Ch 07 fluid and thermal systems

of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien §1.Conservation of Mass For incompressible fluids conservation of mass ⟺ conservation of volume The mass flow rate ??= ??

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7 Fluid and Thermal Systems

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Part 1 Fluid Systems

§1.Conservation of Mass

For incompressible fluids

conservation of mass ⟺ conservation of volume

The mass flow rate

𝑞𝑚= 𝜌𝑞𝑣

𝜌: fluid density,𝑘𝑔/𝑚3

𝑞𝑚: the mass flow rates,𝑘𝑔/𝑠

𝑞𝑣: the volume flow rate,𝑚3/𝑠

1.Density and Pressure

- Density(mass density): mass per unit volume,𝑘𝑔/𝑚3

- Pressure:force per unit area that is exerted by the fluid,𝑁/𝑚2

- Hydrostatic pressure: the pressure that exists in a fluid at rest

System Dynamics 7.03 Fluid and Thermal Systems

The figure is a representation of ahydraulic brake system The piston

in the master cylinder moves inresponse to the foot pedal Theresulting motion of the piston in theslave cylinder causes the brakepad to be pressed against thebrake drum with a force 𝑓3 Theforce𝑓1 depends on the force 𝑓4

applied by the driver’s foot Theprecise relation between𝑓1and𝑓4depends on the geometry of thepedal arm

Obtain the expression for the force𝑓3with the force𝑓1as the inputSystem Dynamics 7.04 Fluid and Thermal Systems

- Conservation of mass

𝑚 = 𝑞𝑚𝑖− 𝑞𝑚𝑜

𝑞𝑚𝑖: the mass inflow rate,𝑘𝑔/𝑠

𝑞𝑚𝑜: the mass outflow rate ,𝑘𝑔/𝑠

𝑞𝑚𝑖= 𝜌𝑞𝑣𝑖

𝑞𝑚𝑜= 𝜌𝑞𝑣𝑜

𝑞𝑣𝑖: total volume inflow rate,𝑚3/𝑠

𝑞𝑣𝑜: total volume inflow rate,𝑚3/𝑠

- The fluid mass𝑚 is related to the container volume 𝑉

𝑚 = 𝜌𝑉 ⟹ 𝑚 = 𝜌 𝑉then 𝜌 𝑉 = 𝜌𝑞𝑣𝑖− 𝜌𝑞𝑣𝑜

⟹ 𝑉 = 𝑞𝑣𝑖− 𝑞𝑣𝑜This is a statement of conservation of volume for the fluidSystem Dynamics 7.06 Fluid and Thermal Systems

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Water is pumped at the mass flow rate

𝑞𝑚𝑜(𝑡) from the tank Replacementwater is pumped from a well at themass flow rate 𝑞𝑚𝑖(𝑡) Determine thewater height ℎ(𝑡), assuming that thetank is cylindrical with a cross section𝐴Solution

From conservation of mass𝑑

Fig.(a): a cylinder and piston connected

to a load mass𝑚Fig.(b): the piston rod connected to arack-and-pinion gear

The pressures𝑝1and𝑝2 are applied toeach side of the piston by two pumps

Neglect the piston rod diameter andassume that the piston and rod masshave been lumped into𝑚

a.Develop a model of the motion of the displacement𝑥 of themass in fig.(a) Also, obtain the expression for the mass flowrate that must be delivered or absorbed by the two pumpsb.Develop a model of the displacement𝑥 in fig.(b) The inertia

of the pinion and the load connected to the pinion is𝐼System Dynamics 7.08 Fluid and Thermal Systems

Solution

a.Model of the motion for figure (a)Assuming that 𝑝1> 𝑝2, the net forceacting on the piston and mass𝑚 is (𝑝1−

𝑝2)𝐴, and thus from Newton’s law

𝑚 𝑥 = (𝑝1− 𝑝2)𝐴Integrate this equation once to obtain the velocity

𝑥 𝑡 = 𝑥 0 +𝐴

𝑚 0

𝑡

[𝑝1𝑢 − 𝑝2(𝑢)]𝑑𝑢The rate at which fluid volume is swept out by the piston is

𝐴 𝑥, and thus if 𝑥 > 0, the pump providing pressure 𝑝1must

supply fluid at the mass rate𝜌𝐴 𝑥, and the pump providing

pressure𝑝2must absorb fluid at the same mass rate

§1.Conservation of MassSolution

b.Model of the motion for figure (b)Firstly, obtain an expression for the equivalent mass of therack, pinion, and load The kinetic energy of the system is

