116 Chapter 4 / Mathematical Modeling of Fluid Systems and Thermal SystemsAs in the case of the force-distance type controller, this controller employs a flapper,
nozzle, and orifices. In Figure 4–10, the drilled opening in the bottom chamber is the
nozzle. The diaphragm just above the nozzle acts as a flapper.
The operation of the force-balance type controller shown in Figure 4–10 may be
summarized as follows: 20-psig air from an air supply flows through an orifice, causing
a reduced pressure in the bottom chamber. Air in this chamber escapes to the atmos-
phere through the nozzle. The flow through the nozzle depends on the gap and the
pressure drop across it. An increase in the reference input pressure while the out-
put pressure remains the same, causes the valve stem to move down, decreasing the
gap between the nozzle and the flapper diaphragm. This causes the control pressure
to increase. Let
(4–20)
Ifpe=0,there is an equilibrium state with the nozzle–flapper distance equal to and
the control pressure equal to At this equilibrium state, and
(4–21)
whereais a constant.
Let us assume that peZ0 and define small variations in the nozzle–flapper distance
and control pressure as xandpc, respectively. Then we obtain the following equation:
(4–22)
From Equations (4–21) and (4–22), we obtain
(4–23)
At this point, we must examine the quantity x. In the design of pneumatic controllers,
the nozzle–flapper distance is made quite small. In view of the fact that x/ais very much
smaller than pc(1-k)A 1 orpeAA 2 - A 1 B—that is, for peZ 0
we may neglect the term xin our analysis. Equation (4–23) can then be rewritten to
reflect this assumption as follows:
and the transfer function between pcandpebecomes
wherepeis defined by Equation (4–20). The controller shown in Figure 4–10 is a
proportional controller. The value of gain Kpincreases as kapproaches unity. Note that
the value of kdepends on the diameters of the orifices in the inlet and outlet pipes of
the feedback chamber. (The value of kapproaches unity as the resistance to flow in the
orifice of the inlet pipe is made smaller.)
Pc(s)
Pe(s)
=
A 2 - A 1
A 1
1
1 - k
=Kp
pc(1-k)A 1 =peAA 2 - A 1 B
x
a
peAA 2 - A 1 B
x
a
pc(1-k)A 1
x=aCpc(1-k)A 1 - peAA 2 - A 1 BD
X
–
+x=aCAP
–
c+pcBA 1 - AP
–
c+pcBkA 1 - peAA 2 - A 1 BD
X
–
=aAP
–
c^ A 1 - P
–
c^ kA 1 B
P 1 =P
–
P c k (where k 6 1)
–
c^.
X
–
pe=Pr-Po
Pc
Po
Pr ,
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