126 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal Systemsvalve is symmetrical,A 1 =A 3 andA 2 =A 4. Assuming the displacement xto be small,
we obtain
wherekis a constant.
Furthermore, we shall assume that the return pressure poin the return line is small
and thus can be neglected. Then, referring to Figure 4–17(a), flow rates through valve
orifices are
where and and gis the specific weight and is given by
g=rg, where ris mass density and gis the acceleration of gravity. The flow rate qto
the left-hand side of the power piston is
(4–25)
The flow rate from the right-hand side of the power piston to the drain is the same as
thisqand is given by
In the present analysis we assume that the fluid is incompressible. Since the valve is
symmetrical, we have q 1 =q 3 andq 2 =q 4 .By equating q 1 andq 3 ,we obtain
or
If we define the pressure difference across the power piston as or
¢p=p 1 - p 2
¢p
ps=p 1 +p 2
ps-p 1 =p 2
q=q 3 - q 2 =C 11 p 2 a
x 0
2
+xb -C 21 ps-p 2 a
x 0
2
- xb
q=q 1 - q 4 =C 11 ps-p 1 a
x 0
2
+xb -C 21 p 1 a
x 0
2
- xb
C 1 =c 1 k 1 2gg C 2 =c 2 k 1 2gg,
q 4 =c 2 A 4
B
2g
g
Ap 1 - p 0 B=C 21 p 1 - p 0 a
x 0
2
- xb =C 21 p 1 a
x 0
2
- xb
q 3 =c 1 A 3
B
2g
g
Ap 2 - p 0 B=C 11 p 2 - p 0 a
x 0
2
+xb =C 11 p 2 a
x 0
2
+xb
q 2 =c 2 A 2
B
2g
g
Aps-p 2 B=C 21 ps-p 2 a
x 0
2
- xb
q 1 =c 1 A 1
B
2g
g
Aps-p 1 B=C 11 ps-p 1 a
x 0
2
+xb
A 2 =A 4 =ka
x 0
2
- xb
A 1 =A 3 =ka
x 0
2
+xb
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