Data-Driven Modeling of Control Valve Stiction Using Revised Binary

Dec 19, 2014 - An accurate stiction model enables the detection, quantification, and compensation of this nonlinear phenomenon in a control valve. Com...
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Data-Driven Modeling of Control Valve Stiction Using Revised Binary-Tree Structure XiaoCong Li,†,‡ Si-Lu Chen,*,†,§ Chek Sing Teo,†,§ Kok Kiong Tan,*,†,‡ and Tong Heng Lee†,‡ †

SIMTech-NUS Joint Lab on Precision Motion Systems, Department of Electrical and Computer Engineering, National University of Singapore (NUS), 4 Engineering Drive 3, Singapore 117583 ‡ National University of Singapore (NUS), 21 Lower Kent Ridge Road, Singapore 119077 § Singapore Institute of Manufacturing Technology (SIMTech), 71 Nanyang Drive, Singapore 638075 ABSTRACT: An accurate stiction model enables the detection, quantification, and compensation of this nonlinear phenomenon in a control valve. Compared with stiction models obtained from physical laws, data-driven models are more popular because of their simplicity in terms of the number of parameters required. In this work, the previously proposed two-layer binary tree datadriven stiction model is revised to overcome its limitations in handling instantaneous input commands on reverse motion. Then, the accuracy of the revised model is validated using the full set of ISA control valve standard tests. From these results, its advantages over other major existing data-driven models are indicated in terms of simplicity and accuracy.

1. INTRODUCTION

the revised binary tree model to demonstrate its accuracy in modeling valve stiction. This article is organized as follows: Section 2 describes valve stiction modeling and the two-layer binary data-driven model. Section 3 points out the weaknesses of the model and proposes and provides a detailed illustration of a revised binary tree model. The ISA standard tests are applied to the revised binary data-driven model to verify its accuracy in section 4. Conclusions are given in section 5.

Stiction widely exists in control valves, and it produces limit cycles in the control loops, which accelerates equipment wear and even affects the stability of closed-loop systems.1,2 Thus, it is important to formulate proper valve stiction models to detect, quantify, and compensate valve stiction. 3−6 A comprehensive literature review of state-of-the-art valve stiction research can be found in ref 7. Intuitively, the behavior of sticky control valves can be described by physical laws incorporating various friction models, such as the Karnopp model,8 the LuGre model,9,10 and the generalized Maxwell slip (GMS) model.11,12 Such physical models have quite a few parameters that are typically hard to determine in practice. To overcome this disadvantage, in recent years, data-driven valve stiction models have been developed that have much fewer parameters. Among them, the most prominent ones are Choudhury et al.’s model,1 Kano et al.’s model,13 and He et al.’s model.14 To examine the effectiveness and accuracy of various data-driven and physical models, a test based on Industry Standard Architecture (ISA) standards15,16 was used by Garcia.17 It was found that Kano et al.’s model was able to pass all of the tests, whereas Choudhury et al.’s and He et al.’s models both failed in some tests. The most recent Xie−Cong−Horch (XCH) model18 used Garcia’s testing results and made modifications based on Choudhury et al.’s model. It is able to reproduce the stiction behavior more closely, but the implementation is relatively complicated. Inspired by He et al.’s binary tree model,14 the two-layer binary tree model19 inherits the simple logic structure and is capable of describing common cases of stiction phenomena and passing various open-loop and closed-loop tests. However, in this work, it is shown that the original version of the two-layer binary tree model fails in handling instantaneous input commands on reverse motion. Thus, a revised binary-tree model is proposed herein to solve this problem. Subsequently, the complete set of 15 ISA tests for control valves is applied to © 2014 American Chemical Society

2. REVIEW OF MODELS OF VALVE STICTION A typical flow control system, including the set point, feedback controller, control valve, and process, is shown in Figure 1,

Figure 1. Block diagram of the flow control system.