𝑚𝑒= 𝑚 + 𝐼

𝑅2

Then, the required model can now be obtained by replacing

𝑚 with 𝑚𝑒in the model developed in part (a)System Dynamics 7.10 Fluid and Thermal Systems

A mixing tank is shown in the figure Pure water

flows into the tank of volume𝑉 = 600𝑚3 at the

constant volume rate of5𝑚3/𝑠 A solution with a

salt concentration of𝑠𝑖𝑘𝑔/𝑚3flows into the tank

at a constant volume rate of2𝑚3/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 concentration𝑠𝑜𝑘𝑔/𝑚3in the outflow is the same as

the concentration in the tank The input is the concentration

𝑠𝑖(𝑡) , whose value may change during the process, thus

changing the value of 𝑠𝑜 Obtain a dynamic model of the

concentration𝑠𝑜

§1.Conservation of MassSolution

Two mass species are conserved here: watermass and salt mass The tank is always full, sothe mass of water𝑚𝑤in the tank is constant, andthus conservation of water mass gives

𝑑𝑚𝑤

𝑑𝑡 = 5𝜌𝑤+ 2𝜌𝑤− 𝜌𝑤𝑞𝑣𝑜= 0 ⟹ 𝑞𝑣𝑜= 7𝑚3/𝑠

𝜌𝑤: the mass density of fresh water

𝑞𝑣𝑜: the volume outflow rate of the mixed solutionThe salt mass in the tank is𝑠𝑜𝑉, and conservation of salt massgives

𝑑

𝑑𝑡 𝑠𝑜𝑉 = 0 5 + 2𝑠𝑖− 𝑠𝑜𝑞𝑣𝑜= 2𝑠𝑖− 7𝑠𝑜⟹𝑑𝑠𝑜

𝑑𝑡 =

2𝑠𝑖− 7𝑠𝑜600

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§2.Fluid Capacitance

- Sometimes it is very useful to think of fluid systems in terms of

electrical circuits

Fluid linear resistance,𝑅 = 𝑝/𝑞𝑚 Electrical resistance, 𝑅 = 𝑣/𝑖

Fluid capacitance, 𝐶 = 𝑚/𝑝 Electrical capacitance,𝐶 = 𝑄/𝑣

Fluid inertance, 𝐼 = 𝑝/(𝑑𝑞𝑚/𝑑𝑡) Electrical inductance, 𝐿 = 𝑣/(𝑑𝑖/𝑑𝑡)

- Fluid resistanceis the relation between pressure and mass

flow rate Fluid resistance relates to energy dissipation

- Fluid capacitanceis the relation between pressure and stored

mass Fluid capacitance relates to potential energy

- Fluid inertancerelates to fluid acceleration and kinetic energy

- Note: the flow is through flexible tubes that can expand and

contract under pressure, then the outflow rate is not the sum ofthe inflow rates This is an example where fluid mass canaccumulate within the system and is analogous to having acapacitor in an electrical circuit

1.Fluid Symbols and Source

- Resistance

Both linear and nonlinear fixed resistances, for

example, pipe flow, orifice flow, or a restriction

- Valve

• manually adjusted valve: faucet

• actuated valve: driven by an electric motor or a

pneumatic device

2.Capacitance Relations

- The figure illustrates the relationbetween stored fluid mass and theresulting pressure caused by the storedmass

-Fluid capacitance 𝐶: the ratio of thechange in stored mass to the change inpressure

𝐶 ≡𝑑𝑚

𝑑𝑝 𝑝=𝑝𝑟𝑚: the stored fluid mass, 𝑘𝑔

𝑝: the resulting pressure, 𝑁/𝑚2

Consider the tank shown in thefigure Assume that the crosssectional area 𝐴 is constant

Derive the expression for thetank’s capacitance

Solution

The total pressure at the bottom of the tank:𝜌𝑔ℎ + 𝑝𝑎Pressure due only to the stored fluid mass: 𝑝 = 𝜌𝑔ℎThe pressure function of the mass𝑚: 𝑝 = 𝑚𝑔/𝐴The capacitance of the tank is given by

