Page 553 - Mechatronics with Experiments

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JWST499-c07
JWST499-Cetinkunt
ELECTROHYDRAULIC MOTION CONTROL SYSTEMS 539
Simulated Case 1: Constant pump displacement, variable main valve opening.
3
D (t) = 4ml∕rev = 4 ⋅ 10 −6 m ∕rev (7.480)
p
w (t) = 2000 rev∕min (7.481)
shaft
A (t) = 10.0; t < 1.0 (7.482)
v
= 10.0 + 40.0(t − 1.0); 1.0 <= t < 2.0 (7.483)
= 50.0; 2.0 <= t < 3.0 (7.484)
= 50.0 − 40.0(t − 1.0); 3.0 <= t < 4.0 (7.485)
= 10.0; 4.0 <= t < 5.0 (7.486)

The fluid compressibility in the volume between pump, relief valve, and main valve is
neglected. Then, we have the net flow rate in and out of this volume as equal to zero,
Q (t) − Q (t) − Q (t) = 0 (7.487)
p
v
r
The relief valve is modeled as a static spool displacement–pressure relationship
(inertial dynamics is not modeled). These equations can be solved over time for the
given simulation conditions above. Notice that if the line pressure does not reach the
maximum pressure setting of the relief valve (if p (t) <= p max ), then the relief valve
s
will be closed and Q (t) = 0.0, then Q (t) = Q (t), which results in the following
p
v
r
equation for the pump pressure,
( ) 2
Q (t)
v
p (t) = p + (7.488)
t
s
C A (x (t))
d v
v
( ) 2
Q (t)
p
= p + (7.489)
t
C A (x (t))
v
d v
( ) 2
D ⋅ w shaft
p
= p + (7.490)
t
C A (x (t))
v
d v
The same relationship can be derived using the rated flow equations,
( ) 2
Q (t) A v,max
v
p (t) = p + ⋅ Δp vr (7.491)
s
t
Q vr A (x (t))
v
v
( ) 2
Q (t) A v,max
p
= p + ⋅ Δp vr (7.492)
t
Q A (x (t))
vr v v
( ) 2
D (t) ⋅ w shaft (t) A v,max
p
= p + ⋅ Δp vr (7.493)
t
Q vr A (x (t))
v
v
The above solution is valid for p (t) < p max .If p (t) ≥ p max , then we must solve
s
s
for a p (t) iteratively from Q (t) and Q (t) equations such that Q (t) − Q (t) − Q (t) =
v
r
s
p
v
r
0. For this case the solution is, for p > p max
s
( ) 2 ( ) 2
Q (t) Q (t)
p
v
p (t) = p + = p + (7.494)
s t t
C A (x (t)) C A (x (t))
d v
d v
v
v
( 2 )
(D ⋅ w shaft )
p
= p + (7.495)
t
2
2
C (A (x (t)) + A (x (t)) 2
d v v r r