wherein standard industrial notation is used to facilitate its reading. PV denotes the process variable or flow rate in this case, OP denotes the operational point or controller output, and MV denotes the manipulated variable or valve position. Here, it is worth pointing out that the control signal generated by the feedback controller has to go through the control valve first before affecting the process. Thus, the existence of stiction in the control valve makes the MV fail to follow the command OP exactly, resulting in inaccurate control of PV. Received: Revised: Accepted: Published: 330

August 7, 2014 December 8, 2014 December 19, 2014 December 19, 2014 DOI: 10.1021/ie5031369 Ind. Eng. Chem. Res. 2015, 54, 330−337

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the actual valve position uv (MV) of a pneumatic valve with stiction is shown in Figure 3. The parameters used here are

Notice that the control valve itself is a motion system governed by physical laws. In earlier years, physical models were used to describe the behavior of control valves impeded by stiction. For example, in a typical pneumatic valve, the force balance equation can be written as1 M

d2x = Fa + Fr + Ff + Fp + Fi dt 2

(1)

where M is the mass of the moving stem; x is the stem position; Fa = Au is the pneumatic actuator force, where u is the actuator air pressure and A is the area of the diaphragm; Fr = −kx is the spring force; Fp is the force due to fluid pressure drop; Fi is the extra force required to force the valve to be in the seat; and Ff is the friction force. Fi and Fp are usually assumed to be zero in simulations and analyses because of their negligible effects. Figure 2 shows a typical pneumatic valve schematic, including the actuator air pressure, spring force, and stiction.

Figure 3. Normalized input−output behavior of a sticky valve.

deadband plus stickband S and slip jump J, and their relationships to static friction fs and dynamic friction fd are given by (3)

fd = (S − J )/2

(4)

As illustrated in Figure 3, if there is no friction in the valve, the valve will move along l0, which makes MV = OP at all times. When dynamic friction fd is present, the curve is shifted to the right (or left) because part of the OP is used to offset fd. As a result, the valve will move along lf in the forward direction and move along lr in the reverse direction. When static friction fs is present, it introduces a stick band J in addition to the dynamic friction, so the MV will stay where it is and jump from B to C when there is an additional increase of J in the MV. Hence, the valve will move along the line ABCDEFGH where stick−slip behavior can be observed. The two-layer binary tree data-driven model19 extends He et al.’s model14 and addresses all possible state transitions, as shown in Figure 4. In this model, the valve movement falls into one of the four possible cases: (1) stick to slipping, (2) keep sticking, (3) slip to sticking, and (4) keep slipping. In each iteration, the change of control input u(k) − u(k − 1) is added to the accumulated value cum_u(k) to check the effects of the incremental control on the change in valve movement. It falls into four different cases according to different conditions on parameters such as dynamic friction fd, static friction fs, and current state (slip or stick). For example, when u(k) increases gradually from 0, cum_u(k) accumulates these changes and falls into case 2 temporarily. It falls into case 1 when the accumulated value cum_u(k) finally exceeds fs.

Figure 2. Pneumatic valve schematic.

The friction force1 can be expressed as follows ⎧−Fc sgn(x)̇ − xḞ v , if x ̇ ≠ 0 ⎪ ⎪ if x ̇ = 0 and |Fa + Fr| ≤ Fs Ff = ⎨−(Fa + Fr), ⎪ ⎪−F sgn(F + F ), if x ̇ = 0 and |F + F | > F ⎩ s a r a r s

fs = (S + J )/2

(2)

where Fc is the Coulomb friction, Fs is the maximum static friction, and vFv is a viscous friction term. The first equation corresponds to the slip state of the valve, and the second equation corresponds to the sticky state of the valve. The third equation corresponds to the friction at the instant of breaking away. By combining the friction force obtained from various friction models and the valve force balance equation, it is possible to obtain a complete physical model for pneumatic control valves. The disadvantage of using such a physical model is that it requires a large set of parameters such as the mass, spring constant, and friction force. Notice that the dynamics of the process parameters, such as pressure, concentration, and liquid level, are much slower than the dynamics of the parameters in the control valve, such as stem position. Thus, except for the stiction nonlinearity, the relationship from the control signal to the valve position can be viewed as a simple static linear relationship. With this observation, one can consider to use equivalent data-driven models with correct descriptions of stiction behavior to replace the physical model, with much fewer parameters required. A data-driven approach was first proposed by Choudhury et al.1 The relationship between the controller output u (OP) and

3. REVISION OF THE TWO-LAYER BINARY TREE MODEL 3.1. Weakness of the Original Two-Layer Binary Tree Model. The effectiveness of this model was demonstrated in open-loop and closed-loop tests in an earlier article,19 where detailed illustrations of this model are explained. However, two cases are absent from the original model: (1) the transition 331

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Figure 4. Original two-layer binary tree of the stiction model.

