĥ r ) =6 ĥ r +3ĥ−1/ 2r (ĥ − ĥ r )

6 ĥ < 24 1 (ĥ 2 16)34

6 ĥ < 12 1 (ĥ 2 4)32

6 ĥ

00

5

10

15

20

25

30

4 8 12

ĥ

16 20 24

Figure 7.4.2 Linearizeapproximations of theresistance relation.


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and is the same as a Taylor series expansion truncated after the rst order term. Noting thatĥ r =h r , this equation becomes

q̂vo =6 h r +3h−1/ 2r (ĥ −h r ) (2)For Case (i), this equation becomesq̂vo

=12

+

3

2( ĥ

−4)

and for Case (ii),

q̂vo =24 +34

( ĥ −16)These are the equations of the straight lines shown in the gure.

Noting that ĥ −h r =h , equation (2) can be expressed in the simpler formq̂vo =6 h r +(3h−1/ 2r )hSubstitute this into equation (1), and note that d ĥ / dt =dh / dt and q̂vi −6√ h r =qvi , to obtain2

dhdt =qvi +(3h−

1/ 2r )h (3)

This is the linearized model that is a good approximation of the nonlinear model (1) near thereference height h r .The time constant of the linearized model (3) is 2 √ h r / 3, and is 4 / 3 sec. for h r = 4 and

8/ 3 sec. for hr = 16. Thus, for Case (i), if the input ow rate is changed slightly from itsequilibrium value of 12, the liquid height will take about 4 (4/ 3) , or 16 / 3, sec to reach its newheight. For Case (ii), if the input ow rate is changed slightly from its value of 24, the liquidheight will take about 4 (8/ 3) , or 32 / 3, seconds to reach its new height.

Note that the model’s time constant depends on the particular equilibrium solution chosenfor the linearization. Because the straight line is an approximation to the ˆ qvo =6 ĥ curve, wecannot use the linearized models to make predictions about the system’s behavior far from theequilibrium point. However, despite this limitation, a linearized model is useful for designinga ow control system to keep the height near some desired value. If the control system works

properly, theheight will stay near theequilibrium value, andthe linearized model will beaccurate.

EXAMPLE 7.4.3 Liquid-Level System with a Flow Source

■ Problem

The cylindrical tank shown in Figure 7.4.3 has a bottom area A. The total mass inow rate fromthe ow source is ˆqmi ( t ) , a given function of time. The total mass outow rate ˆ qmo is not givenand must be determined. The outlet resistance R is the linearized resistance about the referencecondition (h r , qmi r ) . Develop a model of h , the deviation of the liquid height from the constantreference height h r , where ĥ =h r +h .

Figure 7.4.3 A liquid-levelsystem with a ow source.

ĥ

q̂mi

q̂mo

pa

pa R A


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■ Solution

The total mass in the tank is m =ρ Aĥ =ρ A(h +h r ) and from conservation of massdmdt =

d [ρ A(h +h r )]dt =ρ A

dhdt = q̂mi − q̂mo

because ρ , h r , and A are constants. Expressing q̂mi and q̂mo in terms of the deviation variables

qmi and qmo , we have

ρ Adhdt = (qmi +qmi r ) −(qmo +qmor ) = (qmi r −qmor ) +(qmi −qmo )

Because the reference height hr is a constant, the outow rate at equilibrium must equal theinow rate. Thus qmi r −qmor =0, and the model becomes

ρ Adhdt =qmi −qmo (1)

Because R is a linearized resistance, then for small changes h in the height,

qmo

=1

R[(ρ gh

+ pa )

− pa ]

=1

Rρ gh

Substituting this into equation (1) gives the desired model:

ρ Adhdt =qmi −

1 R

ρ gh

which can be rearranged as

R Ag

dhdt +h =

Rρ g

qmi

So the time constant is τ = R A/ g . If qmi = 0 (so the inow rate remains constant at qmi r ),then h will be essentially zero for t > 4τ , which means that the height will return to nearly theequilibrium height h r at that time.

Figure 7.4.4 Electric canalogous to the hydraulsystem shown in Figure

R C is

Some engineers are helped by thinking of a uid system in terms of an analogouselectric circuit, in which pressure difference plays the role of voltage difference, andmass ow rate is analogous to current. A uid resistance resists ow just as an electricalresistor resists current.A uid capacitancestoresuid mass just asanelectrical capacitorstores charge. Figure 7.4.4 shows an electric circuit that is analogous to the tank systemof Figure 7.4.3. The circuit model is

C d vdt = is −

1 R

v

The input current is is analogous to theinow rate qmi , the voltage v across the capacitoris analogous to the uid pressure ρ gh , and the electrical capacitance C is analogous tothe uid capacitance A/ g . It is a matter of personal opinion as to whether such analogieshelp to understand the dynamics of uid systems, and you should decide for yourself.Always keep in mind, however, that we should not get too dependent on analogies fordeveloping models, because they might not always properly represent the underlyingphysics of the original system.


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EXAMPLE 7.4.4 Liquid-Level System with a Pressure Source

■ Problem

The tank shown in cross section in Figure 7.4.5 has a bottom area A. A pressure sourceˆ ps = ps ( t ) + psr is connected through a resistance to the bottom of the tank, where ps (t )

is a given function of time. The resistances R1 and R2 are linearized resistances about the ref-erence condition ( psr , h r ). Develop a model of h, the deviation of the liquid height from theconstant reference height h r , where ĥ =h r +h .

Figure 7.4.5 A liquid-levelsystem with a pressure source.

