Fluid and Thermal Systems Chapter 7 PDF

C H 7 A P T E R Fluid and Thermal Systems CHAPTER OUTLINE Part I. Fluid Systems 397 7.1 Conservation of Mass 397 7.2 Fluid Capacitance 402 7.3 Fluid Resistance 407 7.4 Dynamic Models of Hydraulic Systems 411 7.5 Pneumatic Systems 427 Part II. Thermal Systems 430 7.6 Thermal Capacitance 431 7.7 Thermal Resistance 432 7.8 Dynamic Models of Thermal Systems 441 Part III. MATLAB and Simulink Applications 449 7.9 MATLAB Applications 449 7.10 Simulink Applications 453 7.11 Chapter Review 458 Reference 458 Problems 458 CHAPTER OBJECTIVES When you have ﬁnished this chapter, you should be able to 1. Apply the conservation of mass principle to model simple hydraulic and pneumatic systems. 2. Determine the appropriate resistance relation to use for laminar, turbulent, and oriﬁce ﬂow. 3. Develop a dynamic model of hydraulic and pneumatic systems containing one or more ﬂuid containers. 4. Determine the appropriate thermal resistance relation to use for conduction, convection, and radiation heat transfer. 5. Develop a model of a thermal process having one or more thermal storage compartments. 6. Apply MATLAB and Simulink to solve ﬂuid and thermal system models. ﬂuid system uses one or more ﬂuids to achieve its purpose. The dampers, shock absorbers, and door closer we saw in Chapter 4 are examples of ﬂuid systems because they depend on the viscous nature of a ﬂuid to provide damping. A ﬂuid might be either a liquid or a gas. Part I of this chapter concerns the study of ﬂuid systems, which can be divided into hydraulics and pneumatics. Hydraulics is the study of systems in which the ﬂuid is incompressible, that is, its density stays approximately constant over a range of pressures. Pneumatics is the study of systems in which the ﬂuid is compressible. Hydraulics and pneumatics share a common modeling A 396 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 the temperature and density of a gas can change when its pressure changes. Thus pneumatics provides a bridge to the treatment of thermal systems, which is the subject of Part II of the chapter. Thermal systems are systems that operate due to temperature differences. They thus involve the ﬂow 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 ﬂuid and thermal systems. Fluid and thermal systems are more complicated than most electrical and mechanical systems. While, for example, there are formulas available to compute the spring constant of typical elastic elements, few formulas are available for the coefﬁcients that will appear in our ﬂuid and thermal models, and the coefﬁcients’ values often must be determined experimentally. For this reason, the methods for developing models from data, covered in Chapter 8 and Appendix C, are most important for modeling ﬂuid and thermal systems. ■ PART I. FLUID SYSTEMS In addition to providing damping, other applications of ﬂuid systems include actuators and processes that involve mixing, heating, and cooling of ﬂuids. Active vehicle suspensions use hydraulic and pneumatic actuators to provide forces that supplement the passive spring and damping elements. Water supply, waste treatment, and other chemical processing applications are examples of a general category of ﬂuid systems called liquid-level systems, because they involve regulating the volumes, and therefore the levels of liquids in containers such as tanks. Because all real ﬂuids are compressible to some extent, incompressibility is an approximation. But this approximation is usually sufﬁciently accurate for most liquids under typical conditions, and it results in a simpler model of the system. For this reason we will begin our study of ﬂuid systems with hydraulics. 7.1 CONSERVATION OF MASS We will avoid complex system models by describing only the gross system behavior instead of the details of the ﬂuid motion patterns. The study of such motion belongs to the specialized subject of ﬂuid mechanics and will not be treated here. For incompressible ﬂuids, conservation of mass is equivalent to conservation of volume, because the ﬂuid density is constant. If we know the mass density ρ and the volume ﬂow rate, we can compute the mass ﬂow rate. That is, qm = ρqv , where qm and qv are the mass and volume ﬂow rates. The FPS and SI units for mass ﬂow rate are slug/sec and kg/s, respectively. The units for volume rates are ft3 /sec and m3 /s, respectively. Other common units for volume are the U.S. gallon, which is 0.13368 ft3 , and the liter, which is 0.001 m3 . 7.1.1 DENSITY AND PRESSURE The units for mass density are slug/ft3 and kg/m3 . Sometimes one encounters weight density, whose common symbol is γ . Its units are lb/ft3 or N/m3 , and it is related to the mass density as γ = ρg, where g is the acceleration due to gravity. The mass density of fresh water near room temperature is 1.94 slug/ft3 , or 1000 kg/m3 . The mass 397 398 CHAPTER 7 Fluid and Thermal Systems density of air at sea level and near room temperature is approximately 0.0023 slug/ft3 or 1.185 kg/m3 . Pressure is the force per unit area that is exerted by the ﬂuid. The FPS and SI units of pressure are lb/ft2 and the Pascal (1 Pa = 1 N/m2 ), respectively. Another common unit is psi (lb/in.2 ). At sea level near room temperature, atmospheric pressure, usually abbreviated pa , is 14.7 psi (2117 lb/ft2 ) or 1.0133×105 Pa. Gage pressure is the pressure difference between the absolute pressure and atmospheric pressure, and is often abbreviated 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 ﬂuid at rest. It is caused by the weight of the ﬂuid. For example, the hydrostatic pressure at the bottom of a column of ﬂuid of height h is ρgh. If the atmospheric pressure above the column of liquid is pa , then the total pressure at the bottom of the column is ρgh + pa . E X A M P L E 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 cylinder moves in response to the foot pedal. The resulting motion of the piston in the slave cylinder causes the brake pad to be pressed against the brake drum with a force f 3 . Obtain the expression for the force f 3 with the force f 1 as the input. The force f 1 depends on the force f 4 applied by the driver’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 + ρgh, where h is the height between points 1 and 2. Thus, if h is small, that is, if the pressure ρgh is negligible compared to p2 , then p1 = f 1 /A1 = p2 = f 2 /A2 . The forces are therefore 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 system serves to amplify the pedal force. The force f 3 can be obtained from the lever relation f 3 = f 2 L 1 /L 2 , assuming static equilibrium or negligible lever inertia. x1 Figure 7.1.1 A hydraulic brake system. Pivot f1 Master cylinder p1 A1 f4 A2 f2 p2 L1 x2 Pivot Slave cylinder L2 Drum f3 Brake pad Brake pedal 7.1 