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3.6 Integrating System

We found in the previous chapter that material balances on liquid surge vessels or gas drums often yielded models with the following form:

Equation 3.32

graphics/03equ32.gif


In the Laplace domain, this is

Equation 3.33

graphics/03equ33.gif


Consider an integrating process initially at steady state, with y(0) = 0.

Step Response

If a step input change of Du is made at t = 0,

graphics/03equ33a.gif


and we find the time-domain value

Equation 3.34

graphics/03equ34.gif


That is, the output ramps with a constant slope of kDu.

Impulse Response

If an impulse input of magnitude P is made at t = 0,

graphics/03equ34a.gif


then the output immediately changes to a new steady-state value of

graphics/03equ34b.gif


Example 3.5: Tank-Height Problem

The mathematical model for a liquid surge tank is (see Example 1.3)

graphics/03equ34c.gif


where h is the liquid height, A is the constant cross-sectional area of the tank, F1 is the inlet flow rate, and F2 is the outlet flow rate. Assume that the outlet flow rate remains constant at a steady-state value of F2s. Defining the output and input in deviation variable form as

graphics/03equ34d.gif


For a constant cross-sectional area of 10 m2, the model is

graphics/03equ34e.gif


Step Response

For a step input change of 0.25 m3/min, the output response is

graphics/03equ34f.gif


which is shown in Figure 3-7. If the steady-state height is 2 meters, then the height as a function of time is

graphics/03equ34g.gif


Figure 3-7. Step response of a liquid surge tank. Deviation variables.

graphics/03fig07.gif

Impulse Response

For an impulse input of 1 m3, the output response is

graphics/03equ34h.gif


which makes sense, because the cross-sectional area is 10 m3.

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