1. ME 322: Instrumentation Lecture 39
April 27, 2015
Professor Miles Greiner
Integral Control program, Proportional Control response model

2. Announcements/Reminders
This week: Lab 12 Feedback Control
HW 13 Will accept Wednesday
HW 14 Due Friday, X3 (Last HW assignment)
Review Labs 10, 11, and 12: Wednesday and Friday
Open Lab Practice Session: Saturday and Sunday
Lab Practicum Finals (Starting a week from today)
Schedule on WebCampus
Guidelines,
Drop extra-credit LabVIEW Workshop and Lab 12.1
I will add 6 points to everyone’s Midterm 2 (1.5% of grade)
Wind tunnel CTA
We are working to adjust the CTA feedback system so they will work correctly for the final
You are graded on LabVIEW programming, data acquisition, calculations, clear plots and tables, conclusions (not on equipment failures)

3. Modify Proportional Control
Use Control-U to make block diagram clearer
Shift register, input DTi
Add 𝐹𝑇𝑂 𝑖 = 𝑇 𝑆𝑃 −𝑇 𝐷𝑇𝑖 to FTOp
Display FTOi (bar and numerical indicators)
Add 10log(DTi) and log(DTi) to plots
Add Write to Measurement File VI
Use next available file name
No Headers
One time column
Microsoft Excel

4. Lab 12 Integral Control Block Diagram
Modify proportional VI

5. Figure 1 VI Front Panel
Plots help the user monitor the time-dependent measured and set-point temperatures T and TSP, temperature error T–TSP, and control parameters

6. Process Sample Data Add time scale in minutes Figure 3 Figure 4
Add time scale in minutes
Calculate difference, general format, times 24*60
Figure 3
Plot T, TSP, DT and 10log(DTi) versus time
Figure 4
Plot T-TSP, -DT, 10log(DTi) and 0 versus time
Table 1
Determine time periods when behavior reaches “steady state,” and find 𝑇 𝐴 = 𝑇 and 𝑇 𝑅𝑀𝑆 = 𝜎 𝑇 during those times
Be sure to use an integer number of cycles
Figure 5
Plot 𝑇 𝑅𝑀𝑆 versus DT and DTi
Figure 6
Plot steady state error e= 𝑇 𝐴 − 𝑇 𝑆𝑃 versus DT and DTi

7. Figure 3 Measured, Set-Point, Lower-Control Temperatures and DTi versus Time
Data was acquired for 40 minutes with a set-point temperature of 85°C.
Measure response to each setting for 10 minutes or less
The time-dependent water temperature is shown with different values of the control parameters DT and DTi.
Proportional control is off when DT = 0
Integral control is effectively off when DTi = 107 (10log(DTI) = 70)

8. Figure 4 Temperature Error, DT and DTi versus Time
The temperature oscillates for DT = 0, 5, and 15°C, but was nearly steady for DT = 20°C
Sometimes not steady until DT = 30°C
May dependent on TC location and water level
DTi was set to 100 from roughly t = 25 to 30 minutes, but the systems oscillated, and so it was increased to 1000.
The controlled-system behavior depends on the relative locations of the heater, thermocouple, and side of the beaker, and the amount of water in the beaker. These parameters were not controlled during the experiment (Lab 12.1 investigates these rich behaviors).

9. Table 1 Controller Performance Parameters
This table summarizes the time periods when the system exhibits steady state behaviors for each DT and DTi.
During each steady state period
TA is the average temperature
TA – TSP is an indication of the average controller error
The Root-Mean-Squared temperature TRMS is an indication of controller unsteadiness

10. TRMS is and indication of thermocouple temperature unsteadiness
Figure 5 Controller Unsteadiness versus Proportionality Increment and Set-Point Temperature
TRMS is and indication of thermocouple temperature unsteadiness
Unsteadiness decreased as DT increased, and was not strongly affected by DTi.

11. The average temperature error
Figure 6 Average Temperature Error versus Set-Point Temperature and Proportionality Increment
The average temperature error
Is positive for DT = 0, but decreases and becomes negative as DT increases.
Is significantly improved by Integral control.