Conservation of Mass 399 The tradeoff for force ampliﬁcation is that the master piston must move a distance greater than that of the slave piston. We may see this effect by equating the ﬂuid volume displaced by each piston. volume = A1 x1 = A2 x2 Thus x2 = x1 A1 /A2 , and so x2 < x1 if A1 < A2 . Conservation of mass can be stated as follows. For a container holding a mass of ﬂuid m, the time rate of change ṁ of mass in the container must equal the total mass inﬂow rate minus the total mass outﬂow rate. That is, ṁ = qmi − qmo (7.1.1) where qmi is the mass inﬂow rate and qmo is the mass outﬂow rate. The ﬂuid mass m is related to the container volume V by m = ρV . For an incompressible ﬂuid, ρ is constant, and thus ṁ = ρ V̇. Let qvi and qvo be the total volume inﬂow and outﬂow rates. Thus, qmi = ρqvi , and qmo = ρqvo . Substituting these relationships into (7.1.1) gives ρ V̇ = ρqvi − ρqvo Cancel ρ to obtain (7.1.2) V̇ = qvi − qvo This is a statement of conservation of volume for the ﬂuid, and it is equivalent to conservation of mass, equation (7.1.1), when the ﬂuid is incompressible. A Water Supply Tank E X A M P L E 7.1.2 ■ Problem Water is pumped as needed at the mass ﬂow rate qmo (t) from the tank shown in Figure 7.1.2a. Replacement water is pumped from a well at the mass ﬂow rate qmi (t). Determine the water height h(t), assuming that the tank is cylindrical with a cross section A. Figure 7.1.2 A water supply tank. qmi qmo h A Distribution qmi Well (a) qmo (b) 400 CHAPTER 7 Fluid and Thermal Systems ■ Solution We model the system as shown in part (b) of the ﬁgure. 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. From conservation of mass, we have d (ρ Ah) = qmi (t) − qmo (t) dt Since ρ and A are constant, we have ρA dh = qmi (t) − qmo (t) dt which can be integrated as follows: 1 h(t) = h(0) + ρA t [qmi (u) − qmo (u)] du 0 Once we know the ﬂow rates, we can evaluate the integral. A common hydraulic actuator is the piston-and-cylinder actuator used on many types of heavy equipment, such as the backhoe shown in Figure 7.1.3. When the operator moves a handle, hydraulic ﬂuid under high pressure is sent through the line to the cylinder. The ﬂuid acts on the piston within the cylinder and produces a force that is 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. Piston rod Cylinder Operator controls Hydraulic lines E X A M P L E 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 a frictionless surface. Part (b) of the ﬁgure shows the piston rod connected to a rack-and-pinion gear. The pressures p1 and p2 are applied to each side of the piston by two pumps. Assume the piston rod diameter is small compared to the piston area, so the effective piston area A is the same on both sides of the piston. Assume also that the piston and rod mass have been lumped into m and that any friction is negligible. (a) Develop a model of the motion of the displacement 7.1 Pinion x x m A p1 p2 I p1 401 Figure 7.1.4 A hydraulic cylinder for (a) translating a mass and for (b) rotating a pinion gear. 5 x R R m A Conservation of Mass Rack p2 (a) (b) x of the mass in part (a) of the ﬁgure, assuming that p1 and p2 are given functions of time. Also, obtain the expression for the mass ﬂow rate that must be delivered or absorbed by the two pumps. (b) Develop a model of the displacement x in part (b) of the ﬁgure. The inertia of the pinion 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, and thus from Newton’s law, m ẍ = ( p1 − p2 )A Because p1 and p2 are given functions of time, we can integrate this equation once to obtain the velocity: ẋ(t) = ẋ(0) + A m t [ p1 (u) − p2 (u)] du 0 The rate at which ﬂuid volume is swept out by the piston is A ẋ, and thus if ẋ > 0, the pump providing pressure p1 must supply ﬂuid at the mass rate ρ A ẋ, and the pump providing pressure p2 must absorb ﬂuid at the same mass rate. b. Because we want an expression for the displacement x, we obtain an expression for the equivalent mass of the rack, pinion, and load. The kinetic energy of the system is KE = 1 2 1 2 1 m ẋ + I θ̇ = 2 2 2 m+ I R2 ẋ 2 because R θ̇ = ẋ. Thus the equivalent mass is I R2 The required model can now be obtained by replacing m with m e in the model developed in part (a). me = m + A Mixing Process ■ Problem A mixing tank is shown in Figure 7.1.5. Pure water ﬂows into the tank of volume V = 600 m3 at the constant volume rate of 5 m3 /s. A solution with a salt concentration of si kg/m3 ﬂows into the tank at a constant volume rate of 2 m3 /s. Assume that the solution in the tank is well mixed so that the salt concentration in the tank is uniform. Assume also that the salt dissolves completely so that the volume of the mixture remains the same. The salt concentration so kg/m3 in the outﬂow is the same as the concentration in the tank. The input is the concentration si (t), E X A M P L E 7.1.4 402 CHAPTER 7 Figure 7.1.5 A mixing process. 2 m3/s Solution si Fluid and Thermal Systems 5 m3/s Water V 5 600 m3 so qvo Mixer whose value may change during the process, thus changing the value of so . Obtain a dynamic model of the concentration so . ■ Solution Two mass species are conserved here: water mass and salt mass. The tank is always full, so the mass of water m w in the tank is constant, and thus conservation of water mass gives dm w = 5ρw + 2ρw − ρw qvo = 0 dt where ρw is the mass density of fresh water, and qvo is the volume outﬂow rate of the mixed solution. This equation gives qvo = 5 + 2 = 7 m3 /s. The salt mass in the tank is so V , and conservation of salt mass gives d (so V ) = 0(5) + 2si − so qvo = 2si − 7so dt or, with V = 600, dso (1) = 2si − 7so dt This is the model. The time constant for the mixing process is 600/7 = 85.7 s. Thus, if si is initially zero and then becomes a nonzero constant value S, the salt concentration in the outﬂow will eventually become constant at the value 2S/7 after approximately 4(85.7) = 343 s. 600 7.2 FLUID CAPACITANCE Sometimes it is very useful to think of ﬂuid systems in terms of electrical circuits. Table 7.2.1 gives the ﬂuid quantity, its common nomenclature, its linear relation, and its analogous electrical property. Fluid resistance is the relation between pressure and mass ﬂow rate. Fluid capacitance is the relation between pressure and stored mass. We will concentrate on ﬂuid resistance and capacitance. Fluid resistance relates to energy dissipation while ﬂuid capacitance relates to potential energy. Fluid inertance relates to ﬂuid acceleration and kinetic energy. Fluid systems obey two laws that are analogous to Kirchhoff’s current and voltage laws; these laws are the continuity and the compatibility laws. The continuity law is simply a statement of conservation of ﬂuid mass. This says that the total mass ﬂow into 7.2 Fluid Capacitance 403 Table 7.2.1 