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ProcessControl
Laboratory 3. Mathematical Modelling 3.1 Modelling principles
3.1.1 Model types3.1.2 Model construction3.1.3 Modelling from first principles
3.2 Models for technical systems3.2.1 Electrical systems3.2.2 Mechanical systems3.2.3 Process engineering systems
3.3 Model linearization3.3.1 Motivation3.3.2 Linearization of ODEs
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3. Mathematical Modelling
3.1 Modelling principles
For design and analysis of a control system we need a mathematical model that describes the dynamical behaviour of the system. The dynamics can be described by differential equations for continuous-time dynamics difference equations for discrete-time dynamics
Most processes are time continuous, but some processes are inherently time discrete (e.g. radioactive decay) computer algorithms (i.e. controllers) and many measuring devices produce
outputs at discrete time instantsβ to design such controllers, we sometimes use discrete-time models to
describe continuous-time processes
In this course, we will consider both types of models controller design both in continuous and discrete time
However, the major part of the course deals with continuous time.
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3.1.1 Model types
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3. Mathematical Modelling
3.1.2 Model constructionThere are two main principles for construction of mathematical models Modelling from first principles: we derive models using physical laws and other
known relationships (models). System identification: we use observations (measurements) of the system to
find a model empirically. Usually, designed identification experiments are carried out to generate suitable data.
Often both methods are combined: we derive the basic model from first principles and determine uncertain parameters by system identification.
It is important to realize that all models have a limited validity range, even the physical laws (e.g. Newtonβs laws of motion do not apply close to the speed of light). It is especially important to note that models determined through system identification should not be used outside
the experimental range.
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3.1 Modelling principles
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3. Mathematical Modelling
3.1.3 Modelling from first principlesIn the following we consider modelling from first principles. Because real technical systems tend to be complex we cannot or do not want to include all details of the system in the model.
We try to make a good compromise between the following two require-ments. The model should be sufficiently accurate for its intended purpose simple enough to use e.g. for system analysis and control design
In modelling from first principles, two types of mathematical relationships are used: conservation laws constitutive relationships
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3.1 Modelling principles
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3.1 Modelling principles
Conservation lawsConservation laws apply to additive quantities of the same type in a system. There are two general kinds of conservation laws: flow balances βeffortβ balances
A flow balance for a given quantity in a system has the general formaccumulation / time unit = inflow β outflow + production / time unit
where accumulation and production occur inside the system inflow and outflow cross the system boundaries
Flow balances apply to conserved quantities (under normal conditions). If no chemical or nuclear reactions take place, the production is zero.Examples of flow balance quantities: mass particles (moles) energy current (Kirchhoffβs first law)
Note that volume is not a conserved quantity for compressive fluids.
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3.1.3 Modelling from first principles
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3.1 Modelling principles
An effort balance for a given quantity has the general formchange / time unit = forcing quantity β loading quantity
where change refers to a system property driving and loading refer to interaction with the surrounding
Generally, effort balances are applications of Newtonβs laws of motion and Kirchhoffβs second law.
Examples of effort balance quantities: force momentum angular momentum voltage (Kirchhoffβs second law)
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3.1.3 Modelling from first principles
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3.1 Modelling principles
Constitutive relationshipsConstitutive relationships are static relationships that relate quantities of different kinds in a system.
Examples of constitutive relationships: Ohmβs law: relates the current to the voltage over a resistance valve characteristics: relates the flow rate to the pressure drop over a valve Bernoulliβs law: relates the velocity of the flow out of a tank to the liquid level
in the tank the ideal gas law: relates the temperature to the pressure of a gas in a closed
container
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3.1.3 Modelling from first principles
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3.1 Modelling principles
The general modelling procedure1. Formulate balance equations.2. Introduce constitutive relationships to
β relate variables to each other;β possibly to introduce new variables in the balance equations.
3. Do a correctness check by at least checking thatβ all additive terms in an equation have the same physical unit;β the left and right hand side of an equation have the same unit.
