Preliminary Automatics Course

Chapter 17.
We will analyse the tanks filling process:
-tank without hole
-tank with hole-simplified analysis
-tank with small hole-simplified analysis
-tank with hole-more accurate analysis
The differential equations will occure here. It isn’t a easy subject for beginners but try to fight despite it. But if you don’t understand  , then take it easy and go to the chapter 19. It’s possible to carry on with the next chapters.

Chapter 17.2 The tank without hole filling

Fig. 17-1
Flow Q=1litre/sec it’s 1 water litre in 1 sec.  The tank was 1 water litre “enriched” in this time too!  It’s easy to calculate the water volume V(t) after the time t. When the flow Q is steady of course. We don’t manipulate the valve during the filling process!

Fig. 17-2
Fig. 17-2a The flow formula when Q is steady. We don’t manipulate the valve.
Fig. 17-2b The flow simplified formula when Q isn’t steady. ΔV is the water volume during small Δt time. The smaller is the Δt time–>the more accurate is the Q flow formula.
Fig. 17-2c The accurate flow formula.  Δt aims to 0. The Fig. 17-2b changes to Fig. 17-2c. Flow Q(t) is a  derivative of the volume V(t) and it may be variable in the time now.
Fig. 17-2d The accurate flow formula. Different derivative symbol.
Fig. 17-2c or Fig. 17-2d are the simplest differential equations.

Fig. 17-3
Fig. 17-3a is a differential equation where:
– Flow Q(t) is the function we know
– Volume V(t) is the function we don’t know. Other words – the solution of the V'(t)=Q(t) differential equation.
Fig. 17-3b We integrate the both sides of the equation and the result is Fig. 17-3c the differential equation solution.
By the way. You solved the differential equation in the grammar school in the problems type “The train is going from town A to town B with the speed V, distance is S…” The formula S=V*t was used here. But this is the solution of the differential equation S‘(t)=V where velocity V is constant!
But lets return to our tank without hole.

Fig. 17-4
The tank is a cuboid where  the basis is S=1m2. We open in the time t=0 the valve. The h'(t)=1.25(t) is a version of the general differential equation V'(t)=Q(t). The Fig. 17-4 confirms it.
Let’s integrate both sides of the h'(t)=1.25(t) 

Fig. 17-5
Fig. 17-5a-differential equation of the tank filling – level h(t) is a output
Fig. 17-5b-both sides integration to solve it
Fig. 17-5c
-both sides integration effect-the solution
Fig. 17-5d
-flow step 1.25(t) definition
Fig. 17-5e
-level “ramp” h(t) definition
We solved the differential equation. The result- “ramp” is obviously. The level h(t) increases with the steady speed.
Let’s solve this problem by Xcos.
Call Desktop/PID/04_rownania_rozniczkowe/01_zbiornik_skok.zcos

Fig. 17-6
See Fig. 16-14 chapter 16. The output y(t) is an integral of the input x(t) and vice versa–>the input x(t) is a derivative of the output y(t).
x(t)=y'(t) e.i. 1.25(t)=h'(t)
Click “Start”

Fig. 17-7
Experiment confirms the differential equation solution from the Fig. 17-5. The h(t) level increases when the valve is open.

Chapter 17.3 The tank with hole filling k=1

Fig. 17-8
The accretion speed of the level h(t) will decreases, because there is “hole” outflow Q2(t). The bigger is h(t) level–>The bigger is outflow Q2(t)=k*h(t). What is k? It’s “hole” coefficient. The bigger is hole, the bigger is k. No hole–>k=0–>Fig. 17-4.
Assume k=1 see Fig. 17-8
1.25(t) = h'(t) + h(t)
The differential equation 1.25(t) = h'(t) + h(t) isn’t trivial now as previously. We can’t use “both side integration” method now. There are of course the methods  but we will use our old friend Xcos to solve it!
Call Desktop/PID/04_rownania_rozniczkowe/02_zbiornik_z dziura_skok.zcos
Fig. 17-9
The input x(t) of the integration unit 1/s is a derivative of its output y(t). It isn’t important what’s this input x(t). Single signal, 2 signals sum or difference or  even a product! The cinclusion is. Integration unit is a good tool for differential equations solving.
Click “Start”

Fig. 17-10
There’s seemingly as inertial unit. It’s true! The steady level h=1.25 mm was after 7 sec.
Let’s double minimize   the hole up to k=0.5. Will be the steady level h=2.5 mm?

Chapter 17.4 The tank with small hole filling k=0.5

Fig. 17-11
Call Desktop/PID/04_rownania_rozniczkowe/03_zbiornik_z_mala_ dziura_skok.zcos
Fig. 17-12
Call “Start” –> the differential equation will be solved.

Fig. 17-13
Steady level is h=2.5mm now. But the time constant T is doubled! The tank is more lazy.
The outflow Q2(t)=k*h(t). It’s proportional to the level h(t). But it’s the simplfied model!
The outflow Q2(t) is proportional to the root of the level h(t) as in the next subchapter.

Chapter 17.5 The tank with  hole filling k=1. More accurate model.

Fig. 17-14
It’s an example of the nonliner diferential equations. They are more difficult to solve. But it’s no problem for Xcos.
Call/PID/04_rownania_rozniczkowe/04_zbiornik_z_ dziura_real_skok.zcos

Figs. 17-15
Click “Start”.

Fig. 17-16
Looks like an inertial unit. But it isn’t!

Chapter 17.6 More about differential equations
Tank without hole is the  simplest differential equation example 1.25(t)=h'(t)
Tank with hole is the  more complicacated differential equation example  1.25(t)= h'(t)+h(t).
We can generalise last example as:
x(t) = a1*y'(t) + a0*y(t) when:
x(t) – input signal
x(t) – ouput signal
a0, a1 constant coefficients
We can generalise more for higher derivatives –3 degree here:
x(t) = a3*y”'(t) + a2*y”(t) +a1*y'(t) +a0*y(t)
The derivatives may be on the left equations side too!
b3*x”'(t) + b2*x”(t) + b1*x'(t) + b0*x(t)= a3*y”'(t) + a2*y”(t) + a1*y'(t) + a0*y(t)
There were the examples of the linerar equations examples. All coefficents were constant. Control theory uses mainly  linerar equations. It’s simplification of course! But it works.
The nonlinerar equation example was more acurate tank without hole model Chapter 17.5.
The next chapter – Operational calculus is a comfortable tool for linerar differential equations.

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