Analogous ﬂuid and electrical quantities. Fluid quantity Fluid mass, m Mass ﬂow rate, qm Pressure, p Fluid linear resistance, R R = p/qm Fluid capacitance, C C = m/ p Fluid inertance, I I = p/(dqm /dt) Electrical quantity Charge, Q Current, i Voltage, v Electrical resistance, R R = v/i Electrical capacitance, C C = Q/v Electrical inductance, L L = v/(di/dt) a junction must equal the total ﬂow out of the junction. This is analogous to Kirchhoff’s current law. Flow through two rigid pipes joined together to make one pipe is an example where this applies. If, however, the ﬂow is through ﬂexible tubes that can expand and contract under pressure, then the outﬂow rate is not the sum of the inﬂow rates. This is an example where ﬂuid mass can accumulate within the system and is analogous to having a capacitor in an electrical circuit. The compatibility law is analogous to Kirchhoff’s voltage law, which states that the sum of signed voltage differences around a closed loop must be zero. It is an expression of conservation of energy. The compatibility law states that the sum of signed pressure differences around a closed loop must be zero. 7.2.1 FLUID SYMBOLS AND SOURCES Figure 7.2.1 shows the commonly used symbols for ﬂuid system elements. The resistance symbol is used to represent ﬁxed resistances, for example, due to pipe ﬂow, oriﬁce ﬂow, 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 use ideal pressure and ﬂow sources in our ﬂuid system models. An ideal pressure source is capable of supplying the speciﬁed pressure at any ﬂow rate. An ideal ﬂow source is capable of supplying the speciﬁed ﬂow. These ideal sources are approximations to real devices such as pumps. For example, Figure 7.2.2 shows the steady-state ﬂow-pressure relation for a centrifugal pump, where qm is the mass ﬂow rate produced by the pump when the pressure difference across the pump is p. When the outlet pressure is greater than the inlet pressure, p > 0. Such Resistance p1 ps Manually adjusted valve Actuated valve p2 ps 5 p2 2 p1 qs Ideal pressure source Ideal flow source Pump Figure 7.2.1 Fluid system symbols. 404 CHAPTER 7 Figure 7.2.2 Steady-state ﬂow-pressure relation for a centrifugal pump. Fluid and Thermal Systems s3 s2 s1 qm Dp curves depend on the pump speed, labeled s1 , s2 , and so on in the ﬁgure. To determine the operating condition of the pump for a given speed, we need another relation between qm and p. This relation depends on the load connected to the pump outlet. We will see 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 ﬂuid mass and the resulting pressure caused by the stored mass. Figure 7.2.3 illustrates this relation, which holds for both pneumatic and hydraulic systems. At a particular reference point ( pr , m r ) the slope is C, where dm (7.2.1) C= d p p= pr Thus, ﬂuid capacitance C is the ratio of the change in stored mass to the change in pressure. Figure 7.2.3 General ﬂuid capacitance relation and its linear approximation. m C 1 mr pr p 7.2 Fluid Capacitance 405 Capacitance of a Storage Tank E X A M P L E 7.2.1 ■ Problem Consider the tank shown in Figure 7.2.4. Assume that the sides are vertical so that the crosssectional area A is constant. This is the case, for example, with a cylindrical tank whose horizontal cross section is circular, or with a tank having vertical sides and a rectangular horizontal cross section. Derive the expression for the tank’s capacitance. ■ Solution Because the tank’s sides are vertical, the liquid height h is related to m, the liquid mass in the tank, by m = ρ Ah. The total pressure at the bottom of the tank is ρgh + pa , but the pressure due only to the stored ﬂuid mass is p = ρgh. We can therefore express the pressure as a function of the mass m stored in the tank as p = mg/A. Thus, m= pA g and the capacitance of the tank is given by C= dm A = dp g Figure 7.2.4 Capacitance of a storage tank. pa h h h A A A p (a) (b) When the container does not have vertical sides, such as the one shown in Figure 7.2.5, the cross-sectional area A is a function of the liquid height h, and the relations between m and h and between p and m are nonlinear. In such cases, there is no single value for the container’s capacitance. The ﬂuid mass stored in the container is h m = ρV = ρ A(x)d x, 0 which gives dm = ρA dh Figure 7.2.5 A storage tank of arbitrary shape. A h qmi qmo 406 CHAPTER 7 Fluid and Thermal Systems For such a container, conservation of mass gives dm = qmi − qmo dt (7.2.2) but dm d p dp dm = =C dt d p dt dt Thus C dm dp = = qmi − qmo dt dt (7.2.3) Also dm dh dh dm = = ρA dt dh dt dt So ρA dh = qmi − qmo dt (7.2.4) Equations (7.2.2), (7.2.3), and (7.2.4) are alternative, but equivalent, hydraulic models of a container of ﬂuid. They suggest that either pressure p, mass m, or height h can be chosen as the model’s variable. These variables are all indicators of the system’s potential energy, and as such any one can be chosen as a state variable. If the container’s cross-sectional area is constant, then V = Ah and thus the liquid volume V can also be used as the model variable. E X A M P L E 7.2.2 Capacitance of a V-Shaped Trough ■ Problem (a) Derive the capacitance of the V-shaped trough shown in Figure 7.2.6a. (b) Use the capacitance to derive the dynamic models for the bottom pressure p and the height h. The mass inﬂow rate is qmi (t), and there is no outﬂow. ■ Solution a. From part (b) of the ﬁgure, D = 2h tan θ , and the vertical cross-sectional area of the liquid is h D/2. Thus the ﬂuid mass is given by m = ρV = ρ Figure 7.2.6 A V-shaped trough. A h D h 2 2 L (a) (b) 1 hD 2 L = (ρ L tan θ )h 2 7.3 But p = ρgh and thus, m = (ρ L tan θ) p ρg From the deﬁnition of capacitance, C= b. dm = dp From (7.2.3) with qmo = 0, C ṗ = qmi , or 2L tan θ ρg 2 2 = L tan θ ρg 2 2L tan θ ρg 2 p Fluid Resistance 407 p2 p dp = qmi dt which is a nonlinear equation because of the product p ṗ. We can obtain the model for the height by substituting h = p/ρg. The result is (2ρ L tan θ )h dh = qmi dt 7.3 FLUID RESISTANCE Fluid meets resistance when ﬂowing through a conduit such as a pipe, through a component such as a valve, or even through a simple opening or oriﬁce, such as a hole. We now consider appropriate models for each type of resistance. The mass ﬂow rate q̂m through a resistance is related to the pressure difference p̂ across the resistance. This relation, p̂ = f (q̂m ), is illustrated in general by Figure 7.3.1. We deﬁne the ﬂuid resistance Rr as the slope of f (q̂m ) evaluated at a reference equilibrium condition ( pr , qmr ). That is, d p̂ d p̂ Rr = = (7.3.1) d q̂m q̂m =q̂mr d q̂m r If we need to obtain an approximate linear model of the pressure-ﬂow rate relation, we can use a Taylor series expansion to linearize the expression p̂r = f (q̂m ) near a reference operating