Peter Nachtwey
Member
The purpose of this thread is to use a simple system to show how one can approach systems that are much more complicated. Everyone knows, hopefully, that tank level control can be accomplished with just a proportional gain and or even an simple on off system using limit switches but that isn't the point. I intend this to be a college level 'class' and one doesn't waste good time and money on learning how to kludge a control system together. I feel a college level course should go deeper so excuse me if this seem like over kill but like I said, this is not simply about controlling the level of fluid in a tank but truly understanding what is really happening.
Getting started
If one wants to truly control the level and not use some on-off limits witch then this equation applies
in flow=out flow
If the the in flow is not equal to the out flow then the level will change
d level/dt = ( in flow-out flow ) / area
where the area is the surface area. Right now lets assume the surface are is a constant 1/meter^2. In reality the area may change if the tank is a horizontal cylinder, cone or a sphere so then the more general formula is
d level/dt = ( in flow-out flow ) / area(level)
If the surface area changes as a function of the level then 'complications' are introduce but I will show what to do about his later.
The d level/dt is the rate of change in the level. It should make sense that the greater the flow mismatch the faster the level will change and the greater the surface area the slower the level will change.
The above equations are the first differential equations. My first challenge is to expand the equation for in flow and out flow. There are two cases. In one case the pump controls the out flow and in the other case the pump controls the in flow. In hydraulics Q is used to represent flow so Qin is in flow and Qout is out flow. Just so everyone is using the same terminology lets assume the tank has a volume of 1 cubic meter with a constants surface area of 1 meter squared. Flow is in cubic meter/min. In the case where the pump is controlling the out flow assuming the in flow is Qin(t) which is some function of time we don't have control over but must be in the equation. Do your best, the equations may still seem very general and not good enough for real use but it will be a start.
I helped a college student with a problem like this on LinkedIn about 2 years ago. The difference is that there were two cascaded tanks. I told him I would not give him a passing grade unless he could write the differential equation for his system. He couldn't do it without some help. He still got an A grade from his instructor.
“When you can measure what you are speaking about, and express it in numbers, you know something about it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely, in your thoughts advanced to the stage of science.”
― William Thomson
aka Lord Kelvin
Getting started
If one wants to truly control the level and not use some on-off limits witch then this equation applies
in flow=out flow
If the the in flow is not equal to the out flow then the level will change
d level/dt = ( in flow-out flow ) / area
where the area is the surface area. Right now lets assume the surface are is a constant 1/meter^2. In reality the area may change if the tank is a horizontal cylinder, cone or a sphere so then the more general formula is
d level/dt = ( in flow-out flow ) / area(level)
If the surface area changes as a function of the level then 'complications' are introduce but I will show what to do about his later.
The d level/dt is the rate of change in the level. It should make sense that the greater the flow mismatch the faster the level will change and the greater the surface area the slower the level will change.
The above equations are the first differential equations. My first challenge is to expand the equation for in flow and out flow. There are two cases. In one case the pump controls the out flow and in the other case the pump controls the in flow. In hydraulics Q is used to represent flow so Qin is in flow and Qout is out flow. Just so everyone is using the same terminology lets assume the tank has a volume of 1 cubic meter with a constants surface area of 1 meter squared. Flow is in cubic meter/min. In the case where the pump is controlling the out flow assuming the in flow is Qin(t) which is some function of time we don't have control over but must be in the equation. Do your best, the equations may still seem very general and not good enough for real use but it will be a start.
I helped a college student with a problem like this on LinkedIn about 2 years ago. The difference is that there were two cascaded tanks. I told him I would not give him a passing grade unless he could write the differential equation for his system. He couldn't do it without some help. He still got an A grade from his instructor.
“When you can measure what you are speaking about, and express it in numbers, you know something about it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely, in your thoughts advanced to the stage of science.”
― William Thomson
aka Lord Kelvin