SECTION 1.2. DYNAMIC MODELS

Transcription

1 CHAPTER 1 BY RADU MURESAN Page 1 ENGG4420 LECTURE 5 September :47 PM SECTION 1.2. DYNAMIC MODELS A 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. Obtaining the dynamic model: 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. HEAT AND FLUID FLOW MODELS For 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. 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.

2 CHAPTER 1 BY RADU MURESAN Page 2 HEAT FLOW EQUATIONS Many control systems involve the regulation of temperature the dynamic model of temperature control systems involve the flow and storage of heat energy. Heat energy flows through substances at a rate proportional to the temperature difference across the substance: Where: q = heat energy flow, joules per second [J/sec]; R = thermal resistance, [ o C/J*sec]; T = temperature, [ o C]. The net heat energy flow into a substance affects the temperature of the substance according to the relation: Where C is the thermal capacity. Typically, there are several paths for heat to flow into or out of a substance and q net is the sum of heat flows obeying Eq. (1).

3 CHAPTER 1 BY RADU MURESAN Page 3 EXAMPLE: Equations for Heat Flow A 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. SOLUTION. Application of Eq. (1) and (2) yields: Where: C i = thermal capacity of air within the room T o = temperature outside. T I = temperature inside. R 2 = thermal resistance of the room ceiling. R 1 = thermal resistance of the room wall.

4 CHAPTER 1 BY RADU MURESAN Page 4 SPECIFIC HEAT, THERMAL CONDUCTIVITY Normally the material properties are given in tables as indicated below: a. Specific heat c v at constant volume that is converted to heat capacity by: b. Thermal conductivity k which is related to thermal resistance R by: 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: Where w is the mass flow rate of the fluid at T1 flowing into the reservoir at T2.

5 CHAPTER 1 BY RADU MURESAN Page 5 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. Functionality: 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. Control Objectives: Maintain the temperature of the air at a desired level. Temperature control is achieved by varying the electrical power supplied to the heater grid. The air temperature may be sensed by using a bead thermistor placed in the flow at any of the three positions along the duct. The spatial separation between the thermistor and the heater coil introduces a transport delay into the system. The specific functionality features and settings are presented in the ENGG4420 Lab Manual.

6 CHAPTER 1 BY RADU MURESAN Page 6 PROBLEM: 1) Develop a dynamic model for the PT326 process trainer; 2) derive the transfer function of the process trainer model. Figure below shows the front panel of the PT326 apparatus. See the ENGG4420 Lab Manual for the description of the apparatus. SOLUTION: The physical principle that governs the behaviour of the thermal process in PT326 apparatus is the balance of heat energy.

7 CHAPTER 1 BY RADU MURESAN Page 7 PT326 SYSTEM MODEL Figure 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. The rate at which heat accumulates in a fixed volume V enclosing the heater is: q a = q + (q i q o ) q t (7) Where q is the rate at which heat is supplied by the heater; q i is the rate at which heat is carried into the volume V by the coming air; q o is the rate at which heat is carried out of the volume V by the outgoing air; and q t is the heat lost from the volume V to the surroundings by radiation and conduction.

8 CHAPTER 1 BY RADU MURESAN Page 8 In deriving the model equation for the PT326 apparatus we assume instantaneous heat exchange between the electric heater Re and the air carried into the volume V. 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: Where C is the heat capacity of the air occupying the volume V. Based on the assumption below: Assume instant heat exchange between the electrical resistor Re and the air flowing in volume V Assume that all air coming into volume V leaves volume V instantly. We can conclude that the heat transfer due to Eq. (6) is zero so, q i q o = 0; and the only heat accumulated in volume V is due to heat transferred from Re and the heat lost: As a result, Eq. (5) becomes: q a = q q t (9)

9 CHAPTER 1 BY RADU MURESAN Page 9 ASSUMING a small rise in temperature in volume V that is ΔT = T Ta, then the rate q t 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: 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. Taking Laplace transform for Eq. (11) we get: Where, k 1 = R and τ = RC is the time constant. Note that here R and C relate to thermal resistance and heat capacity, respectively.

10 CHAPTER 1 BY RADU MURESAN Page 10 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: Where, k 2 is the proportionality constant between q and V i. In Eq. (13), ΔT represents the increase in temperature of the air in the volume V. The temperature sensor produces a voltage V o that is proportional to ΔT, that is V o = k 3 ΔT. 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 τ d. The transfer function between the sensor output voltage and the heater input voltage is:

11 CHAPTER 1 BY RADU MURESAN Page 11 BLOCK DIAGRAM OF the PT326 apparatus based on Eq. (14). Where, k = k 1 k 2 k 3 and is the DC gain of the system. The e ( τ ds) term in Eq. (14) arises due to fluid transport and is called a 'transport delay', while term (τs + 1) 1 arises due to the heat transfer dynamics and is called 'transfer delay'. Note the individual transfer block components in the diagram above.

12 CHAPTER 1 BY RADU MURESAN Page 12 SYSTEM STEP RESPONSE The output of the temperature sensor V o and the heater input voltage V i are related by the first order transfer function given by Eq. (14) for small temperature changes from the ambient. 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. Laplace transform of a unit step input is 1/s. Laplace transform of the response of the temperature variation ΔV o to an increase of 1V in the heater input is:

13 CHAPTER 1 BY RADU MURESAN Page 13 DYNAMICS OF MECHANICAL SYSTEMS The equation of motion of Newton's law is basic for obtaining a mathematical model for any mechanical system. Where, F = the sum of all forces applied to a body [N]; a = inertial acceleration [m/sec 2 ]; m = mass of the body [kg]. Application of Newton's law involves: a. Define convenient coordinates to account for the body's motion (position, velocity and acceleration); b. Determine the forces on the body using the freebody diagram; c. Write the equations of motion. Newton's law applied to one dimensional rotational system requires that the above equation be modified to: Where, M = the sum of all external moments [Nm]; I = the mass moment of inertia [kg*m 2 ]; α = angular acceleration [rad/sec 2 ].

14 CHAPTER 1 BY RADU MURESAN Page 14 EXAMPLE: CRUISE CONTROL MODEL Write 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). 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. SOLUTION We make the following assumptions: 1. Rotational inertia of the wheels is negligible; 2. The friction opposing the motion of the car is proportional to speed v. 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.

15 CHAPTER 1 BY RADU MURESAN Page 15 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: Eq. (2) is a first order differential equation in v.

16 CHAPTER 1 BY RADU MURESAN Page 16 TIME RESPONSE USING MATLAB MATLAB can be used to plot the response using the transfer function of the system. The step function in MATLAB calculates the time response of a linear system to a unit step input 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. SOLVE EQ. (2):

17 CHAPTER 1 BY RADU MURESAN Page 17 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. 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]

18 CHAPTER 1 BY RADU MURESAN Page 18 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