Page 551 - Mechatronics with Experiments

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JWST499-c07
JWST499-Cetinkunt
ELECTROHYDRAULIC MOTION CONTROL SYSTEMS 537
pressure setting. Then it moves to fully open position within a small increase of pressure
above the maximum pressure setting.
x (t) = K ⋅ (p (t) − p ); for p (t) > p (7.445)
r px s max s max
= 0.0; else (7.446)
A (t) = K ⋅ x (t); for x (t) > 0 (7.447)
r
ar
r
r
= 0.0; else (7.448)
For a certain value of pressure above maximum pressure, p (t) − p =Δp , the relief
s max ra
valve will be fully open, that is A (t) = A . Given the performance specifications for the
r r,max
relief valve, we would know A r,max and Δp . Then,
ra
A r,max
K ⋅ K px = (7.449)
ar
Δp ra
If we wanted to model the inertial dynamics of the relief valve, then the relationship
is a differential equation,
m ̈ x (t) + c ̇ x (t) + k x (t) = A ⋅ (p (t) − p max ) (7.450)
fb
s
r r
r r
r r
A (t) = K ⋅ x (t); for x (t) > 0 (7.451)
r
ar
r
r
= 0.0 ; else (7.452)
where p max = k ⋅ x preload ∕A , maximum pressure setting is defined in the circuit by the
fb
r
preload on the relief valve spring. In steady-state, the relief valve spool displacement is
A fb
x (∞) = ⋅ (p (t) − p ) (7.453)
r s max
k r
A
A (∞) = K ⋅ fb ⋅ (p (t) − p ); for x (t) > 0 (7.454)
r ar s max r
k
r
In addition, when the relief valve is fully open at the pressure p(t) = p max +Δp , the orifice
ra
area would saturate to A r,max . Hence,
A fb A r,max
K ⋅ = (7.455)
ar
k r Δp ra
The dynamics of the main valve is not modeled as an inertia–force relationship in this
example. Rather, we assume that the spool displacement of the main valve is specified as a
function of time, and the dynamics of the actuation mechanism and the valve are neglected.
Various conditions of the main valve orifice opening (A (t)) are specified below for the
v
simulated conditions. The consideration of the inertial dynamics of the main valve is left
as an exercise for further study of this example.
The governing equations for flow rate across each component (pump, relief valve,
main valve) are as follows (pressure losses are not included since it is discussed above and
their inclusion is very simple)
Q (t) = D (t) ⋅ w (t) (7.456)
p p shaft
√
p (t) − p t
s
Q (t) = Q ⋅ (A (t)∕A r,max ) ⋅ (7.457)
r
r
rr
Δp rr
√
p (t) − p t
s
Q (t) = Q ⋅ (A (t)∕A v,max ) ⋅ (7.458)
v
v
vr
Δp vr