fig IIHHIHIIIHHHUIHIWIHHHIINlllllHlPIHIIIIHIHHHI (HESS IVE itiiiiiiiii '_"I This is to certify that the thesis entitled CLOSED LOOP CONTROL OF A PUMP TEST STAND presented by DAN H. EMMERT has been accepted towards fulfillment of the requirements for ' r professor ‘ Dat 0-7 639 MS U is an Affirmative Action/Equal Opportunity Institution LIBRARY Michigan State University PLACE It RETURN BOXto mouthi- chockomtrom youtncoui. TO AVOID FINES mum on or Moro dd. duo. DATE DUE DATE DUE DATE DUE MSU ioAn Affirmative Action/Ema! Opponunlty Institution Wanna-9.1 CLOSED LOOP CONTROL OF A PUMP TEST STAND By Dan H. Emmert A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE Department Of Mechanical Engineering 1 995 ABSTRACT CLOSED LOOP CONTROL OF A PUMP TEST STAND By Dan H. Emmert The process model transfer functions for system pressure and flow rate, the corresponding closed loop controllers, and the system responses are determined for a pump test stand. The system pressure response to a step change in the control valve position is recorded, and used to determine the process model parameters for a first order transfer function. The poles of the closed loop system are placed to provide an over damped response to step changes in the pressure set point. The flow rate response to a step change in motor voltage exhibits a time delay. Frequency response analysis techniques are used to match a second order with time delay transfer function to the flow process model. Controller parameters for closed loop operation are determined to provide acceptable values for gain margin and phase margin. TABLE OF CONTENTS List Of Tables List Of Figures Test Stand Description Signal Conditioning And Circuitry Pressure Control Flow Control Test Stand Performance Appendix A: Signal Conditioning And Wiring Diagrams Appendix B: Flow Control Appendix C: Pressure Control Appendix D: Flow And Pressure Responses To Step Inputs Bibliography 16 27 30 37 43 49 51 Table 1 Table 2 Table 3 Table 4 LIST OF TABLES Measurement Devices Analog Input Channels Analog Output Channels Frequency Response Data 18 Figure 1 Figure 2 Figure 3 Figure 4 Figure 5 Figure 6 Figure 7 Figure 8 Figure 9 Figure 10 Figure 11 Figure 12 Figure 13 Figure 14 Figure 15 Figure 16 Figure 17 Figure 18 Figure 19 Figure 20 Figure 21 Figure 22 LIST OF FIGURES Block Diagram Test Stand Layout Manifold Cross Section Fluid Path Electrical Cart Electrical Panel Pressure Control Loop Control System Pressure Response To A Step ln Valve Voltage Pressure Response To A Unit Step In Valve Voltage Pressure Closed Loop System Root - Locus Diagram Closed Loop Pressure Response To A Step Input Flow Control Loop Flow Response To A Step In Motor Control Voltage Sine Wave Response Flow: Plant Model Frequency Response Magnitude Flow: Plant Model Frequency Response Phase Flow Closed Loop System Gain Margin vs. Integrator Gain Flow: Closed Loop Frequency Response Magnitude Flow: Closed Loop Frequency Re 10 11 13 14 15 16 17 18 19 19 20 22 24 Figure 23 Figure 24 Figure 25 Figure A1 Figure A2 Figure A3 Figure A4 Figure A5 Figure A6 Figure A7 Figure A8 Figure A9 Figure A10 Figure A11 Figure A12 Figure A13 Figure D1 Figure 02 Figure D3 LIST OF FIGURES Continued Flow: Closed Loop Step Response Pump Section Test Stand Operating range Flow Rate Circuitry Speed Circuitry Pressure Scaling Circuitry Torque Circuitry Power Supply Circuitry Control Relay Circuitry Drive Motor Circuitry Proportional Spool Valve Circuitry Shut Off Valve Circuitry Relay Power Supply Circuitry Flow Meter Zeroing Circuitry Power Supply And Voltage Divider AC Voltage Circuitry Multiple Input - Multiple Output Block Diagram Flow And Pressure Response To A Step In Motor Control Voltage Flow And Pressure Response To A Step In Valve Voltage vi 26 27 28 3O 31 31 32 32 33 33 34 34 35 35 36 49 50 50 Test Stand Description To verify the operating parameters of a new pump design, the pump must be tested in a controlled manner. Repeatable test setups and procedures are required to properly compare the results for different pump designs and design iterations. Manual testing involves the adjustment of the drive motor’s voltage to reach a desired speed, and then adjusting a valve in the discharge line to reach a desired pressure. However, the movement in valve position loads the pump and motor, and this causes the speed to change. The motor voltage must be re-adjusted to correct the speed, but this changes the system pressure again. The iteration between voltage and valve position is time consuming, and it does not easily allow the exact same operating points to be recorded for every pump being tested. Using