C.02 the Common Elements of System Components

8/12/2019 C.02 the Common Elements of System Components

1/25

3/15/2

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

2. THE COMMON

ELEMENTS OF SYSTEM

COMPONENTS

Control System Technology 2.01 The Common Elements of System Components


1. Introduction- Five control system components (phn t): electrical (in), mechanical (c),

liquid flow (lng), gas flow(kh), or thermal (nhit)

- Four behavior elements (c tnh): resistance (tr khng), capacitance (dunglng), inertia(qun tnh), and dead-time delay(thigian tr)

- Four elements are defined in terms of three variables(thng s)

1stvariable: quantity of material (lngvtcht), energy or distance

2ndvariable: driving force or potential that tends to move or change the quantity variable

3rdvariable: time

- The three variable used to define the four elements for each type of components



2/25

3/15/2


1. Introduction

- Resistance

an opposition to the movement of flow of material or energy

being measured in terms of the amount of potential required to produce one unitof electric current, liquid flow rate, gas flow rate, heat flow rate, or velocity

- Capacitance

the amount of material, energy or distance required to make a unit change in potential being measured in terms of the amount of material, energy, or distance required

to make a unit change in potential

- Inertia/Inertance/Inductance

an opposition to change in the state of motion being measured in terms of the amount of potential required to increase electric

current, liquid flow rate, gas flow rate, or velocity by one unit per second

- Dead-time delay

the time interval between the time the signal appears at the input of acomponent and the time the corresponding response appears at the output

the effect of dead time is to delay the input signal by the dead-time delay



2. Electrical Elements1.Electrical resistanceis that property of material which impedes the flow of electric

current. Good conductors have a low resistance; insulators have a very high

resistance


= = () =

=

()


3/25

3/15/2


2. Electrical Elements

- Example: An electrical component is known to have a linear volt-ampere graph.

A test with 24 applied to the terminals of the components resulted in ameasured current of 12. Determine the resistance of the componentSolution: The volt-ampere graph is a straight line, so

= =24

0.012 = 2000

- Example: A light bulb is an example of an electric component with a nonlinear

volt-ampere graph. Determine the resistance of a light bulb at 6 from thefollowing information: 5.95results in 0.5and 6.05results in 0.504Solution: The electrical resistance of a nonlinear component can be

approximated

= = 6.05 5.95 0.504 0.500 = 25


2. Electrical Elements2.Electrical capacitanceis quantity of electric charge, (), required to make a unit

increase in the electrical potential, . The unit of electrical capacitance is ()

= ()

= =

= =

Note if

=

= = sin(+90)

= sin(+90)the current leads the voltage by 90in a capacitor




4/25

3/15/2


- Example: A current pulse with an amplitude of 0.1and a duration of 0.1s isapplied to an electrical capacitor. The voltage across the capacitor is increased

from 0 to +25 by the current pulse. Determine the capacitance, , of thecapacitorSolution:

= = = 0.1 10

0.1 25 0 = 0 . 4 1 0(/) = 0.4

3.Electrical inductanceis voltage required to produce a unit increase in the electric

current each second. The unit of electrical inductance is () = =

()

Note if =

==sin(+90) =sin(+90)

the voltage leads the current by 90in a inductor



2. Electrical Elements4.Electrical dead-time delayis the delay caused by the time it takes a signal to travel

from the source to the destination. The electrical signals travel at tremendous

speeds (2 1 0~ 3 1 0/), therefore in most control systems, the effect ofelectrical dead-time delayed is negligible

= ()




5/25

3/15/2


- Example:

a. Determine the dead-time delay of a

600long transmission line if the velocity

of propagation is 2.310/b. Determine the dead-time delay of a signal from a space vehicle that is located

2000 from the earth station receiving the signal. The signal travels at3 1 0/

Solution:

a. The dead-time delay of a 600long transmission line = 600 2.310/ =2.6110

= 2.61 b. The dead-time delay of a signal from a space vehicle

= 2 1 03 1 0/ =0.6710 = 6.7



2. Electrical ElementsSummary electrical equations

Resistance

= = =

Capacitance

= =

Inductance

= =

Dead-Time Delay

=



resistance, applied voltage, instantaneous current,, change in applied voltage, , change in current, capacitance, change in charge, , interval time, inductance, dead-time delay, distance between the input and output, velocity of travel,/


6/25

3/15/2


3. Liquid Flow Elements

1.Liquid flow resistance is that property of pipes, valves or restrictions which

impedes the flow of a liquid. It is measured in term of the increase of the pressure

required to make a unit increase in flow rate- Liquid resistance is determined by the relationship between the pressure drop

and the flow rate as expressed by a flow equation

- Two types of flow:laminar andturbulent, depending on the Reynolds numbers

= where, fluid density, /

= =

average velocity of fluid flow, /inside diameter of the pipe, absolute viscosity of fluid, .

