Thermal System Closed-Loop Temperature...

MechatronicsThermal System Case Study

Kevin Craig1

Thermal System Closed-Loop Temperature Control

•aluminum plate•thin-film resistive heater•ceramic insulation•conduction and convection

heat transfer•AD590 temperature sensor•microcontroller•on-off closed-loop control

with relay•support analog electronics


Kevin Craig2

Objective of the Case Study

• Control the temperature of a thin aluminum plate, as measured by a temperature sensor positioned in the middle of the top of the plate, by regulating the voltage supplied to a resistive heater positioned under the plate.

• The temperature of the plate is to be regulated to a point 20° C above the temperature of the ambient air.


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How Will We Accomplish The Objective ?

• Apply the general procedure for a dynamic system investigation

• Understand the physical system, develop a physical model on which to base analysis and design, and experimentally determine and/or validate model parameter values

• Develop a mathematical model of the system, analyze the system, and compare the results of the analysis to experimental measurements

• Design a feedback control system to meet performance specifications

• Implement the control system and experimentally validateits predicted performance


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Physical System

Aluminum SupportPlate

Heated ThinAluminum Plate1/32 inch thick

2 in x 2 ink = 177 W/m-K

Ceramic TapeInsulating Layer

1/8 inch thick2 in x 2 in

k = 0.055 W/m-K

Temperature Sensor

Thin-Film Resistive Heaterbetween

Thin Aluminum Plateand

Ceramic Insulation


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Properties of the Resistive Heater

Specification Value

Manufacturer Minco Products

Model Number HK-5169-R185-L12-B

Heater Resistance 185 ohms ± 10%

Heater Area 4 sq. in.

Heater Thickness 0.010 in.


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Specification Value

Rated Temperature Range -55° C to 150° C

Power Supply (min) 4 volts

Power Supply (max) 30 volts

Nominal Output Current @ 298.2 K 298.2 µA

Temperature Coefficient 1 µA/K

Calibration Error ± 2.5° C

Maximum Forward Voltage 44 volts

Maximum Reverse Voltage -20 volts

Case Breakdown Voltage ± 200 volts

Properties of the AD590 Temperature Sensor


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Property Value

Melting Point 775 K

Density, ρ 2770 kg/m3

Specific Heat, cp 875 J/kg-K

Thermal Conductivity, k 177 W/m-K

Material Properties of 6061 Aluminum


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Temperature Sensor Circuit

15 V

1 µA/K

1KΩ

Sensor Voltage

LF 411

Ksensor(volts/K) = (1µA)(Rsensor)


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Physical Model

Ambient air

Thin Aluminum Plate

q co

nve

ctio

n

Heat input q heater

Thin-Film Resistive Heater

Sensor

Ceramic Insulation


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Simplifying Assumptions

• Temperature of the plate is uniform

• No heat loss through the sides of the plate

• Thermal conductivity of the plate is constant

• Heat loss due to radiation is negligible compared to convective heat loss from the plate

• Convection coefficient is constant and is evaluated at the operating temperature of the plate

• Heat loss through the insulative layer is negligible

• Sensor dynamics are negligible

• Ambient air temperature is unaffected by the heat flux from the plate


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Mathematical Modeling

• Define system, system boundary, system inputs and outputs

• Define through and across variables• Write physical relations for each element• Write system relations of equilibrium and/or

compatibility• Combine system relations and physical

relations to generate the mathematical model for the system


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Define System, System Boundary,System Inputs and Outputs

Ambient air

Thin Aluminum Plate

q co

nve

ctio

n

Heat input q heater

Thin-Film Resistive Heater

Sensor

Ceramic Insulation

Input:Voltage supplied to resistive heater

Output:Plate temperature as measuredby sensor on top of plate


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Define Through and Across Variables

Through Variable: heat flow rate q (J/s or W)

Across Variable: temperature θ (K)

Assumption: All points in the body have the same average temperature and temperature deviations from the average at various points do not affect the validity of the single-temperature model.


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Write Physical Relations for Each Element

• Thermal Capacitance

• Thermal ResistanceConductionConvectionRadiation

• Thermal Sources

&θ = −1C

q t q tin outa f a f

q tkA

Lt t

q t hA t t

q t C t t

C t t

( ) [ ( ) ( )]

( ) [ ( ) ( )]

( ) [ ( ) ( )]

[( )( )][ ( ) ( )]

= −

= −

= −

= + + −

θ θ

θ θ

θ θ

θ θ θ θ θ θ

1 2

1 2

14

24

12

22

1 2 1 2

P V i VV

R

V

Rh h hh

h

h

h

= = =2

q tt

R( )

( )=

∆θ


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Write System Relations ofEquilibrium and/or Compatibility

• Select the temperature of each thermal capacitance as a state variable and use: &θ = −

1C

q t q tin outa f a fto obtain the corresponding state-variable equation.