feedback control, a test stand can automatically adjust the motor voltage and valve position with a high degree of repeatability and accuracy. The pump test stand characterizes a pump by measuring the rotational speed and torque required to achieve a desired flow rate and pressure. A data acquisition system measures the pump parameters, and provides the control signals to regulate the motor voltage and valve position. The measured data is displayed on a computer screen, and written to the disk drives if desired. Following a brief discussion of the test stand layout and components, this paper will focus on the determination of the process model transfer functions for pressure and flow rate, the corresponding closed loop controllers, and the systems responses. Figure 1 contains a block diagram for the test stand. The mechanical components are all connected by the rotating drive shaft. The fluid components show the flow path. DRIVE MOTOR —— TORQUE LIMITER ROTATIONAL _ MECHANICAL ENCODER COMPONENTS TORQUE TRANSDUCER PRESSURE I I TRANSDUCER PUMP J 1 SECTION ’ ’ FLU'D PM " . FLOW METER TRANSDUCER SHUT OFF PROPORTIONAL l '— VALVE SPOOL VALVE TANK I l J I FLUID COMPONENTS Block Diagram Figure 1 Figure 2 shows the test stand. The drive motor sits on top Of the structure. It is an one horsepower, universal motor with a DC voltage drive. The DC voltage is supplied by a 0 to 100 volts, 0 to 10 amps power supply. The power supply is controlled by a 0 to 10 volts signal. The next shelf supports the rotational encoder. The encoder provides 60 square pulses (TTL level) per revolution. Hence, the frequency output is a direct reading Of the revolutions per minute (1000 Hz = 1000 rpm). Power for the encoder is from a 5 volt DC power supply. Below the encoder is the torque transducer. This transducer measures the difference in torque between its input and output shafts. Dedicated electronics for the torque transducer convert the measured torque to a 0 to 10 volts signal. The value for the full scale reading (10 volts) is selected with thumb wheels on the electronics front panel, and is set for 100 inch-ounces. The output shaft of the torque transducer is coupled to the pump drive shaft. This Shaft is housed in a manifold which contains the fluid passages. fl ‘fimmm d... l WWMglfirfim/WIWM I g/ '1 L I \Io‘ua mil 1 l T sun writ . 7- le tum mum TWP ixn-R/ mum am»: /' f: '56.)! lMN’SEth mm ”(111? "Ml! gym 7 ‘ / rm um: 17 \\ 7m 4 kn ‘i O ."', _' J.“ . /a air/always | I'Tfi I l i \ ~ VT: ///// /////// SIIPPORI PM um um: m gm 1’4“ a? Test Stand Layout Figure 2 Figure 3 shows a cross section of the manifold. A magnetic shaft seal prevents the fluid from leaking past the pump shaft. The housing for the pump under test is bolted to the underside of the support plate. A small fluid reservoir on a scissors jack can be raised to submerge the pump. W“ MAGNETIC 0m sun I MING SHAH 59x: Iain. /"'//// . l . // iii: .. / ////// iWW‘)// § § .. 1 \ } \s - / % PUMP srAI PlAiE WY////.| _‘%l//1 PUMP HOUSING .,, 7 PUMP StCTIO‘l Manifold Cross Section Figure 3 4 The fluid follows the path from the reservoir through the pump and into the manifold. See Figure 4. A side passage connects the pressure transducer. This transducer outputs a 1 to 6 volts signal corresponding to a gage pressure range of 0 to 200 psi. Power for the pressure transducer is from a 15 volt DC power supply. The fluid is next transferred to the flow meter and back to the manifold. The electronics for the flow meter outputs a frequency signal proportional to the mass flow rate. A flow of 10 grams/second produces a frequency Of 1000 Hz. The flow range is from 0 to 50 grams/second. The fluid then passes through the proportional spool valve that is attached to the side of the manifold. This valve is controlled by a 0 to 10 volts signal. The power for the valve solenoid is from a 24 volts DC power supply. The fluid now leaves the manifold and flows through another valve. This valve is used to completely shut Off the fluid passage since the spool valve will always has some degree of leakage around the spool. This two position solenoid valve is controlled by an 110 volt AC signal through a relay. The fluid then returns to the reservoir. I, HUM PIMP PRISSUPI ’RANSDUCI r / EFT-I‘ll ‘v’Al 'I" i ,4 c/ ./I I 7 r 1 /SH‘U‘OFF ‘\ PRUROWIONFAL VAL“ Fluid Path Figure 4 A cart contains the electrical and electronic components. On the shelves are the computer, torque transducer electronics, and drive motor power supply. The electrical panel is mounted on the rear of the cart. Inside of the panel are the signal conditioning circuits. corresponding power supplies, flow