- 4000:turbulent flow- 2000


7/25

3/15/2



- Laminar flow in round pipe (Hagen-Poiseuille law)

= ()- Turbulent flow in round pipe (Fanning equation)

= () = 2 ./ turbulent flow coefficient

values of friction factor,

interpolation formula = + ( )( )/( )


= 128

(./)

= 8 ./ /


3. Liquid Flow Elements- Example: Oil ( = 880 / , = 0.16 . ) at temperature of 15 flows

through a horizontal tube which is 0.01 in diameter with a flow rate of1.5710/. The line is 10 long. Determine the Reynolds number, theresistance and the pressure drop in the tube

Solution:

The average liquid flow rate

= 4 =4(1.5710/)

(0.01 ) = 2 / The Reynolds number

= =(880 /) (2 /) (0.01 )

0.16 . =109.95 The flow is laminar, the resistance and pressure drop

= 128 =128 (0.16 .) (10 )

(0.01 ) =6.51910./

= =(6.51910 ./)(1.5710/)=1.023510



8/25

3/15/2



- Example: Water (=1000 /, = 0.001 .) at temperature of 15flowsthrough a commercial steel pipe which is 0.01016 (0.4 ) in diameter with aflow rate of 3.785310/ (6 /). The line is 15.240 (50 )long.Determine the Reynolds number, the resistance and the pressure drop in thetube

Solution:

The average liquid flow rate

= 4 =4(3.785310/)

(0.01016 ) =4.669 / The Reynolds number

=

= (1000 /) (4.669 /) (0.01016 )

0.001 .

=47440

the flow is turbulent



3. Liquid Flow Elements Determine the friction factor by interpolation

= + = 0.035 + 0.028 0.035 4744010

10 10

=0.032088



9/25

3/15/2



Turbulent flow coefficient

=8 =

8 (1000 /) (0.032088) (15.24 ) (0.01016 )

=3.661410. /

The resistance of flow and pressure drop

= 2 = 2 (3 . 6 6 1 4 1 0 ./)(3.785310/)=2.771910./

The pressure drop

= =(3.661410 ./)(3.785310/)=5.24610



3. Liquid Flow Elements2.Liquid flow capacitanceis defined in terms of the increase in volume of liquid in a

tank required to make a unit increase in pressure at the outlet of the tank

= (/)

where, (/): liquid flow capacitance (): increase in volume (): corresponding increase in pressure

The increase in pressure in the tank depends on: the increase in level of the liquid

, the acceleration due to gravity

and the density of the liquid

=The increase in level in the tank is equal to the increase in volume divided by the

average surface area, , of the liquid in the tank=/

= =

= =

(

/)



10/25

3/15/2



- Example: A liquid tank has a diameter of 1.83 and a height of 10 .Determine the capacitance of the tank for each of the following fluid:

a. water: =1000/ b. oil: =880/c. kerosene: =800/ c. gasoline: =740/

Solution:

Determine = =

/4 =

(1.83 )/4(9.81 /) =

0.268

Applying for the liquid

a.water: =2.6810/ b.oil: = 3 . 0 5 1 0/

c.kerosene: =3.3510/ c.gasoline: = 3 . 6 2 1 0/



3. Liquid Flow Elements3.Liquid flow inertance is measured in terms of the amount of pressure drop in a

pipe required to increase the flow rate by 1 unit each second

= / (./)

where, (. /): inertance (): pressure drop in the pipe (/): change in flow rate (): time interval

The force will accelerate the fluid in the pipe

= = / with = and =

= (./)- Example: Determine the liquid flow inertance of water ( = 1000 /) in a

pipe that has a diameter of 2.1 and a length of 65 Solution:

= =

/4 =(1000 /) (65 )

(0.021 )/4 =1.8810. /



11/25

3/15/2



4.Dead time occurs whenever liquid is transported from one point to another in a

pipeline

= ()where, (): distance between the input and output , (/):fluid velocity- Example: Liquid flows in pipe that is 200 long and has a diameter of 6 . The

flow rate is 0.0113 /. Determine the dead-time delaySolution:

Average fluid velocity in pipe

= =

/4 =0.0113 /

(0.06 )/4 = 4 /

Determine the dead-time delay

= =200 4 / = 50


3. Liquid Flow ElementsSummary of liquid flow equations

Reynolds Number

= /Averagre Velocity

= =4

Laminar Flow Resistance(4000)

= = = 2 = 8



fluid density, / average fluid velocity in pipe, /

inside diameter of the pipe, absolute viscosity of fluid, . flow rate, / cross-sectional area of pipe, pressure drop from inlet to outlet of pipe, pressure at inlet of pipe, pressure at outlet of pipe, laminar flow resistance, ./ length of the pipe,

turbulent flow coefficient,

.

