Integrating process

Accumulation of fluids in vessels is an integrating process: mass - time-integral of inflow less outflow.

Certainly the dynamic response of an integrating process is different than that of a 1st-order system; and so therefore is the design of an appropriate feedback control system. (Note: a 1st-order lag is the system equivalent of an integrating process with its own integral negative feedback.)

Kelvin B. Hales Kelvin Hales Associates Limited Consulting Process Control Engineers E-mail: snipped-for-privacy@khace.com Web:

This sounds like a question from a college student without much plant experience. And college courses on control theory don't cover practical applications in industry very well.

A key point to start with is that in most industrial processes where automatic control is used, time delays are very important factors that affect how well a process can be controlled. Say, for example, you have a steam line and you need to keep the steam temperature from getting too high. Water to cool the steam sprays from a nozzle in the steam line, and water is evaporated by the steam to prevent excessive temperature. The line carrying the water to the nozzle includes a valve with an actuator. A temperature sensor placed downstream of the nozzle in the steam line is used to measure steam temp., and a controller is tied to the valve actuator to keep the steam temp. at a set point.

If, for example, the volume of steam increases while its raw temperature doesn't change, it will take more water to keep the temperature from rising too high. The temp. sensor will note the higher temp., and the valve's controller will open and let more water pass through to the nozzle to keep steam temperature at set point.

But there is distance between the valve in the water line, and the temperature sensor. When the valve changes position, it takes a while before the temp. sensor notices the changed temperature.

If the valve's controller uses only proportional control (gain), control will be unsatisfactory. If the gain is too high, the valve will open (or close) too much in response to a higher (or lower) steam temperature. The result will be overshoots in steam temperature. You will alternately have either too much or too little cooling water. If the gain is too low, you won't succeed in ever reaching set point, assuming steam flow and steam temp. remain constant.

Integral control makes it much easier to control a process by maintaining the correct set point over time. The integral action is dynamic; as error (measured variable minus set point) changes, the part of the output signal due to integral action modulates smoothly (in a well-tuned process) to keep the steam at the correct temperature, in the above simple example.

I didn't answer all your questions, but perhaps this simple example works a little as a primer. Processes get much more involved than just this, but often the basics of industrial process control are as apparent as they should be.

mh

: I didn't answer all your questions, but perhaps this simple example works a : little as a primer. Processes get much more involved than just this, but : often the basics of industrial process control are as apparent as they : should be. : : mh : Correction: ...often the basics of industrial process control are NOT as apparent as they should be.

Because a first order response simulates flow control. Integrating process is a tank level, pH, conductivity, .....

First order response, X% of action yields Y% flow change

Considering a compressor knock out drum with a requirement to control level. There are 2 outlets from the drum with the top going to compressor suction while bottom liquid is sent to a column via pump. Why there is a need to know the process dynamic of a level (i.e. the integrating dynamic) for designing the level control when it can be simply manipulated with a downstream valve?

This is somewhat the process gain.

What do we call this then? Same as the above?

An integrating process is characterised by a time rather than a gain. A useful number is the hold-up time, defined as the maximum capacity of the volume or "inventory" associated with the integration divided by the maximum value of the conrolled or manipulated flow into or out of that volume.

So a tank with measured volume (0% - 100% level) of 2.5 m^3, and a maximum controlled flow into the tank of 0.05 m^3/s at 100% controller output, has a hold-up time of 2.5 / 0.05 = 2.5 x 20 = 50 s. The level in % can be found by:

Level = 1/50 x integral(inflow - outflow).dt + initial level

Since a fixed flow in gives rise to a steadily increasing or decreasing out, a characteristic must be in terms of 1/time rather than a dimensionless gain.

Bruce.

TheRomanov wrote:

From the glossary page on

Integrating Process: With these loops, making a small change in the controller ouptut, will cause the process variable to ramp until it hits a limit. The larger the change, the faster the ramp. Also the smaller the integral time the faster it will move. It is a common mis-conception that integral time in the controller is not required to hold setpoint with an integrating process. Most control loops are self-regulating. Self- regulating means that with a change in the controller output, the process variable will move and then settle. Integrating loops are also described as non-self-regulating.