meter electronics, and shut Off valve relays. See Figures 5 and 6. Electrical Cart Figure 5 Electrical Panel Figure 6 Signal Conditioning And Circuitry The data acquisition system for the test stand is PC based, and was used for all the initial response testing as well as final control of the stand. The software used was Lab Tech Control from Laboratory Technologies. This package has a quick, simple setup for the HO channels, and built in PID blocks with on-line tuning which allows for easy Operation. The NO card are from Intelligent Instrumentation. Two boards are used. An 8 channel AID board for input, and an 8 channel D/A board for output. The analog input card in the computer is set to read voltages that range from 0 to 10 volts. Of the four signals to be measured, only the torque transducer has this range for its output. Hence, the torque transducer is directly connected to an analog input channel. The remaining signals must be modified to convert them to the desired range. Basic Op-amps circuitry and frequency to voltage converters are used for the signal conditioning. Table 1 lists the measurement devices, their outputs signals, and the signal conditioning required. The schematics for the signal conditioning circuits are shown in appendix A. Table 1 Measurement Devices Flaw Flow Rae 0 to 5000 Hz Fremency to Voltam Vw=(1 volt/1a!) Hz) fin NHer [gins/second] Converter 8. Op-arps Van = -2 (Vm) Rotaiond Speed 0to1tlIDI-lz Freqnncytchlme Van =(1 volt/1a!) Hz) tn Enaxhr [rpni Converter & Op-arps Vw = -V.,1 Preemie Pressue 1t06vclts Op-arps Vm= 2(V-,,)-2 Trmsducer [kPaj Tame Tame 0to10vclts None Vm=Vh Trmsmcer firm-canes] The flow rate and pressure signals provide the feedback to the computer for the control loops. Each Of these lines contains a low pass filter to prevent high frequency noise from contaminating the signals. The cutoff frequency (15 Hz) for the double pole active filters is chosen to be less than half of the 40 Hz sampling frequency. NO filters are used on the speed and torque signals 7 because these values are not used in the control loops. They are displayed as moving averages which does provide some filtering. Table 2 lists the input channels. Table 2 Analog Input Channels to 0 to 10000 0 to 379 The analog output card has its voltage range set to 0 to 5 volts. The drive motor power supply and the spool valve both require a 0 to 10 volt signal, so the two control signals must be doubled. Non-inverting op—amp circuits, each with a gain of 2, modify the signals. A third output channel is used to keep the drive motor Off and the valve completely Open when the data acquisition program is not running. When voltage level of this channel is less than a reference voltage (4 volts), the system is Off. Another channel controls a second valve in the system. This valve completely closes the fluid path. A fifth channel allows the flow meter transducer to be calibrated. Table 3 lists the output channels. Table 3 Analog Output Channels 0 __ System Control > 4 Volts, System On <4 Volts, System Off 1 Motor Voltage 0 to 5 Volts Motor Power Supply Control 2 Valve Voltage 0 to 5 Volts Spool Valve Control 3 Shut Off Valve Relay > 4 Volts, Shut OfT Valve Closed < 4 Volts, Shut Off Valve Open 4 Flow Meter Zeroing > 4 Volts, fem Flow Calibration < 4 Volts, Normal Operation Power is supplied to the test stand by two 20 amp, 110 volt circuits. One is used for the drive motor. The other circuit powers the rest of the equipment. The solenoid valve used to shut Off the 8 fluid flow also take a 110 volt AC signal. This voltage is switched by a relay. Most Of the electronic equipment contains their own internal fuses. For the power supplies that provide the signal conditioning voltages, and for the relay, an external fuse is used for protection. Pressure Control The block diagram for the pressure control loop is shown in Figure 7. The input signal, entered via the keyboard, is the desired pressure in kPa. The actual pressure, also in kPa, is subtracted from the desired pressure, and the error signal is fed into the controller block. The computer outputs a 0 to 5 volt signal that is converted into a 0 to 10 volt signal with an operational amplifier circuit. This voltage controls the position Of the proportional spool valve which Changes the system pressure. The pressure transducer that is located between the pumping section and the valve converts the system pressure into a 1 to 6 volt signal. Another Op-amp circuit scales this to a 0 to 10 volts range, and the signal is read by the computer. The computer scales this value to represent kPa, and this then becomes the actual