engg4420 ‐‐ lecture 5

SECTION 1.2. DYNAMIC MODELSA dynamic model is a mathematical description of the process to be controlled. Specifically, a set of differential equations that describe the dynamic behaviour of the process.

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By using principles of the underlying physics;○By testing a prototype of the device, measuring its response to inputs, and using the data to construct an analytical model ‐‐ called system identification method.

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Obtaining the dynamic model: •

HEAT AND FLUID‐FLOW MODELSFor the purpose of generating dynamic models for use in control systems, the most important aspect of the physics is to represent the dynamic interaction between variables.

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Experiments are usually required to determine the actual values of the parameters and thus to complete the dynamic model for purpose of control systems design.

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ENGG4420 ‐‐ LECTURE 5September‐16‐106:47 PM

CHAPTER 1 BY RADU MURESAN

HEAT FLOW EQUATIONSMany control systems involve the regulation of temperature ‐‐ the dynamic model of temperature control systems involve the flow and storage of heat energy.

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Heat energy flows through substances at a rate proportional to the temperature difference across the substance:

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q = heat energy flow, joules per second [J/sec];○R = thermal resistance, [oC/J*sec];○T = temperature, [oC].○

Where:•

The net heat‐energy flow into a substance affects the temperature of the substance according to the relation:

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Where C is the thermal capacity.Typically, there are several paths for heat to flow into or out of a substance and qnet is the sum of heat flows obeying Eq. (1).

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EXAMPLE: Equations for Heat FlowA room with all but two sides insulated (1/R = 0) is shown in figure below. Find the differential equations that determine the temperature in the room.

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SOLUTION. Application of Eq. (1) and (2) yields:

To = temperature outside.○TI = temperature inside.○R2 = thermal resistance of the room ceiling.○R1 = thermal resistance of the room wall.○

Where: Ci = thermal capacity of air within the room


SPECIFIC HEAT, THERMAL CONDUCTIVITY

Specific heat cv at constant volume that is converted to heat capacity by:

a.

Thermal conductivity k which is related to thermal resistance R by:

b.

Normally the material properties are given in tables as indicated below:

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Where A is the cross sectional area and l is the length of the heat‐flow path.In addition to flow due to transfer as described by Eq. (1), heat can also flow due to warmer mass flowing into a cooler mass, or vice versa:

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Where w is the mass flow rate of the fluid at T1 flowing into the reservoir at T2.


EXAMPLE: MODELING THE PT326 PROCESS TRAINER

The PT326 apparatus models common industrial situations in which temperature control is required in the presence of transport delay and transfer lag.

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Air drawn from the atmosphere by a centrifugal blower.○Air is heated as it passes over a heater grid.○Air is released into the atmosphere through a duct.○

Functionality:•

Maintain the temperature of the air at a desired level.○Temperature control is achieved by varying the electrical power supplied to the heater grid.

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The air temperature may be sensed by using a bead thermistor placed in the flow at any of the three positions along the duct.

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The spatial separation between the thermistor and the heater coil introduces a transport delay into the system.

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Control Objectives:•

The specific functionality features and settings are presented in the ENGG4420 Lab Manual.

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PROBLEM: 1) Develop a dynamic model for the PT326 process trainer; 2) derive the transfer function of the process trainer model.

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Figure below shows the front panel of the PT326 apparatus. See the ENGG4420 Lab Manual for the description of the apparatus.

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SOLUTION: The physical principle that governs the behaviour of the thermal process in PT326 apparatus is the balance of heat energy.

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PT326 SYSTEM MODELFigure below shows a simplified graphical picture of the heat transfer process that takes place in PT326 ‐‐ the volume V around the heater and the heat transfer rates are shown.

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qa = q + (qi ‐ qo) ‐ qt (7)

The rate at which heat accumulates in a fixed volume V enclosing the heater is:

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q is the rate at which heat is supplied by the heater; ○qi is the rate at which heat is carried into the volume V by the coming air;

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qo is the rate at which heat is carried out of the volume V by the outgoing air; and

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qt is the heat lost from the volume V to the surroundings by radiation and conduction.

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Where•


In deriving the model equation for the PT326 apparatus we assume instantaneous heat exchangebetween the electric heater Re and the air carried into the volume V.

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The accumulation of heat in the volume V causes the temperature T of air in V to rise ‐‐ assuming a uniform temperature distribution in the volume V, the rate of heat accumulation is also given based on Eq. (2) as:

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Where C is the heat capacity of the air occupying the volume V.

Assume instant heat exchange between the electrical resistor Re and the air flowing in volume V

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Assume that all air coming into volume V leaves volume V instantly.

