Valve Sizing Calculations (Traditional Method)
626
Te c h n i c a l
Introduction
Fisher® regulators and valves have traditionally been sized using
equations derived by the company. There are now standardized
calculations that are becoming accepted worldwide. Some product
literature continues to demonstrate the traditional method, but
the trend is to adopt the standardized method. Therefore, both
methods are covered in this application guide.
Improper valve sizing can be both expensive and inconvenient.
A valve that is too small will not pass the required ow, and
the process will be starved. An oversized valve will be more
expensive, and it may lead to instability and other problems.
The days of selecting a valve based upon the size of the pipeline
are gone. Selecting the correct valve size for a given application
requires a knowledge of process conditions that the valve will
actually see in service. The technique for using this information
to size the valve is based upon a combination of theory
and experimentation.
Sizing for Liquid Service
Using the principle of conservation of energy, Daniel Bernoulli
found that as a liquid ows through an orice, the square of the
uid velocity is directly proportional to the pressure differential
across the orice and inversely proportional to the specic gravity
of the uid. The greater the pressure differential, the higher the
velocity; the greater the density, the lower the velocity. The
volume ow rate for liquids can be calculated by multiplying the
uid velocity times the ow area.
By taking into account units of measurement, the proportionality
relationship previously mentioned, energy losses due to friction
and turbulence, and varying discharge coefcients for various
types of orices (or valve bodies), a basic liquid sizing equation
can be written as follows
where:
Q = Capacity in gallons per minute
Cv = Valve sizing coefcient determined experimentally for
each style and size of valve, using water at standard
conditions as the test uid
∆P = Pressure differential in psi
G = Specic gravity of uid (water at 60°F = 1.0000)
Thus, Cv is numerically equal to the number of U.S. gallons of
water at 60°F that will ow through the valve in one minute when
the pressure differential across the valve is one pound per square
inch. Cv varies with both size and style of valve, but provides an
index for comparing liquid capacities of different valves under a
standard set of conditions.
Q = CV ∆P / G
(1)
∆P ORIFICE
METER
FLOW
INLET VALVE TEST VALVE LOAD VALVE
Figure 1. Standard FCI Test Piping for Cv Measurement
To aid in establishing uniform measurement of liquid ow capacity
coefcients (Cv) among valve manufacturers, the Fluid Controls
Institute (FCI) developed a standard test piping arrangement,
shown in Figure 1. Using such a piping arrangement, most
valve manufacturers develop and publish Cv information for
their products, making it relatively easy to compare capacities of
competitive products.
To calculate the expected Cv for a valve controlling water or other
liquids that behave like water, the basic liquid sizing equation
above can be re-written as follows
CV = Q
G
(2)
∆P
PRESSURE
INDICATORS
Viscosity Corrections
Viscous conditions can result in signicant sizing errors in using
the basic liquid sizing equation, since published Cv values are
based on test data using water as the ow medium. Although the
majority of valve applications will involve uids where viscosity
corrections can be ignored, or where the corrections are relatively
small, uid viscosity should be considered in each valve selection.
Emerson Process Management has developed a nomograph
(Figure 2) that provides a viscosity correction factor (Fv). It can
be applied to the standard Cv coefcient to determine a corrected
coefcient (Cvr) for viscous applications.
Finding Valve Size
Using the Cv determined by the basic liquid sizing equation and
the ow and viscosity conditions, a uid Reynolds number can be
found by using the nomograph in Figure 2. The graph of Reynolds
number vs. viscosity correction factor (Fv) is used to determine
the correction factor needed. (If the Reynolds number is greater
than 3500, the correction will be ten percent or less.) The actual
required Cv (Cvr) is found by the equation:
C
From the valve manufacturer’s published liquid capacity
information, select a valve having a Cv equal to or higher than the
required coefcient (Cvr) found by the equation above.
= FV CV (3)
vr
627
Te c h n i c a l
Valve Sizing Calculations (Traditional Method)
3000
3,000
4,000
6,000
8,000
10,000
20,000
30,000
40,000
60,000
80,000
100,000
200,000
300,000
400,000
600,000
800,000
C
INDEX
1 2 3 4 6 8 10 20 30 40 60 80 100 200
2000
2,000
1,000
2,000
2,000
3,000
3,000
4,000
4,000
6,000
6,000
8,000
8,000
10,000
10,000
800
600
400
300
200
100
80
60
40
30
20
10
8
6
4
3
2
1
1,000
1,000
2,000
2,000
3,000
3,000
4,000
4,000
6,000
6,000
8,000
8,000
10,000
10,000
20,000
20,000
30,000
30,000
40,000
40,000
60,000
60,000
80,000
80,000
100,000
100,000
200,000
300,000
400,000
800
800
600
600
400
400
300
300
200
200
100
100
80
80
60
60
40
40
35
32.6
30
20
10
8
6
4
3
2
1
0.8
0.6
0.4
0.3
0.2
0.1
0.08
0.06
0.04
0.03
0.02
0.01
1,000
1,000
800
600
800
600
400
300
200
100
80
60
40
30
20
400
300
200
100
80
60
40
30
20
10
10
8
8
6
6
4
4
3
3
2
2
1
0.8
0.6
0.4
0.2
0.1
0.3
1
0.8
0.6
0.4
0.3
0.2
0.1
0.08
0.06
0.04
0.04
0.06
0.08
0.03
0.03
0.02
0.02
0.01
0.01
0.008
0.008
0.006
0.006
0.004
0.004
0.003
0.003
0.002
0.002
0 .001
0.001
0.0008
0.0008
0.0006
0.0006
0.0004
0.0004
0.0003
0.0003
0.0002
0.0002
0.0001
0.0001
1000
800
600
400
300
200
100
80
60
40
30
20
10
8
6
4
3
2
1
0.8
0.6
0.4
0.3
0.2
0.1
0.08
0.06
0.04
0.03
0.02
0.01
3,000
4,000
6,000
8,000
10,000
20,000
30,000
40,000
60,000
80,000
100,000
200,000
300,000
400,000
600,000
800,000
1,000,000
1 2 3 4 6
8 10 20 30 40 60 80 100 200
1 2 3 4 6 8 10 20 30 40 60 80 100 200
2,000
1,000
1,000
2,000
2,000
3,000
3,000
4,000
4,000
6,000
6,000
8,000
8,000
10,000
10,000
20,000
20,000
30,000
30,000
40,000
40,000
60,000
60,000
80,000
80,000
100,000
100,000
200,000
300,000
400,000
800
800
600
600
400
400
300
300
200
200
100
100
80
80
60
60
40
40
35
32.6
30
20
10
8
6
4
3
2
1
1,000
800
600
400
300
200
100
80
60
40
30
20
10
8
6
4
3
2
1
0.8
0.6
0.4
0.2
0.1
0.3
0.04
0.06
0.08
0.03
0.02
0.01
3,000
4,000
6,000
8,000
10,000
20,000
30,000
40,000
60,000
80,000
100,000
200,000
INDEX
1 2 3 4 6 8 10 20 30 40 60 80 100 200
2,000
1,000
1,000
2,000
2,000
3,000
3,000
4,000
4,000
6,000
6,000
8,000
8,000
10,000
10,000
20,000
20,000
30,000
30,000
40,000
40,000
60,000
60,000
80,000
80,000
100,000
100,000
200,000
300,000
400,000
800
800
600
600
400
400
300
300
200
200
100
100
80
80
60
60
40
40
35
32.6
30
20
10
8
6
4
3
2
1
1,000
800
600
400
300
200
100
80
60
40
30
20
10
8
6
4
3
2
1
0.8
0.6
0.4
0.2
0.1
0.3
0.04
0.06
0.08
0.03
0.02
0.01
Q
C
V
V
INDEX
H
R
FOR SELECTING VALVE SIZE
FOR PREDICTING PRESSURE DROP
C
CORRECTION FACTOR, F
V
V
F
V
R
- CENTISTOKES
CS
LIQUID FLOW COEFFICIENT, C
KINEMATIC VISCOSITY V
LIQUID FLOW RATE (SINGLE PORTED ONLY), GPM
LIQUID FLOW RATE (DOUBLE PORTED ONLY), GPM
REYNOLDS NUMBER - N
VISCOSITY - SAYBOLT SECONDS UNIVERSAL
Nomograph Instructions
Use this nomograph to correct for the effects of viscosity. When
assembling data, all units must correspond to those shown on the
nomograph. For high-recovery, ball-type valves, use the liquid
ow rate Q scale designated for single-ported valves. For buttery
and eccentric disk rotary valves, use the liquid ow rate Q scale
designated for double-ported valves.
Nomograph Equations
1. Single-Ported Valves:
2. Double-Ported Valves:
NR = 17250
NR = 12200
Figure 2. Nomograph for Determining Viscosity Correction
Q
C
V νCS
Q
C
V νCS
Nomograph Procedure
1. Lay a straight edge on the liquid sizing coefcient on C
scale and ow rate on Q scale. Mark intersection on index
line. Procedure A uses value of Cvc; Procedures B and C use
value of Cvr.
2. Pivot the straight edge from this point of intersection with
index line to liquid viscosity on proper n scale. Read Reynolds
number on NR scale.
