For dynamics, see ISA 75.25 for the impact of imperfect stem positioning versus control signal to describe control problems caused by valve actuator dynamics.
If your concerns are simple, the valve simulation can be simple. If you are looking for subtle issues or for a difficult application, the simulation will have to be more detailed. A valve is not a simple resistor.
Simulation can be a very powerful tool. The question is always, is it valid? Can it be proven adequate for the questions asked? I know very well how to get the desired answer, but that might not represent the truth.
former chair of ISA 75.25 and 75.07
A: In Simulink simulations, I recall that there is an input function block for step inputs. Please check. (I could be wrong, since I have not used MATLAB, Simulink, Control Toolbox, Signal Processing, etc. for awhile). But having an input function block is of no use. You need to write your own simulation codes for the control valve. This means you need to know the dynamics of your valve—the time constant, the gain, the time lag, etc., of its responses to step changes. More important is the resolution of your valve; typically a diaphragm-operated valve has a resolution of 2%, while a piston-operated valve has a resolution of only 1%. I have not seen any stated resolution for step-motor actuator. (The resolution here means the changes in output fluid pressure caused by the valve for every unit change in valve stroke).
You can study the dynamics of your valve by statistics—clusters analysis, PCA (principle component analysis), DMC (dynamic matrix), etc., from historical data captured with a sampling time at least several times faster than your step motor. (I am assuming that your plant data were captured once every second for a total number of 6000 plant variables). The changes in the output (ideally the flow rate of the valve), if not, the valve stroke, versus the input (the step motor input signal and dynamics of the motor) have to be given.
There is software in the market called Process Doc, written in MATLAB codes, which can help you to pinpoint the key dependent variables as a function of the independent variable (the flow rate of the valve or the valve stroke). But again you need to provide enough statistical data, both for building a model for the valve and for model validation. You can start out by assuming a model, such as Box-Jenkins or ARMA models, etc., and work your way by including white noise, using your Model Identification Toolbox. For reference, please read and examine the codes in IDDEMO. To build a model using MATLAB and Simulink is fast, but garbage in, garbage out, if the dynamics of the valve are unknown or assumed incorrectly.
It is not impossible to build a valve model analytically. But it is very difficult to do so, and basic chemical engineering assumptions have to be used. Sometimes the viscous nature of the fluid can affect the response of your valve to an unknown extent for every change in your step motor input.
I do not believe anyone would, or could, release ready-made software that will simulate your valve for free. I have seen MATLAB + Simulink crashed in non-linear engineering applications.
For any modeling of a control valve, it has to meet the published Cv versus lift in the valve catalog provided by the specific manufacturer. Lift here means the percent opening or travel of the valve plug inside the valve trim. Such published Cv versus lift data is used to validate the valve model. For a given valve size, the Cv versus lift varies from manufacturer to manufacturer because the loss coefficients of their trims are not the same, since their trim geometries are not the same. Even among the various valve types or valve models of the same size made by the same valve manufacturer, the loss coefficient of the trims are not the same. The loss coefficient can be calculated from the geometry of the trim based on knowledge (internal flows) in chemical engineering transport phenomenon or in mechanical engineering fluid mechanics.
For example Crane Technical Publication 410 bases the flow geometry model on a circular flow geometry and suggests the following relationship:
Cv/A = 38/sq.root (k)
Where A = 3.1416d2/4 and is the effective flow area of the valve trim in inches squared,
k is the loss coefficient of the valve trim (dimensionless), and d is the equivalent flow diameter inside the valve trim.
The above equation assumes that the capacity of a control valve (Cv) is a function of the upstream resistance (i.e., loss coefficient) of the valve trim. Knowing the loss coefficient and the Cv, the effective flow area (A) can be calculated. This area is always smaller than the physical flow area of the valve trim, depending on the upstream loss coefficient, k. Cv is called the capacity of the control valve, while Cv/A is called the flow capability of the control valve because it is a function of the trim loss coefficient.
For complicated valve trim geometries, the loss coefficient can be calculated by an iterative method: First assume a numerical value for k; measure the outlet flow area of the trim or from its photograph; calculate the Cv/A at the trim outlet; then calculate the Cv/A at the trim inlet and, therefore, the Cv. If the calculated Cvs match with the published Cv at various strokes to within 5%, then the assumed loss coefficient is the k of the valve trim. Such a reversed engineering method has been published by Joe Steinke of CCI.
There are valve trims with variable loss coefficients as the valve stroke changes. Examples are those control valves used in regulating boiler feedwater flow, where the trim loss coefficient varies to take into account the boiler pump curve for optimized control—low flow rate at small valve opening with high loss coefficient; high flow rate at large valve opening with low loss coefficient.
Gerald Liu, P.Eng.
Control Valve Consultant