# flow data flow factor and orifice size

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```					                                                      ENGINEERING INFORMATION                                                                12
flow data
flow factor and orifice size

Importance of properly sizing valves                Conditions to be known                           The formulas necessary to determine the
Kv are quite complicated and for that
It is important to properly size a valve.           In general, we must obtain as many of the        reason ASCO/JOUCOMATIC has devel-
There are undesirable affects in both               conditions surrounding the application as        oped a series of flow graphs which reduce
undersizing and oversizing.                         possible.                                        that problem to one of a simple multiplica-
tion or division.
Undersizing may result in:                          Flow required in cubic metres per hour           All flow calculation work of a medium has
(m3/h) as used for liquids, Normal Cubic         been simplified to a basic formula:
1) inability to pass desired flow requirements      Metre per hour (nm3/h) as used for gases,
2) flashing of liquids to vapours on the            or Kilogram per hour (kg/h) as used for
Flow required: Q
outlet size of the valve                         steam. This can be obtained by merely            Kv = ---------------------
3) lowering the outlet pressure                     asking the customer's requirements or                 Graph factors: Fgm, Fsg, Fgl
4) creating a substantial pressure loss in          from nameplates on pumping equipment,
a piping system                                  boiler room charts or calculations.              The graph factors Fgm, Fsg, Fgl can be
easily picked out by aligning known pres-
Oversizing may result in:                           Inlet Pressure (p1) - This is usually known      sure conditions on the graphs I to X on the
by knowing the source of the supply or           following pages (for calculations see next
1) unnecessary cost in oversized equipment          readily obtained by placing a gauge near         page).
2) variable flow through the valve or er-           the valve inlet.
ratic control of the flow                                                                         The tables below can be used to estimate
3) shorter life of some valve designs               Outlet Pressure (p2) - This can be ob-           a Kv if the orifice size is known or relate
through oscillating of internal parts            tained by gauge observations but usually         the approximate orifice size if the Kv is
caused by lack of flow to maintain re-           is tied in with specifications regarding         known. The chart is based on the ASCO/
quired internal pressure differentials           allowable system pressure drop. If we            JOUCOMATIC design of in-line globe type
4) erratic operation of some designs such           know the inlet pressure and the pressure         valves. The flow charts must be used for
as failure to shift position due to lack of      drop, we, of course, know the outlet pres-       precise sizing and converting Kv factors
required flow in 3- and 4-way valves             sure.                                            to actual flow terms and the catalogue
5) erosion or wire drawing of seats in                                                               sheet must be consulted for the actual Kv
some designs because they operate at             Pressure Drop (∆p) - In large or compli-         of a particular valve.
nearly closed position                           cated systems, it is desirable to keep the
pressure drop across a valve to a mini-
Definition of Kv                                    mum and often the customer will have
definite specifications concerning the fac-
The flow coefficient Kv in cubic metres per         tor. Of course, if the valve is discharging      Approx.                Approx.
orifice approx. Kv     orifice approx. Kv
hour or litres per minute is a special volu-        to atmosphere, the pressure drop is equal
size                   size
metric flow rate (capacity) through a valve         to the inlet pressure when dealing with           (mm) (m3/h) (l/min)    (mm) (m3/h)(l/min)
at a specified travel and at the following          liquids. With gases and steam, although
conditions:                                         the valve may be discharging to atmos-             0.8   0,02    0,33      13       3 50,0
phere, when sizing a valve, only 50 per-
- the static pressure loss (∆pKv) across            cent of the inlet pressure can be used for                                 16       4    66,7
the valve is 105 Pa (1 bar)                       the pressure drop used in the formulas             1.2   0,05    0,83
- the fluid is water within a temperature           (commonly called critical pressure drop).                                  18      4,5 75,0
range of 278 K to 313 K (5°C to 40°C)             In all other cases the pressure drop is, the
1.6   0,08    1,33      19      6,5   108
- the unit of the volumetric flow rate is the       difference between inlet and outlet pres-
cubic metre per hour or liters per minute         sures.
25      11    183
2.4   0,17    2,83
The value of Kv can be obtained from test           Note: It often is difficult to understand the
32      15    250
results with the help of the following equa-        meaning of the term "minimum operating
tion:                                               pressure differential" (see page V1210).           3.2   0,26    4,33
38      22    366
Certain pilot operated valves operate by
∆pKv . ρ
Kv = Q                                              differential pressures created internally                                  51      41    683
∆p . ρw                                  by "pilot" and "bleed" arrangements. This
3.6   0,31    5,17

where:                                              differential is measured as the difference                                 64      51    850
Q    is the measured volumetric flow rate in        between inlet and outlet conditions on all         4.8   0,45    7,50
cubic metres per hour or liters per minute     valve construction. If pressure conditions                                 76      86 1433
∆pKv is the static pressure loss of 105 Pa (see     are not known, but only flow information,
we can use the graphs or formulas to               6.4   0,60 10,0         80      99 1650
above)
∆p is the measured static pressure loss             solve the resulting pressure drop.
If the drop is less than assigned minimum                                 100     150 2500
across the valve in pascals                                                                             1,5    25,0
differential, the valve is oversized. In these     8
ρ    is the density of the fluid in kilograms per                                                                             125     264 4400
cubic metre                                    situations, offer a valve with a lower min.
ρw is the density of water (see above) in           operating pressure differential, or go to a        9     1,7    28,3      150     383 6375
kilograms per cubic metre (according to        smaller sized valve with a closer Kv.
IEC 534)
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V1215-GB-R2     12
ENGINEERING INFORMATION SECTION 12

