Temperature Control Lab

7/30/2019 Temperature Control Lab

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APPARATUS ....................................................................................... 1

PROCEDURE ....................................................................................... 2

RESULTS/CALCULATIONS ..................................................................... 5

QUESTION 1 ...............................................................................................................5

STEADY- STATE ENERGY BALANCEBETWEENTHE HOTTANKAND PRODUCT.........................................5QUESTION 2 ...............................................................................................................5

STEADY- STATE ENERGY BALANCEFOR HOLDING TUBE...............................................................5

QUESTION 3 ...............................................................................................................6

FOPDT MODELRELATING EXIT TEMPERATURE T4 TO HEATEROUTPUT............................................6

QUESTION 4 ...............................................................................................................8

LINEARITYOF HEATER-OUTPUT-TO-TEMPERATURE DYNAMICS..........................................................8

QUESTION 5 ...............................................................................................................9FOPDT MODELRELATING EXIT TEMPERATURE T1 TO HEATEROUTPUT............................................9

QUESTION 5 .............................................................................................................10

COMPARISONOF FOPDT MODELS....................................................................................10

QUESTION 6 .............................................................................................................10

PROPORTIONAL CONTROL STRENGTHSAND WEAKNESSES.........................................................10

QUESTION 7 .............................................................................................................12

PROPORTIONAL INTEGRAL CONTROL STRENGTHS AND WEAKNESSES 12


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PROPORTIONAL INTEGRAL CONTROL STRENGTHS AND WEAKNESSES 12

GRAPH 1 CHANGE IN TEMPERATURES, T1, T2 AND T4 WITH CHANGES IN

HEATER POWER FOR OPEN LOOP MODE ............................................... A

GRAPH 2 CHANGES IN OUTLET TEMPERATURE, T4 AND IN HEATER

POWER WITH SETPOINT CHANGES FOR CLOSED LOOP MODE ................. A


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PROCEDURE

i. Before turning on the console, the switches were set as follows.

All function switches to MANUAL

All control potentiometers to minimum (fully

counterclockwise)

Valve control switch for SOL1 to Divert

Valve control switch for SOL2 to FEED A

Valve control switch for SOL3 to STOP

Valve control switch for SOL4 to FILL A

Valve control switch for SOL5 to FILL B

ii. Ensuring that the three circuit breakers and the RCCB on the rear of

the console were in the UP position, the console was powered upand the computer turned on.

iii. The water level in the hot water tank was checked, the heaterelement was covered by at least 10 mm of water. The LOW LEVELLED on the console was out.

iv. The flow control valve V1 was fully opened (fully counterclockwise)d th d i l PRV1 l f ll d (f ll


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ix. Double-click the PCT23 icon on the Windows desktop and loadexperiment F - 1 loop (temp T4 to heater PWR). Familiarize

yourself with the basic features of the Armsoft package, especiallythe mimic diagram, graph and table functions available on thetoolbar.

x. The mimic diagram was opened and the values of T1, T2 and T4recorded and then a value of 20 was entered in the Heater Outputbox.

PROCEDURE

xi. Select View graph from the toolbar then Configure the graphdata. Trend variables T1, T2 and T4 on the primary y-axis andHeater Power PWR (kW) on the secondary y-axis. Set the range of

the primary axis to 20-60C and that of the secondary axis to 0.2-0.6 kW (Format, Graph, Axis Scales). Configure the data sampling

for a sample interval of 2 seconds and start the data collection byclicking the green Go icon.

xii. The system was allowed to come to steady-state. When the processtemperatures lined out, the heater output was increased to 30% andwaited for steady-state to be re-established.


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seconds. Click on OK and let the process come to steady-state.

The set point was decreased by 5 C and the process variables once

again allowed to line out.

xviii. When the experiment is complete, stop the data collection and saveyour results (File, Save As) in Formula One and Excel formats asclosedloop.vts and closedloop.xls, respectively. Copy these files toa diskette. Switch the controller to manual, set the heater output to0% and exit the software.

xix. All controls on the console were set to minimum/OFF and all functionswitches to MANUAL. The makeup water valve was closed and theconsole and the computer switched off. The clamps on theperistaltic pump heads were released to prevent permanentdistortion of the flexible tubing.


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RESULTS/CALCULATIONS

Question 1

Steady- state Energy Balance between the Hot tank and Product.

Energy Entering system = Energy Leaving system

i.e. Heat Gained from Flow entering + Heat added via heated vessel =Heat Loss from Flow leaving the system + Heat losses to environment

m*c*T ENTER + Q = m*c*T LEAVE + H

H = m*c*T ENTER + Q m*c*T LEAVE

Now,

Same liquid flowing through system so c is same

Steady state analysis so flow rate entering is same as flow rate leaving

system. , therefore


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Hence heat loss across holding tube = 0.0208 kW

Question 3

FOPDT Model relating Exit Temperature T4 to Heater output

FOPDT model given by equation

Where is time constant, seconds

PV (t) is the process variable, T4 C, in this case

is dead time, seconds

KP is process gain, C / % and

CO (t) is the controller output, heater output, kW

Solving using Laplace transform

Applying Linearity property of Laplace transform


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Question 3

Time Constant,

Time constant is the time taken for a 63.2% in the total steady state PV whena change is made to the CO. For a CO change of 30% to 20%

Initial steady state PV = 33.8000 C

Final steady state PV = 30.8000 C

63.2% PV = 0.632*(33.8000 30.8000) C = 1.8690 C

Value of PV at = 33.8000 1.8690 = 31.9040C

From results, PV(t) = 33.8000 C occurs at sample number 1279 and PV(t) = 31.9040 C occurs at sample number 1420.