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3.1.3 Modelling from first principles
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3. Mathematical Modelling
3.2 Models for technical systems3.2.1 Electrical systemsFig. 3.1 shows three basic components of electrical circuits.Variables π‘π‘ = time π’π’ = voltage [V] ππ = current [A]
Component parameters π π = resistance [Ξ©] πΆπΆ = capacitance [F] πΏπΏ = inductance [H]
Relationships Resistor (Ohmβs law): π’π’ π‘π‘ = π π ππ(π‘π‘) (3.1)
Capacitor: π’π’ π‘π‘ = π’π’ 0 + 1πΆπΆ β«0
π‘π‘ ππ ππ dππ (3.2)
Inductor: π’π’ π‘π‘ = πΏπΏ dππ(π‘π‘)dπ‘π‘
(3.3)
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Fig. 3.1. Basic components in electrical circuits.
resistor capacitor inductor
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3.2 Models for technical systems
Example 3.1. A passive analog low-pass filter.
Notation: π’π’R(π‘π‘) = voltage across the resistor, ππR(π‘π‘) = current through the resistor π’π’C(π‘π‘) = voltage across the capacitor, ππC(π‘π‘) = current through the capacitor
According to Kirchhoffβs second law, the voltages satisfy
π’π’in π‘π‘ = π’π’R π‘π‘ + π’π’C(π‘π‘) (1)
π’π’out π‘π‘ = π’π’C(π‘π‘) (2)
When the output is uncharged, there is no current from the filter. Thus,
ππR π‘π‘ = ππC(π‘π‘) (3)
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3.2.1 Electrical systems
Figure 3.2 shows a passive analog low-pass filter.How does the voltage π’π’out(π‘π‘) on the output side depend on the voltage π’π’in(π‘π‘) on the input side if the circuit is uncharged at the output? Fig. 3.2. A passive analog
low-pass filter.
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3.2.1 Electrical systems
Combination of (1) and (2) and elimination of π’π’R(π‘π‘) by (3.1) giveπ’π’out π‘π‘ = π’π’in π‘π‘ β π π ππR (4)
Elimination of π’π’C(π‘π‘) from (2) by (3.2) yields
π’π’out π‘π‘ = π’π’C π‘π‘ = π’π’C 0 + 1πΆπΆ β«0
π‘π‘ ππC ππ dππ (5)
The derivative of both sides of (5) with respect to time givesdπ’π’out(π‘π‘)
dπ‘π‘= 1
πΆπΆππC π‘π‘ = 1
πΆπΆππR(π‘π‘) (6)
where the last equality is given by (3). Combination of (4) and (6) to eliminate ππR(π‘π‘) gives
π π πΆπΆ dπ’π’out(π‘π‘)dπ‘π‘
+ π’π’out π‘π‘ = π’π’in(π‘π‘) (7)
This is a linear first-order differential equation. The circuit is a low-pass filterthat filters (i.e., reduces the amplitude of) of high-frequency oscillations in π’π’in(π‘π‘).
In practical applications, the output is charged, which violates the assumption of this derivation. However, when an amplifier is used on the output side, (3) still holds (approximatively).
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Example 3.1. A passive analog low-pass filter
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3.2 Models for technical systems
Example 3.2. A simple RLC circuit.Figure 3.3 shows a simple RLC circuitcharged by a current source.
How does the voltage across thecapacitor depend on the currentfrom the current source?
Notation: π’π’R(π‘π‘) = voltage across the resistor, ππR(π‘π‘) = current through the resistor π’π’C(π‘π‘) = voltage across the capacitor, ππC(π‘π‘) = current through the capacitor π’π’L(π‘π‘) = voltage across the inductor, ππL(π‘π‘) = current through the inductor
Kirchhoffβs laws giveπ’π’C π‘π‘ = π’π’R π‘π‘ + π’π’L(π‘π‘) (1)
ππ π‘π‘ = ππR π‘π‘ + ππC(π‘π‘) (2)
ππR π‘π‘ = ππL(π‘π‘) (3)
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3.2.1 Electrical systems
Fig. 3.3. A simple RLC circuit.