SAMPLE PROBLEMS

LIQUIDS (tables I and III)                      AIR AND GASES (tables I and IV - VII)              STEAM (tables VIII - X)

To find Kv: What Kv is required to pass         To find Kv: A valve is required to pass 14         To find Kv: A valve is required to pass 25
22 litres of oil per minute with a specific     nm3/h at an inlet pressure of 4 bar and a          kg/h of saturated steam at an inlet pres-
gravity of 0,9 and a pressure drop of 1,5       pressure drop (∆p) of 0,5 bar.                     sure of 1 bar and a ∆p of 0,2 bar.
bar?                                            Find the Kv if the medium is carbon diox-          What is the Kv?
ide.
The viscosity is less than 9° Engler.

Solution: The formula is:                       Solution: Refer to the 1-10 bar graph.             Solution: Refer to the appropriate Steam
The formula used is:                               Graph.
Use the formula:

Q (m3 /h)                                          Q (Nm3 /h)                                      Q (kg/h)
Kv (m3 /h) =                                     Kv (Nm3 /h) =                                     Kv (m3 /h) =
Fgm . Fsg                                          Fgm . Fsg                                         Fgm

Q (m3 /h)                                         Q (Nm3 /h)                                       Q (kg/h)
Kv (l/min) =                                    Kv (Nl/min) =                                      Kv (l/min) =
Fgl . Fsg                                          Fgl . Fsg                                         Fgl

Locate Fgm at the intersection of 4 bar            Locate Fg on graph corresponding to 1
To find Fg, use the Liquid Flow Graph.          inlet pressure and 0,5 bar pressure drop           bar inlet pressure and a ∆p of 0,2 bar
∆p (along curve). Read down to Fgm =               (along curve).
The Fgm factor is that which corresponds        43,5.                                              Fgm = 13,8 and the Fgl = 0,83
to a pressure drop of 1,5 bar and equals        Fgl factor is 2,61
1,25.
The Fgl factor is 0,075.                        Locate Fsg corresponding to specific grav-
ity of carbon dioxide (= 1,5) on Fsg Chart.
The Fsg factor can be obtained from the         Fsg = 0,81.
Fsg chart and is that which corresponds
to 0,09 specific gravity and equals 1,05.

Insert values into formula :                       Insert values into formula :
Insert values into formula :
−3
Q (Nm3 /h)       14                                Q (kg/h)
Kv = 60.22.10 = 1 m3 /h                         Kv =              =             = 0, 4 Nm3 /h      Kv =            = 25 = 1, 8 m3 /h
1, 25.1, 05                                      Fgm . Fsg    43, 5.0, 81                             Fgm     13, 8
−3
Q (Nm3 /h)       14                                Q (kg/h)
Kv = 60.22.10 = 16, 7 l/min                     Kv =              =             = 6, 62 Nl/min     Kv =            = 25 = 30 l/min
0, 075.1, 05                                       Fgl . Fsg   2, 61.0, 81                             Fgl     0, 83

Formula for liquids                              Formula for gases (with temperature correction)
∆p                                                      ∆p(2P1 − ∆p)
Q (m3 /h) = Kv                                   Q (Nm3 /h) = Kv.18, 9                  ⋅ 293
S.G.                                                        (S.G.)     (273 + t 2 )

∆p                                                ∆p(2P1 − ∆p)
Q (dm3 /min) = Kv1                               Q (Ndm3 /h) = Kv.18, 9                  ⋅ 293
S.G.                                                  (S.G.)     (273 + t 2 )

S.G. : specific gravity related to water for liquids and to air for gases
t2 : fluid temperature (in °C)

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V1215-2
TECHNICAL INFORMATION SECTION 12
Table I : Calculation factor Fsg                                                                       Table II : Calculation factor Ft for temperature correction
Factor Fsg

Factor Ft

FLUID TEMPERATURE t2 (°C)
Specific gravity (S.G.)
The correction for temperature in
OTHER GRAVITIES                                                                                                        OTHER TEMPERATURES                       the range of -7°C to 65°C is very
small and, therefore, can be ignored
specific gravity (for 1 bar absolute and 15°C)                                                            in ordinary applications