𝐶 ≡𝑑𝑚

𝑑𝑝=

𝐴𝑔

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- When the container does nothave vertical sides, the cross-sectional area𝐴 is a function ofthe liquid height ℎ , and therelations between𝑚 and ℎ andbetween𝑝 and 𝑚 are nonlinear

- The fluid mass stored in the container

a Derive the capacitance of the V-shaped

b Derive the dynamic models for the bottompressure𝑝 and the height ℎ The mass inflowrate is𝑞𝑚𝑖(𝑡)

2𝐿𝑡𝑎𝑛𝜃

𝜌𝑔2 𝑝𝑑𝑝

𝑑𝑡= 𝑞𝑚𝑖which is a nonlinear equation because of the

§3.Fluid Resistance

- Fluid meets resistance when flowingthrough a conduit such as apipe, through acomponent such as a valve, or eventhrough a simple opening ororifice, such as

a hole

- The relation between mass flow rate𝑞𝑚andthe pressure difference 𝑝 across theresistance𝑝 = 𝑓(𝑞𝑚) is shown in the figure

- Define thefluid resistance𝑅

𝑅 ≡ 𝑑𝑝

𝑑𝑞𝑚 𝑞=𝑞𝑚𝑟the reference values of𝑞𝑚𝑟 and𝑝𝑟depend on the particularapplication

- The relation𝑝 = 𝑓(𝑞𝑚)

• is linear in a limited number of cases, such as pipe flow

under certain conditions

𝑅𝑟: the linearized resistance at the reference condition(𝑞𝑚𝑟, 𝑝𝑟)

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1.Laminar Pipe Resistance

- Fluid motion is generally divided into two types

𝑅: flow resistance, 𝜇: the fluid viscosity, 𝑁𝑠/𝑚2

𝐿: the length of pipe, 𝑚 𝜌: the fluid density, 𝑘𝑔/𝑚3

𝐷: the diameter of pipe, 𝑚

§3.Fluid Resistance2.System Model

- In liquid-level systems such asshown in the figure, energy is stored

in two ways

• potential energy in the mass ofliquid in the tank

• kinetic energyin the mass of liquid flowing in the pipe

- 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 theliquid in the tank

- Example 7.3.1 Liquid-Level System with a Flow Source

The cylindrical tank shown in the figurehas a bottom area𝐴 The mass inflowrate is𝑞𝑚𝑖(𝑡) The outlet resistance islinear and the outlet discharges toatmospheric pressure 𝑝𝑎 Develop amodel of the liquid heightℎ

𝜌𝐴𝑑ℎ

𝑑𝑡= 𝑞𝑚𝑖−

𝜌𝑔

𝑅ℎ

• pressure difference, 𝜌𝑔ℎ ⟺ voltage difference, 𝑣

• mass flow rate, 𝑞𝑚𝑖 ⟺ current, 𝑖𝑠

• resistance resists flow ⟺ resistor resists current

• capacitance stores fluid mass, 𝐴/𝑔 ⟺ capacitor stores charge, 𝐶System Dynamics 7.28 Fluid and Thermal Systems

3.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

- The mass flow rate𝑞𝑚through the orifice

𝑞𝑚= 𝐶𝑑𝐴𝑜 2𝜌𝑝 = 𝐶𝑑𝐴𝑜 2𝜌 𝑝 = 𝑝/𝑅𝑜

𝐶𝑑: factor, 𝐴0: the area of the orifice,𝑚2

𝑝: the pressure of fluid, 𝑁/𝑚2 𝜌: the fluid density, 𝑘𝑔/𝑚3

𝑅𝑜: orifice resistance

- The volume flow rate𝑞𝑣through the orifice

𝑞𝑣= 𝐶𝑑𝐴𝑜 2𝑔 ℎ0.5

ℎ: the height of fluid, 𝑚

2𝜌𝐶𝑑𝐴𝑜

- Example 7.3.2 Liquid-Level System with an Orifice

The cylindrical tank shown in the figure has

a circular bottom area𝐴 The volume inflowrate from the flow source is𝑞𝑣𝑖(𝑡), a givenfunction of time The orifice in the side wall

