1 Prcess Dynamics Prcess dynamics refer t any number f time elements inherent in each device within the cntrl lp and the prcess. These dynamics are expressed as a dead time and a time lag. Althugh each individual element within a system des have its wn dead time and time lag, these elements cmbine mathematically t prvide a system time lag and a system dead time. It is these tw pieces f data in which we are interested. Dead Time Prcess dead time is easiest f the tw time elements t define, but the mst difficult t cntrl. Dead time is als referred t a transprtatin delay. When a cntrl lp makes a crrective actin t a prcess, the effects f this actin are nt realized within the feedback lp until sensed by the sensr element. The time it takes fr this t happen is called dead time. Cnsider the cnveyr belt system illustrated belw. The belt is mving at a velcity f 15 fps. At the end f the cnveyr belt, 50 ft away, is a scale measuring the weight f prduct released frm the hpper. The cntrller is calibrated t change the psitin f the gate n the hpper, based n input frm the scale, allwing the prduct t drp. Hwever, at any time T, the amunt f material actually released is nt sensed by the scale until the cnveyr belt mves the prduct t a pint abve the scale. Since the cnveyr is mving at a finite velcity, the amunt f time it takes fr the material drpped at T is easily calculated as: T d 50 ft = = 3.33sec 15 fps In many cases, ne can minimize the affects f dead time by placing the sensr prperly. In the case f the cnveyr belt, placing the scale clser t the hpper wuld greatly reduce dead time making the system easier t cntrl. In thery, we culd place the scale directly under the Figure 1: An illustratin f dead time hpper and reduce dead time t nearly zer. Hwever, this wuld nt be prudent. Ding s wuld cause the impact f the prduct t prduce errneus measurements. In ther wrds, prper sensr placement is much mre than a functin f dead time. Based n this analgy, it becmes evident that there will always exist sme amunt f dead time.
2 Multiple dead times in a system are additive. If tw cmpnents f a system are in series, ne with a dead time f ne secnd, the ther with a dead time f tw secnds, then the ttal dead time wuld be three secnds. Using the cnveyr belt example, cnsider the fact the hpper is lcated sme distance abve the cnveyr belt. As such, it takes time fr the prduct t drp frm the hpper t the belt. If this time is ne-half secnd, then the dead time in the abve prcess wuld be 3.83 secnds. First Order Time Lags Ideally, if the engineering value f a prcess variable changes, we wuld like the sensing element t react t this change immediately and synchrnusly. Hwever, this is never the case. There will always be a lag in the sensing device. Similarly, we wuld like t see the mvement f a valve actuatr t exactly and immediately fllw the utput f the cntrller. Althugh there are devices that respnd quickly t an external influence, n device respnds instantaneusly. There is always sme frm f lag. This lag is referred t as a time lag. Cnsider the fllwing scenari. An RTD type f temperature sensr is stabilized at rm temperature, then suddenly immersed in a beaker f water at a temperature f 20 F abve the ambient air temperature. This wuld cnstitute a step change in input signal (temperature) t the RTD. Ideally, the utput f the RTD wuld fllw the input signal. Hwever, the RTD has mass. As such, it takes time fr the temperature f this mass t rise such that the resistance f the device changes. As shwn in the figure belw, the utput signal will nt rise immediately. First, a certain amunt f time will pass befre anything happens. This wuld be a dead time. Once the dead time has passed, the utput f the RTD will increase slwly and at a variable rate. This results in respnse referred t a first rder lag plus dead time (FOLPDT). The terminlgy refers t the type f mathematical equatin, a first rder differential equatin, used t describe this respnse. Figure 2: Graphic determinatin f time cnstant fr first rder lag The shape f this curve is described by a time cnstant. The time cnstant f a first rder respnse is equal t the amunt f time it takes t achieve 63.2% f the ttal utput. The time cnstant can be determined graphically as indicated in the Figure 2. Assume the step functin that generated this respnse ccurred at time t=0. The prcedure fr finding the time cnstant is as fllws.