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3.2.1 Electrical systems
Substitution of (3.1) and (3.3) into (1):
π’π’C π‘π‘ = π π ππR π‘π‘ + πΏπΏ dππL(π‘π‘)dπ‘π‘
(4)
Elimination of ππR(π‘π‘) and ππL(π‘π‘) by (2) and (3):
π’π’C π‘π‘ = π π ππ π‘π‘ β ππC(π‘π‘) + πΏπΏ d ππ π‘π‘ βππC(π‘π‘)dπ‘π‘
(5)
According to eq. (6) in Ex. 3.1:
ππC π‘π‘ = πΆπΆ dπ’π’ππ(π‘π‘)dπ‘π‘
(6)
Substitution of (6) into (5):
π’π’C π‘π‘ = π π ππ π‘π‘ β πΆπΆ dπ’π’ππ(π‘π‘)dπ‘π‘
+ πΏπΏd ππ π‘π‘ βπΆπΆdπ’π’ππ(π‘π‘)
dπ‘π‘dπ‘π‘
(7)
After rearrangement:
πΏπΏπΆπΆ d2π’π’C(π‘π‘)dπ‘π‘2
+ π π πΆπΆ dπ’π’C(π‘π‘)dπ‘π‘
+ π’π’C π‘π‘ = π π ππ π‘π‘ + πΏπΏ dππ(π‘π‘)dπ‘π‘
(8)
where ππ(π‘π‘) is the input signal and π’π’C(π‘π‘) is the output signal.
This is a linear second-order differential equation.
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Example 3.2. A simple RLC circuit
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3. Mathematical Modelling
3.2.2 Mechanical systemsThe modelling of mechanical systems are mainly based on Newtonβs second law
πΉπΉ = ππππ (3.4)πΉπΉ is the force acting on the mass ππ and ππ is the acceleration of the mass.
Example 3.3. An undamped pendulum.Figure 3.4 shows an undamped swinging pendulum. The pendulum can only move in two directions in the plane of the figure. Its point of sus-pension is at a distance π’π’ and its center of mass (the round weight at the lower end of the pendulum) is at a distanceπ¦π¦ from the left-side vertical line.How does the position π¦π¦ depend on π’π’ ?Notation: β = length of pendulum, ππ = weight of mass β = vertical position of the center of mass ππ = angle of swing away from a vertical position πΉπΉ = force acting on the suspension point in the
βnegative directionβ (upwards)
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3.2 Models for technical systems
Fig. 3.4. Swinging pendulum.
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3.2 Models for technical systems
When the pendulum is affected by the suspension force πΉπΉ and the gravitational force ππππ, Newtonβs second law yields horizontal force components: πππ¦ = βπΉπΉ sinππ (1) vertical force components: ππβ = βπΉπΉ cos ππ + ππππ (2)
Here π¦ and β are second-order time derivatives of π¦π¦ and β, respectively, i.e. the acceleration in the respective directions.
Assume that the swing of the pendulum is moderate so that the angle ππ is always small. The pendulum then moves very little in the vertical direction and we can assume that β β 0. Elimination of πΉπΉ then gives
π¦ + ππ tanππ = 0 (3)The angle ππ is given by the trigonometric identity
tanππ = π¦π¦βπ’π’β
β π¦π¦βπ’π’ππ
(4)
Combination of (3) and (4) yields the model
π¦ + ππβπ¦π¦ = ππ
βπ’π’ (5)
Notice that the approximations β β 0 and βππ smallβ limit the validity of themodel.
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3.2.2 Mechanical systems
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3.2 Models for technical systems
Example 3.4. Suspension system in a car.
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3.2.2 Mechanical systems
Fig. 3.5. a) Spring-mounted mass with dampingb) Principle of car suspension system.
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3.2.2 Mechanical systems
a) How does the vertical deviation π¦π¦(π‘π‘) from an equilibrium point depend on a force π’π’(π‘π‘) acting on the spring-mounted mass ππ ?An equilibrium point applies when π¦π¦ = π’π’ = 0 (units omitted). If the downward direction is the positive vertical direction, Newtonβs second law for the spring force and the damping force of the cylinder gives
πππ¦ = βπππ¦ β πππ¦π¦ + π’π’ π‘π‘ , i.e., πππ¦ + πππ¦ + πππ¦π¦ = π’π’ π‘π‘ (1)where ππ and ππ are constants. The gravitational force ππππ is cancelled outwhen deviations from the equilibrium point are considered.
b) How do the deviations π¦π¦1(π‘π‘) and π¦π¦2(π‘π‘) in a car suspension system dependon π’π’(π‘π‘), which denotes the roughness of the ground?ππ1 is the mass of the car, ππ2 is the mass of the wheels and the axles, ππ1 andππ1 describe the dynamics of the car shock absorber, ππ2 denotes the elasticityof the tires. In equilibrium, π¦π¦1 = π¦π¦2 = π’π’ = 0. The model becomes
ππ1π¦1 + ππ1 π¦1 β π¦2 + ππ1 π¦π¦1 β π¦π¦2 = 0 (2)ππ2π¦2 β ππ1 π¦1 β π¦2 β ππ1 π¦π¦1 β π¦π¦2 = ππ2 π’π’ β π¦π¦2 (3)
These are two coupled 2nd order differential equations that describe the vertical position of the car body and the wheels as function of the vertical roughness of the road.
