Page 493 - Mechatronics with Experiments

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JWST499-Cetinkunt
JWST499-c07
ELECTROHYDRAULIC MOTION CONTROL SYSTEMS 479
Even though the commands to each valve do not change, the flow rate to each circuit, and
hence the speed of each cylinder, will vary due to the variation in the load in each circuit.
If it is desired to modify circuit design so that the flow rate distribution between
circuits is independent of the load, the pressure drop across each valve must be same even
if the load pressures are different,
p − p = p − p l2 (7.214)
s
s
l1
Under this condition, flow rate distribution to each circuit is independent of the load
difference between the circuits and would be controlled by specific spool displacement of
each valve (x , x ). When both valves are same size and have the same displacement, flow
v1
v2
rate distribution would be equal, 50% each of total flow, or Q v1 = Q , Q = Q v1 + Q .
s
v2
v2
When load pressures are different (p ≠ p ), the equal pressure drop across the valve for
l2
l1
each circuit can be accomplished by:
1. Adding restriction to the circuit which has the lower load so that the total hydraulic
load seen by the valve is the same as the load on the other circuit. This means adding
“positive hydraulic resistance” to the circuit with lower load.
2. Removing restriction from the circuit which has the higher load so that the total
hydraulic load seen by the valve is same as the load on the other circuit. This means
adding “negative hydraulic resistance” to the circuit with higher load.
The second option is not physically possible. Negative hydraulic resistance cannot be added
to a hydraulic circuit. Therefore, in order to equalize the effective hydraulic resistance
(making the total pressure drop equal in each circuit), we need to add hydraulic resistance
into the circuit which has lower load.
The electrical analogy of this concept is shown in Figure 7.59a. Consider that we
have a constant voltage supply (analogy of pump), and two parallel resistors (analogy of
valve-cylinder of each circuit). The current (analogy of flow rate) is divided between the
two circuits as follows,
i = i + i 2 (7.215)
1
V
i = (7.216)
1
R v1
V
i = (7.217)
2
R
v2
R
v1
i 1
R
i v2
v i 2

(a)

R R
v1 c1
i 1
R R c2
i v2
FIGURE 7.59: (a) Electrical analogy for a
v i 2 two function hydraulic circuit. (b) Same
analogy with adjustable resistors in each
path to control the current (flow rate)
(b) distribution between the two parallel paths.