12. Proportional-Control Thermal AnalysisTW
Heater
QIN = FTO(QMAX)
QOUT = hA(TW -TEnv)
TEnv
TTC
Is TW = TTC? Is TW uniform?
Energy Balance for Water
𝑄−𝑊= 𝑄 𝐼𝑛 − 𝑄 𝑂𝑢𝑡 = 𝑑𝑈 𝑑𝑡 = 𝜌𝑐𝑉 𝑊 𝑑 𝑇 𝑊 𝑑𝑡
𝑄 𝑂𝑢𝑡 =ℎ𝐴( 𝑇 𝑊 − 𝑇 𝐸𝑁𝑉 )
For proportional control: 𝑄 𝐼𝑛 = 𝑄 𝑀𝑎𝑥 𝐹𝑇𝑂 = 𝑄 𝑀𝑎𝑥 𝑇 𝑆𝑃 − 𝑇 𝑇𝐶 𝐷𝑇
𝑄 𝑀𝑎𝑥 𝑇 𝑆𝑃 − 𝑇 𝑇𝐶 𝐷𝑇 −ℎ𝐴 𝑇 𝑊 − 𝑇 𝐸𝑛𝑣 = 𝜌𝑐𝑉 𝑊 𝑑 𝑇 𝑊 𝑑𝑡

13. For Large DT
We observed that TTC is steady when DT is sufficiently large
Under that condition, assume TTC = TW = 𝑇 𝑆𝑆
𝑄 𝑀𝑎𝑥 𝑇 𝑆𝑃 − 𝑇 𝑆𝑆 𝐷𝑇 −ℎ𝐴 𝑇 𝑆𝑆 − 𝑇 𝐸𝑛𝑣 = 𝜌𝑐𝑉 𝑊 𝑑 𝑇 𝑆𝑆 𝑑𝑡
𝑇 𝑆𝑃 𝑄 𝑀𝑎𝑥 𝐷𝑇 + 𝑇 𝐸𝑛𝑣 ℎ𝐴 = 𝑇 𝑆𝑆 𝑄 𝑀𝑎𝑥 𝐷𝑇 +ℎ𝐴
𝑇 𝑆𝑆 = 𝑇 𝑆𝑃 𝑄 𝑀𝑎𝑥 𝐷𝑇 + 𝑇 𝐸𝑛𝑣 ℎ𝐴 𝑄 𝑀𝑎𝑥 𝐷𝑇 +ℎ𝐴
Steady State Error
𝑒 𝑆𝑆 = 𝑇 𝑆𝑆 − 𝑇 𝑆𝑃 = 𝑇 𝑆𝑃 𝑄 𝑀𝑎𝑥 𝐷𝑇 + 𝑇 𝐸𝑛𝑣 ℎ𝐴 − 𝑇 𝑆𝑃 𝑄 𝑀𝑎𝑥 𝐷𝑇 +ℎ𝐴 𝑄 𝑀𝑎𝑥 𝐷𝑇 +ℎ𝐴
𝑒 𝑆𝑆 =− 𝑇 𝑆𝑃 − 𝑇 𝐸𝑁𝑉 𝑄 𝑀𝑎𝑥 𝐷𝑇 ℎ𝐴 +1
How does this prediction compare to measurements?

14. Measured Proportional-Control Steady-State Error
𝑒 𝑆𝑆
𝑇 𝑅𝑀𝑆
The temperature is steady (TRMS becomes small) once DT is sufficiently large
Prediction: 𝑒 𝑆𝑆 = 𝑇 𝑆𝑆 − 𝑇 𝑆𝑃 =− 𝑇 𝑆𝑃 − 𝑇 𝐸𝑁𝑉 𝑄 𝑀𝑎𝑥 𝐷𝑇 ℎ𝐴 +1
Measurements show the error magnitude increases as 𝑇 𝑆𝑃 − 𝑇 𝐸𝑁𝑉 increases
When 𝑇 𝑆𝑃 = 𝑇 𝐸𝑁𝑉 no need for control
Steady state error decreases as 𝑄 𝑀𝑎𝑥 𝐷𝑇 ℎ𝐴 increases (DT decreases)
But when DT is too small the temperature oscillates (TRMS)
Why does this happen? ( 𝑇 𝑊 ≠ 𝑇 𝑇𝐶 )

