Page 613 - Mechatronics with Experiments

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JWST499-c07
JWST499-Cetinkunt
ELECTROHYDRAULIC MOTION CONTROL SYSTEMS 599
−6
2
A (x ) = (40 ⋅ 10 )∕(100 − x )) ⋅ (|x | − x )m ; x ≤ −x (7.818)
AT s db s db s db
2
−6
A (x ) = (10 ⋅ 10 )∕(100 − x )) ⋅ (|x | − x )m ; x ≥ x (7.819)
BT s db s db s db
A (x ) = 0.0 closed-center valve (7.820)
PT s
A = 0.01 m 2 (7.821)
A
A = 0.005 m 2 (7.822)
B
m = 10 000 kg (7.823)
c = 0.0Nt∕(m∕s) (7.824)
l cyl = 1.0 (7.825)
8
= 15.0 ⋅ 10 N∕m 2 (7.826)
V hose,pv = 0.0001 m 3 (7.827)
V = 0.0001 m 3 (7.828)
hose,VA
V = 0.0001 m 3 (7.829)
hose,VB
K relief = 1.0 ⋅ 10 −8 (7.830)
relief = 0.025 (7.831)
6
p relief = 20.0 ⋅ 10 Nt∕m 2 (7.832)
6
p max = 20 ⋅ 10 N∕m 2 (7.833)
6
p min = 15 ⋅ 10 N∕m 2 (7.834)
6
p pre = 15 ⋅ 10 N∕m 2 (7.835)
V disch = 0.005 m 3 (7.836)
K acc = 1.0 ⋅ 10 −6 (7.837)
V disch
2
3
C acc = = 1.0 ⋅ 10 −9 m ∕(N∕m ) (7.838)
p max − p min
The simulated input condition and initial conditions are
w pump (t) = 25 rev∕s (7.839)
2
F load (t) = 10 000.0kg ⋅ 9.81 m∕s = 98.1kN weight (7.840)
x (t) = 0 ; 0 < t <= 1.0 (7.841)
s
= (100∕0.25) ∗ (t − 1) ; 1.0 < t <= 1.25 s (7.842)
= 100 ; 1.25 < t <= 3.0 s (7.843)
= 100 − (100∕0.25) ⋅ (t − 3) ; 3.0 < t <= 3.25 s (7.844)
= 0 ; 3.25 < t < 5.0 (7.845)
p (t) = 0.0; tank pressure (7.846)
T
y(0) = 0.10 ; initial cylinder position (7.847)
̇ y(0) = 0.0; initial cylinder velocity is zero (7.848)
p (0) = p relief (7.849)
P
p (0) = F load (0)∕A ; initial cylinder pressure (7.850)
A
A
p (0) = p = 0.0 (7.851)
T
B
6
p (0) = p min = 15 ⋅ 10 N∕m 2 (7.852)
acc
V (0) = 0.0m 3 (7.853)
acc
Note that the input shaft to the pump runs at a constant speed, the load on the cyclinder is simply a
mass moving against gravity, the initial position of the cylinder is close to the bottom zero position.
The valve opens and closes in a total of 3.0 s in a trapezoidal profile.