3 Draw a hrizntal baseline tangent t the final value f the respnse curve (100%). Dead time ccurs frm time t=0 t the pint f initial respnse (5s). Draw a tangent line t the steepest part f the curve until it intersects the upper baseline. The steepest part f a true first rder respnse always ccurs at the end f the dead time. Frm the intersectin f the tangent line t the upper baseline, draw a vertical line t the time axis. The difference between this time and the end f the dead time represents the time cnstant f this first rder respnse. In Figure 2, the time cnstant is 10 secnds. Als nte the vertical crsses the respnse curve at a pint equal t 62.3% f the ttal change in utput. High Order Time Lags When multiple first rder lags are placed in series, the respnse can be rather cmplex. The multiple time lags may be either interacting r nn-interacting. This can be explained thrugh a hydraulic analgy. Cnsider a grup f series cascading tanks as arranged in Figure 3(a). This represents nn-interacting lags. Assume the valves are partially pen and the system is at steady state. The flw rate frm each tank is dependent nly n the fluid head in each tank and hw much the valve is pened. If we pen valve ne further, the flw frm tanks tw and three are eventually affected. Hwever, if we pen valve tw instead, this will impact tank 3 but nt tank 1. Figure 3: a) Nninteracting Lags b) Interacting Lags Figure 3(b) cnsiders a similar arrangement. Hwever, the flw rate frm each tank is a functin f the differential head between tanks. Any change in flw rate frm ne tank t the dwnstream tank nt nly impacts the dwnstream tank, but als the upstream tanks. Such an arrangement represents interacting lags. Whereas a first rder respnse is typical f a single cmpnent f the cntrl system r prcess, a high rder respnse is the result f several cmpnents f the system r
4 prcess in series. The traces in Figure 4 are representative f such a system. These traces were generated with six nn-interacting time lags and a dead time. The step functin prducing this respnse ccurs at time t=0. Fr ur purpses, we need t analyze the respnse fr an verall time lag and an verall dead time. We are nt usually cncerned with the number f time elements invlved in the respnse r where they ccur within the system. Hwever, as we'll discuss later, it is ften wise t understand where the dminant time lags and the dminant dead times ccur within the system. The analysis f a high rder respnse may be handled in ne f the three ways described in Figure 4. Althugh all three methds described are accepted methds f analysis, they prvide very different results. Nte that the first methd tends t verestimate the time cnstant. In mst cases, this is nt desirable. Hwever, this methd has the advantage f being a very simple methd f graphical analysis. On the ther hand, methds #2 and #3 bth prvide a gd fit t the riginal respnse. Hwever, it is nt necessarily the degree f fit with which we are cncerned. What we need is an accurate apprximatin f the system time cnstant and dead time. Nte in Table 1 that methd #2 apprximates a value f time lag smewhat larger than methd #3 while the dead time is smewhat smaller than that f methd #3. When trying t apprximate these cnstants, it is in fact better t apprximate the dead time n the lw side and the time lag n the high side. This des imply methd #2 is a better methd f finding these cnstants. On the ther hand, research has indicated that perhaps methd #3 prvides mre cnsistent results. In a future chapter regarding lp tuning, we will cmpare hw each f these methds impact the tuning prcess.
