Fisher Manual: Principles of Operations and Regulator Sizing Theory | Fisher Manuals & Guides
Principles of Direct-Operated Regulators
588
Te c h n i c a l
Introduction
Pressure regulators have become very familiar items over the years,
and nearly everyone has grown accustomed to seeing them in
factories, public buildings, by the roadside, and even on the outside
of their own homes. As is frequently the case with such familiar
items, we have a tendency to take them for granted. It’s only
when a problem develops, or when we are selecting a regulator
for a new application, that we need to look more deeply into the
fundamentals of the regulator’s operation.
Regulators provide a means of controlling the ow of a gas or
other uid supply to downstream processes or customers. An
ideal regulator would supply downstream demand while keeping
downstream pressure constant; however, the mechanics of direct-
operated regulator construction are such that there will always be
some deviation (droop or offset) in downstream pressure.
TYPE HSR
The service regulator mounted on the meter outside virtually every
home serves as an example. As appliances such as a furnace or
stove call for the ow of more gas, the service regulator responds
by delivering the required ow. As this happens, the pressure
should be held constant. This is important because the gas meter,
which is the cash register of the system, is often calibrated for a
given pressure.
Direct-operated regulators have many commercial and residential
uses. Typical applications include industrial, commercial, and
domestic gas service, instrument air supply, and a broad range of
applications in industrial processes.
Regulators automatically adjust ow to meet downstream demand.
Before regulators were invented, someone had to watch a pressure
gauge for pressure drops which signaled an increase in downstream
demand. When the downstream pressure decreased, more ow was
required. The operator then opened the regulating valve until the
gauge pressure increased, showing that downstream demand was
being met.
Essential Elements
Direct-operated regulators have three essential elements:
• A restricting element — a valve, disk, or plug
• A measuring element— generally a diaphragm
• A loading element— generally a spring
W1327
133 SERIES
W1934
TYPE 630
Figure 1. Direct-Operated Regulators
Regulator Basics
A pressure reducing regulator must satisfy a downstream demand
while maintaining the system pressure within certain acceptable
limits. When the ow rate is low, the regulator plug or disk
approaches its seat and restricts the ow. When demand increases,
the plug or disk moves away from its seat, creating a larger
opening and increased ow. Ideally, a regulator should provide a
constant downstream pressure while delivering the required ow.
RESTRICTING
ELEMENT
LOADING
ELEMENT
(WEIGHT)
Figure 2. Three Essential Elements
MEASURING
ELEMENT
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589
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Restricting Element
The regulator’s restricting element is generally a disk or plug that
can be positioned fully open, fully closed, or somewhere in between
to control the amount of ow. When fully closed, the disk or plug
seats tightly against the valve orice or seat ring to shutoff ow.
Measuring Element
The measuring element is usually a exible diaphragm that
senses downstream pressure (P2). The diaphragm moves
as pressure beneath it changes. The restricting element is
often attached to the diaphragm with a stem so that when the
diaphragm moves, so does the restricting element.
Loading Element
A weight or spring acts as the loading element. The loading
element counterbalances downstream pressure (P2). The amount of
unbalance between the loading element and the measuring element
determines the position of the restricting element. Therefore, we can
adjust the desired amount of ow through the regulator, or setpoint,
by varying the load. Some of the rst direct-operated regulators used
weights as loading elements. Most modern regulators use springs.
Regulator Operation
To examine how the regulator works, let’s consider these values for
a direct-operated regulator installation:
• Upstream Pressure (P
• Downstream Pressure (P
) = 100 psig
1
) = 10 psig
2
• Pressure Drop Across the Regulator (P) = 90 psi
• Diaphragm Area (A
) = 10 square inches
D
• Loading Weight = 100 pounds
Let’s examine a regulator in equilibrium as shown in Figure 3. The
pressure acting against the diaphragm creates a force acting up to
100 pounds.
Diaphragm Force (FD) = Pressure (P2) x Area of Diaphragm (AD)
or
FD = 10 psig x 10 square inches = 100 pounds
The 100 pounds weight acts down with a force of 100 pounds,
so all the opposing forces are equal, and the regulator plug
remains stationary.
