Page 559 - Mechatronics with Experiments

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JWST499-c07
JWST499-Cetinkunt
ELECTROHYDRAULIC MOTION CONTROL SYSTEMS 545
is not taken into account, the net flow rate into and out of a given section of volume
(i.e., fluid volume in the piping between pump and relief valve and main valve) is
equal to zero.
The pump output pressure will be determined by the dynamic interaction that
involves the compressibility of the fluid. Once we calculate the pump output pressure
from the differential equations, which takes the fluid compressibility into account,
we can further subtract the pressure loss estimations from that in order to determine
the dynamic pressures at the input ports of the two valves. In this case the main
valve is modeled as a variable orifice where the orifice opening area is given as a
function of time, that is the spool position is actuated by a solenoid. The dynamics
of the actuator for the main valve spool motion is not included in the model. The
relief valve is modeled as a variable orifice as a function of the line pressure. A more
accurate model would be to model the inertia–force relationship for the relief valve
spool position, and current–force–inertia relationship for the main valve.

dp (t)
s
= ⋅ (Q (t) − Q (t) − Q (t)); p (t ) = 20 000 kPa (7.502)
p
r
s 0
v
dt V
hose
Simulated input variable conditions are the same as in Case 3, plus finite bulk modulus
of the fluid in the volume between the pump and the two valves.
SimulatedCase5: Variablepumpdisplacement, closedmainvalve, fluidcompressibility
is taken into account, the relief valve activates, where the relief valve is modeled as
a dynamic inertia–force relationship. Simulated conditions are as Case 4, except
that we modeled the relief valve dynamics more accurately. We take into account the
dynamic delay in the relief valve spool position change as a result of the line pressure,
inertia, spring, and damping in the valve (Figure 7.98).
Example: Dynamic Model of a Poppet Valve Consider the poppet valve in
a hydraulic circuit (Figure 7.99). Here, we will develop a dynamic model of this valve. We
will consider the inertial effects of the poppet spool as well as the fluid compressibility
effects inside the valve chamber. The basic operation of the valve is that input flow is in
port 1 and output flow is in port 2 (Figure 7.99b). The poppet gets unseated when the flow
force exceeds the force at the back of the poppet. Similar to sliding spool valves, poppet
valves are also manufactured for multi port applications such as 2, 3, 4 port applications.
Let us express the force–acceleration relationship for the poppet, the spool (Fig-
ure 7.99b),
m ⋅ ̈ x (t) + c ⋅ ̇ x (t) + k ⋅ x (t) =− k ⋅ x preload + F (t) + F shear (t) + p (t) ⋅ A − p (t) ⋅ A 3
p
p
1
3
1
p
sol
p
p
p
p
(7.503)
Notice that the pressure p (t) does not affect the spool motion in this model, because we
2
consider that the pressure inside the valve body in region 2 is uniform. As a result, the net
force applied to the spool by p (t) is zero because it is acting both on the top and bottom
2
surfaces of the poppet.
The relationship between the flow rate through the valve, valve position, pressure in
the valve chamber (assuming no cavitation inside the valve), in-port pressure, and output
(load) pressure is
dp (t)
2
= ⋅ (Q (t) − Q leak (t) − Q (t)) (7.504)
2
4
dt V 2
dp (t) = ⋅ (Q (t) − Q (t)) (7.505)
3
dt V leak 3t
3