Part 1.pdfPart 2.pdfPart 3.pdfPart 4.pdfPart 5.pdfProcess Control Pri nciplesFundamental control conceptsRegulate value of a quantity compared to areference value. Reference called setpointProcess - a collection of equipment andmaterials that takes inputs and has outputsExampleQort depends onV6-lf Qout = Qin, h constantQout t Qin, tank emPtiesQort ( Qin, tank overflowsIHLNo reference value givenet438a-1.pptBasic Control ElementsMeasurementControl decisionSystem modificationIHLcontrollerfinalcontroloout elementtank level, H, is reference (setpoint)h is the control variable.Human is the controller, adjustsQort to maintain h as close as possible to HExample of regulator actionneosurerrentsight gLosset438a-1.pptAutomatic Control SystemsAutomationAutorroticContro[sensorcontnol.elementOor.rtUse sensors and analog or digital electronics tomonitor system.Elements of Automatic Control SystemProcess - single or multiple variableMeasurement - sensors / transducersError detection - compare H to hController - generate corrective actionFinal control element - modify processIHLet438a-1.pptBlock DiagramsReduce complex systems to inputs and outputsSignals form a loop Control loopr = reference setpoint valuec = control variable in the processm = measurement of control variablep = controller outputu = control element changee = r - m (due to signs on summer)Block Diagramof Control SystemControlelementprocesscontrollermeasurementet438a-1.pptPractical System3-15 psipneumaticFinal control elementcurrent topressureconverterSignalConditioningControllerDiff. PressuretransducerSetpointValvemeasurementFlowControllerDiff. pres.Controf System PerformanceSystem signal change over time soe(t)=r-c(t)Where e(t) = error as a function of timer = setpoint valuec(t) = control variable as a function of timeControl System Objectivese(t) = 0 after changes or disturbancesstabfe c(t) after changes or disturbancesStabilitySteady-state regulation - e(t) = 0 or withintolerancesTransient regulation how does systemperform underchange6 et438a-1.pptTypes of system responseuncontrolleprocessprocesscontrolactivatedDamped Responseinstabilityr2c(t)r1Change Setpointr2c(t)r1Transient ResponsesovER s d ool'Setpoint changer1c(t)5e{11,'^ 1 Trtt'eDisturbanceet438a-1.pptAnalog Measurement ErrorError determines accuracyMethods of determing accuracyMeasured Value: Reading +- value1 00 psi +- 2 psiPercent of Full Scale (FS): (FS Value) (%t100)Meter accurate to +-5o/o of FS)10 V scaleerror = 10 V (+-Sotol1 00) -+-0.5VPercentage of Span (Span = max-min)(Span)(%t100)P measurement +-3o/o of span20-50 psierror = (50-20)(+-3%1100) = *-0.9 psiPercentage of Readingreading of 2V +-2o/oerror - 2V (*-Z%1100) = -F- 0.04 V9 et438a-1.pptcSystem AccuracyCummufative ErrorSensorsensor amplifierK - sensor gain G - amplifier gainV - sensor output voltage,AG, av, AK uncertainities in measurementWhat is magnitude of AV?,/ inPutWith no error:With error V+-AV - (K*-AKXG+-AG)cV+-lVMultiple out and=r AV +AG , AK_-- +VGK=* AGv - (KXG)c\\outputsimplify to getWhere + AV = norm a6zedV fractional uncertainity+AKK+-lKG+-AG10 et438a-1.pptG,K= normalized fractional uncertainityCombining the errorsUse RMS (RSS)Notes:AG AK are fractional error.I=rns/ov\\vlRDivide % to getultiply byultiplication&R '''j o/, T ierc.nc€,rth 2% a..c.rosYTM,'F;r(.T?. 3 % ,4rusEFGKAV is fractional RMS error Mrv 100 to get percent.Works on all formulas that are only mtand division.Example: inverting OP AMP T = t[A R+ - A R,i ..