3. Proceed horizontally from intersection on NR scale to proper
curve, and then vertically upward or downward to Fv scale.
Read Cv correction factor on Fv scale.
C
CORRECTION FACTOR, F
V
FOR PREDICTING
FLOW RATE
V
v
Valve Sizing Calculations (Traditional Method)
628
Te c h n i c a l
Predicting Flow Rate
Select the required liquid sizing coefcient (Cvr) from the
manufacturer’s published liquid sizing coefcients (Cv) for the
style and size valve being considered. Calculate the maximum
ow rate (Q
correction required) using the following adaptation of the basic
) in gallons per minute (assuming no viscosity
max
liquid sizing equation:
Q
= Cvr ΔP / G (4)
max
Then incorporate viscosity correction by determining the uid
Reynolds number and correction factor Fv from the viscosity
correction nomograph and the procedure included on it.
Calculate the predicted ow rate (Q
Q
pred
=
) using the formula:
pred
Q
max
(5)
F
V
Predicting Pressure Drop
Select the required liquid sizing coefcient (Cvr) from the published
liquid sizing coefcients (Cv) for the valve style and size being
considered. Determine the Reynolds number and correct factor Fv
from the nomograph and the procedure on it. Calculate the sizing
coefcient (Cvc) using the formula:
C
CVC =
Calculate the predicted pressure drop (∆P
ΔP
pred
vr
(6)
F
v
) using the formula:
pred
= G (Q/Cvc)2 (7)
Flashing and Cavitation
The occurrence of ashing or cavitation within a valve can have a
signicant effect on the valve sizing procedure. These two related
physical phenomena can limit ow through the valve in many
applications and must be taken into account in order to accurately
size a valve. Structural damage to the valve and adjacent piping
may also result. Knowledge of what is actually happening within
the valve might permit selection of a size or style of valve which
can reduce, or compensate for, the undesirable effects of ashing
or cavitation.
P
2
P
2
2
P
2
FLOW
FLOW
P
1
RESTRICTION
Figure 3. Vena Contracta
P
1
P
1
Figure 4. Comparison of Pressure Profiles for
High and Low Recovery Valves
VENA CONTRACTA
P
HIGH RECOVERY
LOW RECOVERY
The “physical phenomena” label is used to describe ashing and
cavitation because these conditions represent actual changes in
the form of the uid media. The change is from the liquid state
to the vapor state and results from the increase in uid velocity at
or just downstream of the greatest ow restriction, normally the
valve port. As liquid ow passes through the restriction, there is a
necking down, or contraction, of the ow stream. The minimum
cross-sectional area of the ow stream occurs just downstream of
the actual physical restriction at a point called the vena contracta,
as shown in Figure 3.
To maintain a steady ow of liquid through the valve, the velocity
must be greatest at the vena contracta, where cross sectional
area is the least. The increase in velocity (or kinetic energy) is
accompanied by a substantial decrease in pressure (or potential
energy) at the vena contracta. Farther downstream, as the uid
stream expands into a larger area, velocity decreases and pressure
increases. But, of course, downstream pressure never recovers
completely to equal the pressure that existed upstream of the
valve. The pressure differential (∆P) that exists across the valve
629
Te c h n i c a l
Valve Sizing Calculations (Traditional Method)
is a measure of the amount of energy that was dissipated in the
valve. Figure 4 provides a pressure prole explaining the differing
performance of a streamlined high recovery valve, such as a ball
valve and a valve with lower recovery capabilities due to greater
internal turbulence and dissipation of energy.
Regardless of the recovery characteristics of the valve, the pressure
differential of interest pertaining to ashing and cavitation is the
differential between the valve inlet and the vena contracta. If
pressure at the vena contracta should drop below the vapor pressure
of the uid (due to increased uid velocity at this point) bubbles
will form in the ow stream. Formation of bubbles will increase
greatly as vena contracta pressure drops further below the vapor
pressure of the liquid. At this stage, there is no difference between
ashing and cavitation, but the potential for structural damage to
the valve denitely exists.
If pressure at the valve outlet remains below the vapor pressure
of the liquid, the bubbles will remain in the downstream system
and the process is said to have “ashed.” Flashing can produce
serious erosion damage to the valve trim parts and is characterized
by a smooth, polished appearance of the eroded surface. Flashing
damage is normally greatest at the point of highest velocity, which
is usually at or near the seat line of the valve plug and seat ring.
However, if downstream pressure recovery is sufcient to raise the
outlet pressure above the vapor pressure of the liquid, the bubbles
will collapse, or implode, producing cavitation. Collapsing of the
vapor bubbles releases energy and produces a noise similar to what
one would expect if gravel were owing through the valve. If the
bubbles collapse in close proximity to solid surfaces, the energy
released gradually wears the material leaving a rough, cylinder
like surface. Cavitation damage might extend to the downstream
pipeline, if that is where pressure recovery occurs and the bubbles
collapse. Obviously, “high recovery” valves tend to be more
subject to cavitation, since the downstream pressure is more likely
to rise above the vapor pressure of the liquid.
Q
(GPM)
Q
(GPM)
K
m
C
v
Figure 5. Flow Curve Showing Cv and K
ACTUAL
FLOW
C
v
Figure 6. Relationship Between Actual ∆P and ∆P Allowable
∆P (ALLOWABLE)
ΔP
PREDICTED FLOW USING
ACTUAL ∆P
ACTUAL ∆P
∆P (ALLOWABLE)
ΔP
PLOT OF EQUATION (1)
CHOKED FLOW
P1 = CONSTANT
m
Choked Flow
Aside from the possibility of physical equipment damage due to
ashing or cavitation, formation of vapor bubbles in the liquid ow
stream causes a crowding condition at the vena contracta which
tends to limit ow through the valve. So, while the basic liquid
sizing equation implies that there is no limit to the amount of ow
through a valve as long as the differential pressure across the valve
increases, the realities of ashing and cavitation prove otherwise.
If valve pressure drop is increased slightly beyond the point where
bubbles begin to form, a choked ow condition is reached. With
constant upstream pressure, further increases in pressure drop (by
reducing downstream pressure) will not produce increased ow.
The limiting pressure differential is designated ∆P
recovery coefcient (Km) is experimentally determined for each
and the valve
allow
valve, in order to relate choked ow for that particular valve to the
basic liquid sizing equation. Km is normally published with other
valve capacity coefcients. Figures 5 and 6 show these ow vs.
pressure drop relationships.
Valve Sizing Calculations (Traditional Method)
630
Te c h n i c a l
1.0
1.0
0.9
c
0.8
0.7
0.6
CRITICAL PRESSURE RATIO—r
0.5
0 500 1000 1500 2000 2500 3000 3500
VAPOR PRESSURE, PSIA
USE THIS CURVE FOR WATER. ENTER ON THE ABSCISSA AT THE WATER VAPOR PRESSURE AT THE
VALVE INLET. PROCEED VERTICALLY TO INTERSECT THE
CURVE. MOVE HORIZONTALLY TO THE LEFT TO READ THE CRITICAL
PRESSURE RATIO, RC, ON THE ORDINATE.
Figure 7. Critical Pressure Ratios for Water Figure 8. Critical Pressure Ratios for Liquid Other than Water
c
0.9
0.8
0.7
0.6
CRITICAL PRESSURE RATIO—r
0.5
0 0.20 0.40 0.60 0.80 1.0
USE THIS CURVE FOR LIQUIDS OTHER THAN WATER. DETERMINE THE VAPOR
PRESSURE/CRITICAL PRESSURE RATIO BY DIVIDING THE LIQUID VAPOR PRESSURE
AT THE VALVE INLET BY THE CRITICAL PRESSURE OF THE LIQUID. ENTER ON THE ABSCISSA AT THE
INTERSECT THE CURVE. MOVE HORIZONTALLY TO THE LEFT AND READ THE CRITICAL
RATIO JUST CALCULATED AND PROCEED VERTICALLY TO
PRESSURE RATIO, RC, ON THE ORDINATE.
VAPOR PRESSURE, PSIA
CRITICAL PRESSURE, PSIA
Use the following equation to determine maximum allowable
pressure drop that is effective in producing ow. Keep in mind,
however, that the limitation on the sizing pressure drop, ∆P
does not imply a maximum pressure drop that may be controlled y
allow
,
the valve.
∆P
= Km (P1 - rc P v) (8)
allow
where:
∆P
= maximum allowable differential pressure for sizing
allow
purposes, psi
Km = valve recovery coefcient from manufacturer’s literature
P1 = body inlet pressure, psia
r
= critical pressure ratio determined from Figures 7 and 8
c
Pv = vapor pressure of the liquid at body inlet temperature,
psia (vapor pressures and critical pressures for
many common liquids are provided in the Physical
Constants of Hydrocarbons and Physical Constants
of Fluids tables; refer to the Table of Contents for the
page number).
After calculating ∆P
equation
Q = CV ∆P / G
actual ∆P is less the ∆P
, substitute it into the basic liquid sizing
allow
to determine either Q or Cv. If the
, then the actual ∆P should be used in
allow
the equation.
The equation used to determine ∆P
calculate the valve body differential pressure at which signicant
should also be used to
allow
cavitation can occur. Minor cavitation will occur at a slightly lower
pressure differential than that predicted by the equation, but should
produce negligible damage in most globe-style control valves.
