Data Sheet
Bulletin 71.1:SR5
7
Capacity Data
The capacity information on the following pages is
based on three offset factors, 10, 20 and 30 percent.
Flow capacity at set point (set ow) was approximately
10% of maximum ow. Offset, or droop, is deviation
from the setpoint of the regulator, and is usually
stated in percentage of set point value. To evaluate
the performance of a regulator, compare the stated
capacities at equivalent operating pressures and
offset factors. Comparing the wide open C
v
does not
consider the overall accuracy.
For example, refer to the air ow rate curve in
Figure 4. The regulator is set for 20 psig (1,4 bar) at
the outlet. Capacity at 10 percent offset is read at the
intersection of a line projected horizontally to the right
from the 18 psig (1,2 bar) outlet pressure point. The
18 psig (1,2 bar) is derived by applying the offset factor
of 10 percent to the regulator setting of 20 psig (1,4 bar).
For the most accurate control, use the lowest range
spring that can be adjusted to the desired set point
(See Table 2 for part numbers of appropriate springs
for each body size). If closer control is necessary,
a regulator of larger capacity should be selected, so
that the necessary ow can be obtained with a smaller
offset factor.
Sometimes it may be necessary to interpolate the
capacity table data to determine capacity for outlet
settings not given. To maintain accuracy, it is important
when interpolating to stay within a spring range if
possible. The following is a procedure for interpolating
the data to calculate ow:
1. Determine which spring is to be used.
2. Find the two outlet settings (P
2a
& P
2b
) that
bracket the actual outlet pressure P
2
.
3. For a given body size and inlet pressure, nd the
capacity Q
a
for P
2a
and Q
b
for P
2b
.
4. Use the following formula to determine the
interpolated capacity (Q).
Q
b
- Q
a
Q
b
- Q
=
P
2b
- P
2a
P
2b
- P
2
Example:
P
1
= 75 psig
1 inch Type SR5 with 5 to 25 psig spring range
with elastomer diaphragm (Table 8)
P
2
= 20 psig
Determine air capacity, Q.
Solution:
Q
a
= 3067 SCFH
Q
b
= 5546 SCFH
P
2a
= 15 psig
P
2b
= 25 psig
5546 - 3067
5546 - Q
=
25 - 15 25 - 20
Q = 4307 SCFH
NOTE
The same interpolation procedure can
be used for different inlet pressures.
An alternative method for interpolating capacities is to
use the C
v
as shown in Table 7.
Contact your Fisher
Representative if you should have any questions about
selecting the proper regulator.
Air Capacities
Regulating capacities at selected pressures and outlet
pressure ows are given in Table 8. Capacities are
provided in SCFH (60°F and 14.7 psia) of air at 60°F
and normal cubic meters per hour at 0°C and 1.01325
bar. To determine the equivalent capacities for other
gases, multiply the table capacities by 1.018 for
nitrogen. For gases of other specic gravities, divide
by the square root of the appropriate specic gravity.
To determine wide-open ow capacity for relief valve
sizing of air at a temperature of 60°F, use one of the
following equations:
For Critical Pressure Drops
Use this equation for critical pressure drops (absolute
outlet pressure equal to one-half or less than one-half
the absolute inlet pressure).
Q = P
1(abs)
x C
g
where,
Q = Gas ow, SCFH (60°F and 14.7 psia)
P
1(abs)
= Absolute inlet pressure psia (add 14.7 psi
to gauge inlet pressure to obtain absolute
inlet pressure)
C
g
=
Wide-open sizing coefcient from Table 6