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Based on the assumption below:•

As a result, Eq. (5) becomes: ○ qa = q ‐ qt (9)

We can conclude that the heat transfer due to Eq. (6) is zero so, qi ‐ qo = 0; and the only heat accumulated in volume V is due to heat transferred from Re and the heat lost:

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ASSUMING a small rise in temperature in volume V that is ΔT = T ‐ Ta, then the rate qt at which heat is lost from the volume V is proportional to the temperature rise ΔT:

As a result, Eq. (9) for small temperature rise ΔT becomes:

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In Eq. (11) term 1 is based on Eq. (2), term 2 is based on Eq. (1) and q represents the heat generated by the resistors Re.

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Taking Laplace transform for Eq. (11) we get:•

Where, k1 = R and τ = RC is the time constant. Note that here R and C relate to thermal resistance and heat capacity, respectively.

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Assuming that the heater supply rate is proportional to the heater input voltage Vi, Eq.(12) yields the transfer function between the temperature rise and the heater input voltage as:

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Where, k2 is the proportionality constant between q and Vi. In Eq. (13), ΔT represents the increase in temperature of the air in the volume V. The temperature sensor produces a voltage Vo that is proportional to ΔT, that is Vo = k3ΔT.

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The sensor is physically located at a certain distance from the heat source and the sensor output responds to a temperature change with a pure time delay τd.

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The transfer function between the sensor output voltage and the heater input voltage is:

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BLOCK DIAGRAM OF the PT326 apparatus based on Eq. (14).

Where, k = k1k2k3 and is the DC gain of the system. •The e(‐τds) term in Eq. (14) arises due to fluid transport and is called a 'transport delay',

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while term (τs + 1)‐1 arises due to the heat transfer dynamics and is called 'transfer delay'.

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Note the individual transfer block components in the diagram above.

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SYSTEM STEP RESPONSEThe output of the temperature sensor Vo and the heater input voltage Vi are related by the first‐order transfer function given by Eq. (14) for small temperature changes from the ambient.

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The transfer function in Eq. (14) is characterized by two parameters, namely, the DC gain k and the time constant τ. Both of these parameters can be determined from the response of the temperature to a step increase in the heater input voltage from a state of thermal equilibrium.

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Laplace transform of a unit step input is 1/s.•Laplace transform of the response of the temperature variation ΔVo to an increase of 1V in the heater input is:

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DYNAMICS OF MECHANICAL SYSTEMSThe equation of motion of Newton's law is basic for obtaining a mathematical model for any mechanical system.

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F = the sum of all forces applied to a body [N];a = inertial acceleration [m/sec2];m = mass of the body [kg].

Where,

Define convenient coordinates to account for the body's motion (position, velocity and acceleration);

a.

Determine the forces on the body using the free‐body diagram;

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Write the equations of motion. c.

Application of Newton's law involves:•

Newton's law applied to one‐dimensional rotational system requires that the above equation be modified to:

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M = the sum of all external moments [Nm];I = the mass moment of inertia [kg*m2];α = angular acceleration [rad/sec2].

Where,


EXAMPLE: CRUISE CONTROL MODELWrite the equations of motion for the speed and forward motion of a car assuming that the engine develops a force u. Take the Laplace transform of the resulting differential equation and find the transfer function between the input u (force) and output v (speed).

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Use MATLAB to find the response of the velocity of the car for the case in which the input jumps from being u = 0 N at time t = 0 to a constant u = 500 N. Assume that the car mass is m = 1000 kg and b = 50 N*sec/m.

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SOLUTIONWe make the following assumptions:

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Rotational inertia of the wheels is negligible;

1.

The friction opposing the motion of the car is proportional to speed v.

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The car can be approximated for modeling purposes by a free body diagram ‐‐ the coordinate of the car's position x is the distance from the reference and is chosen so that positive is to the right.

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FREE BODY DIAGRAM FOR CRUISE CONTROL

In the case of the automotive cruise control the variable of the interest is the speed, v ( ), and the equation of motion becomes:

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Eq. (2) is a first order differential equation in v.•


TIME RESPONSE USING MATLABMATLAB can be used to plot the response using the transfer function of the system.

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The step function in MATLAB calculates the time response of a linear system to a unit step input

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Because the system is linear, the output for this case can be multiplied by the magnitude of the input step to derive a step response of any magnitude. Equivalently numerator can be multiplied by the magnitude of the input step.

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SOLVE EQ. (2):•


Eq. (5) ‐‐ is called "Transfer Function". In order to get the transfer function we substituted d/dt in Eq. (2) with s ‐‐ this is a general rule to obtain the transfer function from a differential equation.

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TIME RESPONSE ‐‐ in order to obtain the time response using MATLAB we divide the transfer function into:Numerator: num = 1/m = 1/1000

Denominator: den = [1 b/m] = [1 50/1000]


MATLAB PROGRAM EXAMPLE

% program to find the time response for the cruise % control system using the transfer function of Eq. (5)>> num = 1/1000; % b/m>> den = [1 50/1000]; % s + b/m>> sys = tf(num*500, den); % step gives unit step response, % so num*500 give u = 500 N>> step(sys); % plots the response>> end


engg4420 ‐‐ lecture 5

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