Increasing Demand
If the downstream demand increases, P2 will drop. The pressure on
the diaphragm drops, allowing the regulator to open further. Suppose
in our example P2 drops to 9 psig. The force acting up then equals
FW = 100 LB
FD = 100 LB
FW = 100 LB
FD = 90 LB
100 LB
P2 = 10 PSIGP1 = 100 PSIG
FD = (P2 x AD) = (10 PSIG)(10 IN2) = 100 LB
AT EQUILIBRIUM
100 LB
FD = (P2 x AD) = (9 PSIG)(10 IN2) = 90 LB
OPEN
Figure 3. Elements
AREA = 10 IN
AREA = 10 IN
P2 = 9 PSIGP1 = 100 PSIG
2
2
90 pounds (9 psig x 10 square inches = 90 pounds). Because of the
unbalance of the measuring element and the loading element, the
restricting element will move to allow passage of more ow.
Decreasing Demand
If the downstream demand for ow decreases, downstream
pressure increases. In our example, suppose P2 increases to 11
psig. The force acting up against the weight becomes 110 pounds
(11 psig x 10 square inches = 110 pounds). In this case, unbalance
causes the restricting element to move up to pass less ow or
lockup.
Weights versus Springs
One of the problems with weight-loaded systems is that they are
slow to respond. So if downstream pressure changes rapidly,
our weight-loaded regulator may not be able to keep up. Always
behind, it may become unstable and cycle—continuously going
from the fully open to the fully closed position. There are other
problems. Because the amount of weight controls regulator
setpoint, the regulator is not easy to adjust. The weight will always
have to be on top of the diaphragm. So, let’s consider using a
spring. By using a spring instead of a weight, regulator stability
increases because a spring has less stiffness.
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2
P1 = 100 PSIG P2 = 10 PSIG
FD = (10 PSIG)(102) = 100 LB
FS = (100 LB/IN)(X) = 100 LB
X = 1-INCH COMPRESSION
Figure 4. Spring as Element
SPRING AS ELEMENT
F
S
FS = (K)(X)
FD = (P2)(AD)
F
D
AT EQUILIBRIUM
AREA = 10 IN
FS = F
D
(TO KEEP DIAPHRAGM
FROM MOVING)
Spring Rate
We choose a spring for a regulator by its spring rate (K). K
represents the amount of force necessary to compress the spring
one inch. For example, a spring with a rate of 100 pounds per inch
needs 100 pounds of force to compress it one inch, 200 pounds of
force to compress it two inches, and so on.
Spring as Loading Element
By using a spring instead of a xed weight, we gain better control
and stability in the regulator. The regulator will now be less likely to
go fully open or fully closed for any change in downstream pressure
(P2). In effect, the spring acts like a multitude of different weights.
Throttling Example
Assume we still want to maintain 10 psig downstream. Consider
what happens now when downstream demand increases and
pressure P2 drops to 9 psig. The diaphragm force (FD) acting up is
now 90 pounds.
FD = P2 x AD
FD = 9 psig x 10 in
FD = 90 pounds
We can also determine how much the spring will move (extend)
which will also tell us how much the disk will travel. To keep the
regulator in equilibrium, the spring must produce a force (FS) equal
to the force of the diaphragm. The formula for determining spring
force (FS) is:
FS = (K)(X)
where K = spring rate in pounds/inch and X = travel or
compression in inches.
2
Equilibrium with a Spring
Instead of a weight, let’s substitute a spring with a rate of
100 pounds per inch. And, with the regulator’s spring adjustor, we’ll
wind in one inch of compression to provide a spring force (FS) of
100 pounds. This amount of compression of the regulator spring
determines setpoint, or the downstream pressure that we want to
hold constant. By adjusting the initial spring compression, we
change the spring loading force, so P2 will be at a different value in
order to balance the spring force.
Now the spring acts down with a force of 100 pounds, and the
downstream pressure acts up against the diaphragm producing
a force of 100 pounds (FD = P2 x AD). Under these conditions
the regulator has achieved equilibrium; that is, the plug or disk is
holding a xed position.