/

turbulent flow resistance, ./ friction factor surface area of liquid in tank, gravity acceleration, / inertance, ./dead-time delay, distance from inlet to outlet of pipe,


12/25

3/15/2


Summary of liquid flow equations

Capacitance

= Inertance

= Dead-Time Delay

=



fluid density, / average fluid velocity in pipe, / inside diameter of the pipe,

flow rate,

/

cross-sectional area of pipe, pressure drop from inlet to outlet of pipe, pressure at inlet of pipe, pressure at outlet of pipe, laminar flow resistance, ./ length of the pipe, absolute viscosity of fluid, . turbulent flow coefficient, ./ turbulent flow resistance, ./ friction factor

surface area of liquid in tank,

gravity acceleration, / inertance, ./dead-time delay, distance from inlet to outlet of pipe,


4. Gas Flow Elements1.Gas flow resistance is that property of pipes, valves or restrictions which impedes

the flow of a gas. It is measured in term of the increase of the pressure required to

produce an increase in gas flow rate of 1 /- Two types of flow:laminar andturbulent, depending on the Reynolds numbers

= 4/ In laminar flow, the pressure drop varies directly with the gas velocity (linear)

In turbulent flow, the pressure drop varies directly with the square of the gasvelocity (nonlinear)

- In practice, gas flow is almost always turbulent, and the common used equations

apply to turbulent flow

- If the pressure drop less than 10%of the initial gas pressure, the equation forincompressible flow gives reasonable accuracy for gas flow

= = = 2 (./)

= 8(./)



13/25

3/15/2


4. Gas Flow Elements

- Example: The graph of mass flow rate ()versus pressure drop (). Determinegas flow resistance as the point where the mass flow rate is 0.6 /

Solution:

Gas flow resistance is the slop of the

tangent line to the curve at the operation

point

= =5 2 1 0

0.72 / =72.222 ./



4. Gas Flow Elements- Example: A smooth tube is supplying 0.03 /of air (=1.8110. , =

1.22 /)at a temperature of 15. The tube is 30 long and has an insidediameter of 4 . Find the pressure drop and the gas flow resistance. The outletpressure ()is 102 , find the inlet pressure ()Solution:

The Reynolds number

= 4 =4 (0.03 /)

(1.8110. ) (0.04 ) = 5 . 2 8 1 0

Friction factor

= + = 0.03 + 0.018 0.030 5.2811 0 1 =0.023Turbulent flow coefficient

= 8 =8(0.023)(30 )

0.04 (1.22 /) =4.46910. /



14/25

3/15/2



The pressure drop

= = 4.469 10. / 0.03 / =4.02 The gas flow resistance

= 2 = 2 4.469 10. / (0.03 /) = 263 . /The inlet pressure

= + = 4.02 + 102 = 106.02



4. Gas Flow Elements2.Gas flow capacitance is defined in terms of the increase in the mass of gas in a

pressure vessel required to produce a unit increase in pressure while the

temperature remains constant

= = 10

where,

absolute pressure of the gas, volume of the gas, number of moles of gas, universal gas constant =8.314/. absolute temperature, mass of the gas,

gram molecular weigh of the gas, / = 10 =

10 Gas flow capacitance

= =

10 (/)



15/25

3/15/2



- Example: pressure tank has a volume of 0.75 . Determine the capacitance ifgas is nitrogen ( = 28.016 /)at 20Solution:

= 10 =( = 28.016 /) (0.75 )