5 Methd #1 a) Fit #1 b) Fit #2 i. Draw a hrizntal baseline at the initial value f the respnse curve (0%) ii. Draw a hrizntal baseline tangent t the final value f the respnse curve (100%) iii. Draw a tangent line t the steepest part f the respnse curve allwing it t intersect with bth baselines iv. The time frm the start f the step input t the lwer pint f intersectin represents the end f the effective system dead time (3.30s) v. The time between the upper and lwer pints f intersectin represents system time lag (4.34s) Methd #2 i. Draw a hrizntal baseline at the initial value f the respnse curve (0%) ii. Draw a tangent line t the steepest part f the respnse curve allwing it t intersect with the baseline iii. Determine that pint at which the respnse reaches 63.2% f its ttal change and draw a vertical line t the crrespnding time. iv. The time frm the start f the step input t the lwer pint f intersectin represents the end f the effective system dead time (3.30s) v. The time between this pint f intersectin and the time fund in step iv represents system time lag (2.90s) Methd #3 i. Determine the time at which the system reaches 63.2% f ttal respnse (6.32s) ii. Determine the time at which the system reaches 28.3% f ttal respnse (4.52s) iii. The time cnstant is 1.5 times the difference between the times fund in steps i and ii (2.70s) iv. The dead time is the difference between the time fund in step i and the time cnstant calculated in step iii (3.62s) c) Fit #3 Figure 4: Analyzing a high rder respnse
6 Methd Time Lag Dead Time # s 3.30 s # s 3.30 s # s 3.62 s Figure 5: Cmparisn f mdels t estimate high rder respnse Table 1: Cmparisn f time elements derived frm the varius methds fr analyzing a high rder respnse Respnse f a valve In a previus sectin, we discussed the perfrmance and selectin characteristics f cntrl valves. Hwever, cntrl valves can als exhibit time delay and dead time. These elements manifest themselves in the frm f valve sticking (sticktin) and deadband. All valves, even new valves, exhibit sme degree f sticking and deadband. It is the quality f the valve (when new), and the maintenance f the valve (when installed), that determines the degree t which these phenmena are manifested. Valve sticking is the result f static frictin frces between the valve stem and valve packing. It is ften referred t as sticktin, shrt fr sticky frictin. Once the valve is placed in service, these frictinal frces may vary fr any number f reasns. If the fluid being cntrlled has lubricating prperties, it is cnceivable ne may see a reductin in frictin. On the ther hand, crrsive fluids may begin t attack the valve stem r packing gland causing an increase in sticktin. Over tightening the packing gland nut will als impact the bserved degree f sticktin. Sticktin will appear as a jerky mvement f the valve stem. When a signal is applied t the valve, it may be inadequate t initially vercme the frictinal frces between the valve stem and packing. This means the cntrller sees n feedback regarding the crrectin it just made. The cntrller will cntinue t increase the signal at the same rate f change. As the signal increases t a pint where it vercmes the frictinal frces, the valve stem jumps. Hwever, it can nly mve a certain amunt befre frictinal frces again exceed the frce applied due t the cntrl signal, thus preventing the valve stem frm mving. Figure 5 illustrates such mvement. This curve shws the result f a valve cycling between 27.5% and 72.5% f ttal valve mvement. The smth curve represents the signal applied t the valve. The staircase
7 trace is the resulting valve mvement. Sticktin is generally stated as a percentage f the ttal valve mvement. The valve in Figure 5 exhibits sticktin f 2%. Valve deadband is the result f dynamic frictinal frces between the valve stem and the valve packing. Deadband shws up when the valve must change directin. It is a measure f the amunt f change in the applied signal that must ccur befre the valve physically changes psitin. Refer again t Figure 5. When the cycle reaches its peak, the signal t the valve reaches a value f 72.5%, hwever, the valve is nly at 69.5%. As the signal reverses, it must drp t a value f 67.5% befre the valve changes psitin. In ther wrds, the valve deadband is = 5%. Nte that bth the deadband and the sticktin will appear t the cntrller as a frm f dead time. In the case f signal reversal, there is apprximately an 8 secnd span ver which the cntrl valve des nt mve even thugh the signal t the cntrl valve is changing. Similarly, each stair-step mvement f the valve stem takes apprximately 1.3 secnds. Ultimately, this affects the tuning f the cntrller. Valves als exhibit a time lag. Figure 6 shws a valve respnse with a valve exhibiting 5% deadband and 2% stick-slip. The slid trace is a first rder apprximatin f the change in valve psitin. If ne were t apply the methds f finding a time cnstant as depicted graphically in Figure 2, ne wuld find a time cnstant fr this valve smewhere between three and fur secnds. Nte that sticktin prevents us finding a mre accurate value f time lag. Figure 6: Valve stick and deadband as may be bserved in a cycling valve. This trace is a screen sht f a valve simulatin frm PC-CntrlLab. Valve stick is 2% and deadband is 5%.