FS = 90 LB
FD = 90 LB
P1 = 100 PSIG P2 = 9 PSIG
0.1-INCH
Figure 5. Plug Travel
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LOCKUP
SETPOINT
DROOP
(OFFSET)
P
2
WIDE-OPEN
FLOW
AS THE FLOW RATE APPROACHES ZERO, P2 INCREASES STEEPLY. LOCKUP IS THE TERM APPLIED TO THE
VALUE OF P2 AT ZERO FLOW.
Figure 6. Typical Performance Curve
We know FS is 90 pounds and K is 100 pounds/inch, so we can
solve for X with:
X = FS ÷ K
X = 90 pounds ÷ 100 pounds/inch
X = 0.9 inch
The spring, and therefore the disk, has moved down 1/10-inch,
allowing more ow to pass through the regulator body.
Regulator Operation and P
Now we see the irony in this regulator design. We recall that the
purpose of an ideal regulator is to match downstream demand
while keeping P2 constant. But for this regulator design to increase
ow, there must be a change in P2.
2
Setpoint
The constant pressure desired is represented by the setpoint. But
no regulator is ideal. The downward sloping line on the diagram
represents pressure (P2) plotted as a function of ow for an actual
direct-operated regulator. The setpoint is determined by the initial
compression of the regulator spring. By adjusting the initial spring
compression you change the spring loading force, so P2 will be
at a different value in order to balance the spring force. This
establishes setpoint.
Droop
Droop, proportional band, and offset are terms used to describe
the phenomenon of P2 dropping below setpoint as ow increases.
Droop is the amount of deviation from setpoint at a given ow,
expressed as a percentage of setpoint. This “droop” curve is
important to a user because it indicates regulating (useful) capacity.
Capacity
Capacities published by regulator manufacturers are given for
different amounts of droop. Let’s see why this is important.
Let’s say that for our original problem, with the regulator set at
10 psig, our process requires 200 SCFH (standard cubic feet per
hour) with no more than a 1 psi drop in setpoint. We need to keep
the pressure at or above 9 psig because we have a low limit safety
switch set at 9 psig that will shut the system down if pressure falls
below this point.
Figure 6 illustrates the performance of a regulator that can do the job.
And, if we can allow the downstream pressure to drop below 9 psig,
the regulator can allow even more ow.
Regulator Performance
We can check the performance of any regulating system by
examining its characteristics. Most of these characteristics can
be best described using pressure versus ow curves as shown in
Figure 6.
Performance Criteria
We can plot the performance of an ideal regulator such that
no matter how the demand changes, our regulator will match
that demand (within its capacity limits) with no change in the
downstream pressure (P2). This straight line performance becomes
the standard against which we can measure the performance of a
real regulator.
The capacities of a regulator published by manufacturers are
generally given for 10% droop and 20% droop. In our example,
this would relate to ow at 9 psig and at 8 psig.
Accuracy
The accuracy of a regulator is determined by the amount of ow it
can pass for a given amount of droop. The closer the regulator is to
the ideal regulator curve (setpoint), the more accurate it is.
Lockup
Lockup is the pressure above setpoint that is required to shut the
regulator off tight. In many regulators, the orice has a knife
edge while the disk is a soft material. Some extra pressure, P2, is
Principles of Direct-Operated Regulators
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required to force the soft disk into the knife edge to make a tight
seal. The amount of extra pressure required is lockup pressure.
Lockup pressure may be important for a number of reasons.
Consider the example above where a low pressure limit switch
would shut down the system if P2 fell below 9 psig. Now consider
the same system with a high pressure safety cut out switch set a
10.5 psig. Because our regulator has a lockup pressure of 11 psig,
the high limit switch will shut the system down before the regulator
can establish tight shutoff. Obviously, we’ll want to select a
regulator with a lower lockup pressure.
Spring Rate and Regulator Accuracy
Using our initial problem as an example, let’s say we now need the
regulator to ow 300 SCFH at a droop of 10% from our original
setpoint of 10 psig. Ten percent of 10 psig = 1 psig, so P2 cannot
drop below 10 to 1, or 9 psi. Our present regulator would not be
accurate enough. For our regulator to pass 300 SCFH, P2 will have
to drop to 8 psig, or 20% droop.