Consequently, initial cavitation and choked ow occur nearly
simultaneously in globe-style or low-recovery valves.
However, in high-recovery valves such as ball or buttery valves,
signicant cavitation can occur at pressure drops below that which
produces choked ow. So although ∆P
predicting choked ow capacity, a separate cavitation index (Kc) is
and Km are useful in
allow
needed to determine the pressure drop at which cavitation damage
will begin (∆Pc) in high-recovery valves.
The equation can e expressed:
∆PC = KC (P1 - PV) (9)
This equation can be used anytime outlet pressure is greater than
the vapor pressure of the liquid.
Addition of anti-cavitation trim tends to increase the value of Km.
In other words, choked ow and incipient cavitation will occur at
substantially higher pressure drops than was the case without the
anti-cavitation accessory.
631
Te c h n i c a l
Valve Sizing Calculations (Traditional Method)
Liquid Sizing Equation Application
EQUATION APPLICATION
1
2 Use to calculate expected Cv for valve controlling water or other liquids that behave like water.
Q = Cv ΔP / G
CV = Q
G
∆P
Basic liquid sizing equation. Use to determine proper valve size for a given set of service conditions.
(Remember that viscosity effects and valve recovery capabilities are not considered in this basic equation.)
3
4 Use to nd maximum ow rate assuming no viscosity correction is necessary.
5 Use to predict actual ow rate based on equation (4) and viscosity factor correction.
6 Use to calculate corrected sizing coefcient for use in equation (7).
7
8
9
C
= FV C
vr
Q
= Cvr ΔP / G
max
Q
Q
=
pred
C
CVC =
F
ΔP
= G (Q/Cvc)
pred
∆P
= Km (P1 - rc P v)
allow
∆PC = KC (P1 - PV)
V
max
F
V
vr
v
2
Liquid Sizing Summary
The most common use of the basic liquid sizing equation is
to determine the proper valve size for a given set of service
conditions. The rst step is to calculate the required Cv by using
the sizing equation. The ∆P used in the equation must be the actual
valve pressure drop or ∆P
step is to select a valve, from the manufacturer’s literature, with a
Cv equal to or greater than the calculated value.
Accurate valve sizing for liquids requires use of the dual
coefcients of Cv and Km. A single coefcient is not sufcient
to describe both the capacity and the recovery characteristics of
the valve. Also, use of the additional cavitation index factor Kc
is appropriate in sizing high recovery valves, which may develop
damaging cavitation at pressure drops well below the level of the
choked ow.
, whichever is smaller. The second
allow
Use to nd actual required Cv for equation (2) after including viscosity correction factor.
Use to predict pressure drop for viscous liquids.
Use to determine maximum allowable pressure drop that is effective in producing ow.
Use to predict pressure drop at which cavitation will begin in a valve with high recovery characteristics.
∆P = differential pressure, psi
∆P
= maximum allowable differential pressure for sizing
allow
purposes, psi
∆Pc = pressure differential at which cavitation damage
begins, psi
Fv = viscosity correction factor
G = specic gravity of uid (water at 60°F = 1.0000)
Kc = dimensionless cavitation index used in
determining ∆P
c
Km = valve recovery coefcient from
manufacturer’s literature
P1 = body inlet pressure, psia
Pv = vapor pressure of liquid at body inlet
temperature, psia
Liquid Sizing Nomenclature
Cv = valve sizing coefcient for liquid determined
experimentally for each size and style of valve, using
water at standard conditions as the test uid
Cvc = calculated Cv coefcient including correction
for viscosity
Q = ow rate capacity, gallons per minute
Q
Q
= designation for maximum ow rate, assuming no
max
viscosity correction required, gallons per minute
= predicted ow rate after incorporating viscosity
pred
correction, gallons per minute
rc = critical pressure ratio
Cvr = corrected sizing coefcient required for
viscous applications
Valve Sizing Calculations (Traditional Method)
632
Te c h n i c a l
Sizing for Gas or Steam Service
A sizing procedure for gases can be established based on adaptions
of the basic liquid sizing equation. By introducing conversion
factors to change ow units from gallons per minute to cubic
feet per hour and to relate specic gravity in meaningful terms
of pressure, an equation can be derived for the ow of air at
60°F. Because 60°F corresponds to 520° on the Rankine absolute
temperature scale, and because the specic gravity of air at 60°F
is 1.0, an additional factor can be included to compare air at 60°F
with specic gravity (G) and absolute temperature (T) of any other
gas. The resulting equation an be written:
520
Q
SCFH
= 59.64 CVP
∆P
1
GT
P
1
The equation shown above, while valid at very low pressure
drop ratios, has been found to be very misleading when the ratio
of pressure drop (∆P) to inlet pressure (P1) exceeds 0.02. The
deviation of actual ow capacity from the calculated ow capacity
is indicated in Figure 8 and results from compressibility effects and
critical ow limitations at increased pressure drops.
Critical ow limitation is the more signicant of the two problems
mentioned. Critical ow is a choked ow condition caused by
increased gas velocity at the vena contracta. When velocity at the
vena contracta reaches sonic velocity, additional increases in ∆P
by reducing downstream pressure produce no increase in ow.
So, after critical ow condition is reached (whether at a pressure
drop/inlet pressure ratio of about 0.5 for glove valves or at much
lower ratios for high recovery valves) the equation above becomes
completely useless. If applied, the Cv equation gives a much higher
indicated capacity than actually will exist. And in the case of a
high recovery valve which reaches critical ow at a low pressure
drop ratio (as indicated in Figure 8), the critical ow capacity of
the valve may be over-estimated by as much as 300 percent.
The problems in predicting critical ow with a Cv-based equation
led to a separate gas sizing coefcient based on air ow tests.
The coefcient (Cg) was developed experimentally for each
type and size of valve to relate critical ow to absolute inlet
pressure. By including the correction factor used in the previous
equation to compare air at 60°F with other gases at other absolute
temperatures, the critical ow equation an be written:
(A)
∆P
= 0.5
P
1
Q
C
∆P
= 0.15
P
1
v
∆P / P
Figure 9. Critical Flow for High and Low Recovery
Valves with Equal C
LOW RECOVERY
HIGH RECOVERY
1
v
Universal Gas Sizing Equation
To account for differences in ow geometry among valves,
equations (A) and (B) were consolidated by the introduction of
an additional factor (C1). C1 is dened as the ratio of the gas
sizing coefcient and the liquid sizing coefcient and provides a
numerical indicator of the valve’s recovery capabilities. In general,
C1 values can range from about 16 to 37, based on the individual
valve’s recovery characteristics. As shown in the example, two
valves with identical ow areas and identical critical ow (Cg)
capacities can have widely differing C1 values dependent on the
effect internal ow geometry has on liquid ow capacity through
each valve. Example:
High Recovery Valve
Cg = 4680
Cv = 254
C1 = Cg/Cv
= 4680/254
= 18.4
Low Recovery Valve
Cg = 4680
Cv = 135
C1 = Cg/C
= 4680/135
= 34.7
v
Q
= CgP1 520 / GT (B)
critical
633
Te c h n i c a l
Valve Sizing Calculations (Traditional Method)
So we see that two sizing coefcients are needed to accurately
size valves for gas ow —Cg to predict ow based on physical size
or ow area, and C1 to account for differences in valve recovery
characteristics. A blending equation, called the Universal Gas
Sizing Equation, combines equations (A) and (B) by means of a
sinusoidal function, and is based on the “perfect gas” laws. It can
be expressed in either of the following manners:
(C)
Q
SCFH
520
=
GT
Cg P1 SIN
59.64
C
1
∆P
P
rad
1
OR
SCFH
520
=
GT
Cg P1 SIN
Q
3417
C
∆P
Deg
P
1
1
(D)
In either form, the equation indicates critical ow when the sine
function of the angle designated within the brackets equals unity.
The pressure drop ratio at which critical ow occurs is known
as the critical pressure drop ratio. It occurs when the sine angle
reaches π/2 radians in equation (C) or 90 degrees in equation
(D). As pressure drop across the valve increases, the sine angle
increases from zero up to π/2 radians (90°). If the angle were
allowed to increase further, the equations would predict a decrease
in ow. Because this is not a realistic situation, the angle must be
limited to 90 degrees maximum.
Although “perfect gases,” as such, do not exist in nature, there are a
great many applications where the Universal Gas Sizing Equation,
(C) or (D), provides a very useful and usable approximation.
General Adaptation for Steam and Vapors
The density form of the Universal Gas Sizing Equation is the most
general form and can be used for both perfect and non-perfect gas
applications. Applying the equation requires knowledge of one
additional condition not included in previous equations, that being
the inlet gas, steam, or vapor density (d1) in pounds per cubic foot.
(Steam density can be determined from tables.)
Then the following adaptation of the Universal Gas Sizing
Equation can be applied:
Q
= 1.06 d1 P1 Cg SIN Deg
lb/hr
3417
C
∆P
P
1
1
(E)
Special Equation Form for Steam Below 1000 psig
If steam applications do not exceed 1000 psig, density changes can
be compensated for by using a special adaptation of the Universal
Gas Sizing Equation. It incorporates a factor for amount of
superheat in degrees Fahrenheit (Tsh) and also a sizing coefcient
(Cs) for steam. Equation (F) eliminates the need for nding the
density of superheated steam, which was required in Equation
(E). At pressures below 1000 psig, a constant relationship exists
between the gas sizing coefcient (Cg) and the steam coefcient
(Cs). This relationship can be expressed: Cs = Cg/20. For higher
steam pressure application, use Equation (E).