pressure. ; ; OP-AMP . ERROR . CIRCUIT : KEY kpa kPa : 338T: 2/589: SPOOL "Pa i BOARD » + CONTROLLER 2 u VALVE S : 1 OP-AMP ; ; CIRCUIT 1 COMPUTER kPa SCAL'NG i 0TO10 ”06 I & IIO BOARD : VOLTS * VOLTS PRESSURE : 137.9 1 - TRANSDUCER ‘— Pressure Control Loop Figure 7 To model the complete plant for pressure control, and to determine the appropriate form for the controller, the simplified block diagram of Figure 8 is used. All external circuitry, the valve, the transducer, and computer scaling have been lumped into the plant expression. The input signal and the actual signal are still in units of kPa. The controller consists of only the PID block Of the data acquisition software. The signal between the controller and the plant is 0 to 5 volts. R(s) Y(s) PID —> PLANT > + Control System Figure 8 9 10 TO determine the plant transfer function, the response of the system pressure to a step change in valve control voltage was measured. The procedure used was to first stabilize the system with a constant voltage applied to the drive motor circuitry to ensure sufficient fluid flow, and 1 volt was applied to the valve circuit. The valve voltage was then changed to 2 volts, and the pressure response recorded. This data is shown in Figure 9. 300 Md )— >- 200 4 Pressure [kPa] ul 8 ------ Step It 100 100 1: -------= ——Response l l 50 + 0 l I l . 3.5 4 4.5 5 5.5 5 Time [seconds] Pressure Response TO A Step In Valve Voltage Step From 1 TO 2 Volts Figure 9 For the pressure control plant, a first order system Of the form Y(s) k R(—s)=rs+1 is assumed with R(s) and Y(s) the input and the output signals respectively. Since the step input did not start from zero, and since the final value of the step is 2 rather than 1, the signals R(s) and Y(s) must be modified to map the above step response to a system which has as its input an unit step with zero steady state conditions. The deviation variables are found by subtracting the 1 1 steady state value of the signals from the physical values for all time after the step has occurred. Let Y*(s) = Y(s) - Yss(s), and R*(s) = R(s) - Rss(s), where the starred values are the deviation variables, Y(s) and R(s) are the physical variables, and the initial values are Yss(s) = 79.7 kPa and Rss(s) = 1 volt. The transfer function in deviation variables is then Y*(s) k R*(s) rs +1 Figure 10 shows the step response for the deviation variables. The values for k and r are determined from the graph. The time constant ‘I.’ is the amount of time required for the pressure to rise to 63 % Of the final value. The final value averages to be 184.9 kPa. 63% Of 184.9 is 116.5 kPa. The time when the pressure equals 116.5 kPa is 0.214 seconds. Hence, the time constant I = 0.214 seconds. The gain k is determined from the ratio of the final value of the pressure deviation variable over the final value of the input deviation variable. For the input, r* = 1 volt. The final value of the output is y* = 184.9 kPa. The gain k = 184.9 kPa/ 1 volt = 184.9 kPa / volt. 200 ..’. -O-—' 175 a} 150 «:— 63 % 116.5 kPa ——————-—————-------_-————-——-———--———--——-—-—--d -l M (II <& .33 Pressure [kPa] ‘2‘ ‘1 0| [I ————— Stepx100 ‘ __ Response 0.214 seconds l — - - "Odd ‘ O p. N” 0| 0.5 0.75 1 1.25 1.5 1.75 2 Time [seconds] Pressure Response To A Unit Step In Valve Voltage [Deviation Variables] Figure 10 12 The transfer function is then Y*(s) _ 184.9 R*(s) ‘ 0.214s+1 The time domain equation for this transfer function is -t. ”t. Y' = kr*[1-e 4) =184.9r*[1_e /0.214] Note that time t* is also a deviation variable Since the step change in the input for the physical variables did not occur at zero seconds, but at 3.79 seconds. The above equation is plotted in figure 10 as the model. Substituting for the deviation variables yields _.t" y—Yss =k(r-rss)(1—e A) y — 79.7 = 184.9(r — 1)[1— e-(t-3.79)/0214] As a check, the physical value for pressure at t = 3.79 seconds (the steady state condition), r = 1 volt is -(3.79-3.79) y—79.7=184.9(1—1)[1-e 4214 = 0 or, y = 0 + 79.7 = 79.7 kPa. The final physical value for the pressure at t = oo, r = 2 volts is y — 79.7 = 184.9(2 — 1)[1— em/O‘ZI") = 184.9 y = 184.9 + 79.7 = 264.6 kPa Wlth an estimate for the for plant transfer function determined, the controller can be specified. For this case, only the integrator portion of the PID block is used. Figure 11 shows the block diagram. 