MV (uv in Figure 4) to a percentage of valve travel. With an increase in the OP, the MV sticks for a while until around 1.75 s, where fs = 30% is overcome successfully and the slip jump happens. Then, the MV follows the OP with an offset equal to fd, which is 15%. The problem occurs at 2 s, where the OP undergoes a huge change to the other direction from 40% to 0%. In this case, the MV is expected to stay at 25% after 2 s because the OP in the opposite direction is not large enough to overcome fs. More specifically, the difference between the OP and the current valve position (|0% − 25%| = 25%) is not larger than fs = 30%. However, as one can observe in Figure 5, the MV obtained using the original model does not stay at 25% but rather jumps to 15%. Because this kind of quick transition of input to try to move the stem in the opposite direction is not taken into account, it cannot be correctly handled by the original model. 3.2. Revised Binary Tree Model. To overcome this weakness, we redefine d(k) as sgn[Δu(k)], which represents the direction of valve motion in the OP (not the actual valve motion MV). Then, we add another condition block immediately after checking whether it is slipping. The condition block checks whether d(k) × d(k−1) ≤ 0 because, if this is true, then the direction of valve motion in the OP is opposite to that of the previous iteration, which is exactly the scenario that is not taken into account in the original two-layer binary tree model. If the condition is false, the algorithm will flow in the original way, and the original model works perfectly in such cases. If this condition is true, which means that the stem is trying to move in the opposite direction, what one needs to do next is to check whether |cum_u(k)| > fs to determine whether the reverse actuator force is large enough to overcome stiction. When the controller commands the valve to move in the opposite direction, the stiction fs still needs to be overcome again to move. The valve still sticks if the condition |cum_u(k)| > fs is not true, even if the controller commands the valve to

from slip in one direction to slip in the opposite direction and (2) the tendency to slip in the opposite direction being stopped by stiction. The original model assumes that slip always goes through the sticking mode before slipping in the opposite direction, but this is not necessarily true because it is possible that the actuator pressure changes very quickly in the opposite direction and is large enough to overcome the valve stiction so that it will not go through the visible sticking mode. Moreover, it is possible that the accumulated force in the opposite direction is between fd and fs, so that the stem sticks, but this situation is falsely categorized in the keep slipping case in the original model. This is quite common when the sampling rate is low. Figure 5 serves as an illustration of this weakness of the original two-layer binary tree model. The input is the required stem position, and the output is the actual stem position affected by stiction. The OP (u in Figure 4) has a scale ranging from 0% to 100%, which represents the percentage of opening with respect to its full stroke, and the stiction forces are converted according to the degree of how much they affect the

Figure 5. Illustration of the weakness of the original two-layer binary tree model. Solid thick line, MV of the revised model; solid thin line, OP; dotted line, MV of the original two-layer binary tree model. 332

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Figure 6. Revised binary tree model.