Q
=
lb/hr
CS P
1
1 + 0.00065T
3417
SIN
sh
C
∆P
1
1
Deg
P
(F)
Gas and Steam Sizing Summary
The Universal Gas Sizing Equation can be used to determine
the ow of gas through any style of valve. Absolute units of
temperature and pressure must be used in the equation. When the
critical pressure drop ratio causes the sine angle to be 90 degrees,
the equation will predict the value of the critical ow. For service
conditions that would result in an angle of greater than 90 degrees,
the equation must be limited to 90 degrees in order to accurately
determine the critical ow.
Most commonly, the Universal Gas Sizing Equation is used to
determine proper valve size for a given set of service conditions.
The rst step is to calculate the required Cg by using the Universal
Gas Sizing Equation. The second step is to select a valve from
the manufacturer’s literature. The valve selected should have a C
which equals or exceeds the calculated value. Be certain that the
assumed C1 value for the valve is selected from the literature.
It is apparent that accurate valve sizing for gases that requires use
of the dual coefcient is not sufcient to describe both the capacity
and the recovery characteristics of the valve.
Proper selection of a control valve for gas service is a highly
technical problem with many factors to be considered. Leading
valve manufacturers provide technical information, test data, sizing
catalogs, nomographs, sizing slide rules, and computer or calculator
programs that make valve sizing a simple and accurate procedure.
g
Te c h n i c a l
Valve Sizing Calculations (Traditional Method)
Gas and Steam Sizing Equation Application
EQUATION APPLICATION
520
A Use only at very low pressure drop (DP/P1) ratios of 0.02 or less.
Q
= 59.64 CVP
SCFH
∆P
1
GT
P
1
B Use only to determine critical ow capacity at a given inlet pressure.
C
D
E
F Use only to determine steam ow when inlet pressure is 1000 psig or less.
Q
SCFH
Q
SCFH
Q
lb/hr
Q
=
lb/hr
Q
= CgP1 520 / GT
critical
520
=
=
GT
520
GT
Cg P1 SIN
Cg P1 SIN
= 1.06 d1 P1 Cg SIN Deg
CS P
1
1 + 0.00065T
59.64
C
OR
3417
C
3417
C
SIN
sh
1
1
1
3417
C
∆P
P
1
∆P
P
1
∆P
P
1
∆P
P
1
1
rad
Deg
Deg
Universal Gas Sizing Equation.
Use to predict ow for either high or low recovery valves, for any gas adhering to the
perfect gas laws, and under any service conditions.
Use to predict ow for perfect or non-perfect gas sizing applications, for any vapor
including steam, at any service condition when uid density is known.
Gas and Steam Sizing Nomenclature
C
C
C
C
d
G = gas specic gravity (air = 1.0)
P
= Cg/C
1
= gas sizing coefcient
g
= steam sizing coefcient, Cg/20
s
= liquid sizing coefcient
v
= density of steam or vapor at inlet, pounds/cu. foot
1
= valve inlet pressure, psia
1
v
∆P = pressure drop across valve, psi
Q
Q
Q
= critical ow rate, SCFH
critical
= gas ow rate, SCFH
SCFH
= steam or vapor ow rate, pounds per hour
lb/hr
T = absolute temperature of gas at inlet, degrees Rankine
T
= degrees of superheat, °F
sh
634
Valve Sizing (Standardized Method)
Introduction
Fisher® regulators and valves have traditionally been sized using
equations derived by the company. There are now standardized
calculations that are becoming accepted world wide. Some product
literature continues to demonstrate the traditional method, but the
trend is to adopt the standardized method. Therefore, both methods
are covered in this application guide.
Liquid Valve Sizing
Standardization activities for control valve sizing can be traced
back to the early 1960s when a trade association, the Fluids Control
Institute, published sizing equations for use with both compressible
and incompressible uids. The range of service conditions that
could be accommodated accurately by these equations was quite
narrow, and the standard did not achieve a high degree of acceptance.
In 1967, the ISA established a committee to develop and publish
standard equations. The efforts of this committee culminated in
a valve sizing procedure that has achieved the status of American
National Standard. Later, a committee of the International
Electrotechnical Commission (IEC) used the ISA works as a basis to
formulate international standards for sizing control valves. (Some
information in this introductory material has been extracted from
ANSI/ISA S75.01 standard with the permission of the publisher,
the ISA.) Except for some slight differences in nomenclature and
procedures, the ISA and IEC standards have been harmonized.
ANSI/ISA Standard S75.01 is harmonized with IEC Standards 5342-1 and 534-2-2. (IEC Publications 534-2, Sections One and Two for
incompressible and compressible uids, respectively.)
In the following sections, the nomenclature and procedures are explained,
and sample problems are solved to illustrate their use.
Sizing Valves for Liquids
Following is a step-by-step procedure for the sizing of control valves for
liquid ow using the IEC procedure. Each of these steps is important
and must be considered during any valve sizing procedure. Steps 3 and
4 concern the determination of certain sizing factors that may or may not
be required in the sizing equation depending on the service conditions of
the sizing problem. If one, two, or all three of these sizing factors are to
be included in the equation for a particular sizing problem, refer to the
appropriate factor determination section(s) located in the text after the
sixth step.
1. Specify the variables required to size the valve as follows:
• Desired design
• Process uid (water, oil, etc.), and
• Appropriate service conditions q or w, P1, P2, or ∆P, T1, Gf, Pv,
Pc, and υ.
The ability to recognize which terms are appropriate for
a specic sizing procedure can only be acquired through
experience with different valve sizing problems. If any
of the above terms appears to be new or unfamiliar, refer
to the Abbreviations and Terminology Table 3-1 for a
complete denition.
2. Determine the equation constant, N.
N is a numerical constant contained in each of the ow equations
to provide a means for using different systems of units. Values
for these various constants and their applicable units are given in
the Equation Constants Table 3-2.
Use N1, if sizing the valve for a ow rate in volumetric units
(GPM or Nm3/h).
Use N6, if sizing the valve for a ow rate in mass units
(pound/hr or kg/hr).
3. Determine Fp, the piping geometry factor.
Fp is a correction factor that accounts for pressure losses due
to piping ttings such as reducers, elbows, or tees that might
be attached directly to the inlet and outlet connections of the
control valve to be sized. If such ttings are attached to the
valve, the Fp factor must be considered in the sizing procedure.
If, however, no ttings are attached to the valve, Fp has a value of
1.0 and simply drops out of the sizing equation.
For rotary valves with reducers (swaged installations),
and other valve designs and tting styles, determine the Fp
factors by using the procedure for determining Fp, the Piping
Geometry Factor, page 637.
4. Determine q
conditions) or ∆P
The maximum or limiting ow rate (q
choked ow, is manifested by no additional increase in ow
rate with increasing pressure differential with xed upstream
conditions. In liquids, choking occurs as a result of
vaporization of the liquid when the static pressure within
the valve drops below the vapor pressure of the liquid.
The IEC standard requires the calculation of an allowable
sizing pressure drop (∆P
of choked ow conditions within the valve. The calculated
∆P
in the service conditions, and the lesser of these two values
is used in the sizing equation. If it is desired to use ∆P
account for the possibility of choked ow conditions, it can
be calculated using the procedure for determining q
Maximum Flow Rate, or ∆P
Drop. If it can be recognized that choked ow conditions will
not develop within the valve, ∆P
5. Solve for required C
• For volumetric ow rate units:
Cv =
• For mass ow rate units:
In addition to Cv, two other ow coefcients, Kv and Av, are
used, particularly outside of North America. The following
relationships exist:
Kv= (0.865) (Cv)
Av= (2.40 x 10-5) (Cv)
6. Select the valve size using the appropriate ow coefcient
table and the calculated Cv value.