13 R*(s> _|_ _, k v19 + s (TS'I'1) Pressure Closed Loop System Figure 11 The Laplace transfer function Of the integrator is US where I is the gain. The closed loop transfer function is then . k| Y _ T R 52+ls+5| T T To determine the value for l, the root locus Of the characteristic equation 52 +1s+5l=q k=184.9, 1:0.214 T T is plotted in figure 12. For values of I between 0 and 0.0063181, the roots are negative and real. This will produce an over damped response. For values greater than 0.0063181, the roots are complex, and the response will be under damped. For the closed loop response to be non- oscillatory, and to account for the uncertainty in the plant model, a conservative value for the integrator gain is chosen to provide over damped response. The value used for the integrator gain is 0.00515. The closed loop transfer function is then 4.44967 s2 + 4.672905 + 4.44967 Y_' _ R! The poles are located at s = -1.33208 and s = -3.34109. Figure 13 shows the actual response of the closed loop system to a step from 200 to 300 kPa. Also plotted is the response of the above closed loop transfer function to the same input. Note, however, that the deviation conditions were again considered in determining the time domain function for this model. The time domain deviation function is 14 y' = r‘[1 + cease-334109" — 1663e'1'33208t'] Substituting r* = r - r35, with r = 300 kPa and r33 = 200 kPa, y* = y - yss, with yss = 200 kPa, and t* = t - 19.18 yields y _ 200 = (300 _ 200)[1 + 0_663e-3.34109(t-19.18) _ 1663 e—1.33208(t-19.18)] Att=oo, r=300 kPa; y=r=300 kPa. Appendix C contains the complete derivation for both the plant and closed loop transfer functions, and the time domain equations. — -L-___ — — e _» -. 2.5 9 Plano 2 \l=o.o1o E‘ i .5 . g 4» 1 5 I = 0.00515 Root - Locus Diagram Figure 12 Pressure [kPa] 325 300 275 250 225 200 175 150 l l I L»_.J__ 17 .PA.._.+ _ A- L 18 F ------ Step ‘ Response - - - - Model Y ‘I 9 20 21 22 23 24 25 Time [seconds] Closed Loop Pressure Response TO A Step Input Step From 200 To 300 kPa Figure 13 l 1.11't 1 1 A 1_+_A_J._..L_.+ A_1LJV 1 L1 441117 L I i I L ._+t__ + .__I___L_.__.L__l 26 27 28 Flow Control The desired flow rate is achieved using the feedback loop shown in Figure 14. The flow produced by the drive motor and pump is measured by the flow meter. This signal is read by the HO cards in the computer, and is compared to the desired flow rate. The resulting error signal is processed, and the control voltage to the motor power supply is adjusted. FLOW CONTROL LOOP : : OP-AMP KEY mm” “W I 3333 C'RCU'T 2,339; MOTOR °J§tg° DRIVE cums/soc ' BOARD °°NTR°LLER ; 2 POWER MOTOR/ e I + ; SUPPLY PUMP : I OP-AMP ' . CIRCUIT : SCALING ' 0T010 ' COMPUTER : VOLTS FREQUENCY ° To 500° HZ FLOW 1 6 I/O BOARDS 5 t voftcites METER : ' CONVERTER TRANSDUCER Flow Control Loop Figure 14 The response of the system to a step input in motor control voltage is shown in Figure 15. The voltage was stepped from 1 to 2 volts, and the response recorded. From the graph, it appears that a time delay exists in the response. Since a time delay results in a non-rational transfer function, the techniques used in frequency response analysis must be used to determine the process model. 16 17 30 TTllTrTf‘T 711 TTTY Flow [grams/sec] ...... Step Input (Volts x 10) Response .+_t_.l_I_J_.L_.L_1_t_1_+_l_L 1.J.J_LA_L_L+_LL_L_LL 1 I 1_A_T_LLLJ_1_L_1_1J_+_L_LLL1_L_I_L_IL‘_LA_A_A_A_A_A_A_J 3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6 Time [seconds] ,_ o $.14 .t_I_.i_.t_t_L.l_+_L_L.t I I-.l_._1.L_+..L_L_l_.LL_LI i_L .+_ Flow Response TO A Step In Motor Control Voltage Step From 1 To 2 Volts Figure 15 The procedure for determining the frequency response of the system involves subjecting the system to sinusoidal input signals with different frequencies. The amplitude ratios Of the output to input signals, and the phase angles between the signals are determined. A typical plot of input and output signals is shown in Figure 16. The amplitude and phase angle of the curve labeled as the model in Figure 16 are determined by using a spread sheet program to fit the curve to the output data. The input signal for all cases is 1.5 + 0.5 sin(mt) where co is varied for each test. 18 30 rfifvfi 10- 7fi' i +Input [Volts] +Response [gramslsec] +Model [gramslsec] J Time [seconds] Sine Wave Response Frequency: 3 radians/second Figure 16 For the response in Figure 16, the model parameters are found to fit the formula 21.615 + 4.8 sin(3t-1.8326). The amplitude ratio is then 4.8/0.5 = 9.6. The phase angle is -1.8326 radians/second, or -105 degrees. The steady state gain is derived from the ratio Of the DC components, 21.615/1.5 = 14.41. Table 4 contains the amplitude ratios and phase angles for all the different frequencies. Table4 Frequency Response Data I—Freqa'qlraymc] 0102030405060708091 2 3 4 5 6 ‘7 8 910] I madam 14.4 14.4 14 138134134132 13 13 13 11.6 96 7 6 4 3 2 1.9 1.5] IMA‘Uflm -18 -Z) -21 -25 -E -Z -& $ -38 4o -75 4% 43) 445 465 -182 JD -215 -%| The amplitude ratios are converted to decibels, and these and the phase angles are plotted versus frequency in Figures 17 and 18 respectively. Also plotted are the equations representing the process model. A second order model with a time delay is assumed. The parameters Of the model are based on trial and error fits with a spread sheet program. The values that best fit the empirical data are as follows: 19 gain k=14.4 grams / second / volt, time constants T1 = 0.24 seconds, 12 = 0.28 seconds, and time delay 9 = 0.15 seconds. 