Electromechanical Dynamics (Part 1).0086

# Electromechanical Dynamics (Part 1).0086 -...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Electromechanical Coupling A-PDF Split3.1.1 DEMO : Purchase from www.A-PDF.com to remove the watermark desired to find the change in stored energy when the independent variables are changed from the point (2a, x,) to the point (2,, x,). To evaluate a line integral we must specify the path of integration; an infinite number of possible paths exist between the two points. A property of a conservative system, however, is that its stored energy is a function of its state (i.e., of the particular values of Aand x that exist) and does not depend on what succession of variable values or what path through variable space was used to reach that state. A consequence of this property is that if the system variables are made to change x along path A from (ZA, a) to (A,, x,) in Fig. 3.1.3 and then along path B back x to (A2, a), the net change in stored energy W, during the process is zero. In a conservative system the change in stored energy between any two points in variable space is independent of the path of integration. Thus we can select the path that makes the integration easiest. As an example, consider the evaluation of the change in energy between points (A, xa) and (A, xb) in Fig. 3.1.3. Along segment 1-2, di = 0; and along segment 2-3, dx = 0. Thus, using path C, integration of (3.1.9) takes on the particular form W,(Ab, xb) W.(Aa,, xa) = -- j ft (A,, x) dx + f i(A,x,)dA. (3.1.10) If, alternatively, we wish to evaluate the integral along path D in Fig. 3.1.3, the result is W.(2b, Xb) Wm(Aa2, Xa) =--- i(A, X,) d2 -J f.(2Ab, x) dx. (3.1.11) The energy difference as evaluated by (3.1.10) and (3.1.11) must, of course, be the same. The integrations given in (3.1.10) and (3.1.11) have a simple physical significance. The integrations of (3.1.10) represent putting energy into the network in two successive steps. First we put the system together mechanically (integrate on x) while keeping Aconstant. In general, this operation requires doing work against the force fe, and this is the contribution of the first integral in (3.1.10) to the energy stored in the coupling network. Then we put energy in through the electrical terminals, keeping the geometry (x) fixed. The second integral is the energy supplied by an electrical source which provides the excitation A. In (3.1.11) these successive steps are reversed in order. We always define electrical terminal pairs that account for the excitation of all electric or magnetic fields in the system. Then, when the electrical terminal variables are zero (A, = 0 in the present example), we can say that there is no force of electrical origin. The difference between (3.1.10) and (3.1.11) with ,= 0 is crucial, for in the first the contribution off' to the integration is zero[f*(0, x) = 0], whereas in the second we must knowf' to ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online