move in the opposite direction. When this condition is true, the stem slips in the opposite direction. Following the new definition of d(k), |fd| in case 4 is also changed to fd accordingly to ensure that cum_u(k) always has the correct sign. The condition fd = 0 and d(k) × d(k − 1) > 0 case in the original model in Figure 4 is separated from the fd > 0 case because it does not work when the valve moves in the opposite direction. Because we have solved this problem by adding another condition block as stated above, we can now combine this condition with the fd > 0 case to make it simpler. The updating rule for cum_u(k) remains the same in the revised model. Another minor improvement is to add saturation conditions on the input when it is greater than 100% or smaller than 0% as has been done in other data-driven models. It should not be greater than 100% or smaller than 0% because the position and control input are represented as a percentage of valve travel. In this case, when the input is smaller than 0%, the output will simply be 0%, and when the input is greater than 100%, the output will be 100%. In the next section, we present a thorough test of the revised binary tree model using ISA standards15,16 to verify its reliability. The whole revised binary tree model is shown in Figure 6, where main changes are highlighted in red. The model can be further illustrated using the normalized input−output behavior plot (OP−MV plot) in Figure 7. The valve movement is indicated by the solid lines with arrows. When the valve first overcomes stiction fs and starts moving, cum_u(k) is immediately reset to d(k)fd as shown in the cases 1 and 4 in Figure 6, and then, all of the increment and decrement of controller output OP will have this value as the reference. In other words, the lines lf and lr will be the reference lines, and the two lines actually show the input−output behavior when

Figure 7. Further illustration using an input−output plot (fd ≥ 0).

stiction is overcome and the valve moves with an offset equivalent to the dynamic friction fd. On line lf, cum_u(k) = fd and d(k) = 1, the region on the right-hand side of line lf is then the region for |cum_u(k)| > fd (gray area). When the input− output falls into this region, it keeps slipping according to the bottom right condition block in Figure 6. If it falls into the white region, the valve will stick again because, in this case, the controller output is not large enough to overcome fd. Similarly, in the overshoot case in Figure 8, the gray area is the region for |cum_u(k)| < −fd. The valve keeps slipping if it falls into this region and sticks again if it falls into the white region. For example, if the valve is moving forward, cum_u(k) = fd ( fd is negative in this case) and d(k) = 1, so any increment in the OP will cause the input−output to fall into the gray area, in which case the valve keeps slipping in the forward direction. Referring back to the undershoot case in Figure 7, the transition from slip to stick or from slip to reverse slip occurs at 333

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and the second test presented in the next subsection provide uniform means of collecting and reporting valve operability data.17 The simulation results for vendor, nominal, and rough valves are shown in Figure 9I,V. Here, we use the rough valve as an illustration. In this case, the valve is supposed to move with a slip jump of around 9%, which is equal to fs − fd (8.62%), after it overcomes fs. The MV should stop moving at t = 4 s when the OP remains at 100% for 1 s. During motion, the offset between the OP and the MV equals fd. When the OP decreases from 100% to 0%, the MV does not move instantaneously with the OP because of fs. It starts moving only when fs is overcome again. Figure 9I shows that the simulation results exactly follow the expected behavior in this case. Similar observations and analyses were performed in the vendor and nominal cases. Thus far, we have shown that the revised binary tree model can follow the expected behavior exactly and pass all three dynamic diagnostic tests, whereas He et al.’s model14 fails in the nominal and rough cases.17 4.2. Test II: Ramp and Pause Test. In this test, the OP is increased by 20% in 1 s, held for 1 s, and then increased by 20% in the next 1 s. The whole process is repeated until the OP reaches 100%, and then the OP is decreased in the same manner until it reaches 0%. The responses of the vendor, nominal, and rough valves are shown in Figure 9II,VI. Here, we use the nominal valve case for illustration. In this case, the MV is supposed to remain at 0% until fs is overcome. After the slip jump, the MV should follow the OP with an offset of fd, and it should stop when the input stops increasing at 40%. The same behavior is observed in the subsequent ramps, and it is easy to verify that the MV matches the expected behavior perfectly. The testing results for the vendor and rough valves also match their expected behavior using our revised binary tree model, whereas Choudhury et al.’s model and He et al.’s model both fail for the nominal and rough valves.17 4.3. Test III: Baseline Test. This test is used to evaluate measurement noise and detect limit cycles. The input starts from 50% and is increased by 2% per second until it reaches a maximum of 90% and is then decreased to 50% in the same manner, as shown in Figure 9III. For the vendor valve, the MV is supposed to follow the OP quite well because the dynamic friction fd is only 0.835% and there is no slip jump. As one can observe, using our model, the MV indeed follows the OP closely, with a difference of only 0.835%. In the nominal case, the MV should start increasing only when the increase of the OP is greater than fd at t = 17 s. The second jump of the MV should happen at t = 20 s when the increment of the OP is greater than J, again after t = 17 s. In the rough valve case, the MV should remain at 50% in view of the fact that the maximum increase of the input is 40%, which is less than fd. As can be observed, the MV indeed follows exactly the expected behavior using our model, whereas, as stated in ref 17, Choudhury et al.’s model fails in the vendor and nominal cases. 4.4. Test IV: Small Step Test. The small step test is used to evaluate the valve deadband and resolution. The initial value of the OP is 50%, and then, at every second, it is increased by 0.1% until it reaches 51.7%. The OP is then decreased in the same manner back to 50%, and the whole process is repeated as shown in Figure 9IV. For the vendor valve, with the OP starting from 50%, the dynamic friction of fd = 0.835% means that the MV starts changing only when the OP is greater than 50.835%, so the MV should start increasing at 9 s when the output is 50.9% . Then, the MV should follow the OP in the same way as in the previous cases. However, in the nominal and rough cases,