10 (8.314/ . ) (293 ) = 8 . 6 1 0/


4. Gas Flow ElementsSummary of gas flow equations

Reynolds Number

= 4Resistance-Low-Pressure Drop

= = = 2 = 8

Capacitance

= 10



gas mass flow rate, / absolute viscosity of fluid, . inside diameter of the pipe, pressure drop from inlet to outlet of pipe, pressure at inlet of pipe, pressure at outlet of pipe, turbulent flow coefficient, ./gas flow resistance, ./ friction factor length of the pipe,

fluid density,

/

gas flow capacitance, / gram molecular weigh of the gas,/

volume of the gas, universal gas constant =8.314/

. absolute temperature,


16/25

3/15/2


5. Thermal Elements

1.Thermal resistanceis that property of a substance that impedes the flow of heat. It

is measured in terms of the different in temperature required to produce a heat

flow rate of 1 (joule per second)Normally heat flow occurs through a wall that separates two fluids (liquid or gas) at

different temperatures. Heat flows through the wall from the hotter fluid to the

cooler fluid

The heat flow rate, , from the outside fluid to theinside fluid

= ()Total resistance of the entire wall

=

(/)

= ()



5. Thermal Elements- The wall separation two fluids is a composite, consisting of several layers, each

contributing to the resistance of the wall

The unit resistance of the film

()= 1 (./)

The unit resistance of the inner layer

( ) = (./)

The unit resistance of the wall

()= 1 + + + 1 (./)The total resistance of the wall

= 11

+ +

+

1 (/)



17/25

3/15/2


5. Thermal Elements

- Film coefficients

+ Natural convection in still air

Horizontal surfaces facing up Horizontal surfaces facing down

= 2.50 . = 1.32 .

Vertical surfaces

= 1.787 .where, is the temperature difference between the main body of the fluidand the wall surface ,



5. Thermal Elements+ Natural convection in still water

= 2 . 2 6 + 34.3 .where, is the water temperature ,

+ Natural convection in oil

= 7.0.

. where, is the absolute viscosity, .

+ Forced convection for air against smooth surfaces and inside straight pipes

= 4.54+4.1 4.6 /7.75. > 4.6 /+ Forced convection for turbulent water flow in straight pipes

= 20.093(68.3+).

. where, is the inner diameter of pipe,



18/25

3/15/2


5. Thermal Elements

- Example: The wall has two inner layers: a steel plate 1 thick (=45/.) and insulation 2 thick (=0.036/.). The inside fluid is still water.The difference between the watertemperature and the surface temperature is

estimated to be = 10. The outside fluidis air, which has a velocity of 6 / . Thewater temperature is = 45 and the airtemperature is = 85 . The walldimensions are = 2 , = 3 . Determinethe unit resistance, the total resistance, the

heat flow, and the direction of the heat flow.

Solution:The area of the wall section

= 2 3 = 6



5. Thermal ElementsThe inside film coefficient

= 2.26 + 34.3 . = 2.26 + 34.3 10. = 566.7 /. The outside film coefficient

=7.75. = 7 . 7 5 6. = 29.71 /. The unit thermal resistance

= 1 + +

+

1

= 129.71 /. +0.01

45 /. +0.02

0.036 /. +1

566.7 /.

= 0.5912 .

/

The total thermal resistance

= =0.5912 ./

6 = 0.09853 /The total heat flow

= =85 45

0.09853 / = 406



19/25

3/15/2


5. Thermal Elements

2.Thermal capacitanceis defined in terms of the increase in heat required to make a

unit increase in temperature

= (/)where, thermal capacitance,/

mass, heat capacity, /.

- Example: Determine the thermal capacitance of 8.31 of water( = 1000 /, = 4190 /. )Solution:

= = 1000 / 8.31 4190/. = 3.48 10/


5. Thermal ElementsSummary of thermal equations

Resistance

= 1 + +

+

1

= = =

Film Coefficients

1. Natural convection in still air

Horizontal surfaces facing up

= 2.50 . Horizontal surfaces facing down

= 1.32 . Vertical surfaces

= 1.787 .



unit thermal resistance, . /outside film coefficient, /. inside film coefficient, /. thickness of layer , thermal conductivity, /. inside film coefficient, /. total thermal resistance, / heat flow rate, outside fluid temperature, inside fluid temperature, area of the wall surface, film coefficient, /. temperature difference between themain body of the fluid and the wall

surface , water temperature , absolute viscosity, .thermal capacitance, / mass,heat capacity, /.


20/25

3/15/2

5. Thermal Elements

Summary of thermal equations

2. Natural convection in still water

=2.26( + 34.3) .

3. Natural convection in oil

= 7.0.

