How should this PID parameter be adjusted?
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The temperature of the hot water tank is regulated by using steam to heat the water. Since it’s difficult to control the liquid level, the current process relies on an overflow mechanism: when the water level rises too high, it overflows. The thermometer is located in the middle of the tank, while the steam enters from the bottom. The green line represents the opening degree of the control valve, the red line represents the set value, and the blue line represents the measured value. After adjusting this parameter, it can fluctuate within a range of two degrees around the set value of 95 degrees. But right now, the control valve keeps opening fully for a while, then closing fully for a while. There’s no need for such large movements when doing it manually. After adding differentiation, the control valve curve exhibits many spikes at the peaks and troughs, resulting in no improvement in temperature control. How should I handle this thing?
Appropriately reduce the ratio Kp (increase the gain), and appropriately increase the differential time Td. Give it a try.
The curve appears to exhibit quasi-sinusoidal oscillations, with the valve opening staying at its maximum and minimum positions for extended periods of time. Please check whether integration is enabled; if so, activate the anti-integration saturation function.
Personally, I think the output response is too fast; the proportion should be reduced gradually and the integral effect should be diminished, with the derivative term removed before making these adjustments.
I'm sharing my personal opinions; please correct me if I'm wrong. This is a typical example of a level or temperature control loop failing due to improperly set PID parameters. Since in PID tuning, the term used for the ‘p’ component of PID varies – some call it the gain coefficient while others refer to it as the proportional gain – I will temporarily use ‘strength’ as a substitute. First, I recommend the following settings for the PID control: 1) Set the strength of the D (derivative) action to 0; there is no need at all to include a derivative term in this circuit. 2) Increase the P strength. 3) The I intensity drops to a lower value. Once the circuit stabilizes, I can be increased appropriately to eliminate the deviation. Second, let’s discuss my personal understanding of PID control. There’s no need to memorize any formulas or tricks; what’s important is to understand the PID formula and the actual curve graphs. Based on the trend chart provided by the poster, when the control parameter reaches its highest or lowest value at a turning point, the control valve should also adjust accordingly (this is the case in manual adjustments; for example, when the liquid level is adjusted manually and it starts to change direction, we should either close or open the valve). However, the automatic control of this circuit did not do so; it took a few minutes before the valve responded. The reason is that there is still a deviation between the adjusted parameter and the set value; as long as there is such a deviation, the process continues. This is due to the I component in PID – I competes with P, and I wins. Therefore, the approach to adjusting this circuit is to increase the intensity of P while reducing the intensity of I. That’s not correct; I welcome criticism as well as discussions. QQ: 23947571.
To add further, on the curve, the output of the control valve starts to change direction from its minimum value; this occurs at the point where there is no deviation in the parameter being controlled (i.e., it matches the set value). It is at this point that the valve begins to change direction, as the deviation is 0 at this stage and therefore the I term has no effect. At this point, the P term and the I term compete with each other, with the P term prevailing, which is why the valve changes direction at this time. This also indicates that the P strength of this loop has decreased, while the I strength has increased. Therefore, this circuit needs to increase the intensity of P and reduce the intensity of I.
Isn’t it only differentiation that enables early action? As long as there is a deviation proportion, it should work; moreover, with a positive deviation proportion, it remains a positive deviation, so it shouldn’t turn around early.
Thank you so much:handshake
The integral effect is too strong. Turn off differentiation, increase the integration time by 1–2 times based on the current value, and appropriately enhance the proportional action. Add differentiation only if the results are satisfactory.