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Example 3.4. Suspension system in a car
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3. Mathematical Modelling
3.2.3 Process engineering systemsProcess engineering systems are typically modeled with flow balances (mass andenergy balances) and constitutive relationships.
Example 3.5. Liquid tank with free outflow.A volumetric flow rate π’π’ is supplied continu-ously to a container and a volumetric flow rateππ flows out freely by gravity as determined bythe height β of the liquid the container.The container has a constant cross-sectionalarea π΄π΄, and the outlet tube has the βeffectiveβcross-sectional area ππ. The liquid has a con-stant density ππ.How does the level of the liquid depend on the inflow π’π’?
Mass balance: ddπ‘π‘
πππ΄π΄β = πππ’π’ β ππππ (1)
Because ππ and π΄π΄ are constants, (1) can be simplified to
π΄π΄ dβdπ‘π‘
= π’π’ β ππ (2)
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3.2 Models for technical systems
Fig. 3.6. Liquid tank with free outflow.
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3.2.3 Process engineering systems
According to Bernoulliβs law, the following constitutive relationship applies for the outflow of a liquid:
π£π£ = 2ππβ (3)
Here π£π£ is the velocity of the outflow and ππ is the acceleration due to gravity. The volumetric flow rate ππ is then given by
ππ = πππ£π£ = ππ 2ππβ (4)
where ππ is the effective cross-sectional area of the outflow tube, which is slightly smaller than the actual cross-sectional area π΄π΄. Combination of (2) and (4) now yields
dβdπ‘π‘
+ ππ 2πππ΄π΄
β = 1π΄π΄π’π’ (5)
This is a nonlinear differential equation that describes how the level β depends on the inflow π’π’.
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Ex. 3.5. Liquid tank with free outflow
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3.2 Models for technical systems
Example 3.6. A mixing tank.Two inflows with the volumetric flow ratesπΉπΉ1 and πΉπΉ2, and concentrations (mass/volume) ππ1 and ππ2 of some componentππ, are mixed continuously in a container.The outflow has the volumetric flow rateπΉπΉ3 and concentration ππ3. The containerhas a constant cross-sectional area π΄π΄ andthe height of liquid is β. The mixing in thecontainer is assumed to be perfect so that the concentration of ππ is ππeverywhere in the container.
How do the level β and the concentration ππ depend on other variables?
If the two inflows have the same constant temperatures and βsmallβ (or not too different) concentrations of ππ , it is reasonable to assume that the liquid density in the different flows is constant and the same.
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3.2.3 Process engineering systems
Fig. 3.7. A mixing tank.
Flow 1 Flow 2
Flow 3
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3.2.3 Process engineering systems
As in Ex. 3.5, the density can be cancelled out yielding a
total mass balance: π΄π΄ dβdπ‘π‘
= πΉπΉ1 + πΉπΉ2 β πΉπΉ3 (1)
Since it is not specified what πΉπΉ3 depends on, it cannot be eliminated now.A mass balance can also be formed for each component ππ:
partial mass balance: ddπ‘π‘
π΄π΄βππ = πΉπΉ1ππ1 + πΉπΉ2ππ2 β πΉπΉ3ππ3 (2)
Because perfect mixing is assumed, the concentration is the same every-where in the container at a given time instant. This means that theconstitutive relationship: ππ3 = ππ (3)must hold. Development of the derivative in (2) according to the product rule and elimination of ππ3 by (3) yield
π΄π΄ππ dβdπ‘π‘
+ π΄π΄β dππdπ‘π‘
= πΉπΉ1ππ1 + πΉπΉ2ππ2 β πΉπΉ3ππ (4)
Combination of (4) with (1) gives
π΄π΄β dππdπ‘π‘
= πΉπΉ1 ππ1 β ππ + πΉπΉ2(ππ2βππ) (5)
This is a linear differential equation with (in general) non-constant parametersπΉπΉ1, πΉπΉ2, ππ1, ππ2.