Spring Rate and Droop
One way to make our regulator more accurate is to change to
a lighter spring rate. To see how spring rate affects regulator
accuracy, let’s return to our original example. We rst tried a
spring with a rate of 100 pounds/inch. Let’s substitute one with a
rate of 50 pounds/inch. To keep the regulator in equilibrium, we’ll
have to initially adjust the spring to balance the 100 pound force
produced by P2 acting on the diaphragm. Recall how we calculate
spring force:
FS = K (spring rate) x X (compression)
Knowing that FS must equal 100 pounds and K = 50 pounds/inch,
we can solve for X, or spring compression, with:
X = FS ÷ K, or X = 2 inches
So, we must wind in 2-inches of initial spring compression to
balance diaphragm force, FD.
Effect on Plug Travel
We saw before that with a spring rate of 100 pounds/inch, when
P2 dropped from 10 to 9 psig, the spring relaxed (and the valve
disk traveled) 0.1 inch. Now let’s solve for the amount of disk
travel with the lighter spring rate of 50 pounds per inch. The force
produced by the diaphragm is still 90 pounds.
To maintain equilibrium, the spring must also produce a force of
90 pounds. Recall the formula that determines spring force:
FS = (K)(X)
Because we know FS must equal 90 pounds and our spring rate (K)
is 50 pounds/inch, we can solve for compression (X) with:
X = FS ÷ K
X = 90 pounds ÷ 50 pounds/inch
X = 1.8 inches
To establish setpoint, we originally compressed this spring 2 inches.
Now it has relaxed so that it is only compressed 1.8 inches, a change
of 0.2-inch. So with a spring rate of 50 pounds/inch, the regulator
responded to a 1 psig drop in P2 by opening twice as far as it did
with a spring rate of 100 pounds/inch. Therefore, our regulator
is now more accurate because it has greater capacity for the same
change in P2. In other words, it has less droop or offset. Using this
example, it is easy to see how capacity and accuracy are related and
how they are related to spring rate.
Light Spring Rate
Experience has shown that choosing the lightest available spring
rate will provide the most accuracy (least droop). For example, a
spring with a range of 35 to 100 psig is more accurate than a spring
with a range of 90 to 200 psig. If you want to set your regulator at
100 psig, the 35 to 100 psig spring will provide better accuracy.
Practical Limits
While a lighter spring can reduce droop and improve accuracy,
using too light a spring can cause instability problems. Fortunately,
most of the work in spring selection is done by regulator
manufacturers. They determine spring rates that will provide good
performance for a given regulator, and publish these rates along
with other sizing information.
Diaphragm Area and Regulator Accuracy
Diaphragm Area
Until this point, we have assumed the diaphragm area to be
constant. In practice, the diaphragm area changes with travel.
We’re interested in this changing area because it has a major
inuence on accuracy and droop.
FD = P2 x AD
Diaphragms have convolutions in them so that they are exible
enough to move over a rated travel range. As they change position,
Principles of Direct-Operated Regulators
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DIAPHRAGM EFFECT ON DROOP
2
A = 10 IN
F
D1
2
A = 11 IN
F
D2
WHEN P2 DROPS TO 9
FD = P2 x A
FD1 = 10 x 10 = 100 LB
FD2 = 9 x 11 = 99 LB
P
2
FLOW
SETPOINT
WIDE-OPEN
CRITICAL FLOW
Figure 7. Changing Diaphragm Area
they also change shape because of the pressure applied to them.
Consider the example shown in Figure 7. As downstream pressure
(P2) drops, the diaphragm moves down. As it moves down, it
changes shape and diaphragm area increases because the centers of
the convolutions become further apart. The larger diaphragm area
magnies the effect of P2 so even less P2 is required to hold the
diaphragm in place. This is called diaphragm effect. The result is
decreased accuracy because incremental changes in P2 do not result
in corresponding changes in spring compression or disk position.
Increasing Diaphragm Area
To better understand the effects of changing diaphragm area, let’s
calculate the forces in the exaggerated example given in Figure 7.
