interacting and non-interacting configurations of liquid tank (model derivation and system response)

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  • 7/28/2019 Interacting and Non-Interacting Configurations of Liquid Tank (Model Derivation and System Response)

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    Example 6.1: Interacting and non-interacting

    configurations of liquid tank (model

    derivation and system response)Contents

    [hide]

    1 Problem Statement 2 Assumptions 3 Noninteracting Configuration 4 Interacting Configuration 5 Comparison

    Problem Statement

    Consider the situation in which the system of Figure 6.1 consists of two identical tanks withidentical time constants 1 minute. The steady-state gains are also identical, and equal

    to 1; 1.

    1. Obtain an expression for the response of (the level in the second tank as a deviation from its

    initial steady-state value) to a unit step change in the inlet flowrate to tank 1.

    2. If these same tanks are now arranged as in Figure 6.2, obtain the unit step response for this

    interacting configuration and compare with the response obtained in (1) for the noninteractingcase.

    Assumptions

    -Linear relation between the liquid level and flowrate through valves.

    - and represent the cross sectional areas of each tank.

    Noninteracting Configuration

    Taking the material balance of each tank ( ),

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    Noninteracting tanks in series.

    Tank 1: :

    Tank 2: :

    Using the system parameters, , , , as well as deviation

    variables, the above expressions change to:

    Taking the Laplace transforms for each equation yields the transfer function for each of the two

    tanks.

    Substituting into the second equation yields the single composite transfer function,

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    Plugging into the equation the values for the time constants, steady-state gains, and using the unit

    step function as the input function,

    Taking the Laplace inversion yields the function;

    Interacting Configuration

    Again starting with a material balance, we yield the following equations for the tanks;

    Interacting tanks in series.

    Tank 1: :

    Tank 2: :

    Substituting the same deviation variables and system parameters used in the noninteractingtanks,

    Plugging in the values for the time constants (1 minute), steady-state gains (1), and

    differentiating the second equation with respect to time yields the following equation:

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    Substituting the differential equation for the first tank into the above equation,

    Taking the Laplace transformation and rearranging,

    Factoring,

    Taking the Laplace inversion,

    Comparison

    Each of the responses were graphed in MATLAB as a function of time.

    clear all; clc;

    t = 0:.25:10;

    y21 = 1-exp(-t)-t.*exp(-t);

    y22 = 1-1.171.*exp(-0.382.*t)+0.171.*exp(-2.618.*t);

    figure(1)

    plot(t,subs(y21),t,subs(y22)) %Plots functions using time domain

    axis([0 10 0 1])

    xlabel('t')

    ylabel('y21,y22')

    legend('Noninteracting','Interacting')

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    As one can see, the noninteracting configuration has a faster response than the interacting

    configuration.

    http://cheg4147.engr.uconn.edu/wiki/index.php/File:Example.jpg