**Chapter.17.1 Introduction
**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 **Q **flow formula when **Q** is steady. We don’t manipulate the valve.

**Fig. 17-2b **The **Q **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.**

**Q****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.**

**Q****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

**-both sides integration effect-the solution**

Fig. 17-5c

Fig. 17-5c

**-flow step**

Fig. 17-5d

Fig. 17-5d

**1.25(t)**definition

**-level**

Fig. 17-5e

Fig. 17-5e

**“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)**

generally

**x(t)=y'(t)+y(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**

**1.25(t)=h'(t)+0.5*h(t)**

generally

**x(t)=y'(t)+0.5*y(t)**

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.

**Note**

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**.

Thanks for the excellent advice, it really is useful.