atmospheric pressure𝑝𝑎 Develop a model

of the liquid heightℎ, assuming that ℎ1> 𝐿Solution

From conservation of mass and the orifice flow relation

𝜌𝐴𝑑ℎ

𝑑𝑡= 𝜌𝑞𝑣𝑖− 𝐶𝑑𝐴𝑜 2𝑝𝜌where 𝑝 = 𝜌𝑔ℎ Thus the model becomes

𝐴𝑑ℎ

𝑑𝑡= 𝑞𝑣𝑖− 𝐶𝑑𝐴𝑜 2𝑔ℎSystem Dynamics 7.30 Fluid and Thermal Systems

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4.Turbulance and Component Resistance

- 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

𝑞𝑚= 𝑝/𝑅

• turbulent flow is described by the nonlinear relation

𝑞𝑚= 𝑝/𝑅1

- Components, such as valves, elbow bends, couplings,

porous plugs, and changes in flow area resist flow and

usually induce turbulent flow at typical pressures, and𝑞𝑚=

𝑝/𝑅1is often used to model them

- Experimentally determined values of 𝑅 are available for

common types of components

§4.Dynamic Models of Hydraulic Systems

- Example 7.4.1 Liquid-Level System with a Pressure Source

Consider the system shown in thefigure The linear resistance 𝑅represents the pipe resistancelumped at the outlet of the pressuresource The bottom of the water tank

is a height𝐿 above the pressure source Develop a model ofthe water heightℎ with the supply pressure 𝑝𝑠and the flowrate𝑞𝑚𝑜(𝑡) as the inputs

SolutionThe total mass in the tank 𝑚 = 𝜌𝐴ℎSince𝜌 and 𝐴 are constants, from conservation of mass𝑑𝑚

𝑑𝑡= 𝜌𝐴

𝑑ℎ

𝑑𝑡= 𝑞𝑚𝑖− 𝑞𝑚𝑜System Dynamics 7.32 Fluid and Thermal Systems

Because the outlet resistance is linear

The desired model

Consider the system shown in thefigure, the input was the specified flowrate 𝑞𝑚𝑖 The linear resistance 𝑅represents at the outlet of the pressuresource The bottom of the water tank is

a height𝐿 above the pressure source

Develop a model of the water height ℎ with the supplypressure𝑝𝑠and the flow rate𝑞𝑚𝑜(𝑡) as the inputs

The cylindrical tanks have bottomareas𝐴1and𝐴2 The mass inflowrate𝑞𝑚𝑖(𝑡) from the flow source is

a given function of time Theresistances are linear and theoutlet discharges with pressure𝑝𝑎

a.Develop a model of the liquid heights ℎ1and ℎ2b.Suppose𝑅1= 𝑅2= 𝑅, and 𝐴1= 𝐴 and 𝐴2= 3𝐴 Obtain thetransfer function𝐻1(𝑠)/𝑄𝑚𝑖(𝑠)

c.Use the transfer function to solve for the steady stateresponse forℎ1if the inflow rate𝑞𝑚𝑖is a unit-step function,and estimate how long it will take to reach steady state Is itpossible for liquid heights to oscillate in the step response?

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Solution

a.Assume thatℎ1> ℎ2so that themass flow rate𝑞𝑚 1is positive ifflowing from tank 1 to tank 2

Conservation of mass applied totank 1 gives

c.The characteristic equation is3𝑠2+ 5𝐵𝑠 + 𝐵2= 0, with roots

𝑠 = −5 ± 13 𝐵/6 = −1.43𝐵, −0.232𝐵

⟹ 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 with 𝑄𝑚𝑖𝑠 = 1/𝑠

ℎ1𝑠𝑠= lim

𝑠→0𝑠𝐻1(𝑠) = lim

𝑠→0

𝑅𝐵2/𝜌𝑔3𝑠2+ 5𝐵𝑠 + 𝐵2

1

𝑠=

𝑅𝜌𝑔The time constants are

4.32𝐵The largest time constant is𝜏2and thus it will take a time

equal to approximately4𝜏2= 17.2𝐵 to reach steady state

§4.Dynamic Models of Hydraulic Systems5.Hydraulic Damper

Dampers oppose a velocity difference across them, and thusthey are used to limit velocities The most common application

of dampers is in vehicle shock absorbers

Consider a shock absorber:

a piston of diameter𝑊 andthickness𝐿 has a cylindricalhole of diameter𝐷

The piston rod extends out of the housing, which is sealedand filled with a viscous incompressible fluid Assuming thatthe flow through the hole is laminar and that the entrancelength𝐿𝑒 is small compared to𝐿, develop a model of therelation between the applied force 𝑓 and 𝑥 , the relativevelocity between the piston and the cylinder