point ( pr , qmr ) as follows (after dropping the second order and higher terms in the expansion): d p̂ p̂ = pr + (qm − qmr ) = pr + Rr (qm − qmr ) (7.3.2) d q̂m r where Rr is the linearized resistance deﬁned by (7.3.1). p^ Figure 7.3.1 General ﬂuid resistance relation and its linear approximations. Rr 1 p pr qm qmr q^m 408 CHAPTER 7 Fluid and Thermal Systems Referring to Figure 7.3.1, we deﬁne a new set of variables p and qm , called deviation variables, that represent small but ﬁnite changes in p̂ and q̂m from their reference values pr and qmr . From Figure 7.3.1, we see that p = p̂ − p̂r (7.3.3) qm = q̂m − q̂mr (7.3.4) In terms of these deviation variables, we can rewrite (7.3.2) as p̂ − p̂r = Rr (q̂m − q̂mr ) or p = Rr qm , or p (7.3.5) Rr 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 = resulting small change in mass ﬂow rate Rr 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 (q̂m ), which depends on the application. In a limited number of cases, such as pipe ﬂow under certain conditions, the relation of p̂ versus q̂m is linear so that p̂ = R q̂m , or p̂ (7.3.6) q̂m = R where R is the linear resistance. In some other applications the relation is a square-root relation. p̂ q̂m = (7.3.7) B where B is a constant that often must be determined empirically. The relation (7.3.7) gives p̂ = B q̂m2 and thus d p̂ Rr = = 2Bqmr d q̂m r and p qm = (7.3.8) 2Bqmr This is the linearized model corresponding to the relation (7.3.7). This relation can be √ expressed instead in terms of pr as follows. Because qmr = pr /B, we obtain Rr = 2B pr /B = 2 Bpr qm = and thus 1 √ p (7.3.9) 2 Bpr When only the curve of p̂ versus q̂m is available, we can obtain a linearized model by graphically computing the slope S of the tangent line that passes through the reference point ( p̂r , q̂mr ). The equivalent, linearized resistance Rr is the slope S. The resistance symbol shown in Figure 7.3.2 represents all types of ﬂuid resistance, whether linear or not. Although the symbol looks like a valve, it can represent ﬂuid resistance due to other causes, such as pipe wall friction and oriﬁces. qm = 7.3 qm p1 R1 R2 p2 qm qm p3 p1 Fluid Resistance R Figure 7.3.2 Combination of (a) series resistances and (b) parallel resistances. p3 R 5 R1 1 R2 (a) R1 qm1 qm qm R qm 1 51 11 R R1 R2 R2 qm2 qm 5 qm1 1 qm 2 (b) As with electrical resistances, linear ﬂuid resistance elements obey the series and parallel combination rules. These are illustrated in Figure 7.3.2. Series ﬂuid resistances carry the same ﬂow rate; parallel ﬂuid resistances have the same pressure difference across them. 7.3.1 LAMINAR PIPE RESISTANCE Fluid motion is generally divided into two types: laminar ﬂow and turbulent ﬂow. Laminar ﬂow can be described as “smooth” in the sense that the average ﬂuid particle velocity is the same as the actual particle velocity. If the ﬂow is “rough,” the average particle velocity will be less than the actual particle velocity, because the ﬂuid particles meander while moving downstream. This is turbulent ﬂow. You can see the difference between laminar and turbulent ﬂow by slightly opening a faucet; the ﬂow will be smooth. As you open the faucet more, eventually the ﬂow becomes rough. If the pipe ﬂow is laminar, the following linear relation applies. q̂m = p̂ R (7.3.10) or equivalently p (7.3.11) R The laminar resistance for a level pipe of diameter D and length L is given by the Hagen-Poiseuille formula qm = R= 128μL πρ D 4 (7.3.12) where μ is the ﬂuid viscosity. The viscosity is a measure of the “stickiness” of the ﬂuid. Thus molasses has a higher value of μ than that of water. Not all pipe ﬂow is laminar. A useful criterion for predicting the existence of laminar ﬂow is the Reynolds number Ne , the ratio of the ﬂuid’s inertial forces to the viscosity forces. For a circular pipe, Ne = ρv D μ 409 (7.3.13) 410 CHAPTER 7 Fluid and Thermal Systems where v = qv /(π D 2 /4), the average ﬂuid velocity. For Ne > 2300 the ﬂow is often turbulent, while for Ne < 2300 laminar ﬂow usually exists. The precise value of Ne above which the ﬂow becomes turbulent depends on, for example, the ﬂow conditions at 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 proﬁle no longer changes with increasing distance, is much less than 0.06D Ne . Because laminar ﬂow can be expected only if Ne < 2300, L e 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 the pipe length, the more reliable will be our resistance calculations. 7.3.2 TORRICELLI'S PRINCIPLE An oriﬁce can simply be a hole in the side of a tank or it can be a passage in a valve. We saw an example of oriﬁce ﬂow in Example 1.5.2 in Chapter 1, in which we analyzed the ﬂow rate of water through a small hole in the side of a plastic milk bottle. We found that the ﬁtted function is f = 9.4h 0.558 , where f is the outﬂow rate in ml/s and the water height h is in centimeters. It turns out that the empirically determined exponent 0.558 is close to its theoretical value of 0.5, as we will now demonstrate. Around 1640 Torricelli discovered that the ﬂow rate through an oriﬁce is proportional to the square root of the pressure difference. This observation can be simply derived by considering a mass m of ﬂuid a height h above the oriﬁce (see Figure 7.3.3). The potential energy of the mass is mgh. As the mass falls toward the oriﬁce its potential energy is converted to kinetic energy mv 2 /2. If all the potential energy is converted to kinetic energy at the oriﬁce, then mgh =√mv 2 /2, and the maximum speed the ﬂuid mass can attain through the oriﬁce is v = 2gh. Because the pressure √ drop across the oriﬁce is p = ρgh, we can express the maximum speed as v = 2 p/ρ. Thus the mass√ﬂow rate qm √through the oriﬁce of area Ao can be no greater than Ao ρv = Ao ρ 2 p/ρ = Ao 2 pρ. The actual ﬂow rate will be less than this value because of friction effects. To account for these frictional effects, we introduce a factor Cd in the ﬂow rate equation as follows: q̂m = Cd Ao 2 pρ (7.3.14) The factor Cd is the discharge coefﬁcient, which must lie in the range 0 < Cd ≤ 1. A typical value for water is 0.6. Because p = ρgh, (7.3.14) can be expressed in terms of the volume ﬂow rate q̂v and the height h as follows: q̂v = Cd Ao 2g h 0.5 Figure 7.3.3 Derivation of Torricelli's principle. m PE 5 mgh, KE 5 0 h 2 PE 5 0, KE 5 m v 2 m Ao 7.4 Dynamic Models of Hydraulic Systems 411 Thus the theoretical value of the exponent (0.5) is close to the value obtained in the bottle experiment. Equation (7.3.14) depends on the oriﬁce area being small enough so that the pressure variation over the oriﬁce area is negligible compared to the average pressure at the oriﬁce. For a liquid-level system with a circular oriﬁce, this implies that the liquid height above the oriﬁce must be large compared to the oriﬁce diameter. The oriﬁce relation (7.3.14) can be rearranged in the form of (7.3.7). q̂m = Cd Ao 2ρ p̂ = p̂ Ro (7.3.15) where the oriﬁce resistance is deﬁned as Ro = 1 2ρCd2 A2o (7.3.16) 7.3.3 TURBULENT AND COMPONENT RESISTANCE For us, the practical importance of the difference between laminar and turbulent ﬂow lies in the fact that laminar ﬂow can be described by the linear relation (7.3.5), while turbulent ﬂow is described by the nonlinear relation (7.3.7). Components, such as valves, elbow bends, couplings, porous plugs, and changes in ﬂow area resist ﬂow and usually induce turbulent ﬂow at typical pressures, and (7.3.7) is often used to model them. Experimentally determined values of B are available for common types of components. 