|1 Crore+ students have signed up on EduRev. Have you?|
Second Order Process
A second order process is a process whose output is modeled by a second order differential equation.
where, u(t) and y(t) are input and output of the process respectively. If , then define the following:
Hence, the second order differential equation takes the following form:
At steady state condition , the equation can be re-written as
Subtracting eq. (68) from eq. (67), we obtain
Where, and are respectively the deviation forms of the output and input of the process around the steady state, whose initial conditions are assumed to be the following:
Taking Laplace Transform of the eq. (70) we obtain,
Rearranging the above we obtain,
Kp is called the gain of the process.
Example of a second order process
Consider the U tube manometer as in Fig.6. The liquid inside the manometer has been shown in a pressurized state. Initially mercury levels at both the legs were at the same height. The present pressurized state is obtained upon exerting a pressure of on Leg I.
Applying force balance on both the legs of the manometer across plane of initial pressurized state, we obtain:
Where, cross-sectional area of manometer leg(s), P= density of manometer liquid, f =Fanning' friction factor, v = velocity of manometer liquid, D = diameter of manometer leg(s), L =length of manometer liquid in the tube, m = mass of manometer liquid. Assuming laminar flow inside the manometer, the friction factor can be expressed as , where is the Reynold's number. Hence the force balance equation takes the form:
The velocity of manometer liquid is rate of change of h . Hence,
Comparing eq.(77) with eq.(67), the following can be obtained: and and .
Dynamic Response of a Second Order Process to a Step Change in the Input
For a step input of magnitude A , the Laplace Transform of u(t) would be,
Hence, second order process takes the following form,
The process response will grossly depend upon the value of ξ and there can be three distinguished cases of ξ, i.e. ξ >1; ξ = 1 and ξ <1 .
Case A: ξ = 1
In this case the process response equation in the Laplace domain takes the following form:
Using the following:
in eq. (80), we obtain
For ξ ≠ 1, using the following:
in eq. (79), we obtain
Case B: When ξ >1
In the above equations, the following trigonometric identities have been used: and
Hence we get the final expression for process response when ξ >1,
In the above equations, the following trigonometric identities have been used:
One can also use the following trigonometric identity for the above expression:
Hence we get the final expression for process response for ξ < 1,
he frequency of oscillation is
whereas the phase lag is