Figure 8. Further illustration using an input−output plot ( fd < 0).

point R. In the input−output graph, we notice that it goes horizontally to the left and then starts reverse slip, but this transition period might be very short, and the valve will directly start its reverse motion without sticking. The original two-layer binary tree model fails to capture this transition, whereas the revised binary tree model will take this case into account. During reverse slip, the valve keeps slipping when it falls into the region |cum_u(k)| > fd, which is on the left-hand side of line lr. It sticks when the input−output falls into the white region. Similarly, in the overshoot case in Figure 8, at the turning point R, it is possible that the valve goes directly from R to line lf, and then the valve keeps slipping in the reverse direction when it falls into the gray region |cum_u(k)| < −fd. We now examine the previous case, and its results are shown in Figure 5. One can observe that the issue has been resolved because the output stays at 25% without jumping to 15%. This is just one example of the issue, and this issue can arise whenever a rapid change of inputs to move the stem to the other direction occurs. The improved revised binary tree model takes this issue into account, and it is a more reliable datadriven model.

4. ISA STANDARD TESTS ON THE REVISED BINARY TREE MODEL The ISA standards basically consist of five different tests, and their detailed descriptions can be found in refs 15−17. On the basis of the friction levels on the valve from small to large, valves can be categorized as “vendor”, “nominal”, and “rough” valves, respectively. The parameters fs and fd are not the actual values of the static and dynamic frictions, but the percentages of valve travel that these frictions will affect. As defined in Garcia’s article,17 the parameters used in vendor valves are listed in Table 1. In the following ISA standard tests, the input is the Table 1. Valve Stiction Parameters valve

fs (%)

fd (%)

vendor nominal rough

0.835 32.22 50.42

0.835 26.72 41.82

command stem position (OP in Figure 3), and the output is the actual stem position (MV in Figure 3). The input and output should coincide if no stiction exists. 4.1. Test I: Dynamic Test. In this test, the input signal increases linearly from 0% to 100% in 4 s, remains at 100% for 1 s, and decreases linearly to 0% in the following 4 s. This test 334

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Figure 9. ISA standard test for the revised binary tree model: (I) dynamic test, (II) ramp and pulse test, (III) baseline test, (IV) small step test, (V) magnified dynamic test, (VI) magnified ramp and pulse test. Black line, OP; green line, MV of the vendor valve; gray line, MV of the nominal valve; blue line, MV of the rough valve.