*JX s:ro,o.s 1: q,,n*, '-Ra -R; - /sq ui" '(cc'd w I41t.; =i&- =*o, o z14'" too Bui =t\m6t 1r" J ^r,o: * /Co os;e+(r"'s)11 et438a-1.ppt -16. o Z3 oR +'/rc\' /m\\a) . \^/Se nsor Ch a racte ri sticsSensitivity - Change in output for changein input. Equals slope in linear deviceHysteresis - output different for increasing ordecreasing inputResolution - Smallest measurement a sensorcan make.Linearity - how close to a line is the llOrelationshipcm=m(c)+coWhere c = measured control variablem = slopeGo = offset (y intercept)G,n = sensor output12 et438a-1.pptExampleFinding m and co from data pointsY Y1=m\x xl) lrt=Y2-Y1x2 x1Sensor has a linear resistance change of 100 to 195ohms, as temperature changes from 20 - 120 C.Find the sensor l/O relationshipbe{'^e 't{,..t= ,xr rl,) = (zo t,, loosr) X= 1r-,prJ(x, y.,. (tzo"r,l9ssz) t= o''-fP'-.iS l"pe. ,,.\.- l?Sra - /oo ;i _ gS sr_frl: lZoo(- 2,:oc = m= o.gs-n/r.3 - too.o- = o.9s f".(x - Z""c)J= o.9sx o9s(2") + lao iY=O.?sX+ 8t -./"t yPl"-l *u cleq/- po,,.t-s " l .* /d b, e one_I '''' e' '13 et438a-1.pptSensor ResponseFirst Order Response - idealc(t)cfStep change inPractical sensorc(t) b(t; =measured variableresponsesensor response functioncfsrr PTNCAtAsfSTf Pb Ec. ATAT,Ebr = final valuebi = initial valueno time delayet438a-1.pptModef ing 1st Order ResponeFor step increase:b(t)=bi. (0,z* (a -z)( ,- u*/" o"")Ztz(r-e*/"'o"o)9oo/o o{ 4.oV = r,d Vz(r - s %'*zs;Wherebr = final sensor valuebi = initial sensor valuet = timet = time constant of sensorFor step decreaseExample: Step increasebr-4.0V b, -2.0V r=0.0025/sFind time it takes for sensor to reach g0o/o of final valueo ,) (,,-Tb(t)=(0, o+)." rt\- r)-e Ib{t1 =b(t)=b (-r) or3.6= z +3+3 = t - e */o.aozs-O.8-( = -e-tl".ooZs_t/6.oo zf+o.Z:t€ il"Go,z)= l^(dt/"'^{-/.eo9: -t/o.oozr1.6o7(,:.ao2s) = 6o.Qo4 S = t4"4S: { /nrs,,,.>-F- il15et438a-1.pptExample: Step decreaseThe sensor initial output is 1.0 V how long does ittake to change to 0 .2 V if the time constant of thesensor is 0. 1/s. -t lrbi = l.oV [4"o.oV ?"st/s b(+)=(['-b{) -- t/"' 'btt) " o.Z O. Z' I eIn(o.r1 1 Llot+/- l.c o14 = - '/o' I/.(o94G,i) -. t,-.#r ! +"*Significant Digits in measurement anddesignIn meaurement: readable output of instrumentsresolution of sensors andtransducers.Calulation using meaurements: Trunicate calculatoranswers to match significant digits of measurementsand readings,16 et438a-1.pptSignificant digit examplescompute power based on the following measuredvalues. Use correct number of significant digits.| = 3.25 A V - 117.8 V p = V(t)3.25A 3 significantdigits117.8 V 4 significantdigitscalculator 382.85 WTrunciate to 3 significant digits p - 383 WSignificant digits not factor in design calculationsDevice values assumed to have no uncertainity.Compute the current flow through a resistor that hasa measured R of 1.234 ko and a voltage drop of1.344 Vdc.ft = 1.234 kA 4 significant digitsV - 1.344V 4 significant digitsl= (1.344)l(1 .234x 103) = 1.089 mA 4 digits17 et438a-1.pptBasic Statistics*=(l{WhereVaridnce (nrr)di/-J