15. Predict time-dependent TTC(t) for TCTW
TTC
QIN = hA(TW -TTC)
TC Energy Balance (assuming 𝑇 𝑊 ≠ 𝑇 𝑇𝐶 )
𝑄−𝑊= 𝜌𝑐𝑉 𝑇𝐶 𝑑 𝑇 𝑇𝐶 𝑑𝑡
𝑄= ℎ𝐴 𝑇𝐶 𝑇 𝑊 − 𝑇 𝑇𝐶 = 𝜌𝑐𝑉 𝑇𝐶 𝑑 𝑇 𝑇𝐶 𝑑𝑡
𝑯 𝑻𝑪 𝑻 𝑾 − 𝑻 𝑻𝑪 = 𝑴 𝑻𝑪 𝒅 𝑻 𝑻𝑪 𝒅𝒕 (Eqn. A)
𝐻 𝑇𝐶 = ℎ𝐴 𝑇𝐶
𝑀 𝑇𝐶 = 𝜌𝑐𝑉 𝑇𝐶
Constants
Dynamic relationship between 𝑇 𝑇𝐶 (t) and 𝑇 𝑊 (𝑡)

16. Water Energy-Balance TW
Heater
QIN = FTO(QMAX)
QOUT = hA(TW -TENV)
TENV
TTC
𝑄 𝐼𝑛 − 𝑄 𝑂𝑢𝑡 = 𝜌𝑐𝑉 𝑊 𝑑 𝑇 𝑊 𝑑𝑡 = 𝑀 𝑊 𝑑 𝑇 𝑊 𝑑𝑡
For proportional control:
𝑄 𝐼𝑛 = 𝑄 𝑀𝑎𝑥 𝑇 𝑆𝑃 − 𝑇 𝑇𝐶 𝐷𝑇 =𝐺 𝑇 𝑆𝑃 − 𝑇 𝑇𝐶
Where the controller gain is 𝐺= 𝑄 𝑀𝑎𝑥 𝐷𝑇
𝑄 𝑂𝑢𝑡 = ℎ𝐴 𝐴 𝑇 𝑊 − 𝑇 𝐸𝑁𝑉 + ℎ𝐴 𝑇𝐶 ( 𝑇 𝑊 − 𝑇 𝑇𝐶 )
𝐺 𝑇 𝑆𝑃 − 𝑇 𝑇𝐶 − 𝐻 𝐴 𝑇 𝑊 − 𝑇 𝐸𝑁𝑉 − 𝐻 𝑇𝐶 ( 𝑇 𝑊 − 𝑇 𝑇𝐶 )= 𝑀 𝑊 𝑑 𝑇 𝑊 𝑑𝑡
𝐻 𝐴 = ℎ𝐴 𝐴 , 𝑀 𝑊 = 𝜌𝑐𝑉 𝑊 , 𝑇 𝐸𝑁𝑉 and 𝑇 𝑆𝑃 are constants
In addition to 𝐺= 𝑄 𝑀𝑎𝑥 𝐷𝑇 , 𝐻 𝑇𝐶 = ℎ𝐴 𝑇𝐶 , 𝑀 𝑇𝐶 = 𝜌𝑐𝑉 𝑇𝐶

17. Collect Terms Couple with Eqn. A: 𝑯 𝑻𝑪 𝑻 𝑾 − 𝑻 𝑻𝑪 = 𝑴 𝑻𝑪 𝒅 𝑻 𝑻𝑪 𝒅𝒕
𝐺 𝑇 𝑆𝑃 − 𝑇 𝑇𝐶 − 𝐻 𝐴 𝑇 𝑊 − 𝑇 𝐸𝑁𝑉 − 𝐻 𝑇𝐶 ( 𝑇 𝑊 − 𝑇 𝑇𝐶 )= 𝑀 𝑊 𝑑 𝑇 𝑊 𝑑𝑡
𝑴 𝑾 𝒅 𝑻 𝑾 𝒅𝒕 + 𝑯 𝑻𝑪 + 𝑯 𝑨 𝑻 𝑾 = 𝑻 𝑻𝑪 𝑯 𝑻𝑪 −𝑮 + 𝑮 𝑻 𝑺𝑷 + 𝑯 𝑨 𝑻 𝐸𝑁𝑉
(Eqn. B)
Dynamic relationship between 𝑇 𝑊 (𝑡) and 𝑇 𝑇𝐶 (𝑡)
Couple with Eqn. A: 𝑯 𝑻𝑪 𝑻 𝑾 − 𝑻 𝑻𝑪 = 𝑴 𝑻𝑪 𝒅 𝑻 𝑻𝑪 𝒅𝒕
What do we have?
Two, 1st-order, coupled, constant-coefficient liner-differential equations for 𝑇 𝑊 (t) and 𝑇 𝑇𝐶 (t)

