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MAY-JUN 2017

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38 INTECH MAY/JUNE 2017 WWW.ISA.ORG SPECIAL SECTION: PROCESS INDUSTRY TRENDS Combining the parameters and cal- culated values, the loop at sample six is solved as follows, where red is the pro- portional, blue is the integral, and green is the derivative term: M n = K c * e n + Ki * ∑ e i + Kr * (e n – e n-1 ) + M o M(6) = (3 * 20) + (1 * 20) + (0 * (20 – 0)) + 30 M(6) = 60 + 20 + 0 + 30 M(6) = 110 The result is 80 more than the ini- tial control output of 30, with the pro- portional term providing most of the increase. The controller converts this result into an analog output to the con- trolled device—a heating element in this example—in the form of a 4–20 mA or 0–10 VDC signal. As the control output drives the tem- perature toward the set point, at sample seven the error is decreasing, so e(7) = SP – PV = 350 – 340 = 10. At this time, the sum of all the sample errors is ∑ e i = ((e(1) + e(2) + e(3) + e(4) + e(5) + e(6) + e(7)) = (0+0+0+0+0+20+10) = 30). At this moment in time, the result of the equation is: M(7) = (3 * 10) + (1 * 30) + (0 * (10 – 20)) + 30 M(7) = 30 + 30 + 0 + 30 M(7) = 90 With the error corrected, the out- put decreases due to the drop in the proportional term, even though the integral term increased. At sample eight, the PV temperature recovers, to 349.5°F, making the sum of all errors 30.5. At this moment in time, the con- trol output is: M(8) = (3 * 0.5) + (1 * 30.5) + (0*(0.5 – 10)) + 30 M(8) = 1.5 + 30.5 + 0 + 30 M(8) = 61.5 Now the proportional term is ap- proaching zero, and the integral bias is having more effect on the control out - put. While the proportional term has de- creased with the error from 60 to 30 to 1.5, the integral term has increased from 20 to 30 to 30.5. This highlights a loop tun - ing tip: with large error, the proportional term drives the output, and with small error, the integral term takes control. Add a little D for stability The derivative term adds stability to the control loop. When a rapid rate of change occurs, such as the 20°F drop at sample six, there is a risk of instabil- ity, which the derivative term reduces, while adding correction. For example, if the derivative coeffi- cient is changed to Kr = 1 by setting the derivative time (Td) to 20 seconds (Kr = Kc * (derivative time/sample rate) or Kr = 3*20/60 = 1), the control output at sam- ples six, seven, and eight would become: M(6) = (3 * 20) + (1 * 20) + (1 * (20 – 0)) + 30 M(6) = 60 + 20 + 20 + 30 M(6) = 130 M(7) = (3 * 10) + (1 * 30) + (1 * (10 – 20)) + 30 M(7) = 30 + 30 – 10 + 30 M(7) = 80 M(8) = (3 * 0.5) + (1 * 30.5) + (1 * (0.5 – 10)) + 30 M(8) = 1.5 + 30.5 – 9.5 + 30 M(8) = 52.5 The results above show the derivative term increased the control output at l Kr = 0 (This sets the derivative time to zero, making this a PI controller, which is a good starting point.) l K c = 3 (proportional gain) l Ts = 60 second sample rate l Ki = 1 (set integral time to 180 sec- onds as Ki = K c * (sample rate/inte- gral time) or Ki = 3*60/180 = 1 l M(0) = 30 (initial control output) The graph (figure 2) charts tem- perature fluctuation over the past 9 minutes. The graph shows that the temperature is stable for the first five samples before dropping 20°F at sam - ple six. The error at sample six is: e(6) = SP – PV = 350 – 330 = 20. The sum of all errors can also be calculated: ∑ e i = (e(1) + e(2) + e(3) + e(4) + e(5) + e(6)) = (0+0+0+0+0+20) = 20. i=1 n Figure 2. This chart shows how the process variable changes in a PID temperature control application for an industrial oven. Figure 1. Temperature in this industrial oven can be accurately controlled using the PID algorithm. i=1 n i=1 n

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