Solution

Assume that therod’s cross-sectional area and the hole area

𝜋(𝐷/2)2are small compared to the piston area𝐴 Let 𝑚 be the

combined mass of the piston and rod Then the force𝑓 acting

on the piston rod creates a pressure difference(𝑝1− 𝑝2) across

the piston such that

𝑚 𝑦 = 𝑓 − 𝐴(𝑝1− 𝑝2)

If𝑚 or 𝑦 is small, then 𝑚 𝑦 ≈

0 ⟹ 𝑓 = 𝐴(𝑝1− 𝑝2)For laminar flow through the hole

𝑞𝑣=1

𝜌𝑅(𝑝1− 𝑝2)The volume flow rate𝑞𝑣can be expressed as

𝑞𝑣= 𝐴 𝑦 − 𝑧 = 𝐴 𝑥

§4.Dynamic Models of Hydraulic SystemsCombining the above equations, we obtain

𝑓 = 𝐴 𝜌𝑅𝐴 𝑥 = 𝜌𝑅𝐴2 𝑥 = 𝑐 𝑥, 𝑐 ≡ 𝜌𝑅𝐴2

From the Hagen-Poiseuille formula, for a cylindrical conduit

𝑅 =128𝜇𝐿𝜋𝜌𝐷4 ⟹ 𝑐 =128𝜇𝐿𝐴

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6.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

The figure shows a double-acting piston and cylinder The

device moves the load mass 𝑚 inresponse to the pressure sources 𝑝1

and𝑝2 Assume the fluid is incompressible,the resistances are linear, and thepiston mass is included in𝑚

Derive the equation of motion for𝑚

§4.Dynamic Models of Hydraulic SystemsSolution

Define the pressures𝑝3and𝑝4to be the pressures on the and right-hand sides of the piston

left-The mass flow rates through the resistances are

𝑝1+ 𝑝𝑎− 𝑝3= 𝑅1𝜌𝐴 𝑥

𝑝4− 𝑝2− 𝑝𝑎= 𝑅2𝜌𝐴 𝑥

⟹ 𝑝4− 𝑝3= 𝑝2− 𝑝1+ (𝑅1+ 𝑅2)𝜌𝐴 𝑥System Dynamics 7.44 Fluid and Thermal Systems

FromNewton’s law

- Example 7.4.6 Hydraulic Piston with Negligible Load

Develop a model for the motion of theload mass𝑚 in the figure, assumingthat the product of the load mass 𝑚and the load acceleration 𝑥 is verysmall

Solution

If𝑚 𝑥 is very small, from 𝑚 𝑥 + 𝑅1+ 𝑅2𝜌𝐴2 𝑥 = 𝐴(𝑝1− 𝑝2),

we obtain the model

𝑅1+ 𝑅2𝜌𝐴2 𝑥 = 𝐴(𝑝1− 𝑝2)which can be expressed as

𝑥 = 𝑝1− 𝑝2(𝑅1+ 𝑅2)𝜌𝐴

if𝑝1− 𝑝2is constant, the mass velocity 𝑥 will also be constantSystem Dynamics 7.46 Fluid and Thermal Systems

The implications of the approximation 𝑚 𝑥 = 0 can be seen

fromNewton’s law

𝑚 𝑥 = 𝐴(𝑝3− 𝑝4)

If𝑚 𝑥 = 0, the above equation impliesthat 𝑝3= 𝑝4 ⟹ the pressure is thesame on both sides of the pistonFrom 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 onNewton’s law The resulting

model will be first order rather than second order

The pilot valve controls the flowrate of the hydraulic fluid fromthe supply to the cylinder Whenthe pilot valve is moved to theright of its neutral position, thefluid enters the right-handpiston chamber and pushes thepiston to the left The fluiddisplaced by this motion exitsthrough the left-hand drain port