R2 R1

pa pa

ĥ

A

pa

p̂ s

■ Solution

The total mass in the tank is m =ρ Aˆh =ρ A(h +h r ) , and from conservation of massdm

dt =d [ρ A(h +h r )]

dt =ρ Adhdt = q̂mi − q̂mo

or

ρ Adhdt = (qmi +qmir ) −(qmo +qmor ) = (qmi −qmo ) +(qmi r −qmor )

Because at the reference equilibrium, the outow rate equals the inow rate, qmi r −qmor = 0,and we have

ρ Adhdt =qmi −qmo (1)

This is a linearized model that is valid for small changes around the equilibrium state.Because the outlet resistance has been linearized,

qmo = 1 R2 [(ρ gh + pa ) − pa ] = ρ gh R2Similarly for the mass inow rate, we have

qmi =1

R1[( ps + pa ) −(ρ gh + pa )] =

1 R1

( ps −ρ gh )Substituting into equation (1) gives

ρ Adhdt =

1 R1

( ps −ρ gh ) −ρ gh R2 =

1 R1

ps −ρ g R1 + R2

R1 R2h

This can be rearranged as R1 R2 A

g( R1 + R2)dhdt +h =

R2ρ g( R1 + R2)

ps

The coefcient of d h / dt gives the time constant, which is τ = R1 R2 A/ g( R1 + R2) .

When a uid system contains more than one capacitance, you should apply the con-servation of mass principle to each capacitance, and then use the appropriate resistancerelations to couple the resulting equations. To do this you must assume that some pres-sures or liquid heights are greater than others and assign the positive-ow directionsaccordingly. If you are consistent, the mathematics will handle the reversals of owdirection automatically.


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Two Connected Tanks EXA MPL E

■ Problem

The cylindrical tanks shown in Figure 7.4.6a have bottom areas A1 and A2 . The total massinow rate from the ow source is ˆ qmi ( t ) , a givenfunction of time. The resistances are linearizedresistances about the reference condition h 1r , h 2r , qmi r . (a) Develop a model of the liquid heightsh 1 and h 2 . (b) Suppose the resistances are equal: R1 = R2 = R, and the areas are A1 = A and A2 =3 A. Obtain the transfer function H 1(s )/ Q mi (s ) . (c) Use the transfer function to solve forthe steady-state response for h 1 if the inow rate qmi is a unit-step function, and estimate howlong it will take to reach steady state. Is it possible for liquid heights to oscillate in the stepresponse?

■ Solution

a. Using deviation variables as usual, we note that

ĥ 1 =h 1r +h 1ĥ 2 =h 2r +h 2

q̂mi =qmi r +qmiFor convenience, assume that h1 > h 2 . This is equivalent to assuming that the ow ratefrom tank 1 to tank 2 increases. From conservation of mass applied to tank 1, we obtain

d (ρ A1 ĥ 1)dt =

d [ρ A1(h 1 +h 1r )]dt =ρ A1

dh 1dt = −q̂1m = −(q1m +q1mr )

From physical reasoning we can see that the two heights must be equal at equilibrium, andthus q1mr =0. Therefore,

ρ A1dh 1dt = −q1m

Because R1 is a linearized resistance,

q1m = ρ g R1 (h 1 −h 2)

(a)

R2 R1

q̂mi

ĥ1 ĥ2 A2 A1

(b)

v 1 v 2

R2 C 2

R1 i1

i2

C 1 is

Figure 7.4.6 (a) Twoconnected tanks.(b) Analogous electric ci


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So, after canceling ρ on both sides, the model for tank 1 is

A1dh 1dt = −

g R1

(h 1 −h 2) (1)Similarly for tank 2,

d (ρ A2 ĥ 2)

dt =d [ρ A2(h 2 +h 2r )]

dt =ρ A2

dh 2

dt Conservation of mass gives

ρ A2dh 2dt = q̂mi + q̂1m − q̂mo =(qmi +qmi r ) +(q1m +q1mr ) −(qmo +qmor )

Recalling that q1mr =0, we note that this implies that qmi r =qmor , and thusρ A2

dh 2dt =qmi +q1m −qmo

Because the resistances are linearized, we have

ρ A2dh 2dt =qmi +

ρ g R1

(h 1 −h 2) −ρ g R2

h 2 (2)

The desired model consists of equations (1) and (2).b. Substituting R1 = R2 = R, A1 = A, and A2 =3 A into the differential equations and

dividing by A, and letting B =g / R A we obtainḣ 1 = − B(h 1 −h 2)

3ḣ 2 =qmiρ A + B(h 1 −h 2) − Bh 2 =

qmiρ A + Bh 1 −2 Bh 2

Apply the Laplace transform of each equation, assuming zero initial conditions, andcollect terms to obtain

(s + B) H 1(s ) − B H 2(s ) = 0 (3)

− B H 1(s ) +(3s +2 B) H 2(s ) =1

ρ A Q mi (s ) (4)

Solve equation (3) for H 2(s ) , substitute the expression into equation (4), and solve for H 1(s ) to obtain

H 1(s )Q mi (s ) =

R B2 /ρ g3s 2 +5 Bs + B2

(5)

c. The characteristic equation is 3 s2 +5 Bs + B2 =0 and has the two real rootss =−

5 ±√ 136

B = −1.43 B, −0.232 BThus the system is stable, and there will be a constant steady-state response to a step input.The step response cannot oscillate because both roots are real. The steady-state height canbe obtained by applying the nal value theorem to equation (5) with Qmi (s ) =1/ s .

h 1ss = lims→0s H 1(s ) = lims→0

s R B2 /ρ g

3s2 +5 Bs + B21s =

Rρ g

The time constants are

τ 1 =1

1.43 B =0.699

Bτ 2 =

10.232 B =

4.32 B

modelamento de sistemas fluidicos

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