The first model for the tank that we developed was:
Let's set the cross-section area of the tank (A) to 10 sq metres and the initial value for the level in the tank to 1 sq metre. Also let's set the initial values of the two flows to 1 cubic metre/minute (it is important that all control analysis starts from steady-state - if it doesn't then the system will show a dynamic response prior to any input changes being made).
Now let's see what happens when the inlet flowrate increases by 0.5 m3.min-1. Have a go at solving the problem yourself before looking at the solution below:
The VisSim solution to the problem is here.
The analytical and numerical solutions in this case give identical results. The reason for this is that the process is linear and no approximation was necessary in forming the analytical solution. The response of the tank is for the level to grow linearly with time (until it overflows - but we haven't modelled that). This makes sense - if you increase the flow in, but don't change the outlet flow then the level should rise.
We shall now look at the alternative model for the tank, when the tank discharge is free draining and dependant on the level in the tank, i.e.
This model contains a non-linearity (the square root term on the level) which needs to be linearised. Before linearisation remember to convert everything to deviation variables. Have a go at the linearisation, but don't solve the equation just yet. The solution is given below:
Prior to solving this problem we need to sort out the value of the constant terms. We know that the initial value of the level is 1m - we'll take this to be the nominal steady-state value. For the system to be at steady-state the volumetric flow (in this case - because we have assumed constant and equal densities!) out has to equal the flow in. The only way this can happen is if the discharge coefficient has a value of one, so the linearised model with the problem related parameters becomes:
You can see that I have rearranged the equation into the standard form for a linear first-order differential equation. We already know the standard solution to this equation and can write down the response we expect to see when the flow in changes by a step of 0.5 cubic metres/min:
As t tends to infinity the level in the tank will tend to increase by 1m. For practical purposes the response should be complete after 100 minutes (5 time constants).
Look at the VisSim solution to this problem here.
In this case you'll see that there is quite a bit of difference between the analytical, linearised, solution and the results of the dynamic simulation. The 'real' dynamics in the simulation finish with the tank level greater than 2.2m (a change of 1.2m) and the response hasn't completed in 150 minutes. The reason for this difference is that the linearised solution is only approximately correct - the further away from the point of linearisation the worse the error becomes (you can try the simulation with much smaller steps in flow - you should find that it and the analytical solution now correspond more closely). If the process was linearised around a nominal flowrate of 1.5 cu metres/min (the flow after the step) and a level of 2.2m, then the time constant would be approximately 29 minutes (c.f. 100 minutes) and the steady-state gain 2.9 (c.f. 2) - always remember that linearised solutions are approximations!