Te c h n i c a l
(the maximum ow rate at given upstream
max
(the allowable sizing pressure drop).
max
), commonly called
max
), to account for the possibility
max
value is compared with the actual pressure drop specied
max
max
, the
max
Cv=
N
N6F
1Fp
w
p
, the Allowable Sizing Pressure
max
need not be calculated.
max
, using the appropriate equation:
v
q
(P1-P2) γ
P1 - P
G
2
f
to
635
Valve Sizing (Standardized Method)
636
Te c h n i c a l
SYMBOL SYMBOL
C
v
d Nominal valve size P
Valve sizing coefcient P
D Internal diameter of the piping P
F
d
F
F
F
k
F
L
F
LP
F
P
G
f
G
g
Valve style modier, dimensionless P
Liquid critical pressure ratio factor,
dimensionless
Ratio of specic heats factor, dimensionless ∆P
Rated liquid pressure recovery factor,
dimensionless
Combined liquid pressure recovery factor and
piping geometry factor of valve with attached
ttings (when there are no attached ttings, FLP
equals FL), dimensionless
Piping geometry factor, dimensionless q
Liquid specic gravity (ratio of density of liquid at
owing temperature to density of water at 60°F),
dimensionless
Gas specic gravity (ratio of density of owing
gas to density of air with both at standard
conditions
to molecular weight of air), dimensionless
k Ratio of specic heats, dimensionless x
K Head loss coefcient of a device, dimensionless x
M Molecular weight, dimensionless Y
N Numerical constant
1. Standard conditions are dened as 60°F and 14.7 psia.
Table 3-1. Abbreviations and Terminology
(1)
, i.e., ratio of molecular weight of gas
1
2
c
v
Upstream absolute static pressure
Downstream absolute static pressure
Absolute thermodynamic critical pressure
Vapor pressure absolute of liquid
at inlet temperature
∆P Pressure drop (P1-P2) across the valve
max(L)
∆P
max(LP)
Maximum allowable liquid sizing
pressure drop
Maximum allowable sizing pressure
drop with attached ttings
q Volume rate of ow
max
T
1
Maximum ow rate (choked ow conditions)
at given upstream conditions
Absolute upstream temperature
(deg Kelvin or deg Rankine)
w Mass rate of ow
Ratio of pressure drop to upstream absolute
static pressure (∆P/P1), dimensionless
T
Z Compressibility factor, dimensionless
1
γ
υ
Rated pressure drop ratio factor, dimensionless
Expansion factor (ratio of ow coefcient for a
gas to that for a liquid at the same Reynolds
number), dimensionless
Specic weight at inlet conditions
Kinematic viscosity, centistokes
kPa
bar
psia
- - - -
- - - -
- - - -
- - - kPa
bar
psia
kPa
bar
kPa
bar
kPa
bar
psia
kPa
bar
kPa
bar
(1)
(2)
γ
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
kg/m
kg/m
pound/ft
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
T d, D
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
3
3
3
- - - -
- - - -
- - - -
- - - -
deg Kelvin
deg Kelvin
deg Kelvin
deg Kelvin
deg Kelvin
deg Kelvin
deg Rankine
deg Kelvin
deg Kelvin
deg Kelvin
deg Kelvin
- - - -
- - - -
- - - mm
inch
mm
inch
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
N w q p
Table 3-2. Equation Constants
N
1
N
2
N
5
N
6
Normal Conditions
TN = 0°C
Standard Conditions
(3)
N
7
(3)
N
9
1. Many of the equations used in these sizing procedures contain a numerical constant, N, along with a numerical subscript. These numerical constants provide a means for using
different units in the equations. Values for the various constants and the applicable units are given in the above table. For example, if the ow rate is given in U.S. GPM and
the pressures are psia, N1 has a value of 1.00. If the ow rate is Nm3/h and the pressures are kPa, the N1 constant becomes 0.0865.
2. All pressures are absolute.
3. Pressure base is 101.3 kPa (1,01 bar) (14.7 psia).
Ts = 16°C
Standard Conditions
Ts = 60°F
N
8
Normal Conditions
TN = 0°C
Standard Conditions
TS = 16°C
Standard Conditions
TS = 60°F
0.0865
0.865
1.00
0.00214
890
0.00241
1000
2.73
27.3
63.3
3.94
394
4.17
417
1360 - - - - SCFH psia - - - - deg Rankine - - - -
0.948
94.8
19.3
21.2
2120
22.4
2240
7320 - - - - SCFH psia - - - - deg Rankine - - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - kg/hr
kg/hr
pound/hr
- - - -
- - - -
- - - -
- - - -
kg/hr
kg/hr
pound/hr
- - - -
- - - -
- - - -
- - - -
Nm3/h
Nm3/h
GPM
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
Nm3/h
Nm3/h
Nm3/h
Nm3/h
- - - -
- - - -
- - - -
Nm3/h
Nm3/h
Nm3/h
Nm3/h
Valve Sizing (Standardized Method)
637
Te c h n i c a l
Determining Piping Geometry Factor (Fp)
Determine an Fp factor if any ttings such as reducers, elbows, or
tees will be directly attached to the inlet and outlet connections
of the control valve that is to be sized. When possible, it is
recommended that Fp factors be determined experimentally by
using the specied valve in actual tests.
Calculate the Fp factor using the following equation:
-1/2
2
C
∑K
v
2
d
N
2
Fp=
1 +
where,
N2 = Numerical constant found in the Equation Constants table
d = Assumed nominal valve size
Cv = Valve sizing coefcient at 100% travel for the assumed
valve size
In the above equation, the ∑K term is the algebraic sum of the
velocity head loss coefcients of all of the ttings that are attached to
the control valve.
∑K = K1 + K2 + KB1 - K
B2
where,
K1 = Resistance coefcient of upstream ttings
K2 = Resistance coefcient of downstream ttings
KB1 = Inlet Bernoulli coefcient
KB2 = Outlet Bernoulli coefcient
The Bernoulli coefcients, KB1 and KB2, are used only when the
diameter of the piping approaching the valve is different from the
diameter of the piping leaving the valve, whereby:
4
KB1 or K
B2
= 1-
d
D
• For an outlet reducer:
K2= 1.0
• For a valve installed between identical reducers:
K1 + K2= 1.5
Determining Maximum Flow Rate (q
Determine either q
develop within the control valve that is to be sized. The values can
be determined by using the following procedures.
Values for FF, the liquid critical pressure ratio factor, can be
obtained from Figure 3-1, or from the following equation:
Values of FL, the recovery factor for rotary valves installed without
ttings attached, can be found in published coefcient tables. If the
given valve is to be installed with ttings such as reducer attached to
it, FL in the equation must be replaced by the quotient FLP/FP, where:
d
1-
D
N
q
=
1FLCV
max
FF = 0.96 - 0.28
2
2
2
1-
max
2
2
d
2
D
)
max
or ∆P
if it is possible for choked ow to
max
P1 - FF P
V
G
f
P
V
P
C
where,
d = Nominal valve size
D = Internal diameter of piping
If the inlet and outlet piping are of equal size, then the Bernoulli
coefcients are also equal, KB1 = KB2, and therefore they are dropped
from the equation.
The most commonly used tting in control valve installations is
the short-length concentric reducer. The equations for this tting
are as follows:
• For an inlet reducer:
2
2
d
K1= 0.5
1-
2
D
-1/2
2
1
C
K
V
1
FLP =
+
2
2
F
d
N
2
L
and
K1 = K1 + K
B1
where,
K1 = Resistance coefcient of upstream ttings
KB1 = Inlet Bernoulli coefcient
(See the procedure for Determining Fp, the Piping Geometry Factor,
for denitions of the other constants and coefcients used in the
above equations.)
Valve Sizing (Standardized Method)
638
Te c h n i c a l
ABSOLUTE VAPOR PRESSURE-bar
FACTOR—F
LIQUID CRITICAL PRESSURE RATIO
1.0
0.9
F
0.8
0.7
0.6
34
69
103
138
172 207
241
A2737-1
0.5
0
500
USE THIS CURVE FOR WATER, ENTER ON THE ABSCISSA AT THE WATER VAPOR
PRESSURE AT THE VALVE INLET, PROCEED VERTICALLY TO INTERSECT THE CURVE,
MOVE HORIZONTALLY TO THE LEFT TO READ THE CRITICAL PRESSURE RATIO, FF, ON
THE ORDINATE.
1000 1500
Figure 3-1. Liquid Critical Pressure Ratio Factor for Water
Determining Allowable Sizing Pressure Drop (∆P
∆P
(the allowable sizing pressure drop) can be determined from
max
the following relationships:
max
)
For valves installed without ttings:
∆P
max(L)
= F
L
2
(P1 - FF PV)
For valves installed with ttings attached:
2
F
∆P
max(LP)
=
LP
F
P
(P1 - FF PV)
where,
P1 = Upstream absolute static pressure
P2 = Downstream absolute static pressure
Pv = Absolute vapor pressure at inlet temperature
Values of FF, the liquid critical pressure ratio factor, can be obtained
from Figure 3-1 or from the following equation:
P
FF = 0.96 - 0.28
V
P
c
An explanation of how to calculate values of FLP, the recovery
factor for valves installed with ttings attached, is presented in the
preceding procedure Determining q
Once the ∆P
equation, it should be compared with the actual service pressure
value has been obtained from the appropriate
max
differential (∆P = P1 - P2). If ∆P
(the Maximum Flow Rate).
max
is less than ∆P, this is an
max
2000
ABSOLUTE VAPOR PRESSURE-PSIA
indication that choked ow conditions will exist under the service
conditions specied. If choked ow conditions do exist (∆P
< P1 - P2), then step 5 of the procedure for Sizing Valves for
Liquids must be modied by replacing the actual service pressure
differential (P1 - P2) in the appropriate valve sizing equation with
the calculated ∆P
Once it is known that choked ow conditions will
develop within the specied valve design (∆P
calculated to be less than ∆P), a further distinction
can be made to determine whether the choked
ow is caused by cavitation or ashing. The
choked ow conditions are caused by ashing if
the outlet pressure of the given valve is less than
the vapor pressure of the owing liquid. The
choked ow conditions are caused by cavitation if
the outlet pressure of the valve is greater than the
vapor pressure of the owing liquid.