The action is reversed for a pilot valve displacement to theleft Both return lines are connected to a sump from which apump draws fluid to deliver to the supply line Derive a model

of the system assuming that𝑚 𝑥 = 0 is negligibleSystem Dynamics 7.48 Fluid and Thermal Systems

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𝐴𝑜: uncovered area of the port

𝐴𝑜≈ 𝑦𝐷, 𝐷: the port depth

𝐶𝑑: the discharge coefficient𝜌: mass density of the fluid

If𝐶𝑑,𝜌, 𝑝, and 𝐷 are taken to be constant

𝑞𝑣= 𝐶𝑑𝐷𝑦 2∆𝑝/𝜌 = 𝐵𝑦

where,𝐵 = 𝐶𝑑𝐷 2∆𝑝/𝜌

§4.Dynamic Models of Hydraulic SystemsThe rate at which the piston pushes fluid out of the cylinder is

𝐴𝑑𝑥/𝑑𝑡 From conservation of volume

𝑞𝑣= 𝐴𝑑𝑥𝑑𝑡Combining the last two equations givesthe model for the servomotor

𝑑𝑥

𝑑𝑡=

𝐵

𝐴𝑦This model predicts a constant pistonvelocity𝑑𝑥/𝑑𝑡 if 𝑦 is held fixed

The same pressure drop𝑝 across both the inlet and outlet valves

∆𝑝 = 𝑝𝑠+ 𝑝𝑎 − 𝑝1= 𝑝2− 𝑝𝑎⟹ 𝑝1− 𝑝2= 𝑝𝑠− 2∆𝑝FromNewton’s law 𝑚 𝑥 = 𝐴(𝑝1− 𝑝2), 𝑚 𝑥 ≈ 0 ⟹ 𝑝1= 𝑝2, andthus𝑝 = 𝑝𝑠/2 Therefore 𝐵 = 𝐶𝑑𝐷 𝑝𝑠/𝜌

7.Pump Models

- Pump behavior, especially dynamic response, can be quite

complicated and difficult to model

- Here, based on the steady-state performance curves to

obtain linearized models for pump

- Typical performance curves for a centrifugal pump which

relates the mass flow rate𝑞𝑚

through the pump to thepressure increase𝑝 in goingfrom the pump inlet to itsoutlet, for a given pumpspeed𝑠𝑗

- For a given speed and given equilibrium values(𝑞𝑚)𝑒 and

(𝑝)𝑒, we can obtain a linearizeddescription of the figure

be found as a function of𝑝System Dynamics 7.52 Fluid and Thermal Systems

- Example 7.4.8 A Liquid-Level System with a Pump

The figure shows a liquid-levelsystem with a pump input and a drainwhose linear resistance is 𝑅2 Theinlet from the pump to the tank has alinear resistance 𝑅1 Obtain alinearized model of the liquid heightℎSoution

Let𝑝 ≡ 𝑝1− 𝑝2 Denote the mass flow rates through each

resistance as𝑞𝑚1and𝑞𝑚2These flow rates are

(1)(2)

§4.Dynamic Models of Hydraulic SystemsFrom conservation of mass

𝑞𝑚 2= 1

𝑅1+ 𝑅2∆𝑝This is simply an expression of the series resistance law,which applies here because ℎ = 0 at equilibrium and thus thesame flow occurs through𝑅1and𝑅2

(3)

(4)

(5)

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Plotted the flow rate𝑞𝑚 2on the same plot as the pump curve,

the intersection gives the equilibriumvalues of𝑞𝑚1and𝑝 A straight linetangent to the pump curve andhaving the slope−1/𝑟 then givesthe linearized model

𝛿𝑞𝑚 1= −1

𝑟𝛿(∆𝑝)

𝛿𝑞𝑚 1,𝛿(𝑝): the deviations from

the equilibrium valuesFrom eq.(4) and eq.(6)

𝑅1+ 𝑅2

𝑅2 𝜌𝑔𝛿ℎSystem Dynamics 7.55 Fluid and Thermal Systems

(6)

(7)