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Example 3.6. A mixing tank
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3.2 Models for technical systems
Example 3.7. A water heater.The inflow of water to the water heaterhas the mass flow rate π1 and tempera-ture ππ1 whereas the outflow has the massflow rate π2 and temperature ππ2. Themass of water in the heater is ππ and it isheated to a temperature ππ with a heatingpower π. The mixing of water in theheater is assumed to be perfect.How do the amount of water and the temperature in the heater depend on other variables?
Mass balance: dππdπ‘π‘
= π1 β π2 (1)
Energy balance: dπΈπΈdπ‘π‘
= πΈ1 β πΈ2 + π (2)
Here, πΈ1 and πΈ2 are energy flows associated with the inflow and the outflow, respectively.
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3.2.3 Process engineering systems
Fig. 3.8. A water heater.
Flow 1
Flow 2
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3.2.3 Process engineering systems
The energy in a substance is proportional to its mass or mass flow rate. For liquids it applies with good accuracy that the energy is also proportional to itstemperature. This results in theconstitutive relationships: πΈπΈ = ππpππππ, πΈ1 = ππpππ1π1, πΈ2 = ππpππ2π2 (3)Here ππp is the specific heat capacity for water, which in this case is assumed to be constant independently of the water temperature. Combination of (2) and (3) and development of the derivative according to the product rule give
ππ dππdπ‘π‘
+ ππdππdπ‘π‘
= ππ1π1 β ππ2π2 + πππp
(4)
Because of the assumption of perfect mixing, there is also aconstitutive relationship: ππ2 = ππ (5)Elimination of βdππ dπ‘π‘ from (4) by (1) and substitution of (5) give
ππdππdπ‘π‘
= π1 ππ1 β ππ + πππp
(6)
Equation (1) and (6) show how the mass and the temperature in the heater depend on the inflow and the heating power π.
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Example 3.7. A water heater
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3.2.3 Process engineering systems
If we want to use volumetric units instead of mass units in the model, this can easily be accomplished by the substitutions
ππ = πππ΄π΄β, π1 = ππ1πΉπΉ1 (7)
which applied to (6) yield
πππ΄π΄β dππdπ‘π‘
= ππ1πΉπΉ1 ππ1 β ππ + πππp
(8)
Note that the water density is not assumed to be constant in equation (8).
Equation (1) expressed in volumetric units becomes more complicated when the water density is non-constant., i.e.,
π΄π΄ dππβdπ‘π‘
= ππ1πΉπΉ1 β ππ2πΉπΉ2 = ππ1πΉπΉ1 β πππΉπΉ2 (9)
It is possible to show that even if ππ β ππ1 due to the fact that ππ β ππ1, the effects tend to cancel out in such a way that
π΄π΄ dβdπ‘π‘β πΉπΉ1 β πΉπΉ2 (10)
becomes a good approximation of (1) and (9).
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Example 3.7. A water heater
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3.2 Models for technical systems
Example 3.8. Gas flow through a tank.Figure 3.9 illustrates a closed gastank with the volume ππ, amountof substance ππ (number of moles),pressure ππ, and temperature ππ.The inflow to the tank is the molarflow rate π1 at pressure ππ1 beforeValve 1, the outflow is the molar flow rate π2 at pressure ππ2 after Valve 2. Valve 2 can be used for control by adjustment of the valve position π’π’.
How does the pressure ππ in the tank depend on other variables?
Molar balance: dππdπ‘π‘
= π1 β π2 (1)
The molar flow rate through a valve with a given opening position can be assumed to be proportional to the square root of the pressure difference across the valve. It is also reasonable to assume that the proportionality factor is proportional to the square of the βlinearβ valve position π’π’. Thus:
Constitutive relationships: π1 = ππ1 ππ1 β ππ , π2 = ππ2π’π’2 ππ β ππ2 (2)
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3.2.3 Process engineering systems
Fig. 3.9. Gas flow through a tank.
Valve 1 Valve 2
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3.2.3 Process engineering systems
In the absence of other information, we can assume that the ideal gas law holds. Thus, we have
Constitutive relationship: ππππ = πππ π ππ (3)
Here, π π is the general gas constant and ππ is the temperature expressed in Kelvin.