7.4 DYNAMIC MODELS OF HYDRAULIC SYSTEMS In this section we consider a number of hydraulic system examples dealing with liquidlevel 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 mass of liquid in the tank, and as kinetic energy in the mass of liquid ﬂowing in the pipe. In many systems, the mass of the liquid in the pipes is small compared to the liquid mass in the tanks. If the mass of liquid in a pipe is small enough or is ﬂowing at a small enough velocity, the kinetic energy contained in it will be negligible compared to the potential energy stored in the liquid in the tank. If the kinetic energy of the liquid is signiﬁcant, more advanced ﬂuid-ﬂow theory is required. This is usually not the case for the scope of applications considered here. Liquid-Level System with an Oriﬁce ■ Problem The cylindrical tank shown in Figure 7.4.1 has a circular bottom area A. The volume inﬂow rate from the ﬂow source is q̂vi (t), a given function of time. The oriﬁce in the side wall has an area Ao and discharges to atmospheric pressure pa . Develop a model of h, the deviation of the liquid height from a reference equilibrium value h r , assuming that h 1 > L. E X A M P L E 7.4.1 412 CHAPTER 7 Fluid and Thermal Systems q^vi Figure 7.4.1 A liquid-level system with an oriﬁce. pa h h1 hr A h^ L ■ Solution If the inﬂow rate q̂vi is held constant at the rate qvir , the liquid level eventually becomes constant at the height h r . Using the oriﬁce ﬂow relation (7.3.14), this height can be found from ρqvir = Cd Ao 2ρ(ρgh r ) or hr = 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 Ah 1 ) d(h + h r + L) dh = ρA = ρA dt dt dt Conservation of mass implies that ρA dh = ρ q̂vi − Cd Ao 2 p̂ρ dt (1) where the pressure drop across the oriﬁce is p̂ = pa + ρg(h + h r ) − pa = ρg(h + h r ) Therefore, equation (1) becomes ρA dh = ρ q̂vi − Cd Ao 2gρ 2 (h + h r ) dt Canceling ρ gives the desired model. A dh = q̂vi − Cd Ao 2g(h + h r ) dt or, because ĥ = h + h r , A d ĥ = q̂vi − Cd Ao dt 2g ĥ (2) Note that the height L does not appear in the model because the liquid below the oriﬁce does not affect the pressure at the oriﬁce. 7.4 Dynamic Models of Hydraulic Systems 413 Linearizing a Model E X A M P L E 7.4.2 ■ Problem Consider the liquid-level system with an oriﬁce, treated in Example 7.4.1. The model is given by equation (2) of that example. d ĥ = q̂vi − q̂vo = q̂vi − Cd Ao 2g ĥ dt √ Consider the case where A = 2 ft2 and Cd Ao 2g = 6. Estimate the system’s time constant for two cases: (i) the inﬂow rate is held constant at q̂vi = 12 ft3 /sec and (ii) the inﬂow rate is held constant at q̂vi = 24 ft3 /sec. A ■ Solution Substituting the given values, we obtain 2 d ĥ = q̂vi − q̂vo = q̂vi − 6 ĥ dt (1) When the inﬂow 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 ĥ = h r and d ĥ/dt equal to 2 zero. This gives 36h r = qvir . 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 outﬂow ﬂow rate q̂vo = 6 ĥ through the oriﬁce as a function of the height 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. The general equation for these lines is q̂vo = 6 ĥ = 6 ĥ r + 30 d q̂vo d ĥ (ĥ − ĥ r ) = 6 ĥ r + 3ĥ r−1/2 (ĥ − ĥ r ) r Figure 7.4.2 Linearized approximations of the resistance relation. 3 6 h^ < 24 1 4 (h^ 2 16) 25 20 3 6 h^ < 12 1 2 (h^ 2 4) 6 h^ 15 10 5 0 0 4 8 12 h^ 16 20 24 414 CHAPTER 7 Fluid and Thermal Systems and is the same as a Taylor series expansion truncated after the ﬁrst order term. Noting that ĥ r = h r , this equation becomes q̂vo = 6 h r + 3h r−1/2 (ĥ − h r ) (2) For Case (i), this equation becomes 3 q̂vo = 12 + (ĥ − 4) 2 and for Case (ii), 3 q̂vo = 24 + (ĥ − 16) 4 These are the equations of the straight lines shown in the ﬁgure. Noting that ĥ − h r = h, equation (2) can be expressed in the simpler form q̂vo = 6 h r + (3h r−1/2 )h √ Substitute this into equation (1), and note that d ĥ/dt = dh/dt and q̂vi − 6 h r = qvi , to obtain dh (3) = qvi + (3h r−1/2 )h dt This is the linearized model that is a good approximation of the nonlinear model (1) near the reference 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 h r = 16. Thus, for Case (i), if the input ﬂow rate is changed slightly from its equilibrium value of 12, the liquid height will take about 4(4/3), or 16/3, sec to reach its new height. For Case (ii), if the input ﬂow rate is changed slightly from its value of 24, the liquid height 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 chosen for the linearization. Because the straight line is an approximation to the q̂vo = 6 ĥ curve, we cannot use the linearized models to make predictions about the system’s behavior far from the equilibrium point. However, despite this limitation, a linearized model is useful for designing a ﬂow control system to keep the height near some desired value. If the control system works properly, the height will stay near the equilibrium value, and the linearized model will be accurate. 2 E X A M P L E 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 inﬂow rate from the ﬂow source is q̂mi (t), a given function of time. The total mass outﬂow rate q̂mo is not given and must be determined. The outlet resistance R is the linearized resistance about the reference condition (h r , qmir ). Develop a model of h, the deviation of the liquid height from the constant reference height h r , where ĥ = h r + h. Figure 7.4.3 A liquid-level system with a ﬂow source. q^mi pa h^ A R pa q^mo 7.4 Dynamic Models of Hydraulic Systems 415 ■ Solution The total mass in the tank is m = ρ Aĥ = ρ A(h + h r ) and from conservation of mass d[ρ A(h + h r )] dh dm = = ρA = q̂mi − q̂mo dt dt dt because ρ, h r , and A are constants. Expressing q̂mi and q̂mo in terms of the deviation variables qmi and qmo , we have ρA dh = (qmi + qmir ) − (qmo + qmor ) = (qmir − qmor ) + (qmi − qmo ) dt Because the reference height h r is a constant, the outﬂow rate at equilibrium must equal the inﬂow rate. Thus qmir − qmor = 0, and the model becomes ρA dh = qmi − qmo dt (1) Because R is a linearized resistance, then for small changes h in the height, qmo = 1 1 [(ρgh + pa ) − pa ] = ρgh