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How to adjust the D value of temperature PID?

2009-03-07View Original

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How to adjust the D value of temperature PID?
Reply #22009-03-07
The empirical PID tuning parameters are as follows when the medium is fluid (gas, liquid). The empirical PID tuning parameters are as follows: 1. Flow adjustment (F): Generally P=120~200%, I=50~100S, D=0S ; 2. For anti-surge system: Generally P=120~200%, I=20~40S, D=15~40S ; 3. Pressure adjustment (P): Generally P=120~180%, I=50~100S, D=0S ; 4. For venting system: Generally P=80~160%, I=20~60S, D=15~40S ; 5. Adjust the liquid level (L): 1] , large container (diameter 4 meters, height more than 2 meters tower tank): Generally P=80~120%, I=200~900S, D=0S ; 2] , medium container (diameter 2--4 meters, height 1.5--2 meters tower tank): Generally P=100~160%, I=80~400S, D=0S ; 3] , small container (diameter 2 meters, height less than 1.5 meters tower tank): Generally P=120~300%, I=60~200S, D=0S ; 6. Temperature adjustment (T): Generally P=120~260%, I=50~200S, D=20~60S ; The above parameters are empirical and not absolute. In addition, in practice, sometimes there are problems with a process object or valve (positioner) of the regulating system, which can be overcome by changing the PID parameters to enable automatic operation. Automatic betting requires patient observation and constant correction. In practice, whether it can be put into automation, the most critical thing is that the valve (positioner) and actuator are easy to use and the movement is flexible. In a cascade regulation system (e.g.: There are 2 regulators), the entire inner ring (secondary tune, its Ko * Kv * Kc > 0) is equivalent to the Kv of the main ring, which is always positive. Results of PID parameter tuning: Observe the curve, it is generally a first-order attenuation characteristic (of course it is a second-order attenuation characteristic in theory).
Reply #32009-03-07
D is often unnecessary. If it is really adjusted, it should be adjusted according to the two items of P and I. If the effect of adjusting the first two items is not good, then adjust D. This is mainly for control objects with large inertia.
Reply #42009-03-07
It is generally not recommended to use D, because D participates in control by judging the upward or downward trend of the data. Personally, it feels like it is controlling the acceleration of an object's movement. It doesn't matter how big the gap between PV and SV is. As long as you change too much per unit time, it will suppress the trend of data changes. Sometimes it causes adjustment lag.
Reply #52009-03-07
I'd like to ask, doesn't D mean "Diagram"? Why is it still worth it?
Reply #62009-03-09
ls really misleads everyone that pid adjustment depends more on experience.
Reply #72009-03-10
The PID mentioned here refers to: Proportional differential integral is used on conventional regulators. The PID you are referring to is the pipeline instrumentation diagram (also called the process flow diagram with control points). This post was last edited by zhaohh3211 on 2009-3-10 07:40 ]
Reply #82009-03-10
It is generally not recommended to use D, because D participates in control by judging the upward or downward trend of the data. Personally, it feels like controlling the acceleration of an object's movement. It sometimes suppresses the trend of data changes and sometimes causes adjustment lag.
Reply #92009-03-10
Temperature control objects generally have large time constants. PID can appropriately increase the D value based on the PI adjustment, that is, increasing the differential effect, which can reduce the control time constant and make the control timely. Generally, PID parameters are adjusted based on experience, and relevant information can also be consulted.
Reply #102009-03-10
The temperature link can generally be regarded as a first-order large inertia link (K/(T* s+1)), from the perspective of the transfer function, the coefficient of the denominator is relatively large. From a mathematical point of view, the PID controller can also be regarded as a transfer function (KP(TD) composed of three numbers KP, TI, and TD. * s+1)/(TI * s+1)), then the system exists (TD * s+1)/(T * s+1), making the inertia of the entire system weaker.
Reply #112009-03-10
Temperature adjustment is relatively lagging, just like pouring a pot of boiling water into cold water. It is impossible to adjust the temperature to the set value immediately. Therefore, the temperature adjustment needs to add integration time. The specific method is as follows: Without adding integration, observe the time for the temperature curve to approach the set value from top to bottom, which is what we often call a cycle, and then divide it by 2. Then fill in the value of D with the number you calculated. If it still doesn't work, just gradually add differential time.
Reply #122022-01-24
Whether you should use D or not, I think it depends on the adjustment situation, just add it if necessary, generally add less!

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