Liquid Sizing Sample Problem
Assume an installation that, at initial plant startup, will not be
operating at maximum design capability. The lines are sized
for the ultimate system capacity, but there is a desire to install a
control valve now which is sized only for currently anticipated
requirements. The line size is 8-inch (DN 200) and an ASME
CL300 globe valve with an equal percentage cage has been
specied. Standard concentric reducers will be used to install the
valve into the line. Determine the appropriate valve size.
2500 3000
value.
max
Note
3500
max
max
is
Valve Sizing (Standardized Method)
639
Te c h n i c a l
1.0
0.9
F
0.8
FACTOR—F
0.7
LIQUID CRITICAL PRESSURE RATIO
0.6
0.5
0
0.10
0.20 0.30
USE THIS CURVE FOR LIQUIDS OTHER THAN WATER. DETERMINE THE VAPOR PRESSURE/
CRITICAL PRESSURE RATIO BY DIVIDING THE LIQUID VAPOR PRESSURE AT THE VALVE INLET
BY THE CRITICAL PRESSURE OF THE LIQUID. ENTER ON THE ABSCISSA AT THE RATIO JUST
CALCULATED AND PROCEED VERTICALLY TO INTERSECT THE CURVE. MOVE HORIZONTALLY
TO THE LEFT AND READ THE CRITICAL PRESSURE RATIO, FF, ON THE ORDINATE.
0.40
ABSOLUTE VAPOR PRESSURE
ABSOLUTE THERMODYNAMIC CRITICAL PRESSURE
Figure 3-2. Liquid Critical Pressure Ratio Factor for Liquids Other Than Water
1. Specify the necessary variables required to size the valve:
• Desired Valve Design—ASME CL300 globe valve with
equal percentage cage and an assumed valve size of 3-inches.
• Process Fluid—liquid propane
• Service Conditions—q = 800 GPM (3028 l/min)
P1 = 300 psig (20,7 bar) = 314.7 psia (21,7 bar a)
P2 = 275 psig (19,0 bar) = 289.7 psia (20,0 bar a)
∆P = 25 psi (1,7 bar)
T1 = 70°F (21°C)
Gf = 0.50
Pv = 124.3 psia (8,6 bar a)
Pc = 616.3 psia (42,5 bar a)
2. Use an N1 value of 1.0 from the Equation Constants table.
3. Determine Fp, the piping geometry factor.
Because it is proposed to install a 3-inch valve in
an 8-inch (DN 200) line, it will be necessary to determine
the piping geometry factor, Fp, which corrects for losses
caused by ttings attached to the valve.
0.50 0.60
Fp =
1 +
0.70
P
P
∑K
N
v
c
0.80
2
C
v
2
d
2
0.90
-1/2
1.00
where,
N2 = 890, from the Equation Constants table
d = 3-inch (76 mm), from step 1
Cv = 121, from the ow coefcient table for an ASME CL300,
3-inch globe valve with equal percentage cage
To compute ∑K for a valve installed between identical
concentric reducers:
∑K = K1 + K
= 1.5
= 1.5
1 -
1 -
2
D
(3)
(8)
2
2
d
2
2
2
2
= 1.11
Te c h n i c a l
Valve Sizing (Standardized Method)
where,
D = 8-inch (203 mm), the internal diameter of the piping so,
-1/2
2
1.11
Fp =
1 +
890
121
3
2
= 0.90
4. Determine ∆P
(the Allowable Sizing Pressure Drop.)
max
Based on the small required pressure drop, the ow will not
be choked (∆P
max
> ∆P).
5. Solve for Cv, using the appropriate equation.
Cv =
=
q
P1 - P
N1F
p
G
800
(1.0) (0.90)
2
f
25
0.5
= 125.7
6. Select the valve size using the ow coefcient table and the
calculated Cv value.
The required Cv of 125.7 exceeds the capacity of the assumed
valve, which has a Cv of 121. Although for this example it
may be obvious that the next larger size (4-inch) would
be the correct valve size, this may not always be true, and a
repeat of the above procedure should be carried out.
Assuming a 4-inches valve, Cv = 203. This value was
determined from the ow coefcient table for an ASME
CL300, 4-inch globe valve with an equal percentage cage.
Recalculate the required Cv using an assumed Cv value of
203 in the Fp calculation.
where,
∑K = K1 + K
= 1.5
= 1.5
1 -
1 -
2
2
2
d
2
D
2
16
64
= 0.84
and
-1/2
2
C
∑K
Fp =
=
1.0 +
1.0 +
N
0.84
890
v
2
d
2
-1/2
2
203
2
4
= 0.93
and
Cv =
N1F
=
(1.0) (0.93)
q
P1 - P
p
G
800
2
f
25
0.5
= 121.7
This solution indicates only that the 4-inch valve is large enough to
satisfy the service conditions given. There may be cases, however,
where a more accurate prediction of the Cv is required. In such
cases, the required Cv should be redetermined using a new Fp value
based on the Cv value obtained above. In this example, Cv is 121.7,
which leads to the following result:
-1/2
2
C
∑K
Fp =
=
1.0 +
1.0 +
N
0.84
890
2
v
2
d
121.7
4
-1/2
2
2
= 0.97
The required Cv then becomes:
Cv =
N1F
=
(1.0) (0.97)
q
P1 - P
800
2
G
f
p
25
0.5
= 116.2
Because this newly determined Cv is very close to the Cv used
initially for this recalculation (116.2 versus 121.7), the valve sizing
procedure is complete, and the conclusion is that a 4-inch valve
opened to about 75% of total travel should be adequate for the
required specications.
640
Valve Sizing (Standardized Method)
Te c h n i c a l
Gas and Steam Valve Sizing
Sizing Valves for Compressible Fluids
Following is a six-step procedure for the sizing of control valves
for compressible ow using the ISA standardized procedure.
Each of these steps is important and must be considered
during any valve sizing procedure. Steps 3 and 4 concern the
determination of certain sizing factors that may or may not
be required in the sizing equation depending on the service
conditions of the sizing problem. If it is necessary for one or
both of these sizing factors to be included in the sizing equation
for a particular sizing problem, refer to the appropriate factor
determination section(s), which is referenced and located in the
following text.
1. Specify the necessary variables required to size the valve
as follows:
• Desired valve design (e.g. balanced globe with linear cage)
• Process uid (air, natural gas, steam, etc.) and
• Appropriate service conditions—
q, or w, P1, P2 or P, T1, Gg, M, k, Z, and γ
The ability to recognize which terms are appropriate
for a specic sizing procedure can only be acquired through
experience with different valve sizing problems. If any
of the above terms appear to be new or unfamiliar, refer to
the Abbreviations and Terminology Table 3-1 in Liquid
Valve Sizing Section for a complete denition.
2. Determine the equation constant, N.
N is a numerical constant contained in each of the ow
equations to provide a means for using different systems of
units. Values for these various constants and their applicable
units are given in the Equation Constants Table 3-2 in
Liquid Valve Sizing Section.
Use either N7 or N9 if sizing the valve for a ow rate in
volumetric units (SCFH or Nm3/h). Which of the two
constants to use depends upon the specied service
conditions. N7 can be used only if the specic gravity, Gg,
of the following gas has been specied along with the other
required service conditions. N9 can be used only if the
molecular weight, M, of the gas has been specied.
Use either N6 or N8 if sizing the valve for a ow rate in mass
units (pound/hr or kg/hr). Which of the two constants to use
depends upon the specied service conditions. N6 can
be used only if the specic weight, γ1, of the owing gas has
been specied along with the other required service
conditions. N8 can be used only if the molecular weight, M,
of the gas has been specied.
3. Determine Fp, the piping geometry factor.
Fp is a correction factor that accounts for any pressure losses
due to piping ttings such as reducers, elbows, or tees that
might be attached directly to the inlet and outlet connections
of the control valves to be sized. If such ttings are attached
1
to the valve, the Fp factor must be considered in the sizing
procedure. If, however, no ttings are attached to the valve, Fp
has a value of 1.0 and simply drops out of the sizing equation.
Also, for rotary valves with reducers and other valve designs
and tting styles, determine the Fp factors by using the procedure
for Determining Fp, the Piping Geometry Factor, which is located
in Liquid Valve Sizing Section.
4. Determine Y, the expansion factor, as follows:
Cv =
x
3Fk x
T
x
3Fk x
N7 FP P1 Y
Note
T
q
x
Gg T1 Z
Y = 1 -
where,
Fk = k/1.4, the ratio of specic heats factor
k = Ratio of specic heats
x = P/P1, the pressure drop ratio
xT = The pressure drop ratio factor for valves installed without
attached ttings. More denitively, xT is the pressure
drop ratio required to produce critical, or maximum, ow
through the valve when Fk = 1.0
If the control valve to be installed has ttings such as reducers
or elbows attached to it, then their effect is accounted for in the
expansion factor equation by replacing the xT term with a new
factor xTP. A procedure for determining the xTP factor is
described in the following section for Determining xTP, the
Pressure Drop Ratio Factor.
