Section 4–4 / Hydraulic Systems 127
then
For the symmetrical valve shown in Figure 4–17(a), the pressure in each side of the
power piston is (1/2)pswhen no load is applied, or As the spool valve is dis-
placed, the pressure in one line increases as the pressure in the other line decreases by
the same amount.
In terms of psand we can rewrite the flow rate qgiven by Equation (4–25) as
Noting that the supply pressure psis constant. the flow rate qcan be written as a func-
tion of the valve displacement xand pressure difference or
By applying the linearization technique presented in Section 3–10 to this case, the lin-
earized equation about point is
(4–26)
where
Coefficientsaandbhere are called valve coefficients. Equation (4–26) is a linearized
mathematical model of the spool valve near an operating point
The values of valve coefficients aandbvary with the operating point. Note that
is negative and so bis negative.
Since the normal operating point is the point where near the
normal operating point Equation (4–26) becomes
(4–27)
where
K 2 =AC 1 +C 2 B
x 0
4121 ps
70
K 1 =AC 1 +C 2 B
A
ps
2
70
q=K 1 x-K 2 ¢p
x–=0,¢p–=0, q–=0,
0 f0¢p
x=x–,¢p=¢p–,q=q–.
+
C 2
2121 ps+¢p–
ax 0
2
- x–bR 60
b=
0 f
0¢p
2
x=x–,¢p=¢p–=-c
C 1
2121 ps-¢p–
ax 0
2
+x–b
a=
0 f
0 x
2
x=x–,¢p=¢p–=C 1
A
ps-¢p–
2
+C 2
A
ps+¢p–
2
q–=f(x–,¢p–)
q-q–=a(x-x–)+b(¢p-¢p–)
x=x–,¢p=¢p–,q=q–
q=C 1
A
ps-¢p
2
ax 0
2
+xb -C 2
A
ps+¢p
2
ax 0
2
- xb =f(x,¢p)
¢p,
q=q 1 - q 4 =C 1
A
ps-¢p
2
ax 0
2
+xb -C 2
A
ps+¢p
2
ax 0
2
- xb
¢p,
¢p=0.
p 1 =
ps+¢p
2
, p 2 =