18. System Solution Solve Eqn. A for 𝑇 𝑊
𝑇 𝑊 = 𝑀 𝑇𝐶 𝐻 𝑇𝐶 𝑑 𝑇 𝑇𝐶 𝑑𝑡 + 𝑇 𝑇𝐶
𝑑 𝑇 𝑊 𝑑𝑡 = 𝑀 𝑇𝐶 𝐻 𝑇𝐶 𝑑 2 𝑇 𝑇𝐶 𝑑 𝑡 2 + 𝑑 𝑇 𝑇𝐶 𝑑𝑡
Plug into Eqn. B and collect terms
𝑀 𝑊 𝑑 𝑇 𝑊 𝑑𝑡 + 𝐻 𝑇𝐶 + 𝐻 𝐴 𝑇 𝑊 = 𝑇 𝑇𝐶 𝐻 𝑇𝐶 −𝐺 + 𝐺 𝑇 𝑆𝑃 + 𝐻 𝐴 𝑇 𝐸𝑁𝑉
𝑀 𝑊 𝑀 𝑇𝐶 𝐻 𝑇𝐶 𝑑 2 𝑇 𝑇𝐶 𝑑 𝑡 2 + 𝑑 𝑇 𝑇𝐶 𝑑𝑡 + 𝐻 𝑇𝐶 + 𝐻 𝐴 𝑀 𝑇𝐶 𝐻 𝑇𝐶 𝑑 𝑇 𝑇𝐶 𝑑𝑡 + 𝑇 𝑇𝐶 = 𝑇 𝑇𝐶 𝐻 𝑇𝐶 −𝐺 + 𝐺 𝑇 𝑆𝑃 + 𝐻 𝐴 𝑇 𝐸𝑁𝑉

19. Collect Terms
𝐴 𝑑 2 𝑇 𝑇𝐶 𝑑 𝑡 2 +𝐵 𝑑 𝑇 𝑇𝐶 𝑑𝑡 + 𝐺+ 𝐻 𝐴 𝑇 𝑇𝐶 =𝐺 𝑇 𝑆𝑃 + 𝐻 𝐴 𝑇 𝐸𝑁𝑉
More Constant
𝐴= 𝑀 𝑊 𝑀 𝑇𝐶 𝐻 𝑇𝐶
B= 𝑀 𝑊 + 𝑀 𝑇𝐶 + 𝐻 𝐴 𝑀 𝑇𝐶 𝐻 𝑇𝐶
2nd order, linear, Constant-coefficient differential-equation, non-homogeneous (RHS= Constant)
Solution 𝑇 𝑇𝐶 = 𝑇 𝑃 + 𝑇 𝐻
Particular Solution for RHS = Constant, 𝑇 𝑃 = 𝑇 𝑆𝑆 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐺+ 𝐻 𝐴 𝑇 𝑆𝑆 =𝐺 𝑇 𝑆𝑃 + 𝐻 𝐴 𝑇 𝐸𝑁𝑉
𝑇 𝑆𝑆 = 𝐺 𝑇 𝑆𝑃 + 𝐻 𝐴 𝑇 𝐴 𝐺+ 𝐻 𝐴 = 𝑄 𝑀𝑎𝑥 𝐷𝑇 𝑇 𝑆𝑃 + ℎ𝐴 𝐴 𝑇 𝐸𝑁𝑉 𝑄 𝑀𝑎𝑥 𝐷𝑇 + ℎ𝐴 𝐴
Same as for simple 1st order system

