Page 494 - Mechatronics with Experiments

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In order to make current division to each path equal, i = i , the resistance in each path
1
2
must be the same, R v1 = R . This can be accomplished by adding an adjustable resistor in
v2
each path (Figure 7.59b), where R , R are adjustable resistors. If R v1 > R , we can make
v2
c1
c2
R c1 = 0.0, R c2 = R v1 − R v2 to make the current in each path equal. Similarly, if R v1 < R ,
v2
we can make R c1 = R v2 − R , R c2 = 0.0 to make the current in each path equal,
v1
i = i + i 2 (7.218)
1
V
i = (7.219)
1
R + R
v1 c1
V
i = (7.220)
2
R v2 + R c2
If we make, R + R = R + R , then i = i ,evenif R ≠ R .
v1 c1 v2 c2 1 2 v1 v2
The same concept can be accomplished in hydraulic circuits by adding an adjustable
orifice (“hydraulic resistor”) in series with each circuit. There are two methods for this:
1. pre-pressure compensated valves,
2. post-pressure compensated valves.
In multi function circuits as shown in Figure 7.60, both pre- and post-pressure compensated
circuits provide the same behavior until the pump saturates or relief pressure is reached.
Figure 7.60a shows a circuit with pre-pressure compensation. Figure 7.60b shows the
post-pressure compensated circuit. We assume that both circuits have the same pump,
same valves and cylinders and pressure compensator valves, except that the location of the
pressure compensator valves and their feedback control signals are different. The pump is
controlled in load-sensing mode and has the same margin pressure for both circuits.
Consider the pre-pressure compensated case, and the pump is not saturated. Let us
assume that the preload setting in the pressure compensator valves are equal, p = p .
sp1 sp2
The pre-pressure compensator valve spools move to control the restriction (adding hydraulic
resistor from zero to infinity by opening and closing the compensator valve orifice by spool
movement) so that the pressure difference across the main valve is constant. In other words,
the pressure compensator valves work to maintain the following relationship as long as
they are not saturated (fully closed or fully open),
p cp1 − p = p sp1 (7.221)
l1
p cp2 − p = p sp2 (7.222)
l2
Since flow rate across the main valves is determined by the following relationships,
√ √
Q v1 = C A (x ) p cp1 − p = C A (x ) p sp1 (7.223)
v1
v
l1
d
v1
d
v1
√ √
Q v2 = C A (x ) p cp2 − p = C A (x ) p sp2 (7.224)
l2
v2
v
d
d
v2
v2
which shows that the flow rates are independent of the load pressures (p , p ). Instead, the
l1 l2
flow rates Q and Q are functions of p and p , which are the pressure settings by the
v1 v2 sp1 sp2
preloaded springs in the pressure compensator valves, respectively. Notice that p − p is
c1 l1
the pressure drop across main flow control valve number one, and p − p is the pressure
c2 l2
drop across main flow control valve number two, and they are equal to the preload pressure
setting of the pre-pressure compensator valves, p sp1 , p sp2 , respectively. If they are equal by
design, p sp1 = p sp2 , then the pressure drop across each main valve is regulated to that same
value as long as saturation is not reached. Hence, the flow rate distribution between the
circuit is load independent. The flow rate distribution is controlled by the spool position of
the main flow control valves.