24 _ _ ____ 20 l 15 m 12 -‘ —o—Responee_1 1’ L - - Model a . 4 D o .. . e- 0.1 1 10 Frequency (redleneleec) Flow: Plant Model Frequency Response Magnitude Figure 17 o - . f E .A_L_J 04""--... 1 1b so . g .120 .- Deg —o—Reeponee - - Model Frequency (redlenslsec) Flow: Plant Model Frequency Response Phase Figure 18 20 The transfer function for the model of the plant for flow then has the form ke—OS T18 +1)(125 +1) 69(5) = ( or with the values substituted 1446-0158 69(5) = (0.24s +1)(o.28s + 1) The equation plotted in figure f4 for the magnitude frequency response is 1 4.4 IGprII = (mym) The equation plotted in figure f5 for the phase angle frequency response is zepuo) = —o.15o — tan“[o.24o] — tan‘1[0.28w] With the plant specified, the controller, Gc(s), can be chosen. A simple integrator with gain, Gc(s) = y , is used for closed loop control. Figure 19 shows the closed loop block diagram. The integrator term eliminates any Off set error between the set point and process variable. It also reduces the effect of high frequency signals by attenuating the amplitude by a factor proportional to the reciprocal of the signal frequency. The integrator acts as a low pass filter. TO ensure that the system remains stable, the open loop transfer function frequency response is analyzed. The Bode Stability Criterion requires that the amplitude ratio Of the open loop transfer function, GOL(S), be less than unity when the phase angle equals -180 degrees. The open loop transfer function is derived from the closed loop system when the feed back leg is disconnected. G,(s) G,(s) R(S)+ I ke-Os Y(S) s (t.s+1) ' Flow Closed Loop System Figure 19 21 The open loop transfer function is then I 14.461153 Gods) = Gc(S)GP(s) = (§)[(0.24s + 1)(o.23s + 1)] The magnitude of this transfer function when s = jw is 14.4l (ob/c0 2(0.24)2 +1)(‘/m 2(0.28)2 +1) |G0L(l°3)l = and the phase angle is zoom) = g - 0.150) - tan'1[0.24m] — tan“[0.28o] Note that the Open loop magnitude equation, IGOLUCDH, is simply the process model magnitude equation, |Gp(jm)l, multiplied by I la). . I . IGOL(Ith D (D I. I515?“ iI—T (:) ® I; S '99” ‘ <:> “FIRM. ~ !‘ [ -1591]; HUI-RN J ,6 I; 15m: LAWN" \JV LAWN Power Supply Circuitry Figure A5 Analog Output Channels: (0 to 5 Volts) #0: System Control. A voltage comparator is used to switch a transistor to control a relay. When the channel output voltage exceeds 4 volts, the comparator output turns on the Darlington transistor. The collector is pulled to ground, and current flows through the relay coil. Computer controlled relay #1 (ccr1) prevents the motor and spool valve from Operating when the data acquisition program is not running. Figure A6 contains this schematic. 33 ——-— A _ V1201, ~— m v i (HP {7\‘\.0 3 awn? ‘ massif ‘ + ‘L mm 3 /73\ W“ . , l‘f’?‘ \' _Hr l ‘A ~ W4 \JW‘ANNM ‘2 /- ‘ I “(”1 )9 go '1’“. y Ip; ‘ b l‘COI‘H‘QOL) . \jl] M {WC 6/4/ 1 3n? ARA'I‘ Control Relay Circuitry Figure A6 #1: Motor Speed. The l/O board outputs a 0 to 5 vdc signal. Since the motor power supply requires a O to 10 vdc range, a non-inverting op-amp circuit with gain of 2 is used. The op-amp output is switched by the first set of normally open contacts in ccr1. The motor power supply output 0 to 100 vdc to drive the universal motor. Figure A7 contains this schematic. some will)? _ DC mm 110 vac 01)le E ”M J" 601”?! CHANNEl I - m L] _4 20m 1 E comm 3 MOTOR LL @ (ROTOR) ‘ ' 9cm pgng1 (aging? 3'31“ _ - l 93 dfl mm mm 090.110 Drive Motor Circuitry Figure A7 #2: V Pressure. The IIO board outputs a 0 to 5 vdc signal. Since the spool valve requires a 0 to 10 vdc range, a non-inverting op-amp circuit with gain of 2 is used. The op-amp output is switched by the second set of normally open contacts in ccr1. The spool valve is powered by a +24 vdc power supply. Figure A8 contains this schematic. i’RfSSU‘h’ CO‘i‘ROl VAZVY ClRCUl'RY ._ ll:l VAC moo ———— 50,"; 0 3:66 l — L .l {—l___, ‘ ‘91”. J [LL—(Ll) Lama # ‘ ‘ "j PTFWER lmn . ; ‘ :] (Al‘v't v $333.6?) 