If the temperature ππ is constant, substitution of (2) and (3) in (1) givesdππdπ‘π‘
= π π ππππdππdπ‘π‘
= π π ππππ
ππ1 ππ1 β ππ β ππ2π’π’2 ππ β ππ2 (4)
which is a relatively complex nonlinear differential equation, even if it is of first order.
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Example 3.8. Gas in closed tank
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3.3 Model linearization
We have in a number of examples shown how to derive dynamic models for many types of technical systems. In all cases, the obtained models are ordinary differential equations. We note that the differential equations (DEs) are often nonlinear even if they are linear, the coefficients are generally not constant because
they depend on some physical time-varying quantity it is difficult, maybe impossible, to find general solutions to these kinds of DEs
Therefore, we need to study special cases and/or do simplifying assumptions
Frequently used simplifications is to assume that some quantities are constant, even if they are (slightly) time-varying input signals change in some ideal (but reasonable) way
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3.3.1 Motivation
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3.3 Model linearization
In practice, it is often enough to know the system behaviour in a limited region close to a known operating point. Then, the model simplification may be to linearize the model equations at the operating point.
The advantage of this is that efficient analysis, synthesis, and design methods based on linear algebra can
be used.
If the system is very nonlinear, or the operating region very large, one can use several linear models that are linearized at different operating points.
Because of the reasons mentioned above, modelling from first principles is often followed by a linearization of the
model.
In this course, we are only considering models obtained from ordinary differential equations, not partial DEs.
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3.3.1 Motivation
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3. Mathematical Modelling
3.3.2 Linearization of ODEsA general ODEConsider an ππ:th order ODE, which we can formally write as
ππ π¦π¦ ππ , β¦ , π¦, π¦π¦,π’π’ = 0 (3.5)
For simplify, time derivatives of π’π’ are not included; they can be handled in the same way as the time derivatives of π¦π¦.
Usually the time derivatives appear linearly in (3.5); however, the linearization applies also when they do not appear linearly.
Equation (3.5) can be linearized by a first-order Taylor series expansion at the nominal operating point π¦π¦ ππ , β¦ , π¦π¦, π¦π¦, π’π’ , denoted by ππ:
ππ π¦π¦ ππ , β¦ , π¦, π¦π¦,π’π’ = ππ π¦π¦ ππ , β¦ , π¦π¦, π¦π¦, π’π’ + πππππππ¦π¦(ππ) ππ
π¦π¦ ππ β π¦π¦ ππ +
β―+ πππππππ¦ ππ
π¦ β π¦π¦ + πππππππ¦π¦ ππ
π¦π¦ β π¦π¦ + πππππππ’π’ ππ
π’π’ β π’π’ (3.6)
Usually the operating point is a static (steady-state) point with all time derivatives equal to zero, but (3.6) holds even if this is not so.
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3.3 Model linearization
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3.3 Model linearization
We introduce the variables
βπ¦π¦(ππ) β‘ π¦π¦ ππ β π¦π¦ ππ , β¦ , βπ¦ β‘ π¦ β π¦π¦, βπ¦π¦ β‘ π¦π¦ β π¦π¦, βπ’π’ β‘ π’π’ β π’π’ (3.7)
which denote deviations from the nominal operating point. We call such variables deviation variables, or simply, β-variables.
Combination of (3.5), (3.6) and (3.7) with the fact that the nominal operating point has to satisfy (3.5) yields
πππππππ¦π¦(ππ) ππ
βπ¦π¦(ππ) + β―+ πππππππ¦ ππ
βπ¦ + πππππππ¦π¦ ππ
βπ¦π¦ + πππππππ’π’ ππ
βπ’π’ = 0 (3.8)
This is a linear ππ:th order ODE with constant coefficients.
Note: If the ODE contains time derivatives of π’π’, they appear as β-variables in (3.8) in the same way as the time derivatives of π¦π¦ appear.