. 4. Forced convection for air against smooth

surfaces and inside straight pipes

Air velocity, 4.6 / = 4 . 5 4 + 4 . 1 Air velocity, > 4.6 /

=7.75.5. Forced convection for turbulent water

flow in straight pipes

= 20.093(68.3+).. Capacitance

=



unit thermal resistance, . /outside film coefficient, /. inside film coefficient, /.

thickness of layer

,

thermal conductivity, /. inside film coefficient, /. total thermal resistance, / heat flow rate, outside fluid temperature, inside fluid temperature, area of the wall surface, film coefficient, /. temperature difference between the

main body of the fluid and the wall

surface ,

water temperature ,

absolute viscosity, .thermal capacitance, / mass,heat capacity, /.

6. Mechanical Elements1.Mechanical resistanceor friction(ma st) is that property of a mechanical system

that impedes motion. It is measured in terms of the increase in force required to

produce an increase in velocity of 1/ = = =

(./)

where, mechanical resistance, ./ mechanical resistance, ./ force, velocity, / change in force, change in velocity, /




21/25

3/15/2

6. Mechanical Elements

- Example: Mechanical resistance



A mechanical part with resistance provided by a dashpot Resistance due to laminar flow in a dashpot-viscous friction

Resistance due to turbulent flow in a dashpot Resistance due to both viscous friction and coulomb friction

6. Mechanical Elements- Example: A dashpot is used to provide mechanical resistance in a packaging

machine. A test was conducted in which a force of 98 produced a velocity of24 /. Determine the mechanical resistance

Solution:

The mechanical resistance = =

98 24 / = 4.08 . /




22/25

3/15/2


- Example: A mechanical system consists of a sliding load (Coulomb friction) and

a shock absorber (viscous friction). The force

versus velocity is shown in the figure. The

following data were obtained from the system

Determine the resistance, , and the Coulombfriction force, () . Write an equation for theforce, (), in terms of the velocity, ()

Solution:

The resistance

= = = =

9.607.1015.759.60 = 0.476 . /

The Coulomb friction force = + = = 7.1 0.476 . / 10.5/ = 2.1 Equation for the force

= + =2.1+0.476 ()



6. Mechanical Elements2.Mechanical capacitance (c dung/th nng) is defined as the increase in the

displacement of a spring required to make a unit increase in spring force

= =1 (/)

where, mechanical capacitance, / change in displacement, change in force, spring constant, /

- Example: A spring is used to provide mechanical capacitance in a system. A

force of 100 compresses the spring by 30 . Determine the mechanicalcapacitance and the spring constant

Solution:

The mechanical capacitance and the spring constant

= =0.30 100 =0.003 /

= 1 = 333.3 /




23/25

3/15/2


3.Mechanical inertia (qun tnh) is measured in terms of the force required to

produce a unit increase in acceleration

= ()

= ()where, average force, mass,

change in velocity, / increment of time, / rate of change of velocity, /



6. Mechanical Elements3.Mechanical inertia (qun tnh) is measured in terms of the force required to

produce a unit increase in acceleration

= () = ()

where, average force, mass, change in velocity, / increment of time, force, /rate of change of velocity, /

- Example: Automobile has a mass of 1500 . Determine the average forcerequired to accelerate A from 0 27.5 / in 6 . Automobile required anaverage force of 8000 to accelerate from 0 27.5 /in 6 . Determine mass of Solution:The required average force for

= = 1500 27.5 /

6 = 6875 The mass of

= = 8000 6

27.5 / = 1745




24/25

3/15/2


- Example: Automobile has a mass of 1500 . Determine the average forcerequired to accelerate A from 0 /to 27.5 /in 6 . Automobile required anaverage force of 8000 to accelerate from 0 /to 27.5 /in 6 . Determinemass of

- Solution:

Automobile = = 1500

27.5 /6 = 6875

Automobile = 8 0 0 0 =

27.506 = 800

627.5 = 1745



6. Mechanical Elements4.Mechanical dead time(thigian tr) is the time required to transport material from

one place to another

= ()where, : dead time delay, ; : distance, ; : velocity, /- Example: A belt conveyor is 30 long and has a belt speed of 3 /. Determine

the dead-time delay between the input and output ends of the belt. Write the

equation for the output mass flow rate ()in terms of the input mass flow rate ()Solution:

The dead-time delay

= =30 3 / = 10

The equation for the output mass

flow rate

= = 1 0




25/25

3/15/2


Summary of mechanical equations

Resistance

= = = Capacitance

= =1

Inertia

=

= Dead Time

=



mechanical resistance, ./ mechanical resistance, ./ force, velocity, / change in force, change in velocity, / mechanical capacitance, / change in displacement, spring constant, / average force, mass, increment of time, / rate of change of velocity, / dead time delay, distance,

C.02 the Common Elements of System Components

Documents