0f. PM? NW7 C‘U-"UY SUL‘ULY SCML _ _ . l' l ,1 —© ' N? J“ " , [M'H ’ 33023.0 Proportional Spool Valve Circuitry Figure A8 #3: Shut Off Valve. A voltage comparator is used to switch a transistor to control a relay. When the channel output voltage exceeds 4 volts, the comparator output turns on the Darlington transistor. The collector is pulled to ground, and current flows through the relay coil. Computer controlled relay #2 (ccr2) switches a 24 vdc line that energizes control relay #1 (cr1). Cr1 switches a 110 vac line that powers the shut off valve. Figure A9 contains the dc voltage schematic, and Figure A10 shows the 24 vdc power supply and relay for the 110 vac line. SlilllOFF VAL Vi CONTROL ClR‘CUllR‘l R2335? (Infill r CHANW L fl l (6,»... NWT q "11 W.) Q Shut Off Valve Circuitry Figure A9 323:6 33236 I "i ll WWW ‘;1’7“‘,'°E__,_+ POIIR U POWER LmamR (‘ “’6” SUPPLY 0qu cm" le ) _ “($3 I] I 2 C 2mm '1 $3 CROWD ‘5— no w; 60 H: I r _. ®—°—l l—« ”‘23? 4515‘ CRl-l “—JV (,7? i an? Relay Power Supply Circuitry Figure A10 35 #4: Flow Meter Zeroing Circuitry. A voltage comparator is used to switch a transistor to control a relay. When the channel output voltage exceeds 4 volts, the comparator output turns on the Darlington transistor. The collector is pulled to ground, and current flows through the relay coil. Computer controlled relay #3 (ccr3) connects two terminals on the flow meter. This connection enables the calibration and zero flow level adjustment on the flow meter. Figure A11 contains this schematic. Power Supply. ROW ”[39. ZEROLNS ClFfilel‘RY l /’ 7 u [R “I '\/\ P; (\ MR) 31 Air1 man; (A? I \ / , L ragténn; “ "'l - , . \ ‘s —I*J ’l‘- ‘ T’ ‘3 “WW '3" ‘ _‘U" q \ mm lg . ,yg, _ L 791211024 {4332' t, "‘ ” L. N A, LJ‘J‘m 31L unill \ 3 <1 . s >— ,; :3 . " ,x’ M. [\l‘u. NOW 343th is H a) . s /l~’3ll:‘£i k , z’l’ 312132.613) "’ “ a, 34me 5/ ““49”" f _- .. ,. ._ \/ t '\ V \ {—2 - m ~~——o—{ IELCW HE [R ‘ I 4 ‘ELLCTRO‘llCS ll 1 Flow Meter Zeroing Circuitry Figure A11 Figure A12 shows the dual power supply used to power the op-amps and comparators for the analog output channels. comparators. A voltage divider is used to provide the 4 volt reference for the no w. H) M l .r 20 AMPS ’_— ' "‘1 .2 fl\ 3 _‘ , —< ‘ ‘.)lrL'C f"‘\ Qy—Jm .J L U "3’le 13 WC ill-3f) , y {ovum or, raw 6P1] DUAL [54 em 0 swim Pit-322R - SUPPLY -2 4000 l— -—@ > a 45m ‘ ’ Dill” \é/ K57 CHUNG Power Supply And Voltage Divider Figure A12 36 Figure A13 shows the schematic for the 110 vac lines that power all the devices. A separate line is provided for the motor power supply. Each line requires a 20 amp circuit. An external fuse is located before the power supplies used for the op-amps and the shut off solenoid valve. 'u (in 6‘ K at: (0 3:3" T LL m 3 ‘2 3 - 3‘23 —3 3—3;3_mi 2;; 3 a _. .3 3 L; 33173 — -3 l 3 l , l l CH1} '— t' .n. l' 1 .32. 6‘ L “>3. W @:l_::fl3a__l _:::: :: I 363 *— 3 , /‘\ — r——fi:3—J L——a: : 2.1 ”j: +3 3"}: ' __ 3 w _d H —_JT;:“ r3 ————< @345?" l {,3 :1: J l 37;: 3 —< » f: : “‘3 3'35" 1 33‘. ‘1‘_3 l —_:T; ___:2 L 3,, ## 3L-._ I I 33 7'3 ‘ 2 _ 3:7 '3 3,1 i' 3. 33 p___- {‘IM 76] a—# .91 “3'7 _ AC Voltage Circuitry Figure A13 Appendix B Flow Control A 2nd order transfer function with time delay has the form _ ke—es G(S) - ( T1S+1ths+ 1) Let G1 5 G2 s 6(8) = 333536432) where (31(5) = k, 62(5) = e‘es, 63(5) = 1,5 +1, and 64(3) = 125 +1. With s=jw, the magnitudes of the above equations are as follows: |G1(J‘°)| = " 3G2(j‘°)| = 1 |G3(ja))| = 3112002 +1 |G4(jo))| = ‘39sz +1 Then, _ 330°33'64”» k ’ l63(jw)||G4(jw)| -‘/112c02 +1J122c02 +1 |G(iw)| 14.4 032 +1JO.282m2 +1 or, I600)» = J0 242 The phase angle for the model is 16(5) = AG1(s) + 462(5) - 4G3(s) - 464(5). VWth s = jco, the above angles are as follows: 46,000) = 0 162(k)) = —6m , 46300)) = tan—111w] 46400)) = tan-1120)] 37 38 Then, 4600)) = —9w - tan-111m] - tan—1120)] = —0.150) — tan“‘[0.24m] — tan-10.2803] The open loop transfer function is GOL(S) = Gc(S)GP(5) = [an ke‘°’ 3 T18 + 1X128 + 1) .3..- 22:33:33. .3.-. .23-.. W 64(5) = s ' 65(5) = T19> +1. and 66(8) = 125 +1. The magnitudes of these function are as follows; 36100» = k |Gz(jm)l =l 363003 =1 |G4(iw)l =w 365(ij = W IGe(jo))| = W Then, . 14.4l IGOL (10))3 = k' = a)(\/1:12a)2 +1)( 122(02 +1) w(Jo.242m2 +1)(Jo.282m2 +1) The phase angle for the open loop transfer function is 46(3) = 461(5) + 162(s) + 463(5) - 464(5) - 465(5) — 166(5). With 5 = jw, the phase angles for the individual functions are as follows: 46100)) = 0 46200)) = 0 4G3(j0)) = —90) , [(340m) = % 4G5(jm) = tan’1[r1m] 1G5(j(o) = tan-11200] or 460(1)) = —9m — g- - tan-1110)] — tan-1[12m] 39 4600)) = —0.15(o - g - tan-10.248] - tan-10280)] The critical frequency is found by setting the phase angle equation for the open loop transfer function equal to -n. _ 7! —1 —1 -1t — -0.15(o — E —- tan [0.24m] — tan [0.28m] w = 25875 radians/second. Substituting this into the corresponding magnitude equation yields: |Go._(j2.5875)| = 14"" - = 3.8286 I (2.5875)(J(0.24)2(2.5875)2 +1)(J(0.28)2(2.5875)2 +1) For stability, the open loop gain must be less than unity at the critical frequency. 