8 Figure 7: An apprximatin f the first rder respnse f a valve exhibiting sticktin and deadband. The time cnstant fr this valve is between three and fur secnds. Time cnstants fr physical systems Physical systems can take many frms. In this discussin, we will address electrical systems, fluid systems, and thermdynamic systems. All elemental devices have a time cnstant assciated with it. Mathematically, we can shw such a time cnstant is a functin f a resistance t energy flw and the capacity t stre energy. At this pint, it becmes a matter f hw we define these tw quantities fr individual systems. Electrical Systems A typical RTD temperature transmitter is shwn in Figure 8. At the risk f being verly simplistic, such devices can ften be reduced t an equivalent RLC (resistance, inductance, capacitance) circuits. Althugh such circuits may be series, parallel r series - parallel, circuits, ur discussin will cncentrate n a series circuit as indicated in Figure 9. Such circuits cntain tw energy strage devices, a capacitr and an inductr, in series with a resistance, and will exhibit a secnd rder respnse similar t that shwn in Figure 4. T understand the time cnstant f a high rder system, we must first understand first rder electrical systems in the frm f an elementary RC circuit, as shwn in Figure 10, and an elementary RL circuit as shwn in Figure 11.
9 Figure 8: Temperature Transmitter (Curtesy KMC Cntrls) Figure 9: An electrical circuit can be mdeled as ne with Resistance, Capacitance and Inductance Frm a curse in basic electrical circuits, we knw the fllwing: Quantity f electrical energy (q) is measured in Culmbs as an electrical charge. The flw f electrical energy is measured in amperes (I). An ampere is the flw f ne culmb (q) f electrical charge per secnd. The flw f current is the result f an electrmtive frce measured as vltage (E). It is this electrical 'pressure' that frces an electrical current t flw thrugh an electrical resistance. Ohm's law states that vltage is the prduct f current and resistance. If we rearrange this equatin, we can define resistance as vltage divided by current. A capacitr is used t stre electrical charge. The ttal capacitance f a capacitr, which is measured in farads, is a functin f vltage and ttal stred charge. The greater the vltage acrss the capacitr, the mre charge it will hld. Mathematically, charge is expressed as the prduct f capacitance and vltage. Rearranging, we can express capacitance as charge per unit f vltage. An inductr is a device cmprised f several cils f wire that ppses a change in current. This ppsitin is in the frm f an induced EMF. In ther wrds, as current in a circuit increases, the inductr generates an ppsing EMF. The ttal inductance f an inductr, which is measured in henries, is a functin f the generated vltage and the bserved change in current. Mathematically, the EMF generated is expressed as the prduct f inductance and the change in current. Rearranging, we can express inductance as vltage per unit change f current. Assume the RC circuit in Figure 10 is unpwered and the capacitr has n stred charge. A digital vltmeter is used t measure the vltage acrss the capacitr as shwn. Clsing the switch cnstitutes a step input t the circuit; an instantaneus increase frm zer vlts t battery vltage. The capacitr prevents the vltmeter frm seeing the full battery
10 vltage immediately. Rather, the capacitr begins t stre charge and the vltage acrss the capacitr will increase in a predictable manner. The respnse f this system will be identical t the first rder respnse shwn in Figure 2. It was stated earlier a system time cnstant is the prduct f resistance and capacitance. Using the basic definitins described abve, we can write: E = I * R therefre R=E/I q = C * E therefre C=q/E Since τ = R*C, then: τ = E/I * q/e = q/i = q / (q/sec) = sec Figure 10: A simple RC circuit. The respnse f such a system is shwn in the graph. Nte the time cnstant is the prduct f resistance and capacitance. Thus an RC circuit with a capacitance f 10,000 µf and a resistance f 100 Ω will have a time cnstant f 0.01 F * 100 Ω = 1 secnd. The time cnstant f a RL circuit is analyzed in a similar fashin. Assume the circuit in Figure 11 is unpwered. A digital vltmeter is used t measure the vltage acrss the inductr. By clsing the switch, we impse a step vltage input t the system. Since an inductr ppses a change in current, initial current will be zer while the initial vltage acrss the inductr is equal t system vltage. As current begins t increase, the vltage acrss the inductr decreases. This ccurs predictably in accrdance with the system time cnstant. The respnse f this system will be a first rder respnse similar t that shwn in Figure 11.