R R Substituting this into equation (1) gives the desired model: ρA dh 1 = qmi − ρgh dt R which can be rearranged as R A dh R +h = qmi g dt ρg So the time constant is τ = R A/g. If qmi = 0 (so the inﬂow rate remains constant at qmir ), then h will be essentially zero for t > 4τ , which means that the height will return to nearly the equilibrium height h r at that time. Some engineers are helped by thinking of a ﬂuid system in terms of an analogous electric circuit, in which pressure difference plays the role of voltage difference, and mass ﬂow rate is analogous to current. A ﬂuid resistance resists ﬂow just as an electrical resistor resists current. A ﬂuid capacitance stores ﬂuid mass just as an electrical capacitor stores charge. Figure 7.4.4 shows an electric circuit that is analogous to the tank system of Figure 7.4.3. The circuit model is C 1 dv = is − v dt R The input current i s is analogous to the inﬂow rate qmi , the voltage v across the capacitor is analogous to the ﬂuid pressure ρgh, and the electrical capacitance C is analogous to the ﬂuid capacitance A/g. It is a matter of personal opinion as to whether such analogies help to understand the dynamics of ﬂuid systems, and you should decide for yourself. Always keep in mind, however, that we should not get too dependent on analogies for developing models, because they might not always properly represent the underlying physics of the original system. Figure 7.4.4 Electric circuit analogous to the hydraulic system shown in Figure 7.4.3. is R C v 416 CHAPTER 7 Fluid and Thermal Systems E X A M P L E 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 source p̂ s = 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 reference condition ( psr , h r ). Develop a model of h, the deviation of the liquid height from the constant reference height h r , where ĥ = h r + h. pa Figure 7.4.5 A liquid-level system with a pressure source. h^ pa p^s A R1 pa R2 ■ Solution The total mass in the tank is m = ρ Aĥ = ρ A(h + h r ), and from conservation of mass dm d[ρ A(h + h r )] dh = = ρA = q̂mi − q̂mo dt dt dt or dh ρA = (qmi + qmir ) − (qmo + qmor ) = (qmi − qmo ) + (qmir − qmor ) dt Because at the reference equilibrium, the outﬂow rate equals the inﬂow rate, qmir − qmor = 0, and we have dh ρA (1) = qmi − qmo dt This is a linearized model that is valid for small changes around the equilibrium state. Because the outlet resistance has been linearized, 1 ρgh qmo = [(ρgh + pa ) − pa ] = R2 R2 Similarly for the mass inﬂow rate, we have 1 1 qmi = [( ps + pa ) − (ρgh + pa )] = ( ps − ρgh) R1 R1 Substituting into equation (1) gives ρgh 1 R1 + R2 1 dh ( ps − ρgh) − = ps − ρg h = ρA dt R1 R2 R1 R1 R2 This can be rearranged as R1 R2 A dh R2 +h = ps g(R1 + R2 ) dt ρg(R1 + R2 ) The coefﬁcient of dh/dt gives the time constant, which is τ = R1 R2 A/g(R1 + R2 ). When a ﬂuid system contains more than one capacitance, you should apply the conservation of mass principle to each capacitance, and then use the appropriate resistance relations to couple the resulting equations. To do this you must assume that some pressures or liquid heights are greater than others and assign the positive-ﬂow directions accordingly. If you are consistent, the mathematics will handle the reversals of ﬂow direction automatically. 7.4 Dynamic Models of Hydraulic Systems 417 Two Connected Tanks E X A M P L E 7.4.5 ■ Problem The cylindrical tanks shown in Figure 7.4.6a have bottom areas A1 and A2 . The total mass inﬂow rate from the ﬂow source is q̂mi (t), a given function of time. The resistances are linearized resistances about the reference condition h 1r , h 2r , qmir . (a) Develop a model of the liquid heights h 1 and h 2 . (b) Suppose the resistances are equal: R1 = R2 = R, and the areas are A1 = A and A2 = 3A. Obtain the transfer function H1 (s)/Q mi (s). (c) Use the transfer function to solve for the steady-state response for h 1 if the inﬂow rate qmi is a unit-step function, and estimate how long it will take to reach steady state. Is it possible for liquid heights to oscillate in the step response? ■ Solution a. Using deviation variables as usual, we note that ĥ 1 = h 1r + h 1 ĥ 2 = h 2r + h 2 q̂mi = qmir + qmi For convenience, assume that h 1 > h 2 . This is equivalent to assuming that the ﬂow rate from tank 1 to tank 2 increases. From conservation of mass applied to tank 1, we obtain d(ρ A1 ĥ 1 ) d[ρ A1 (h 1 + h 1r )] dh 1 = = ρ A1 = −q̂1m = −(q1m + q1mr ) dt dt dt From physical reasoning we can see that the two heights must be equal at equilibrium, and thus q1mr = 0. Therefore, ρ A1 dh 1 = −q1m dt Because R1 is a linearized resistance, q1m = ρg (h 1 − h 2 ) R1 q^mi h^1 h^2 A1 A2 R1 R2 (a) v1 C1 R1 i1 v2 i2 R2 C2 (b) is Figure 7.4.6 (a) Two connected tanks. (b) Analogous electric circuit. 418 CHAPTER 7 Fluid and Thermal Systems So, after canceling ρ on both sides, the model for tank 1 is A1 dh 1 g = − (h 1 − h 2 ) dt R1 (1) Similarly for tank 2, d[ρ A2 (h 2 + h 2r )] dh 2 d(ρ A2 ĥ 2 ) = = ρ A2 dt dt dt Conservation of mass gives dh 2 = q̂mi + q̂1m − q̂mo = (qmi + qmir ) + (q1m + q1mr ) − (qmo + qmor ) dt Recalling that q1mr = 0, we note that this implies that qmir = qmor , and thus ρ A2 dh 2 = qmi + q1m − qmo dt Because the resistances are linearized, we have ρ A2 ρ A2 b. dh 2 ρg ρg (h 1 − h 2 ) − h2 = qmi + dt R1 R2 (2) The desired model consists of equations (1) and (2). Substituting R1 = R2 = R, A1 = A, and A2 = 3A into the differential equations and dividing by A, and letting B = g/R A we obtain ḣ 1 = −B(h 1 − h 2 ) 3ḣ 2 = qmi qmi + B(h 1 − h 2 ) − Bh 2 = + Bh 1 − 2Bh 2 ρA ρA Apply the Laplace transform of each equation, assuming zero initial conditions, and collect terms to obtain (s + B)H1 (s) − B H2 (s) = 0 −B H1 (s) + (3s + 2B)H2 (s) = 1 Q mi (s) ρA (3) (4) Solve equation (3) for H2 (s), substitute the expression into equation (4), and solve for H1 (s) to obtain H1 (s) R B 2 /ρg (5) = 2 Q mi (s) 3s + 5Bs + B 2 c. The characteristic equation is 3s 2 + 5Bs + B 2 = 0 and has the two real roots √ −5 ± 13 s= B = −1.43B, −0.232B 6 Thus 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 can be obtained by applying the ﬁnal value theorem to equation (5) with Q mi (s) = 1/s. h 1ss = lim s H1 (s) = lim s s→0 s→0 R B 2 /ρg R 1 = 3s 2 + 5Bs + B 2 s ρg The time constants are τ1 = 1 0.699 = 1.43B B τ2 = 1 4.32 = 0.232B B 7.4 Dynamic Models of Hydraulic Systems 419 The largest time constant is τ2 and thus it will take a time equal to approximately 4τ2 = 17.2/B to reach steady state. Figure 7.4.6b shows an electrical circuit that is analogous to the hydraulic system shown in part (a) of the ﬁgure. The currents i s , i 1 , and i 2 are analogous to the mass ﬂow rates qmi , qm 1 , and qmo . The voltages v1 and v2 are analogous to the pressures ρgh 1 and ρgh 2 , and the