3.8286 I < 1 I < 0.2612 Let I = 0.07. The gain at the critical frequency is then |G(j25875)| = (3.8286)(0.07) = 0.2680 Then gain margin is then 20 Log(0.2680) = -11.4 dB To determine the phase margin, the frequency where the open loop gain equals 1 is required. (14.4)(0.07) (m)(\/O.242m 2 +1)(\/0.282m2 +1) a) = 0.9497 radians/second The phase angle at this frequency is 46( 10.9497) = —(o.15)(o.9497) — 323 — tan"[(o.24)(o.9497)] - tan“1[(0.28)(0.9497)] 46009497) = -2.197 radians/second = -125.9 ° The phase margin is then 180° + (-125.9°) = 54.1 °. 40 Closed Loop Response. From figure f6, the transfer function is Dropping the “(s)”, Y = (9po02 — Y) Y = GpoR — GpoY Y + GpoY = GpoR Y(1+ Gpo) = GpoR Y ('3po R ke-GS Vlfith Gc = US, and GD = (115 + ”(128 + 1) . mike-es G _ (11$ +1)(1.'2$ +1) d — 1+ (%)ke‘95 (11$ +1)(tzs +1) Simplifying: (gym—es (115 + 1)(rzs + 1) + ()é)ke‘95 kle—93 8(118 +1)(12s +1) + kle—95 Gcl = Gd: For frequency response, let s=jm, lac-rim9 <3cl = _. 9 (jco)(1:1jm +1)(12j(o +1) + kle 1“) 41 VWth e"“’°=cos(- Choose over damped response, and let p1 and p2 be the poles, Wlth p1, p2 < 0: Note: 52+ES+§'=(S-p1)(s_p2) 1 kl 82 +¥S+?=sz —(p1+92)5+9192 1 T = ’(91 + P2) 1 kl -"'“=P1pz I Partial fraction expansion: kl . kl" :r : A B C 1 kl = = + + {sz+—s+_) S(S‘F’1)($"l32) s S—p1 s—p2 1 1 kl ,, kl * _r __r A: I = T =r* P1P2 E T kl ,, —r _ T P1(P1-Pz) kl * —r C_ 1.’ p (92-91) 51". Era .-,Y*=.r_.+ T 1 ]+ T 1 j s 91031-92) s-p1 92(92-p1) s—p2 47 Inverse Laplace transform: kl ( mt‘) kl [ p2?) —l' e —r e I ‘t y* t" =r*+ —— + I ) P1(P1-P2) P2(P2-P1) Multiply second term on the right side by p2 / P2. and the third term by p1 / p1: kl ,, [ mt') kl , ( pzt') —r P2 e —r P1 e 1: T + P1P2(P1 - P2) P1P2(P2 - P1) y*(t*)=r*+ Recall: fl = p1p2 T kl p1t' kl 921* —P2 e —P1 e 1' T + kl kl (P1-P2) :(P2 'P1) T mt. pzta P2 e P1 e (91-92) + (P2 -p1) y*(t*)=r* 1+ y*(t*)=r* 1+ Rearranging yields P1" 92" P2 e -P1 e (P1-P2) y*(t*)=r* 1+ Check: y*(o) = r*[1+——p2(1)'p‘(1)) = r*[1+&—'—pl) = |1*[1_1]= 0 P1-P2 P1-P2 Check: Y*(oo) = r*[1+____p2(0)—p1(0)] = r*[1+0]= r*[1]= r* P1-P2 48 Converting back to the physical variables yields l- — 1.1.131)31:34:») (P1-P2) Y(t)- Yss =(r- rss)1+ _ J Wlth p1 = -3.34109, p2 = -1 .33208, yss =200 kPa, rss = 200 kPa, tstep = 19.18 seconds, y(t) — 200 = (r — 200)(1+ 0.663633410904938) _ 16538-133208(t-19-18)) Check: Wlth r = 200 kPa, t= 19.18 seconds, y(19.18) = (200 - 200)(1+ 0.663(1) — 1663(1)) + 200 = (0)(1- 1) + 200 = 200 Wlth r =300 kPa, t = 00, y(oo) = (300 — 200)(1+ 0.663(0) — 1663(0)) + 200 = (100)(1— 0) + 200 = 300 Appendix D Flow And Pressure Responses To Step Inputs The response of the flow to a step change in motor voltage, and the response of the pressure to a step change in valve voltage was used to determine the process models for the control loops. Note that the flow rate also changes when the valve position is altered, and the pressure changes when the motor speed is modified. Hence, The two control loops used on the test stand are not completely independent. A multiple input - multiple output (MIMO) system could be developed for the test stand. Figure D1 show a block diagram for such a system. The determination of all the process models and the corresponding controllers for the MIMO system is beyond the scope of this project; however, the flow and pressure responses to step changes in motor and valve voltages are presented. Figure 02 shows the responses to a step change in motor voltage, and Figure 03 shows the responses to a step change in valve voltage. R1*(S) - + Y1*(S) + Controller1 —?-H G1(s) + b G3(s) + G4(s) + R *(S) Y *(S) 2+ _ Controller2 H 62(3) + 2 ~ Multiple Input - Multiple Output Block Diagram Figure D1 49 50 30 200 E 14 1" 'b 180 25 «3 P R t nsssurs ssponss T1” 3 + 140 20 + . 1 Stop Input [Volts x 10] 4L 120 {rY’frWr-fi Flow Response i. 8 Pressure [kPa] Flow [gramslssc] a 1 YY LIJLLJLALAALAA 3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 5 11m [seconds] Flow And Pressure Responses To A Step In Motor Control Voltage Step From 1 To 2 Volts Figure DZ 300 , 20 Promo Rsmonss 250 i Flow Response 3» 15 1 r 14 Stop Input Watts )1 - 3" 13 100] «j» 12 AALALALLAAIAA r V I s .-: Flow [gumslssoond] . IIO- . . 4» 1 a . «- s 4 ’ 4 <— 5 4 I 4 4 b