20. Homogeneous Solution 𝐴 𝑑 2 𝑇 𝐻 𝑑 𝑡 2 +𝐵 𝑑 𝑇 𝐻 𝑑𝑡 + 𝐺+ 𝐻 𝐴 𝑇 𝐻 =0
𝐴 𝑑 2 𝑇 𝐻 𝑑 𝑡 2 +𝐵 𝑑 𝑇 𝐻 𝑑𝑡 + 𝐺+ 𝐻 𝐴 𝑇 𝐻 =0
Assumed solution 𝑇 𝐻 𝑡 =𝐶 𝑒 𝑏𝑡 , 𝑏= ?
Characteristic Equation 𝐴 𝑏 2 +𝐵𝑏+𝐶 𝑒 𝑏𝑡 =0
𝑏= −𝐵± 𝐵 2 −4𝐴 𝐺+ 𝐻 𝐴 2𝐴
If DT is small enough, then
G= 𝑄 𝑀𝑎𝑥 𝐷𝑇 will be large enough so that
𝐵 2 −4𝐴 𝐺+ 𝐻 𝐴 <0,
b will be imaginary and
𝑇 𝐻 will be oscillatory 𝑇 𝐻 𝑡 =𝐷 𝑒 −𝑎𝑡 sin⁡(𝜔𝑡+𝜙)
We observed oscillations for small DT

21. What is the largest minimum value of DT that will have a steady behavior?
𝐵 2 −4𝐴 𝐺+ 𝐻 𝐴 = 𝐵 2 −4𝐴 𝑄 𝑀𝑎𝑥 𝐷𝑇 + 𝐻 𝐴 <0
𝐵 2 4𝐴 − 𝐻 𝐴 < 𝑄 𝑀𝑎𝑥 𝐷𝑇
𝐷𝑇 < 𝐷𝑇 𝑀𝑖𝑛 = 𝑄 𝑀𝑎𝑥 𝐵 2 4𝐴 − 𝐻 𝐴 = 𝑄 𝑀𝑎𝑥 𝑀 𝑊 + 𝑀 𝑇𝐶 + 𝐻 𝐴 𝑀 𝑇𝐶 𝐻 𝑇𝐶 𝑀 𝑊 𝑀 𝑇𝐶 𝐻 𝑇𝐶 − 𝐻 𝐴 = 𝑄 𝑀𝑎𝑥 𝐻 𝑇𝐶 𝑀 𝑊 + 𝑀 𝑇𝐶 1+ 𝐻 𝐴 𝐻 𝑇𝐶 𝑀 𝑊 𝑀 𝑇𝐶 − 𝐻 𝐴 𝐻 𝑇𝐶
𝐻 𝐴 = ℎ𝐴 𝐴 , 𝑀 𝑊 = 𝜌𝑐𝑉 𝑊
𝐻 𝑇𝐶 = ℎ𝐴 𝑇𝐶 , 𝑀 𝑇𝐶 = 𝜌𝑐𝑉 𝑇𝐶

22. Problem X3
Problem X3: A 200-Watt heater and a 1.5-mm-diameter thermocouple are placed in a water-filled beaker of diameter 3 cm and height 5 cm. If the heat transfer coefficient between the beaker and air is 5 W/m2K, and between the water and thermocouple is 1000 W/m2K, estimate the lowest proportional control temperature increment DTMin, for which the control system will be steady (not oscillatory). Assume the thermocouple properties to be that of iron, and evaluate water properties at 30°C.
Heater: Q = 200 W
Beaker: D = 3 cm, H = 5 cm, hAir = 5 W/m2K
TC: D = 1.5 mm, hTC = 1000 W/m2K, iron

26. Proportional Control
1st Law

27. Proportional Control Find
Want DT to be small, but that leads to oscillation.

28. Proportional Control
No way for a 1st order constant coefficient deferential equation to give oscillation.
How to predict/model oscillations?
(Twater ≠ TTC)
Better Model

29. 1st Law TC

30. Water
Heater HA HT MW
Unknown: T, T­W

31. A B
Homogenous Solutions:

32. Characteristic EquationIf
Then set complex
If DT is small enough then get oscillations.