• The net heat flow rate into a thermal capacitance depends on the heat sources and heat flow rates through thermal resistances. Use

q tR

t ta f a f a f= −1

1 2θ θ

to express the heat flow rates through the resistances in terms of the system’s state variables.


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Combine System Relations and Physical Relations to Obtain the Mathematical Model

q tV

R

q tR

hA

Cq t q t

Cq t

R

RC Cq t

RC

inh

h

out ambient ambient

in out

in ambient

in ambient

a f

a fa f

a f a f

a f

= =

= − = −

= −

= − −LNM

OQP

+ = +

heater input2

1

1

1 1

1 1 1

θ θ θ θ

θ

θ θ θ

θ θ θ

& )

&

&


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Predicted Dynamic Response

Solve the mathematical equation both analytically and numerically to predict the dynamic response of the physical system and gain physical insight.

&& ( )

θ θ θ

τθ θ θ τ

+ = +

+ = + =

1 1 1RC C

q tRC

Rq t RC

in ambient

in ambient

a f

τ

τω

ωτωτ

τdqdt

q Kq q Kq e

QQ

sKs

QQ

iK

oo i o is

t

o

i

o

i

+ = = −FHG

IKJ

=+

=+

∠ −

−FH

IK

−

1

1 121a f a f a f tan


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Experimental Testing:Model Parameter Identification

Two physical parameters need to be determined, one by calculation and the other by experiment.

Thermal Capacitance C = Mcp = ρVcp = 4.96 J/K

Thermal Resistance R = 1/hAEither measure τ = RC and solve for R ormeasure the steady-state temperature of the plate and use

q qVR Rin out

h

h

ambient= =−

or 2 θ θa f


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Control System Design

Error

To Workspace4

Qin

To Workspace2

RC.s+1

R(s)

System

u[1]^2/Rh

Electrical-Thermal

ConversionController: Relay

+

-

SumTemperature

Reference

Temp

To Workspace

Clock

time

To Workspace1Matlab / Simulink Block DiagramThermal Closed-Loop Control System


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Control System Simulation

Simulation of Control about 20° C Setpoint with a 2° C Deadband

0 20 40 60 80 1000

5

10

15

20

25

time (sec)

tem

pera

ture

(de

gree

s C

)


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Experimental Set-Up & Control Implementation


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Microcontroller Software Design

SetpointRead

SensorSignal

Deadband

Sensor SignalLess Than

Deadband Bottom

Sensor SignalGreater ThanDeadband Top

Sensor SignalIn

Deadband

HeaterOn

HeaterOff

IssuePrevious

Command

YesYes Yes

Decision Logic

InitalizeVariables

Read Sensorand Setpoint

Signals

Compute ErrorSignal

ImplementLogic

CommandActuator

Flow Diagram


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Microcontroller Real-Time Basic Program100 REM set up Port A as an output, reset Port B110 xby(0fd03h)=128:xby(0fd01h)=6120 PRINT USING(####)130 REM calibrate the A to D converters140 XBY(0ff03h)=2150 DO: SFR=XBY(0ff03h):WHILE SFR.AND.2160 IF SFR.AND.40h THEN GOTO 140165 MEMX=0170 REM read current temperature180 XBY(0ff00h)=1190 CURTEMP=256*XBY(0ff01h)200 CURTEMP=CURTEMP+XBY(0ff00h)210 REM read setpoint temperature220 XBY(0ff00h)=0230 SETEMP=256*XBY(0ff01h)240 SETEMP=SETEMP+XBY(0ff00h)250 ERROR=SETEMP-CURTEMP260 IF (ERROR<-82).AND.(MEMX<-82) THEN COMMAND=0270 IF (ERROR>82).AND.(MEMX>82) THEN COMMAND=255275 MEMX=ERROR280 IF (ERROR>-82).AND.(ERROR<82) THEN COMMAND=COMMAND290 XBY(0fd00h)=COMMAND300 PRINT "CURRENT TEMP=",CURTEMP,"SET TEMP=",SETEMP310 GOTO 170


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Comparison of Predicted Dynamic Behaviorwith

Actual, Measured Dynamic Behavior

0 50 100 150 2000

5

10

15

20

25

time (sec)

tem

pera

ture

(de

gree

s C

)

Predicted: red curve (-)Measured: blue curve (--)

Thermal System Closed-Loop Temperature...

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