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3.3.2 Linearization of ODEs
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3.3 Model linearization
ODEs linear in the time derivativesIf the time derivatives appear linearly in (3.5), we can formally write
ππππ π¦π¦,π’π’ π¦π¦ ππ + β―+ ππ1 π¦π¦,π’π’ π¦ + ππ0 π¦π¦,π’π’ = 0 (3.9)
Equation (3.6) can now be used to linearize every term separately giving
ππ0 π¦π¦,π’π’ = ππ0 π¦π¦, π’π’ + ππππ0πππ¦π¦ ππ0
βπ¦π¦ + ππππ0πππ’π’ ππ0
βπ’π’
The term with the ππ:th derivative results in
ππππ π¦π¦,π’π’ π¦π¦ ππ = ππππ π¦π¦, π’π’ π¦π¦ ππ + ππππ π¦π¦, π’π’ βπ¦π¦ ππ + πππππππππ¦π¦ ππππ
π¦π¦ ππ βπ¦π¦ + πππππππππ’π’ ππππ
π¦π¦ ππ βπ’π’
Substitution into (3.9) and cancelling out constant terms give
ππππ π¦π¦, π’π’ βπ¦π¦ ππ + β―+ ππ1 π¦π¦, π’π’ βπ¦ + ππππ0πππ¦π¦ ππ0
βπ¦π¦ + ππππ0πππ’π’ ππ0
βπ’π’ = β ππ (3.10)
where β ππ = ββππ=1ππ π¦π¦ ππ πππππππππ¦π¦ ππππ
βπ¦π¦ + πππππππππ’π’ ππππ
βπ’π’ (3.11)
Note that β ππ = 0 if the operating point is a steady-state with all π¦π¦ ππ = 0.
KEH Process Dynamics and Control 3β31
3.3.2 Linearization of ODEs
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Laboratory
3.3 Model linearization
Constitutive relationshipsNonlinear constitutive relationships also need to be linearized. Such a relationship can be formally written
ππ π§π§, π¦π¦,π’π’ = 0 (3.12)
where π§π§ is a new variable that is related to π¦π¦ and π’π’ according to (3.12). Linearization using a first-order Taylor series expansion as in (3.6) gives
πππππππ§π§ ππ
βπ§π§ + πππππππ¦π¦ ππ
βπ¦π¦ + πππππππ’π’ ππ
βπ’π’ = 0 (3.13)
If the nominal operating point is a steady state with all time derivatives zero, differentiation with respect to time gives for the ππ:th time derivative
πππππππ§π§ ππ
βπ§π§(ππ) + πππππππ¦π¦ ππ
βπ¦π¦(ππ) + πππππππ’π’ ππ
βπ’π’(ππ) = 0 (3.14)
If desired, the variable βπ§π§ can be introduced as dependent variable instead of βπ¦π¦in (3.8) or (3.10) by means of (3.13) and (3.14).
KEH Process Dynamics and Control 3β32
3.3.2 Linearization of ODEs
ProcessControl
Laboratory
3.3 Model linearization
Example 3.9. Linearization of a first-order DE.In Example 3.5 we derived the nonlinear DE
π΄π΄β + ππ 2ππβ β π’π’ = 0 (1)
We want to linearize this DE at the steady-state operating point β, π’π’ . Application of (3.10) gives
π΄π΄ββ + ππππβ
ππ 2ππβ β π’π’β,π’π’
ββ + πππππ’π’
ππ 2ππβ β π’π’β,π’π’
βπ’π’ = 0
βΉ π΄π΄ββ + ππ 2ππ d βdβ β
ββ β dπ’π’dπ’π’ π’π’
βπ’π’ = 0 (2)
which gives
π΄π΄ββ + ππ 2ππ
2 βββ = βπ’π’ (3)
KEH Process Dynamics and Control 3β33
3.3.2 Linearization of ODEs
ProcessControl
Laboratory
3.3 Model linearization
Exercise 3.1. Linearization of constitutive relationship.Consider a control valve ina pipeline, schematicallyillustrated in the figure.At a given pressure, theflow rate ππ through thevalve depends on the position π₯π₯ of the valve plug (stem) according to the valve characteristic
ππ = πΆπΆ(πΌπΌπ₯π₯ β 1)/(πΌπΌ β 1)where πΆπΆ and πΌπΌ are constant parameters depending on the size and construction of the valve. The valve is closed when π₯π₯ = 0 and fully open when π₯π₯ = 1.The position π₯π₯ is adjusted by an actuator responding to a control signal π’π’. Because of inertia, π₯π₯ follows π’π’ according to the dynamic relationship
πππ₯ + π₯π₯ = πΎπΎπ’π’where ππ and πΎπΎ are constant parameters (time constant and static gain).
Derive a linear dynamic model that shows how ππ depends on π’π’ close to an operating point ππ = ππ.
KEH Process Dynamics and Control 3β34
3.3.2 Linearization of ODEs