11 The time cnstant f an RL circuit is the qutient f inductance and resistance. We can shw this thrugh the fllwing mathematical relatinships. E = L* I = L*(I/sec) therefre L = (E*sec)/I Frm abve, R = E/I Since τ = L/R, Then τ = [(E*sec)/I] / (E/I) = sec Thus an RL circuit with an inductance f 100,000 mη and a resistance f 100 Ω has a time cnstant f 100 Η / 100 Ω = 1 sec. The respnse f an RL circuit is shwn in Figure 11. Figure 11: A simple RL circuit. Nte the time cnstant is the qutient f inductance and resistance. Since an inductr resists a change in current, the plt used t determine the time cnstant is %current vs. time. We can determine the time variable vltage acrss the inductr by applying hms law. Fluid Systems The cntrl f fluid systems is cmmnplace in HVAC and prcess cntrl. Analgus t electrical systems, the time cnstant f a fluid system is the prduct f hydraulic resistance and hydraulic capacitance. T determine these quantities, let us investigate the flw f fluid thrugh a prcess tank. Cnsider the prcess tank shwn in Figure 12. This tank has a crss sectinal area f 'A' and a fluid head equal t 'H' Frm a basic curse in fluid mechanics, yu will remember the cntinuity equatin states: Q = V xa Q = Vlume flw rate (ft 3 /sec) V = Flw velcity (ft/sec) A = Area f flw cnduit (ft 2 ) Energy is required fr fluid t flw thrugh the nzzle at the base f the tank. This energy is a cnversin f the ptential Figure 12 Prcess Tank
12 energy stred in the tank t flw energy. Ptential energy is equal t the height f the fluid in the tank. Flw energy is a functin f the flw velcity. Thus: H v 2 V = 2g Rearranging: H v = Velcity head. In this case, it is equal t ttal fluid head, H V = 2gH Substituting int the cntinuity equatin yields: Q= Ax 2gH This equatin is knwn as Trcelli's therem. It is the theretical flw rate thrugh a nzzle f area 'A' given a fluid ptential f 'H'. Hwever, certain phenmena will result in a smewhat lesser flw rate. In rder t accunt fr these lsses, we will multiply the theretical quantity by a discharge cefficient, C d. Q= C xax 2gH d Nw let us cnsider the pressure drp thugh any passive hydraulic element such as a valve r nzzle. Such pressure drp is defined as: n H = RxQ xsg.. H = Pressure drp (in feet f fluid head) Q = Vlume flw rate R = Hydraulic resistance n = Parameter t handle turbulent flw S.G. = Fluid specific gravity Cnsidering the fluid level in the tank f Figure 12 and making the fllwing assumptins: Then: Cnsider the variatin in fluid level in the tank as a flw. This flw is laminar, thus the 'n' parameter in the abve equatin is equal t 1 The H term is the differential head between sme value f fluid head, H, and zer head, H 0, which ccurs with an empty tank. Thus H = H H = H 0 = H Assume the fluid has a specific gravity f 1 H = RxQ
13 Nte the similarity t Ohm's law. Rearranging, we can define the general frm fr hydraulic resistance as: R = H Q Hwever, there is ne significant difference between a fluid system and an electrical system. Since head (H) and flw (Q) can vary with time, hydraulic resistance will als vary with time. As such, we must determine the hydraulic resistance at any instant in time. We can d s by differentiating head with respect t flw. dh R = dq 2 Q H = C xa x 2 g 2 2 d dh Q R = = dq C x A x g 2 2 d By substituting Tricelli's equatin fr Q: Cd x A x 2gH 2gH R = = C xa xg C xaxg 2 2 d d Multiply the tp and bttm by 2H R = C xax 2 gh d 2gH and simplify: Nte the denminatr is equal t Q, s instantaneus hydraulic resistance can is defined as: 2H R = Q Nw that resistance is defined, we need t define capacitance in a hydraulic system. Referring again t the tank in Figure 12, we knw the ttal capacity f the tank is the vlume f the tank. Taking the area f the tank in square feet and fluid head in feet, the vlume f the tank, in cubic feet, is calculated as: V = H xa By making an analgy t electrical systems, tank vlume can be cnsidered the [quantity f] charge n a capacitr and fluid head is analgus t vltage [ptential]. We