First, assume that the regulator is in equilibrium with a downstream
pressure P2 of 10 psig. Also assume that the area of the diaphragm
in this position is 10 square inches. The diaphragm force (FD) is:
FD = (P2)(AD)
FD = (10 psi) (10 square inches)
FD = 100 pounds
Now assume that downstream pressure drops to 9 psig signaling the
need for increased ow. As the diaphragm moves, its area increases
to 11 square inches. The diaphragm force now produced is:
FD = (9 psi) (11 square inches)
FD = 99 pounds
The change in diaphragm area increases the regulator’s droop.
While it’s important to note that diaphragm effect contributes to
Figure 8. Critical Flow
droop, diaphragm sizes are generally determined by manufacturers
for different regulator types, so there is rarely a user option.
Diaphragm Size and Sensitivity
Also of interest is the fact that increasing diaphragm size can result
in increased sensitivity. A larger diaphragm area will produce
more force for a given change in P2. Therefore, larger diaphragms
are often used when measuring small changes in low-pressure
applications. Service regulators used in domestic gas service are
an example.
Restricting Element and Regulator Performance
Critical Flow
Although changing the orice size can increase capacity, a
regulator can pass only so much ow for a given orice size and
inlet pressure, no matter how much we improve the unit’s accuracy.
Shown in Figure 8, after the regulator is wide-open, reducing P2
does not result in higher ow. This area of the ow curve identies
critical ow. To increase the amount of ow through the regulator,
the owing uid must pass at higher and higher velocities. But, the
uid can only go so fast. Holding P1 constant while decreasing P2,
ow approaches a maximum which is the speed of sound in that
particular gas, or its sonic velocity. Sonic velocity depends on the
inlet pressure and temperature for the owing uid. Critical ow is
generally anticipated when downstream pressure (P2) approaches a
value that is less than or equal to one-half of inlet pressure (P1).
Principles of Direct-Operated Regulators
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Orifice Size and Capacity
One way to increase capacity is to increase the size of the orice.
The variable ow area between disk and orice depends directly
on orice diameter. Therefore, the disk will not have to travel
as far with a larger orice to establish the required regulator
ow rate, and droop is reduced. Sonic velocity is still a limiting
factor, but the ow rate at sonic velocity is greater because more
gas is passing through the larger orice.
Stated another way, a given change in P2 will produce a larger
change in ow rate with a larger orice than it would with a smaller
orice. However, there are denite limits to the size of orice that
can be used. Too large an orice makes the regulator more sensitive
to uctuating inlet pressures. If the regulator is overly sensitive, it
will have a tendency to become unstable and cycle.
Orifice Size and Stability
One condition that results from an oversized orice is known as
the “bathtub stopper” effect. As the disk gets very close to the
orice, the forces of uid ow tend to slam the disk into the orice
and shutoff ow. Downstream pressure drops and the disk opens.
This causes the regulator to cycle—open, closed, open, closed. By
selecting a smaller orice, the disk will operate farther away from
the orice so the regulator will be more stable.
Performance Limits
The three curves in Figure 9 summarize the effects of spring rate,
diaphragm area, and orice size on the shape of the controlled
pressure-ow rate curve. Curve A is a reference curve representing
a typical regulator. Curve B represents the improved performance
from either increasing diaphragm area or decreasing spring rate.
Curve C represents the effect of increasing orice size. Note that
increased orice size also offers higher ow capabilities. But
remember that too large an orice size can produce problems that
will negate any gains in capacity.
P
2
SETPOINT
C
B
A
FLOW
Figure 9. Increased Sensitivity
SETPOINT
Orifice Size, Lockup, and Wear
A larger orice size also requires a higher shutoff pressure,
or lockup pressure. In addition, an oversized orice usually
produces faster wear on the valve disk and orice because it
controls ow with the disk near the seat. This wear is accelerated
with high ow rates and when there is dirt or other erosive
material in the ow stream.
Orifice Guideline
Experience indicates that using the smallest possible orice is
generally the best rule-of-thumb for proper control and stability.
Increasing P1
Regulator capacity can be increased by increasing inlet pressure (P1).
Factors Affecting Regulator Accuracy
As we have seen, the design elements of a regulator—the spring,
diaphragm, and orice size—can affect its accuracy. Some
of these inherent limits can be overcome with changes to the
regulator design.
P
2
FLOW INCREASED
TIME
Figure 10. Cycling
Cycling
The sine wave in Figure 10 might be what we see if we increase
regulator sensitivity beyond certain limits. The sine wave indicates
instability and cycling.