§4.Dynamic Models of Hydraulic SystemsThe linearized form of eq.(3) 𝜌𝐴𝑑(𝛿ℎ)/𝑑𝑡 = 𝛿𝑞𝑚 1− 𝛿𝑞𝑚 2From eq.(2) and eq.(7)

𝜌𝐴𝑑

𝑑𝑡𝛿ℎ = −

1𝑟

𝑅1+ 𝑅2

𝑅2 𝑔𝛿ℎThis is the linearized model, and it is of the form𝑑

𝑑𝑡𝛿ℎ = −𝑏𝛿ℎ, 𝑏 =

1𝑟

𝑅1+ 𝑅2

𝑅2

𝑔𝐴The equation has the solution 𝛿ℎ(𝑡) = 𝛿ℎ(0)𝑒−𝑏𝑡 Thus ifadditional liquid is added to or taken from the tank so that𝛿ℎ(0) = 0 , the liquid height will eventually return to itsequilibrium value The time to return is indicated by the timeconstant, which is1/𝑏

8.Nonlinear System

- Common causes of nonlinearities in hydraulic system models

are a nonlinear resistance relation, such as due to orifice flow

or turbulent flow, or a nonlinear capacitance relation, such as

a tank with a variable cross section

- If the liquid height is relatively constant, say because of a

liquid-level controller, we can analyze the system by

linearizing the model

- In cases where the height varies considerably, we must solve

the nonlinear equation numerically

- Example 7.4.9 Liquid-Level System with an Orifice

Consider the liquid-level system with anorifice as in the figure The model is

𝐴𝑑ℎ

𝑑𝑡= 𝑞𝑣𝑖− 𝑞𝑣𝑜

= 𝑞𝑣𝑖− 𝐶𝑑𝐴𝑜 2𝑔ℎwhere𝐴 = 2𝑓𝑡2and𝐶𝑑𝐴𝑜 2𝑔 = 6Estimate thesystem’s time constant for two cases(i) the inflow rate is held constant at𝑞𝑣𝑖= 12𝑓𝑡3/𝑠𝑒𝑐(ii)the inflow rate is held constant at𝑞𝑣𝑖= 24𝑓𝑡3/𝑠𝑒𝑐System Dynamics 7.58 Fluid and Thermal Systems

Solution

Substituting the given values, we obtain

2𝑑ℎ

𝑑𝑡= 𝑞𝑣𝑖− 𝑞𝑣𝑜= 𝑞𝑣𝑖− 6 ℎWhen the inflow rate 𝑞𝑣𝑒= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, theliquid height reaches an equilibrium value

ℎ𝑒that can be found by setting𝑑ℎ/𝑑𝑡 = 0

The two cases of interest to us are (i)ℎ𝑒= 122/36 = 4𝑓𝑡 and

(ii)ℎ𝑒= 242/36 = 16𝑓𝑡 The graph is a plot of the flow rate

6 ℎ through the orifice as a function of the height ℎ The two

points corresponding toℎ𝑒= 4 and ℎ𝑒= 16 are indicated on

the plot

(1)

In the figure two straight lines areshown, each passing through one

of the points of interest (ℎ𝑒= 4andℎ𝑒= 16), and having a slopeequal to the slope of the curve atthat point

The general equation for theselines is

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2𝑑ℎ

𝑑𝑡= 𝑞𝑣𝑖− 6 ℎ𝑒− 3ℎ𝑒

−1

ℎ − ℎ𝑒 = 𝑞𝑣𝑖− 3 ℎ𝑒− 3ℎ𝑒−1/2ℎThe time constant of this linearized model isτ = 2 ℎ𝑒/3

𝜏

ℎ 𝑒 =4=4

3, 𝜏ℎ 𝑒 =16=8

3

- If the input rate𝑞𝑣𝑖is changed slightly from its equilibrium

value of𝑞𝑣𝑖= 12, the liquid height will take about 4(4/3) =

16/3sec to reach its new height

- If the input rate𝑞𝑣𝑖is changed slightly from its value of𝑞𝑣𝑖=

24, the liquid height will take about 8(4/3) = 32/3sec to

reach its new height

§4.Dynamic Models of Hydraulic Systems9.Fluid Inertance

Fluid inertance𝐼: the ratio of the pressure difference over therate of change of the mass flow rate