14 als knw that electrical capacitance is charge divided by vltage. Rearranging the abve equatin: V A= H Thus the area f a tank in a fluid system is analgus t electrical capacitance. Thus, the time cnstant f a tank is equal t hydraulic resistance multiplied by hydraulic capacitance. Dimensinally: 2H feet τ = = = Q feet secnd 3 x A x feet secnds 3 Example: A prcess tank experiences an inflw f 80 gpm and an utflw f 160 gpm. The tank has a diameter f 8 feet and a fluid head f 5 feet. Determine the time cnstant. The net utflw is: Q = = 80 gpm 3 3 gal min ft ft 80 x x = min 60sec 7.48 gal sec The area f the tank is: A= π 4 = 50.2 ft 2 2 The system time cnstant is then equal t: τ = 2 x 5 ft = = 2 x50.2 ft 2823secnds 47minutes 3 ft sec Thermal Systems Anther cmmn prcess in HVAC cntrl is that f heat transfer. As with electrical and fluid systems, the time cnstant f a heat transfer prcess is als the prduct f a resistance and a capacitance. Frm a basic curse in heat transfer, we knw that the heat absrbed by any material is defined by: Q= mcp T Q = Ttal energy transfer in BTU (Quantity) m = Mass f material in lb m c p = Specific heat f material in BTU/(lb m F) (Heat Capacity) T = Temperature difference (Ptential)
15 Similarly, the heat transfer thrugh a slid material is: 1 Q= A T R If we let Q = Heat transfer rate in BTU per unit time (Quantity) A = Area f surface thrugh which heat is transferred R = Resistance t heat transfer T = Temperature difference (Ptential) R rearrange and write: R T = Q = RA, where R is the verall resistance t heat transfer, then we can Thermal resistance f many materials is well knwn and tabulated. Fr example, the ASHRAE Handbk f Fundamentals tabulates the thermal resistance f many cmmn building materials. These material prperties may be expressed as ne f fur values: Resistance Cmmnly referred t as R-value, this prperty represents the number f hurs it takes fr ne BTU f energy t pass thrugh ne square ft f area with a ne-degree temperature difference fr a given thickness f material. R 2 ft Fhr BTU Resistance per inch This is R-value per inch r 2 ft Fhr BTU in Cnductance Thermal cnductance is the reciprcal f thermal resistance. It represents the number f BTU's that will pass thrugh ne square ft f area within ne hur given a temperature differential f ne degree fr a given thickness f material. C BTU ft 2 Fhr Cnductivity Thermal cnductivity is the reciprcal f R-value per inch. BTU in k 2 ft Fhr
16 Given all f the abve definitins, we can nw define ttal thermal resistance and ttal thermal capacitance f a system. Since verall resistance t heat transfer, R, is simply equal t R-value divided by area, then: R 2 ft F hr R = = BTU = 2 A ft BT Fhr U Ttal capacitance is equal t the specific heat f the material multiplied by the ttal quantity (mass) f material. BTU BTU C = c xmass= xlb = p m lbm F F Then the time cnstant f a thermal system is: τ = Fhr BTU R xc = x hur BTU F = Example: A finned-tube ht water cil is fabricated f aluminum tubes and fins. The ttal external heat transfer area