fails in the vendor and nominal cases, and He et al.’s model fails in the rough valve case.17 4.6. Discussion. According to previous studies,17 He et al.’s model can pass 10 of 15 ISA standard tests, whereas Choudhury et al.’s model can pass only 8 of 15 ISA standard tests. To further illustrate and compare the different performances,20 we provide one example here in Figure 11. It shows the ramp and pause test in the rough valve case. The revised binary tree model shows the expected behavior, whereas the MV of Choudhury et al.’s model is stuck at 0%. Details regarding the other tests where Choudhury et al.’s model and He et al.’s model fail can be found in ref 17. The original binary tree model actually has the same response as the revised binary tree model because the five ISA standard tests do not include any instantaneous reverse motion (in which the valve is originally slipping in one direction when the command signal suddenly changes to the reverse direction). This is a potential weakness that cannot be detected using only the ISA tests, but it does affect the response severely when encountered. The five tests provided by ISA are not sufficient to comprehensively assess data-driven models. The ISA tests are conducted mainly to show the advantages of

the OP is not large enough to overcome the valve stiction, so the MV should remain at 50%. As one can observe in Figure 9IV, the simulation results exactly follow the expected behavior as stated above. We can conclude that the revised binary tree model can also pass all three small step tests, whereas, as reported in ref 17, Choudhury et al.’s model fails in the vendor case. 4.5. Test V: Response Time Test. In the response time test, the OP is moved away from 50% gradually with step sizes of 0.1%, 0.2%, 0.5%, 1%, 2%, 5%, 10%, 20%, and 50%, as shown in Figure 10. In the vendor valve case, the MV should start changing when fs is overcome and should then follow the OP closely with an offset of fd. In the nominal case, the MV should remain at 50% for a long time before the OP finally exceeds 82.22%, because fs is now 32.22%, whereas in the rough valve case, the MV is not supposed to move at all in view of the fact that the OP is never large enough to overcome the strong stiction. As shown in Figure 10, the testing results obtained using our model exactly follow the expected behavior. Notice that this is a difficult case to pass because the input changes violently after around 20 s. In fact, Choudhury et al.’s model 335

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Figure 10. Response time test for the vendor, nominal, and rough valves.

fail some of the ISA standard tests,17 whereas the XCH model can pass all of these tests but is too complicated to implement, considering its complex algorithm and long flowchart.18 On the basis of the ISA standard tests, the revised binary tree model is better than the above-mentioned three models in view of its reliability and simpler algorithm.

5. CONCLUSIONS In this article, the two-layer binary tree data-driven model is reviewed and improved, especially in handling the case of instantaneous reverse motion. Then, the revised binary tree model is tested using 15 different test cases according to the ISA standard, showing that it is able to pass all ISA standard tests in the vendor, nominal, and rough valve cases. Thus, it should be considered as one of the best data-driven models. Its advantages come from its accuracy in the ISA standard tests compared with Choudhury et al.’s model and He et al.’s model and its simplicity compared with the XCH model and Choudhury et al.’s model. Future work can be conducted on testing the revised binary tree model as well as other existing data-driven models and physical models based on real valve parameters under operating conditions to compare their behaviors in various plants. New approaches to stiction detection, quantification, and compensation could be developed based on this revised binary tree model.

Figure 11. Comparison between Choudhury et al.’s model and the revised binary tree model under the ramp and pause test. Black solid line, OP; blue dashed line, MV of the revised binary tree model; green solid line, MV of Choudhury et al.’s model.

the revised binary tree model compared to Choudhury et al.’s model, He et al.’s model, and the recently proposed XCH model. However, in terms of ISA standard tests only, the revised binary tree model is as satisfactory as Kano et al.’s model. Further tests could be conducted using data and parameters from real plants to further assess the performance of these two models, possibly other data-driven, and physical models as well. To summarize, the revised binary tree model can pass all 15 ISA standard tests with respect to vendor, nominal, and rough valve situations. Thus, it is reliable for use in modeling control valve stiction. Choudhury et al.’s model and He et al.’s model

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AUTHOR INFORMATION

Corresponding Authors

*E-mail: [email protected]. 336

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Industrial & Engineering Chemistry Research *E-mail: [email protected].

(20) Xu, Q. Digital Sliding Mode Control of Piezoelectric Micropositioning System Based on Input−Output Model. Ind. Electron., IEEE Trans. 2014, 61, 5517−5526.

Notes

The authors declare no competing financial interest.

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ACKNOWLEDGMENTS This work was supported by SIMTech-NUS Joint Lab on Precision Motion Systems U12-R-024JL

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REFERENCES

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