𝑑𝑞𝑚/𝑑𝑡

Consider fluid flow (either liquid or gas) in anonaccelerating pipe Derive the expressionfor the inertance of a slug of fluid of length𝐿Solution

The mass of the slug: 𝜌𝐴𝐿, 𝜌: the fluid mass densityThe net force acting on the slug due to the pressures𝑝1and𝑝2

𝐴(𝑝2− 𝑝1)

ApplyingNewton’s law to the slug

𝑑𝑞𝑚

𝑑𝑡 = 𝑝2− 𝑝1With𝑝 = 𝑝2− 𝑝1, we obtain

Inertance is larger for longer pipes and for smaller cross section pipes

§5.Pneumatic Systems

- Working fluid: a compressible fluid, most commonly air

- The response of pneumatic systems can be slower and moreoscillatory than that of hydraulic systems because of thecompressibility of working fluid

- The inertance relation is not usually needed to develop amodel because the kinetic energy of a gas is usuallynegligible Instead, capacitance and resistance elements formthe basis of most pneumatic system models

- Theperfect gas law

𝑝𝑉 = 𝑚𝑅𝑔𝑇𝑝: the absolute pressure, 𝑁/𝑚2 𝑉: gas volume, 𝑚3

- The perfect gas law enables us to solve for one of the

variables𝑝,𝑉,𝑚, or 𝑇 if the other three are given Additional

information is usually available in the form of a

pressure-volume or“process” relation

- The following process models are commonly used

𝑊 = 𝑚𝑐𝑣(𝑇1− 𝑇2) 𝑊:the external work

If𝑚 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, this reduces to the previous processes if 𝑛

is chosen as0,∞,1, 𝛾, and if the perfect gas law is usedSystem Dynamics 7.66 Fluid and Thermal Systems

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1.Pneumatics Capacitance

Fluid capacitance𝐶 is the ratio of the change in stored mass,

𝑚, to the change in pressure, 𝑝

If the gas undergoes a polytropic process

𝐶 =𝑚𝑉

𝑛𝑝𝑉=

𝑉

𝑛𝑅𝑔𝑇System Dynamics 7.67 Fluid and Thermal Systems

Obtain the capacitance of air in a rigid cylinder of volume0.03𝑚3, if the cylinder is filled by an isothermal process

Assume the air is initially at room temperature,293𝐾Solution

The filling of the cylinder can be modeled as an isothermalprocess if it occurs slowly enough to allow heat transfer tooccur between the air and its surroundings

In this case,𝑛 = 1 in the polytropic process equation, and weobtain

Air at temperature𝑇 passes through a valveinto a rigid cylinder of volume𝑉, as shown

in the figure The mass flow rate through

difference ∆𝑝 = 𝑝𝑖− 𝑝, and is given by anexperimentally determined function

𝑞𝑚𝑖= 𝑓(𝑝)Assume the filling process is isothermal Develop a dynamic

model of the gage pressure𝑝 in the container as a function

of the input pressure𝑝𝑖

§5.Pneumatic SystemsSolution

From conservation of mass, if𝑝𝑖= 𝑝 > 0

𝑞𝑚𝑖= 𝑓(𝑝) ⟹𝑑𝑚

𝑑𝑡 = 𝑞𝑚𝑖= 𝑓(∆𝑝)But

𝑑𝑚

𝑑𝑡 =

𝑑𝑚𝑑𝑝

𝑑𝑝

𝑑𝑡= 𝐶

𝑑𝑝𝑑𝑡

𝐶𝑑𝑝

𝑑𝑡= 𝑓 ∆𝑝 = 𝑓(𝑝𝑖− 𝑝)where the capacitance𝐶 is given with 𝑛 = 1

𝑅𝑔𝑇

If the function 𝑓 is nonlinear, then the dynamic model isnonlinear

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tienand thus

Part 2 Thermal Systems

- A thermal system is one in which energy is stored andtransferred as thermal energy, commonly called heat

- Thermal systems operate because of temperature differences,

as heat energy flows from an object with the highertemperature to an object with the lower temperature

- Thermal systems are analogous to electric circuits

• conservation of charge ⟺ conservation of heat

• voltage difference ⟺ temperature differenceSystem Dynamics 7.72 Fluid and Thermal Systems

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