is 60 ft 2. The ttal mass f aluminum f which the cil is fabricated is 200 lb m. The verall cnductivity f the cil is 8.45 BTU/(hr ft 2 F). What is the time cnstant fr this cil? Frm handbk data, c p fr aluminum is BTU/lb m F BTU BTU C = x200lbm = 42.8 lb F F R m 1 BTU 8.45 = = ft BTU 2 hr ft F Fhr Fhr BTU τ = R xc = * 42.8 = hur = 5.1 minutes BTU F Putting it tgether It is nt unusual t get lst in the mathematics f the abve material and nt realize the physical and practical imprtance f what it means t cntrl system design, peratin, and maintenance. Althugh there are definitely situatins in which the calculatin f the time cnstants f each cmpnent f a prcess system is necessary, the majrity f situatins d nt require ne t d s. Hwever, it is quite useful t be able t determine which cmpnent f the prcess lp is likely t be the dminant time lag r
17 dminant dead time. The abve mathematical presentatins are prvided t aid yu in identifying thse cmpnents that may be dminant. In an effrt t prvide such insight, cnsider the fllwing scenaris. All t frequently, the applicatin f any single cntrl device t a given prcess is ften cnsidered nly in the case f steady-state peratin. Fr example, a previus sectin discussed the applicatin f temperature sensrs. Assume an applicatin is cntrlled by sensing the temperature f sme flwing fluid. If the system is at steady-state cnditins, des it really matter if the temperature sensr is a highly respnsive device with a time cnstant f nly secnds, r a massive element in an il-filled pipe well with a time cnstant measured in minutes? In the case f steady-state peratin, it really desn t matter. But we dn t prvide cntrl t handle the steady-state scenari. Cntrl is necessary t maintain the prcess variable at a fairly cnstant value during therwise dynamic cnditins. It becmes apparent that if the cntrl prcess had t be very respnsive, then we wuld want t take the time t select a respnsive (shrt time cnstant) sensing element. In a similar vein, certain types f liquid-t-liquid heat exchangers are mre respnsive than thers. Fr example, plate and frame heat exchangers tend t have a smewhat shrter time cnstant than a shell and tube exchanger. Such a cnsideratin may r may nt be imprtant t the design f the system. On the ther hand, cnsider tw prcess tanks as shwn in Figure 13. Each tank has the same utflw, Q, and the same liquid head, H, thus each has the same hydraulic resistance. Hwever, they each have a different capacitance as measured by their areas. Assume yu are t cntrl the level in each tank. Which tank wuld yu suppse t be mre sensitive t variatin in inflw and utflw? If yu said the smaller tank, yu wuld be crrect. This is rather self-evident when ne realizes that the smaller area means there is a smaller vlume f liquid fr every ft f liquid head. Thus, the fluid level will drp further Figure 13 Tw prcess tanks f equal utflw, equal head, but unequal capacitance and faster fr a given utflw than with the larger tank. This is brn ut by the fact the smaller capacitance als means a smaller time cnstant fr the tank. This means the tank is mre respnsive t variatin in flw. As will be seen in the chapters n tuning, this will have definite impact n cntrl lp tuning.
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Physics 3204 Public Exam Questins Unit 1: Circular Mtin NAME: August 2009---------------------------------------------------------------------------------------------------------------------- 12. Which
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