capacitances C1 and C2 are analogous to the ﬂuid capacitances A1 /g and A2 /g. 7.4.2 HYDRAULIC DAMPERS Dampers oppose a velocity difference across them, and thus they are used to limit velocities. The most common application of dampers is in vehicle shock absorbers. Linear Damper E X A M P L E 7.4.6 ■ Problem A damper exerts a force as a result of a velocity difference across it. Figure 7.4.7 shows the principle used in automotive shock absorbers. A piston of diameter W and thickness L has a cylindrical hole of diameter D. The piston rod extends out of the housing, which is sealed and ﬁlled with a viscous incompressible ﬂuid. Assuming that the ﬂow through the hole is laminar and that the entrance length L e is small compared to L, develop a model of the relation between the applied force f and ẋ, the relative velocity between the piston and the cylinder. ■ Solution Assume that the rod’s cross-sectional area and the hole area π(D/2)2 are small compared to the piston area A. Let m be the combined mass of the piston and rod. Then the force f acting on the piston rod creates a pressure difference ( p1 − p2 ) across the piston such that m ÿ = f − A( p1 − p2 ) (1) If the mass m or the acceleration ÿ is small, then m ÿ ≈ 0, and we obtain f = A( p1 − p2 ) (2) 1 1 qm = ( p1 − p2 ) ρ ρR (3) For laminar ﬂow through the hole, qv = The volume ﬂow rate qv is the rate at which the piston sweeps out volume as it moves, and can be expressed as qv = A( ẏ − ż) = A ẋ x· 5 y· 2 z· z· (4) Figure 7.4.7 A damper. y· p2 A p1 f L W D 420 CHAPTER 7 Fluid and Thermal Systems because the ﬂuid is incompressible. Combining equations (2), (3), and (4), we obtain f = A(ρ R A ẋ) = ρ R A2 ẋ = c ẋ where the damping coefﬁcient c is given by c = ρ R A2 . From the Hagen-Poiseuille formula (7.3.7) for a cylindrical conduit, R= 128μL πρ D 4 and thus the damping coefﬁcient can be expressed as c= 128μL A2 π D4 The approximation m ÿ ≈ 0 is commonly used for hydraulic systems to simplify the resulting model. To see the effect of this approximation, use instead equation (1) with equations (3) and (4) to obtain qv = 1 1 ( p1 − p2 ) = ( f − m ÿ) = A ẋ ρR ρRA Thus, f = m ÿ + ρ R A2 ẋ Therefore, if m ÿ cannot be neglected, the damper force is a function of the absolute acceleration as well as the relative velocity. 7.4.3 HYDRAULIC ACTUATORS Hydraulic actuators are widely used with high pressures to obtain high forces for moving large loads or achieving high accelerations. The working ﬂuid may be liquid, as is commonly found with construction machinery, or it may be air, as with the air cylinder-piston units frequently used in manufacturing and parts-handling equipment. E X A M P L E 7.4.7 Hydraulic Piston and Load ■ Problem Figure 7.4.8 shows a double-acting piston and cylinder. The device moves the load mass m in response to the pressure sources p1 and p2 . Assume the ﬂuid is incompressible, the resistances are linear, and the piston mass is included in m. Derive the equation of motion for m. pa Figure 7.4.8 A double-acting piston and cylinder. p2 A pa p1 R1 p3 R2 p4 x m 7.4 Dynamic Models of Hydraulic Systems 421 ■ Solution Deﬁne the pressures p3 and p4 to be the pressures on the left- and right-hand sides of the piston. The mass ﬂow rates through the resistances are 1 ( p1 + pa − p3 ) (1) R1 1 qm 2 = ( p4 − p2 − pa ) (2) R2 From conservation of mass, qm 1 = qm 2 and qm 1 = ρ A ẋ. Combining these four equations we obtain qm 1 = p1 + pa − p3 = R1 ρ A ẋ (3) p4 − p2 − pa = R2 ρ A ẋ (4) Adding equations (3) and (4) gives p4 − p3 = p2 − p1 + (R1 + R2 )ρ A ẋ (5) m ẍ = A( p3 − p4 ) (6) From Newton’s law, Substitute equation (5) into (6) to obtain the desired model: m ẍ + (R1 + R2 )ρ A2 ẋ = A( p1 − p2 ) (7) Note that if the resistances are zero, the ẋ term disappears, and we obtain m ẍ = A( p1 − p2 ) which is identical to the model derived in part (a) of Example 7.1.3. Hydraulic Piston with Negligible Load ■ Problem Develop a model for the motion of the load mass m in Figure 7.4.8, assuming that the product of the load mass m and the load acceleration ẍ is very small. ■ Solution If m ẍ is very small, from equation (7) of Example 7.4.7, we obtain the model (R1 + R2 )ρ A2 ẋ = A( p1 − p2 ) which can be expressed as ẋ = p1 − p 2 (R1 + R2 )ρ A (1) From this we see that if p1 − p2 is constant, the mass velocity ẋ will also be constant. The implications of the approximation m ẍ = 0 can be seen from Newton’s law: m ẍ = A( p3 − p4 ) (2) If m ẍ = 0, equation (2) implies that p3 = p4 ; that is, the pressure is the same on both sides of the piston. From this we can see that the pressure difference across the piston is produced by a large load mass or a large load acceleration. The modeling implication of this fact is that if we neglect the load mass or the load acceleration, we can develop a simpler model of a hydraulic system—a model based only on conservation of mass and not on Newton’s law. The resulting model will be ﬁrst order rather than second order. E X A M P L E 7.4.8 422 CHAPTER 7 Fluid and Thermal Systems E X A M P L E 7.4.9 Hydraulic Motor ■ Problem A hydraulic motor is shown in Figure 7.4.9. The pilot valve controls the ﬂow rate of the hydraulic ﬂuid from the supply to the cylinder. When the pilot valve is moved to the right of its neutral position, the ﬂuid enters the right-hand piston chamber and pushes the piston to the left. The ﬂuid displaced by this motion exits through the left-hand drain port. The action is reversed for a pilot valve displacement to the left. Both return lines are connected to a sump from which a pump draws ﬂuid to deliver to the supply line. Derive a model of the system assuming that m ẍ is negligible. ■ Solution Let y denote the displacement of the pilot valve from its neutral position, and x the displacement of the load from its last position before the start of the motion. Note that a positive value of x (to the left) results from a positive value of y (to the right). The ﬂow through the cylinder port uncovered by the pilot valve can be treated as ﬂow through an oriﬁce. Let p be the pressure drop across the oriﬁce. Thus, from (7.3.14) with p replaced by p, the volume ﬂow rate through the cylinder port is given by 1 1 qv = qm = Cd Ao 2pρ = Cd Ao 2p/ρ (1) ρ ρ where Ao is the uncovered area of the port, Cd is the discharge coefﬁcient, and ρ is the mass density of the ﬂuid. The area Ao is approximately equal to y D, where D is the port depth (into the page). If Cd , ρ, p, and D are taken to be constant, equation (1) can be written as qv = Cd Dy √ 2p/ρ = By (2) where B = Cd D 2p/ρ. Assuming that the rod’s area is small compared to the piston area, the piston areas on the left and right sides are equal to A. The rate at which the piston pushes ﬂuid out of the cylinder is A d x/dt. Conservation of volume requires