Number of samples taken during = 1420-1279 = 141

Sample time is 2s, therefore = 141*2 = 282s

Dead time,

From Graph 1, when 30% to 20% change in CO was made

Sample number at point of step change = 1279

S l b h PV (t) b i t h 1298


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temperature than for warm/water to lose heat and decrease in temperature

in the same surroundings.

Question 5

FOPDT Model relating Exit Temperature T1 to Heater output

Again using Graph 1, from approximately sample number 1279 till end for T1

Process Gain, KP

Time Constant,

Time constant is the time taken for a 63.2% in the total steady state PV whena change is made to the CO. For a CO change of 30% to 20%

Initial steady state PV = 33.0000 C

Fi l t d t t PV 30 3000 C


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Question 5

Comparison of FOPDT Models

The final equations for T4 and T1 are, respectively,

.....Eq 1

..Eq 2

From these equations, we see that

- Process gain is great for T4 model than T1, (0.3>0.27) this means that

T4 value changed more T1 for the same change in heater output

- Dead time is smaller for T4 than T1, (38


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The main weakness of proportional only control is offset, i.e. the difference

between steady-state value and set point after a change was made. From

graph 2 we see that the set point was given a value of 38 C, when steady-

state was established, T4 had a value of 35 C. This is a characteristic of all P-

only control loops.


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Question 7

Proportional Integral Control Strengths and Weaknesses

Again we consider Graph 2, from sample number 500 when the last

change is made. We see the advantage of PI control right away:

Process variable goes to the set point The set point was 31 C and the

last temperature measured was 30.9 C, this is close considering the noise

in the system.

The disadvantage that is visible is the length of time taken to return to a

steady-state. The change was made at sample number 512 but steady-

state was not established till sample number 612, (512-612 * 2s = 200).

Another major problem with PI control loops is oscillations. Usually the

process variable oscillates a few times about the set point before it finally

stays at the set point value. This is not visible in the data collected since

data collection was stopped as soon as T4 reach the set point value, if

however the experiment was continued we may have seen the

oscillations.

Question 8


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applied to this system, there were no oscillations observed for thetime data was collected.

- Another major reason not to use PID control for this system is theamount of noise in the system. Derivative control is based on rate ofchange, in a noisy system such as in this experiment, derivativeaction amplifies noise and makes the controller output unstable.

- Additionally, properly tuning the system would be a problem as it isnot always certain which term, P, PI or PID in affecting the system.


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DISCUSSION

FOPDT Model

Valve Action

AO, air to open and AC, air to close are the two type of valve action for

pneumatic valves. Air-to-open means that air is required to move the

valve stem upwards so that the fluid moving through the valve can pass

freely. These types of valve are fail close meaning that if a failure occurs

in the system, the air supply to the system will be stopped and the valve

will close completely.

Time Delays / Dead Time

Dead time is the delay from when a controller output (CO) signal is issueduntil when the measured process variable (PV) first begins to respond.

Dead time is usually not wanted in industry as it can decrease the

efficiency of many processes and increase the level of difficulty of

controlling a plant/process. The only advantage of having dead time is for

the situation were an error may have been made in changing a set point

t ll t t hi h ld l d t h d it ti A ti


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DISCUSSION

Safety Analysis

It was ensured that the initial water level in the hot water tank was

10mm higher than the heating element to prevent the element from

begin damaged.

The opening and closing of valves were done as specified to prevent

water flowing into the wrong places.

The main valve controlling the water supply was opened slowly to

prevent the water pounding on the pressure sensor L1 at the

bottom of the tank else it would have been damaged.

The level in the feed tanks was monitored and the supply valve

adjusted to ensure the tanks did not overflow or empty out

Sources of Error

O it hi f P l t PI t l th t i t t t


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CONCULSION

From the results, it can be concluded that:

Heat loss between the heater and product was 0.6274 kW

Heat loss in the holding tube was 0.0208 kW.

The transfer functions for the system between exchanger and, T4 and T1,in seconds are, respectively:

In this experiment where it was necessary to control the temperature the

Proportional Integral controller was better suited than the Proportional

l t ll i it i t i d th t i t l ft t h


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REFERENCES

www.controlguru.com

Cooper, Douglas

Validating Our Heat Exchanger FOPDT Model (2007)

PI Control of the Heat Exchanger (2008)

A Discrete Time Linear Model of the Heat Exchanger By Peter

Nachtwey (2006)

Cooper, Douglas. Practical Process Control using Control Sataion. 2004.

Luben, Luben and. Essentials of Process Controls. McGraw Hill, 1997.
http://www.controlguru.com/http://www.controlguru.com/


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APPENDIX A

GRAPH 1 CHANGE IN TEMPERATURES, T1, T2 AND T4 WITH CHANGES IN HEATER POWER FOR OPEN LOOP MODE

A1


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APPENDIX A

GRAPH 2 CHANGES IN OUTLET TEMPERATURE, T4 AND IN HEATER POWER WITH SETPOINT CHANGES FOR CLOSED LOOPMODE

A2

Documents

Temperature Control Lab