Conditions of critical pressure drop are realized
when the value of x becomes equal to or exceeds the
appropriate value of the product of either Fk xT or
Fk xTP at which point:
y = 1 - = 1 - 1/3 = 0.667
Although in actual service, pressure drop ratios can, and often
will, exceed the indicated critical values, this is the point where
critical ow conditions develop. Thus, for a constant P1,
decreasing P2 (i.e., increasing P) will not result in an increase in
the ow rate through the valve. Values of x, therefore, greater
than the product of either FkxT or FkxTP must never be substituted
in the expression for Y. This means that Y can never be less
than 0.667. This same limit on values of x also applies to the ow
equations that are introduced in the next section.
5. Solve for the required Cv using the appropriate equation:
For volumetric ow rate units—
• If the specic gravity, Gg, of the gas has been specied:
641
Valve Sizing (Standardized Method)
642
Te c h n i c a l
• If the molecular weight, M, of the gas has been specied:
Cv =
N7 FP P1 Y
q
x
M T1 Z
For mass ow rate units—
• If the specic weight, γ1, of the gas has been specied:
N6 FP Y
w
x P1 γ
1
Cv =
• If the molecular weight, M, of the gas has been specied:
Cv =
N8 FP P1 Y
w
x M
T1 Z
In addition to Cv, two other ow coefcients, Kv and Av, are
used, particularly outside of North America. The following
relationships exist:
Kv = (0.865)(Cv)
Av = (2.40 x 10-5)(Cv)
6. Select the valve size using the appropriate ow coefcient table
and the calculated Cv value.
Determining xTP, the Pressure Drop Ratio Factor
If the control valve is to be installed with attached ttings such as
reducers or elbows, then their effect is accounted for in the expansion
factor equation by replacing the xT term with a new factor, xTP.
-1
2
xT
xTP = 1 +
F
p
xT K
2
i
N
5
where,
N5 = Numerical constant found in the Equation Constants table
d = Assumed nominal valve size
Cv = Valve sizing coefcient from ow coefcient table at
100% travel for the assumed valve size
Fp = Piping geometry factor
xT = Pressure drop ratio for valves installed without ttings attached.
xT values are included in the ow coefcient tables
In the above equation, Ki, is the inlet head loss coefcient, which is
dened as:
Ki = K1 + K
B1
where,
K1 = Resistance coefcient of upstream ttings (see the procedure
for Determining Fp, the Piping Geometry Factor, which is
contained in the section for Sizing Valves for Liquids).
Cv
d2
KB1 = Inlet Bernoulli coefcient (see the procedure for Determining
Fp, the Piping Geometry Factor, which is contained in the
section for Sizing Valves for Liquids).
Compressible Fluid Sizing Sample Problem No. 1
Determine the size and percent opening for a Fisher® Design V250
ball valve operating with the following service conditions. Assume
that the valve and line size are equal.
1. Specify the necessary variables required to size the valve:
• Desired valve design—Design V250 valve
• Process uid—Natural gas
• Service conditions—
P1 = 200 psig (13,8 bar) = 214.7 psia (14,8 bar)
P2 = 50 psig (3,4 bar) = 64.7 psia (4,5 bar)
P = 150 psi (10,3 bar)
x = P/P1 = 150/214.7 = 0.70
T1 = 60°F (16°C) = 520°R
M = 17.38
Gg = 0.60
k = 1.31
q = 6.0 x 106 SCFH
2. Determine the appropriate equation constant, N, from the
Equation Constants Table 3-2 in Liquid Valve Sizing Section.
Because both Gg and M have been given in the service
conditions, it is possible to use an equation containing either N
or N9. In either case, the end result will be the same. Assume
that the equation containing Gg has been arbitrarily selected for
this problem. Therefore, N7 = 1360.
3. Determine Fp, the piping geometry factor.
Since valve and line size are assumed equal, Fp = 1.0.
4. Determine Y, the expansion factor.
Fk =
=
k
1.40
1.31
1.40
= 0.94
It is assumed that an 8-inch Design V250 valve will be adequate
for the specied service conditions. From the ow coefcient
Table 4-2, xT for an 8-inch Design V250 valve at 100% travel
is 0.137.
x = 0.70 (This was calculated in step 1.)
7
Valve Sizing (Standardized Method)
643
Te c h n i c a l
Since conditions of critical pressure drop are realized
when the calculated value of x becomes equal to or
exceeds the appropriate value of FkxT, these values
should be compared.
FkxT = (0.94) (0.137)
= 0.129
Because the pressure drop ratio, x = 0.70 exceeds the
calculated critical value, FkxT = 0.129, choked ow
conditions are indicated. Therefore, Y = 0.667, and
x = FkxT = 0.129.
5. Solve for required Cv using the appropriate equation.
Cv =
N7 FP P1 Y
The compressibility factor, Z, can be assumed to be 1.0 for
the gas pressure and temperature given and Fp = 1 because
valve size and line size are equal.
So,
Cv = = 1515
(1360)(1.0)(214.7)(0.667)
6. Select the valve size using the ow coefcient table and
the calculated Cv value.
The above result indicates that the valve is adequately sized
(rated Cv = 2190). To determine the percent valve opening,
note that the required Cv occurs at approximately 83 degrees
for the 8-inch Design V250 valve. Note also that, at
83 degrees opening, the xT value is 0.252, which is
substantially different from the rated value of 0.137 used
initially in the problem. The next step is to rework the
problem using the xT value for 83 degrees travel.
The FkxT product must now be recalculated.
x = Fk x
= (0.94) (0.252)
= 0.237
The required Cv now becomes:
Cv =
=
= 1118
The reason that the required Cv has dropped so dramatically
is attributable solely to the difference in the xT values at rated
and 83 degrees travel. A Cv of 1118 occurs between 75 and
80 degrees travel.
q
x
Gg T1 Z
6.0 x 10
T
q
N7 FP P1 Y
(1360)(1.0)(214.7)(0.667)
Gg T1 Z
6
x
6.0 x 10
0.129
(0.6)(520)(1.0)
6
0.237
(0.6)(520)(1.0)
The appropriate ow coefcient table indicates that xT is higher at
75 degrees travel than at 80 degrees travel. Therefore, if the
problem were to be reworked using a higher xT value, this should
result in a further decline in the calculated required Cv.
Reworking the problem using the xT value corresponding to
78 degrees travel (i.e., xT = 0.328) leaves:
x = Fk x
= (0.94) (0.328)
= 0.308
and,
=
= 980
The above Cv of 980 is quite close to the 75 degree travel Cv. The
problem could be reworked further to obtain a more precise predicted
opening; however, for the service conditions given, an 8-inch
Design V250 valve installed in an 8-inch (203 mm) line
will be approximately 75 degrees open.
T
Cv =
N7 FP P1 Y
(1360)(1.0)(214.7)(0.667)
q
x
Gg T1 Z
6.0 x 10
6
0.308
(0.6)(520)(1.0)
Compressible Fluid Sizing Sample Problem No. 2
Assume steam is to be supplied to a process designed to operate at
250 psig (17 bar). The supply source is a header maintained at
500 psig (34,5 bar) and 500°F (260°C). A 6-inch (DN 150) line
from the steam main to the process is being planned. Also, make
the assumption that if the required valve size is less than 6-inch
(DN 150), it will be installed using concentric reducers. Determine
the appropriate Design ED valve with a linear cage.
1. Specify the necessary variables required to size the valve:
a. Desired valve design—ASME CL300 Design ED valve
with a linear cage. Assume valve size is 4 inches.
b. Process uid—superheated steam
c. Service conditions—
w = 125 000 pounds/hr (56 700 kg/hr)
P1 = 500 psig (34,5 bar) = 514.7 psia (35,5 bar)
P2 = 250 psig (17 bar) = 264.7 psia (18,3 bar)
P = 250 psi (17 bar)
x = P/P1 = 250/514.7 = 0.49
T1 = 500°F (260°C)
γ1 = 1.0434 pound/ft3 (16,71 kg/m3)
(from Properties of Saturated Steam Table)
k = 1.28 (from Properties of Saturated Steam Table)
Valve Sizing (Standardized Method)
644
Te c h n i c a l
2. Determine the appropriate equation constant, N, from the
Equation Constants Table 3-2 in Liquid Valve Sizing Section.
Because the specied ow rate is in mass units, (pound/hr), and
the specic weight of the steam is also specied, the only sizing
equation that can be used is that which contains the N6 constant.
Therefore,
N6 = 63.3
3. Determine Fp, the piping geometry factor.
Fp = 1 +
K
Σ
N
2
Cv
d2
-1/2
2
where,
N2 = 890, determined from the Equation Constants Table
d = 4 inches
Cv = 236, which is the value listed in the ow coefcient
Table 4-3 for a 4-inch Design ED valve at 100%
total travel.
K = K1 + K
Σ
= 1.5 1 -
= 1.5 1 -
2
2
d2
D2
2
42
62
= 0.463
Finally,
-1/2
2
Fp = 1 +
0.463
890
(1.0)(236)
(4)
2
= 0.95
4. Determine Y, the expansion factor.
Y = 1 -
3Fkx
x
TP
where,
k
Fk =
1.40
1.28
=
1.40
= 0.91
x = 0.49 (As calculated in step 1.)
Because the 4-inch valve is to be installed in a 6-inch line, the xT
term must be replaced by xTP.