Design Variations
All direct-operated regulators have performance limits that result
from droop. Some regulators are available with features designed
to overcome or minimize these limits.
Principles of Direct-Operated Regulators
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E0908
Type HSR
Decreased Droop (Boost)
The pitot tube offers one chief advantage for regulator accuracy, it
decreases droop. Shown in Figure 12, the diaphragm pressure, PD,
must drop just as low with a pitot tube as without to move the disk
far enough to supply the required ow. But the solid curve shows
that P2 does not decrease as much as it did without a pitot tube. In
fact, P2 may increase. This is called boost instead of droop. So
the use of a pitot tube, or similar device, can dramatically improve
P2 = 10 PSIGP1 = 100 PSIG
SENSE PRESSURE HERE
Figure 11. Pitot Tube
Improving Regulator Accuracy with a Pitot Tube
droop characteristics of a regulator.
In addition to the changes we can make to diaphragm area, spring
rate, orice size, and inlet pressure, we can also improve regulator
accuracy by adding a pitot tube as shown in Figure 11. Internal to
the regulator, the pitot tube connects the diaphragm casing with a
low-pressure, high velocity region within the regulator body. The
pressure at this area will be lower than P2 further downstream.
PIVOT POINT
By using a pitot tube to measure the lower pressure, the regulator
will make more dramatic changes in response to any change in
P2. In other words, the pitot tube tricks the regulator, causing it to
respond more than it would otherwise.
P
2
P
D
PRESSURE
FLOW
Figure 12. Performance with Pitot Tube
SETPOINT
P2 = DOWNSTREAM PRESSURE
PD = PRESSURE UNDER DIAPHRAGM
Numerical Example
For example, we’ll establish setpoint by placing a gauge
downstream and adjusting spring compression until the gauge
reads 10 psig for P2. Because of the pitot tube, the regulator might
INLET PRESSURE
OUTLET PRESSURE
ATMOSPHERIC PRESSURE
E0908
Figure 13. Lever Style Regulator
Improving Performance with a Lever
The lever style regulator is a variation of the simple direct-operated
regulator. It operates in the same manner, except that it uses a lever
to gain mechanical advantage and provide a high shutoff force.
In earlier discussions, we noted that the use of a larger diaphragm
can result in increased sensitivity. This is because any change in P2
will result in a larger change in diaphragm force. The same result
is obtained by using a lever to multiply the force produced by the
diaphragm as shown in Figure 13.
The main advantage of lever designs is that they provide increased
force for lockup without the extra cost, size, and weight associated
with larger diaphragms, diaphragm casings, and associated parts.
actually be sensing a lower pressure. When P2 drops from 10 psig
to 9 psig, the pressure sensed by the pitot tube may drop from
8 psig to 6 psig. Therefore, the regulator opens further than it
would if it were sensing actual downstream pressure.
Principles of Pilot-Operated Regulators
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Pilot-Operated Regulator Basics
In the evolution of pressure regulator designs, the shortcomings of
the direct-operated regulator naturally led to attempts to improve
accuracy and capacity. A logical next step in regulator design is to
use what we know about regulator operation to explore a method
of increasing sensitivity that will improve all of the performance
criteria discussed.
PILOT
REGULATOR
INLET PRESSURE, P
OUTLET PRESSURE, P
ATMOSPHERIC PRESSURE
LOADING PRESSURE, P
1
2
L
Figure 1. Pilot-Operated Regulator
Regulator Pilots
To improve the sensitivity of our regulator, we would like to be
able to sense P2 and then somehow make a change in loading
pressure (PL) that is greater than the change in P2. To accomplish
this, we can use a device called a pilot, or pressure amplier.
The major function of the pilot is to increase regulator sensitivity.
If we can sense a change in P2 and translate it into a larger change
in PL, our regulator will be more responsive (sensitive) to changes
in demand. In addition, we can signicantly reduce droop so its
effect on accuracy and capacity is minimized.
Gain
The amount of amplication supplied by the pilot is called
“gain”. To illustrate, a pilot with a gain of 20 will multiply
the effect of a 1 psi change on the main diaphragm by 20. For
example, a decrease in P2 opens the pilot to increase PL 20 times
as much.
