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Hydraulic Oils and Theor etical Backgr ound 55

system, the differential equation (DE), can be transformed into Laplace
domain, just by replacing the term (d/dt) by (s). The resulting transfer
function takes the form

Gs() = Ps() where P(s) and Q(s) are polynomials in s. (2A.1)
Qs()

For the system to be physically realizable, the order of P(s) should
not exceed that of Q(s).
Consider the linear system of input x(t) and output y(t), described
by the following differential equation of zero initial conditions:

2
3
4
2
dy dy dy dy dx dx
a + a + a + a + ay = b + b + bx (2A.2)
4 4 3 3 2 2 1 0 2 2 1 0
dt dt dt dt d dt dt
The application of Laplace transform to Eq. (2A.2) yields:
4 (
as + a s + as + a s a Y s =( b s + b s b X ) (s) (2A.3)
+
+
4
3
2
2
0) ()
s
2
3
1
0
2
1
+
2
Ys () bs + b s b
then, Gs () = = 2 1 0 (2A.4)
+
4
2
3
Xs () as + a s + as + a s + a
4 3 2 1 0
The transfer function can be deduced as follows:
• Write down the equations governing the system behavior.
• If there are any nonlinear relations, a linearization procedure
or simplifying assumptions should be considered to obtain
linearized equations.
• By substitution, eliminate the variables that are not of direct
interest, leaving the relation between the input and output
variables.
• If there are any variables of nonzero initial conditions, x(t = 0)
= x and x ≠ 0 , substitute it by a new variable, x = x(t) − x .
o o 1 o
The new variable x is of zero initial value.
1
• Apply Laplace transform and find the required transfer
function.
Appendix 2B Laminar Flow in Pipes
In the laminar flow, the paths of individual particles of fluid do not
cross. So, the flow in a pipe may be considered as a series of concentric
cylinders sliding over each other. Consider a cylinder of fluid of
length dx and radius r, flowing steadily in the center of a pipe.
Consider the cylindrical fluid element illustrated by Fig. 2B.1,
moving in a pipe as shown in Fig. 2B.2. In the steady state, the fluid