-1
2
xT
xTP = 1 +
2
F
p
xT K
N
5
Cv
i
d2
where,
N5 = 1000, from the Equation Constants Table
d = 4 inches
Fp = 0.95, determined in step 3
xT = 0.688, a value determined from the appropriate
listing in the ow coefcient table
Cv = 236, from step 3
and
Ki = K1 + K
= 0.5 1 - + 1 -
= 0.5 1 - + 1 -
B1
2
2
d
D2 D
2
2
4
62
4
d
4
4
6
= 0.96
where D = 6-inch
so:
-1
2
xTP = 1 + = 0.67
0.69
0.952
(0.69)(0.96)
1000
236
42
Finally:
Y = 1 -
= 1 -
x
3 F
k xTP
0.49
(3) (0.91) (0.67)
= 0.73
5. Solve for required Cv using the appropriate equation.
Cv =
=
w
N6 FP Y
x P1 γ
(63.3)(0.95)(0.73)
1
125,000
(0.49)(514.7)(1.0434)
= 176
Valve Sizing (Standardized Method)
645
Te c h n i c a l
Table 4-1. Representative Sizing Coefcients for Type 1098-EGR Regulator
LINEAR CAGE
BODY SIZE,
INCHES (DN)
1 (25) 16.8 17.7 17.2 18.1 0.806 0.43
2 (50) 63.3 66.7 59.6 62.8 0.820 0.35
3 (80) 132 139 128 135 0.779 0.30
4 (100) 202 213 198 209 0.829 0.28
6 (150) 397 418 381 404 0.668 0.28
BODY SIZE,
INCHES (DN)
1 (25) 16.7 17.6 15.6 16.4 0.753 0.10
2 (50) 54 57 52 55 0.820 0.07
3 (80) 107 113 106 110 0.775 0.05
4 (100) 180 190 171 180 0.766 0.04
6 (150) 295 310 291 306 0.648 0.03
Line Size Equals Body Size 2:1 Line Size to Body Size
C
v
Regulating Wide-Open Regulating Wide-Open
Line Size Equals Body Size Piping 2:1 Line Size to Body Size Piping
C
v
Regulating Wide-Open Regulating Wide-Open
C
v
WHISPER TRIMTM CAGE
C
v
X
T
X
T
F
D
F
D
F
0.84
F
0.89
L
L
VALVE SIZE,
INCHES
1
1-1/2
2
3
4
6
8
10
12
16
Table 4-2. Representative Sizing Coefcients for Rotary Shaft Valves
VALVE STYLE
V-Notch Ball Valve 60
V-Notch Ball Valve 60
V-Notch Ball Valve
High Performance Buttery Valve
V-Notch Ball Valve
High Performance Buttery Valve
V-Notch Ball Valve
High Performance Buttery Valve
V-Notch Ball Valve
High Performance Buttery Valve
V-Notch Ball Valve
High Performance Buttery Valve
V-Notch Ball Valve
High Performance Buttery Valve
V-Notch Ball Valve
High Performance Buttery Valve
V-Notch Ball Valve
High Performance Buttery Valve
DEGREES OF
VALVE OPENING
90
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
60
90
C
15.6
34.0
28.5
77.3
59.2
132
58.9
80.2
120
321
115
237
195
596
270
499
340
1100
664
1260
518
1820
1160
2180
1000
3000
1670
3600
1530
3980
2500
5400
2380
8270
3870
8600
V
F
0.86
0.86
0.85
0.74
0.81
0.77
0.76
0.71
0.80
0.74
0.81
0.64
0.80
0.62
0.69
0.53
0.80
0.58
0.66
0.55
0.82
0.54
0.66
0.48
0.80
0.56
0.66
0.48
0.78
0.63
- - - -
- - - -
0.80
0.37
0.69
0.52
L
X
0.53
0.42
0.50
0.27
0.53
0.41
0.50
0.44
0.50
0.30
0.46
0.28
0.52
0.22
0.32
0.19
0.52
0.20
0.33
0.20
0.54
0.18
0.31
0.19
0.47
0.19
0.38
0.17
0.49
0.25
- - - -
- - - -
0.45
0.13
0.40
0.23
T
F
- - - -
- - - -
- - - -
- - - -
- - - -
- - - -
0.49
0.70
0.92
0.99
0.49
0.70
0.92
0.99
0.49
0.70
0.91
0.99
0.49
0.70
0.91
0.99
0.49
0.70
0.91
0.99
0.49
0.70
0.92
0.99
0.49
0.70
0.92
1.00
- - - -
- - - -
D
Te c h n i c a l
Valve Sizing (Standardized Method)
Table 4-3. Representative Sizing Coefcients for Design ED Single-Ported Globe Style Valve Bodies
VALVE SIZE,
INCHES
1/2 Post Guided Equal Percentage 0.38 (9,7) 0.50 (12,7) 2.41 0.90 0.54 0.61
3/4 Post Guided Equal Percentage 0.56 (14,2) 0.50 (12,7) 5.92 0.84 0.61 0.61
1
1-1/2
2 Cage Guided
3 Cage Guided
4 Cage Guided
6 Cage Guided
8 Cage Guided
VALVE PLUG
STYLE
Micro-Form
Cage Guided
Micro-Form
Cage Guided
TM
TM
FLOW
CHARACTERISTICS
Equal Percentage
Linear
Equal Percentage
Equal Percentage
Linear
Equal Percentage
Linear
Equal Percentage
Linear
Equal Percentage
Linear
Equal Percentage
Linear
Equal Percentage
Linear
Equal Percentage
PORT DIAMETER,
INCHES (mm)
3/8
(9,5)
1/2
(12,7)
3/4
(19,1)
1-5/16
(33,3)
1-5/16
(33,3)
3/8
(9,5)
1/2
(12,7)
3/4
(19,1)
1-7/8
(47,6)
1-7/8
(47,6)
2-5/16
(58,7)
2-5/16
(58,7)
3-7/16 (87,3) 1-1/2 (38,1) 148
- - - - - - - -
4-3/8 (111) 2 (50,8) 236
- - - - - - - -
7 (178) 2 (50,8) 433
- - - - - - - -
8 (203) 3 (76,2) 846
- - - - - - - -
RATED TRAVEL,
INCHES (mm)
3/4
(19,1)
3/4
(19,1)
3/4
(19,1)
3/4
(19,1)
3/4
(19,1)
3/4
(19,1)
3/4
(19,1)
3/4
(19,1)
3/4
(19,1)
3/4
(19,1)
1-1/8
(28,6)
1-1/8
(28,6)
C
3.07
4.91
8.84
20.6
17.2
3.20
5.18
10.2
39.2
35.8
72.9
59.7
136
224
394
818
V
F
0.89
0.93
0.97
0.84
0.88
0.84
0.91
0.92
0.82
0.84
0.77
0.85
0.82
0.82
0.82
0.82
0.84
0.85
0.87
0.86
L
X
0.66
0.80
0.92
0.64
0.67
0.65
0.71
0.80
0.66
0.68
0.64
0.69
0.62
0.68
0.69
0.72
0.74
0.78
0.81
0.81
T
F
0.72
0.67
0.62
0.34
0.38
0.72
0.67
0.62
0.34
0.38
0.33
0.31
0.30
0.32
0.28
0.28
0.28
0.26
0.31
0.26
D
6. Select the valve size using ow coefcient tables and the
calculated C
value.
v
Refer to the ow coefcient Table 4-3 for Design ED valves with
linear cage. Because the assumed 4-inch valve has a Cv of 236 at
100% travel and the next smaller size (3-inch) has a Cv of
only 148, it can be surmised that the assumed size is correct. In
the event that the calculated required Cv had been small enough
to have been handled by the next smaller size, or if it had been
larger than the rated Cv for the assumed size, it would have been
necessary to rework the problem again using values for the new
assumed size.
7. Sizing equations for compressible uids.
The equations listed below identify the relationships between
ow rates, ow coefcients, related installation factors, and
pertinent service conditions for control valves handling
compressible uids. Flow rates for compressible uids may
be encountered in either mass or volume units and thus equations
are necessary to handle both situations. Flow coefcients may be
calculated using the appropriate equations selected from the
following. A sizing ow chart for compressible uids is given in
Annex B.
The ow rate of a compressible uid varies as a function of the
ratio of the pressure differential to the absolute inlet pressure
( P/P1), designated by the symbol x. At values of x near zero,
the equations in this section can be traced to the basic Bernoulli
equation for Newtonian incompressible uids. However,
increasing values of x result in expansion and compressibility
effects that require the use of appropriate factors (see Buresh,
Schuder, and Driskell references).
7.1 Turbulent ow
7.1.1 Non-choked turbulent ow
7.1.1.1 Non-choked turbulent ow without attached ttings
[Applicable if x < FγxT]
The ow coefcient shall be calculated using one of the
following equations:
Eq. 6
N6Y
W
N8P1Y
W
xP1ρ
1
T1Z
xM
C =
Eq. 7
C =
Eq. 8a
C =
Q
N9P1Y
MT1Z
x
Eq. 8b
C =
NOTE 1 Refer to 8.5 for details of the expansion factor Y.
Q
N7P1Y
GgT1Z
x
NOTE 2 See Annex C for values of M.
7.1.1.2 Non-choked turbulent ow with attached ttings
[Applicable if x < FγxTP]
646
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