the volume ﬂow rate into the cylinder be equal to the volume ﬂow rate out. Therefore, qv = A Figure 7.4.9 A hydraulic motor. y Pilot valve Return dx dt Supply (3) Return Spool Port x m Hydraulic lines A Piston 7.4 Dynamic Models of Hydraulic Systems pa Figure 7.4.10 Pressures in a hydraulic motor. ps pa p2 423 pa p1 Combining the last two equations gives the model for the servomotor: dx B = y (4) dt A This model predicts a constant piston velocity d x/dt if the pilot valve position y is held ﬁxed. The pressure drop p can be determined as follows. We assume that because of geometric symmetry, the pressure drop is the same across both the inlet and outlet valves. From Figure 7.4.10 we see that p = ( ps + pa ) − p1 = p2 − pa and thus p1 − p2 = ps − 2p (5) where p1 and p2 are the pressures on either side of the piston. The force on the piston is A( p1 − p2 ), and from Newton’s law, m ẍ = A( p1 − p2 ). Using the approximation m ẍ = 0, we √ see that p1 = p2 , and thus equation (5) shows that p = ps /2. Therefore B = Cd D ps /ρ. Equation (4) is accurate for many applications, but for high accelerations or large loads, this model must be modiﬁed because it neglects the effects of the inertia of the load and piston on the pressures on each side of the piston. 7.4.4 PUMP MODELS Pump behavior, especially dynamic response, can be quite complicated and difﬁcult to model. At our level of treatment, we will conﬁne ourselves to obtaining linearized models based on the steady-state performance curves. Typical performance curves for a centrifugal pump are shown in Figure 7.4.11a, which relates the mass ﬂow rate qm through the pump to the pressure increase p in going from the pump inlet to its outlet, for a given pump speed s j . For a given speed and given equilibrium values (qm )e and (p)e , we can obtain a linearized description as shown in part (b) of the ﬁgure. This linearized model consists of a straight line tangent to the pump curve at the equilibrium point, and can be expressed as 1 δqm = − δ(p) r where δqm and δ(p) are the deviations of qm and p from their equilibrium values. Thus, δqm = qm − (qm )e and δ(p) = p − (p)e . Identiﬁcation of the equilibrium values depends on the load connected downstream of the pump. Once this load is known, the resulting equilibrium ﬂow rate of the system 424 CHAPTER 7 Figure 7.4.11 (a) Performance curves for a centrifugal pump. (b) Linearized model. Fluid and Thermal Systems s3 s2 qm s1 Dp (a) qm qm 5 2 1r (Dp) (qm)e Pump curve Dp (Dp)e (b) can be found as a function of p. When this function is graphed on the same plot as the pump curve, the intersection of the two curves will establish the equilibrium point. The required procedure is illustrated by the following example. E X A M P L E 7.4.10 A Liquid-Level System with a Pump ■ Problem Figure 7.4.12 shows a liquid-level system with a pump input and a drain whose linear resistance is R2 . The inlet from the pump to the tank has a linear resistance R1 . The resistances were linearized about the reference height h = hr . Obtain a linearized model of the liquid height h. 7.4 Dynamic Models of Hydraulic Systems Figure 7.4.12 A liquid-level system with a pump. pa h A p1 R1 R2 pa qm2 qm1 pa ■ Solution Let p = p1 − pa . Denote the mass ﬂow rates through each resistance as qm 1 and qm 2 . These ﬂow rates are 1 1 qm 1 = ( p1 − ρgh − pa ) = (p − ρgh) (1) R1 R1 1 1 qm 2 = (ρgh + pa − pa ) = ρgh (2) R2 R2 From conservation of mass, ρA 425 dh 1 1 (p − ρgh) − ρgh = qm 1 − qm 2 = dt R1 R2 (3) At equilibrium, qm 1 = qm 2 , so from equation (3), 1 1 (p − ρgh) = ρgh R1 R2 which gives ρgh = R2 p R1 + R2 (4) Substituting this into expression (2) we obtain an expression for the equilibrium value of the ﬂow rate qm 2 as a function of p: qm 2 = 1 p R1 + R2 (5) This is simply an expression of the series resistance law, which applies here because ḣ = 0 at equilibrium and thus the same ﬂow occurs through R1 and R2 . When equation (5) is plotted on the same plot as the pump curve, as in Figure 7.4.13, the intersection gives the equilibrium values of qm 1 and p. A straight line tangent to the pump curve and having the slope −1/r then gives the linearized model: 1 δqm 1 = − δ(p) r where δqm 1 and δ(p) are the deviations from the equilibrium values. From equation (4), p = R1 + R2 ρgh R2 and thus δ(p) = R1 + R2 ρg δh R2 (6) 426 CHAPTER 7 Fluid and Thermal Systems qm Figure 7.4.13 Graphical solution of the pump model. qm1 5 2 1r (Dp) qm2 5 1 Dp R1 1 R2 (qm1)e Pump curve Dp (Dp)e and from equation (6), 1 1 R1 + R2 ρg δh δqm 1 = − δ(p) = − r r R2 (7) The linearized form of equation (3) is ρA d δh = δqm 1 − δqm 2 dt From equations (2) and (7), ρA ρg 1 R1 + R2 d ρg δh − δh δh = − dt r R2 R2 or d 1 R1 + R2 1 A δh = − + dt r R2 R2 g δh This is the linearized model, and it is of the form d δh = −b δh dt where b= 1 R1 + R2 1 + r R2 R2 g A The equation has the solution δh(t) = δh(0)e−bt . Thus if additional liquid is added to or taken from the tank so that δh(0) = 0, the liquid height will eventually return to its equilibrium value. The time to return is indicated by the time constant, which is 1/b. 7.4.5 FLUID INERTANCE We have deﬁned ﬂuid inertance I as I = p dqm /dt 7.5 Pneumatic Systems 427 which is the ratio of the pressure difference over the rate of change of the mass ﬂow rate. The inertance is the change in pressure required to produce a unit rate of change in mass ﬂow rate. Thus inertance relates to ﬂuid acceleration and kinetic energy, which are often negligible either because the moving ﬂuid mass is small or because it is moving at a steady rate. There are, however, some cases where inertance may be signiﬁcant. The effect known as water hammer is due partly to inertance. Also, as we will see in the following example, inertance can be signiﬁcant in conduits that are long or that have small cross sections. Calculation of Inertance E X A M P L E 7.4.11 ■ Problem Consider ﬂuid ﬂow (either liquid or gas) in a nonaccelerating pipe (Figure 7.4.14). Derive the expression for the inertance of a slug of ﬂuid of length L. ■ Solution The mass of the slug is ρ AL, where ρ is the ﬂuid mass density. The net force acting on the slug due to the pressures p1 and p2 is A( p2 − p1 ). Applying Newton’s law to the slug, we have dv = A( p2 − p1 ) dt where v is the ﬂuid velocity. The velocity v is related to the mass ﬂow rate qm by ρ Av = qm . Using this to substitute for v, we obtain ρ AL L dqm = A( p2 − p1 ) dt or L dqm = p2 − p 1 A dt With p = p2 − p1 , we obtain L p = A dqm /dt Thus, from the deﬁnition of inertance I , L A Note that the inertance is larger for longer pipes and for pipes with smaller cross section. I = The signiﬁcance of inertance in a given application is often difﬁcult to assess. Often

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