MAIN
REGULATOR
Identifying Pilots
Analysis of pilot-operated regulators can be simplied by viewing
them as two independent regulators connected together. The
smaller of the two is generally the pilot.
Setpoint
We may think of the pilot as the “brains” of the system. Setpoint
and many performance variables are determined by the pilot. It
senses P2 directly and will continue to make changes in PL on
the main regulator until the system is in equilibrium. The main
regulator is the “muscle” of the system, and may be used to control
large ows and pressures.
Spring Action
Notice that the pilot uses a spring-open action as found in directoperated regulators. The main regulator, shown in Figure 1, uses
a spring-close action. The spring, rather than loading pressure, is
used to achieve shutoff. Increasing PL from the pilot onto the main
diaphragm opens the main regulator.
Pilot Advantage
Because the pilot is the controlling device, many of the
performance criteria we have discussed apply to the pilot. For
example, droop is determined mainly by the pilot. By using very
small pilot orices and light springs, droop can be made small.
Because of reduced droop, we will have greater usable capacity.
Pilot lockup determines the lockup characteristics for the system.
The main regulator spring provides tight shutoff whenever the pilot
is locked up.
Gain and Restrictions
Stability
Although increased gain (sensitivity) is often considered an
advantage, it also increases the gain of the entire pressure
regulator system. If the system gain is too high, it may become
unstable. In other words, the regulator might tend to oscillate;
over-reacting by continuously opening and closing. Pilot gain
can be modied to tune the regulator to the system. To provide
a means for changing gain, every pilot-operated regulator system
contains both a xed and a variable restriction. The relative size
of one restriction compared to the other can be varied to change
gain and speed of response.
Principles of Pilot-Operated Regulators
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PILOT
FLOW
P
2
P
P
1
VARIABLE
RESTRICTION
L
SMALLER FIXED
RESTRICTION
TO MAIN
REGULATOR
P
2
Figure 2. Fixed Restrictions and Gain (Used on Two-Path Control Systems)
Restrictions, Response Time, and Gain
Consider the example shown in Figure 2 with a small xed
restriction. Decreasing P2 will result in pressure PL increasing.
Increasing P2 will result in a decrease in PL while PL bleeds out
through the small xed restriction.
If a larger xed restriction is used with a variable restriction, the
gain (sensitivity) is reduced. A larger decrease in P2 is required to
increase PL to the desired level because of the larger xed restriction.
PILOT
FLOW
P
P
P
1
L
2
P
2
PILOT
P
2
P
1
VARIABLE
RESTRICTION
VARIABLE
RESTRICTION
FLOW
INLET PRESSURE, P
OUTLET PRESSURE, P
ATMOSPHERIC PRESSURE
LOADING PRESSURE, P
FLOW
P
L
TO MAIN
FIXED RESTRICTION
1
2
L
REGULATOR
LARGER FIXED
RESTRICTION
P
2
FIXED
RESTRICTION
TO MAIN
REGULATOR
VARIABLE
RESTRICTION
Figure 3. Unloading Systems
Loading and Unloading Designs
A loading pilot-operated design (Figure 2), also called two-path
control, is so named because the action of the pilot loads PL onto
the main regulator measuring element. The variable restriction, or
pilot orice, opens to increase PL.
An unloading pilot-operated design (Figure 3) is so named because
the action of the pilot unloads PL from the main regulator.
Figure 4. Two-Path Control
Two-Path Control (Loading Design)
In two-path control systems (Figure 4), the pilot is piped so that
P2 is registered on the pilot diaphragm and on the main regulator
diaphragm at the same time. When downstream demand is
constant, P2 positions the pilot diaphragm so that ow through the
pilot will keep P2 and PL on the main regulator diaphragm. When
P2 changes, the force on top of the main regulator diaphragm
and on the bottom of the pilot diaphragm changes. As P2 acts on
the main diaphragm, it begins repositioning the main valve plug.
This immediate reaction to changes in P2 tends to make two-path
designs faster than other pilot-operated regulators. Simultaneously,
P2 acting on the pilot diaphragm repositions the pilot valve and
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