Page 554 - Mechatronics with Experiments

JWST499-Cetinkunt
JWST499-c07
540 MECHATRONICS Printer: Yet to Come October 9, 2014 8:41 254mm×178mm
The same relationship can be derived using the rated flow equations,
[ ]
( ) 2 ( ) 2
A v 1 A r 1 2
p (t) = p + Q (t) ⋅ + Q (t) ⋅ ⋅ Q (t) (7.496)
s t vr rr p
A Δp A Δp
v⋅max vr r,max rr
In this case, there is a constant supply of flow from the pump. Since the main
valve is initially open a small amount, the line pressure increases quickly in order to
force the incoming flow rate through the valve. As the main valve gradually opens
more, the line pressure gradually drops since a lower pressure differential is needed
to pass the same flow rate through a larger orifice area. If the line pressure had to
increase beyond the maximum pressure setting of the relief valve in order to pass the
supplied flow rate through the main valve, then the relief valve would have opened.
®
In Simulink , we simply impose the conservation of flow condition equation,
and let the iterative algorithm solve for p (t) using algebraic constraint equation solver,
s
Q (t) − Q (t) − Q (t) = 0, instead of evaluating these explicit solution equations for
r
v
p
p (t) (Figure 7.94).
s
Simulated Case 2: Variable pump displacement, variable main valve opening
(Figure 7.95). The same conditions as in the previous case except that the pump
displacement is specified as function of time as follows. The dynamics of the pump
displacement actuation mechansims and the pump response is not modeled.
D (t) = 0.0; t < 1.0 (7.497)
p
= 4.0(t − 1.0); 1.0 <= t < 2.0 (7.498)
= 4.0; 2.0 <= t < 3.0 (7.499)
= 4.0 − 4.0(t − 1.0); 3.0 <= t < 4.0 (7.500)
= 0.0; 4.0 <= t < 5.0 (7.501)
Same equations as in Case 1 apply, only the input variables are different for the
simulation. Notice that since the displacement of the pump matches the main valve
opening, the line pressure does not fluctuate as high as the first case.
Simulated Case 3: Variable pump displacement, closed main valve, relief valve
activates – relief valve modeled as a static relation between pressure and valve
opening (Figure 7.96). Since in this case the only path the flow can go through is
the relief valve, the line pressure will quickly build up to maximum line pressure to
open the relief valve. Same equations as in Case 1 apply. The pump displacement as
a function of time is the same as in Case 2. The main valve orifice area is different:
A (t) = 0.0 for the whole simulation. The relief valve orifice opening is determined
v
instantaneously by the line pressure. In other words, the inertial dynamics of the
relief valve is not modeled, rather it is the orifice area (or spool position) versus line
pressure that is modeled as a static relationship, as defined above.
Simulated Case 4: Fluid compressibility in the piping between components is taken
into account (Figure 7.97). Let us take into account the finite stiffness (bulk modulus),
that is the compressible nature of the fluid in the volume between the pump and the
two valves. There is not much interest in taking into account the compressibility of
the fluid between the output port of the valves and the tank connection since the
overall pressure in the return lines is rather small and the fluid compressibility does
not make much difference in the pressure in that line.
Due to the compressibility of the fluid, the net flow rate change in a volume
results in a change (time derivative) of the pressure scaled by the bulk modulus divided
by the total fluid volume in the section considered. When the fluid compressibility