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
MechatronicsThermal System Case Study
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.
MechatronicsThermal System Case Study
Kevin Craig3
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
MechatronicsThermal System Case Study
Kevin Craig4
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
MechatronicsThermal System Case Study
Kevin Craig5
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.
MechatronicsThermal System Case Study
Kevin Craig6
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
MechatronicsThermal System Case Study
Kevin Craig7
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
MechatronicsThermal System Case Study
Kevin Craig8
Temperature Sensor Circuit
15 V
1 µA/K
1KΩ
Sensor Voltage
LF 411
Ksensor(volts/K) = (1µA)(Rsensor)
MechatronicsThermal System Case Study
Kevin Craig9
Physical Model
Ambient air
Thin Aluminum Plate
q co
nve
ctio
n
Heat input q heater
Thin-Film Resistive Heater
Sensor
Ceramic Insulation
MechatronicsThermal System Case Study
Kevin Craig10
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
MechatronicsThermal System Case Study
Kevin Craig11
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
MechatronicsThermal System Case Study
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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
MechatronicsThermal System Case Study
Kevin Craig13
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.
MechatronicsThermal System Case Study
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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
MechatronicsThermal System Case Study
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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.
MechatronicsThermal System Case Study
Kevin Craig16
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
& )
&
&
MechatronicsThermal System Case Study
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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
sKs
iK
oo i o is
t
o
i
o
i
+ = = −FHG
IKJ
=+
=+
∠ −
−FH
IK
−
1
1 121a f a f a f tan
MechatronicsThermal System Case Study
Kevin Craig18
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
MechatronicsThermal System Case Study
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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
MechatronicsThermal System Case Study
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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
)
MechatronicsThermal System Case Study
Kevin Craig21
Experimental Set-Up & Control Implementation
MechatronicsThermal System Case Study
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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
MechatronicsThermal System Case Study
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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
MechatronicsThermal System Case Study
Kevin Craig24
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 (--)