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JWST499-Cetinkunt
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It is of interest to simulate the motion of the hydraulic system and plot the results:
1. Q (t) – pump flow rate
s
2. Q (t) – flow rate from pump to A-side of the cylinder
PA
3. Q (t) – flow rate from pump to B-side of the cylinder
PB
4. Q (t) – flow rate from A-side of the cylinder to the tank
AT
5. Q (t) – flow rate from B-side of the cylinder to the tank
BT
6. Q (t) – flow rate through the relief valve
r
7. Q (t) – flow rate into or out of the accumulator
acc
8. x (t) – main valve spool displacement
s
9. F load (t) – external load
10. p (t) – pump pressure
P
11. p (t) – pressure in the A-side of the cylinder
A
12. p (t) – pressure in the B-side of the cylinder
B
13. p (t) – pressure in the accumulator
acc
14. V (t) – fluid volume in the accumulator
acc
15. ̇ y(t) – cylinder velocity
16. y(t) – cylinder position, which is obtained from initial condition plus the integtration of cylinder
t
velocity, y(t) = y(t ) + ∫ ̇ y( )d
0 t 0
Remark: The actual pressure developed at the outlet of the pump is determined by the load pressure,
valve size, and relief pressure. Note that if the valve is undersized for the application, A (x (t)) is
v s
small, then p will be equal to p relief most of the time to support the most flow rate it can. If, on the
P
other hand, the valve is oversized, then A (x (t)) is too large, then p will not need to be as large as
v s P
p relief most of the time to support the flow, and the relief valve will not need to open, hence the system
will be more efficient. However, the larger the A (x (t)) gain is, the worse the control accuracy of the
s
v
flow rate, and hence the control accuracy of the EH motion control system.
It is instructive for the reader to change the valve size by changing the A (x ) and observe
v
s
the effect of it on the p , that is what happens if we have sized the valve too small or too large
P
for the application. As the valve size gets larger, the control accuracy gets poorer, while efficiency
improves. The fundamental variable we control is the valve spool displacement, x , through which
s
the valve orifice area A (x ) is determined. The smallest resolution that the valve is controllable is
v
s
determined by the position control of the valve. Let us assume that it is a fixed quantity by the valve
control accuracy, depending on the controller, amplifier, and valve position feedback accuracy (if any
closed loop control is used on the valve position). For a given Δx , that is the smallest increment the
s
valve position can be controlled, the change in flow rate, hence the cylinder speed is proportionally
affected by the A (x ). The larger that gain is (the larger the valve size), the larger the smallest change
s
v
in the cylinder velocity that we can affect, hence the worse the control resolution is (or control
accuracy). Conversely, the smaller the valve size (A (x )), the smaller the flow rate change we can
s
v
affect, hence the smaller cylinder velocity change we can affect, the better the control resolution
(control accuracy). In contrast, energy efficiency requires a larger valve size, since energy efficiency
requires us to minimize the power loss. It is desirable to support the largest flow rate with as little
pressure drop as possible across the valve. In other words, as the valve size gets larger, the pressure
drop needed to support a given desired flow rate gets smaller. Hence, the power loss
Q ⋅ (p − p ) + Q ⋅ (p − p ) or (7.854)
PA P A BT B T
Q PB ⋅ (p − p ) + Q AT ⋅ (p − p ) (7.855)
B
P
T
A
across the valve (due to pressure drop across the valve) is smaller. This is a fundamental design
conflict between control resolution (accuracy) and energy efficiency in hydraulic systems. A good
design must find a balance between these two conflicting requirements.
Case 1 – Simplified Model: We consider the following simplified version of the model
neglect the cylinder and load inertial dynamics,
neglect fluid compressibility throughout the hydraulic circuit,
assume we have an ideal relief valve,
assume we do not have any accumulator.