Table III : Flow calculation Fgm and Fgl for liquids

0,54
Factor Fgm (m3/h)
Factor Fgl (l/min)

0,48

0,42

0,36

0,30

0,24

0,18

0,12

0,06

0,03

0

Pressure drop ∆p (bar)

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V1215-3       12
ENGINEERING INFORMATION SECTION 12
Table IV : Flow calculation Fgm and Fgl for air/gas

Pressure drop ∆p (bar)
Inlet pressure 0,01 to 0,1 bar gauge)

Limiting flow curve - Do not read
beyond this curve

Factor Fgm (m3/h)
0,17     0,18           0,21                   0,24                   0,27               0,30               0,36              0,42              0,48              0,54
Factor Fgl (l/min)

Table V : Flow calculation Fgm and Fgl for air/gas
Pressure drop ∆p (bar)

Limiting flow curve - Do not read
Inlet pressure of 0,1 to 1 bar (gauge)

beyond this curve

Factor Fgm (m3/h)
0,24    0,30     0,36   0,42    0,48          0,6          0,72       0,84     0,96      1,08     1,2      1,32      1,44     1,56      1,68     1,8      1,92     2,04
0,54         0,66          0,78     0,9      1,02      1,14    1,26      1,38      1,5      1,62      1,74    1,86     1,98      2,1
Factor Fgl (l/min)
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V1215-4
ENGINEERING INFORMATION SECTION 12
Table VI : Flow calculation Fgm and Fgl for air/gas
Pressure drop ∆p (bar)
Inlet pressure of 1 to 10 bar (gauge)

Limiting flow curve - Dot not read
beyond this curve

Factor Fgm (m3/h)
0,6       1,2       1,8   2,4        3,0        3,6   4,2    4,8    5,4        6         6,6      7,2        7,8        8,4    9         9,6        1,02   1,08
Factor Fgl (l/min)

Table VII : Flow calculation Fgm and Fgl for air/gas
Pressure drop ∆p (bar)
Inlet pressure 10 to 100 bar (gauge)

Limiting flow curve - Do not read
beyond this curve

Factor Fgm (m3/h)
0         6               12         18           24         30     36       42     48    54      60         66   72    78    84         90    96    102
Factor Fgl (l/min)
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V1215-5   12
ENGINEERING INFORMATION SECTION 12
Table VIII : Flow calculation Fgm and Fgl for steam
Pressure drop ∆p (bar)
Inlet pressure of 0,1 to 1 bar (gauge)

Limiting flow curve - Do not read
beyond this curve

Factor Fgm (m3/h)
0,18    0,3     0,42   0,54  0,66   0,78    0,9          1,02     1,08     1,14     1,2         1,26     1,32    1,38    1,44    1,5         1,56     1,62     1,68
0,24    0,36    0,48 0,6    0,72    0,84    0,96
Factor Fgl (l/min)

Table IX : Flow calculation Fgm and Fgl for steam                                                                                                                              Pressure drop ∆p (bar)
Inlet pressure 1 to 10 bar (gauge)

Limiting flow curve - Do not read
beyond this curve

Factor Fgm (m3/h)
0               0,6          1,2           1,8     2,4      3,0      3,6      4,2         4,8      5,4     6,0     6,6     7,2         7,8      8,4      9,6
Factor Fgl (l/min)

Table X : Flow calculation Fgm and Fgl for steam                                                                                                                               Pressure drop ∆p (bar)
Inlet pressure of 10 to 100 bar (gauge)

Limiting flow curve - Do not read
beyond this curve

Factor Fgm (m3/h)
0       6                12            18              24       30       36       42           48      54      60      66      72          78       84
Factor Fgl (l/min)
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V1215-6
ENGINEERING INFORMATION SECTION 12

ADDITIONAL FLOW FORMULAS AND                    Definition of Kv- (or Cv-) coefficient           expressed in units volume "A" per time
PHYSICAL DATA                                                                                    unit "B", that will pass through a valve
Valve flow coefficient Kv (or Cv) is the         with a pressure drop equal to pressure
flow of water (specific gravity = 1),            unit "C".
(See table below)

Kv- and Cv-conversion table
units
symbol                     conversion formulas
volume "A"                time "B"             press. "C"
litre                        min.                bar                         Kvl           16,7 Kv       = 17,3 Cve = 14,4             Cv
cubic metre                  hour                bar                         Kv            0,06 Kvl      = 1,04 Cve = 0,865            Cv
imp. gallon                  min.                psi                         Cve           0,058 Kvl     = 0,963 Kv = 0,833            Cv
U.S. gallon                  min.                psi                         Cv            0,069 Kvl     = 1,16 Kv = 1,2               Cve

Flow calculation                                Air and Gases                                    Steam and vapours (e.g. refrigerants)

General: Pressure drop values for which                                                          For steam:
no curves are shown, may be determined          Fgm = 18, 9 ∆p (2p1 − ∆p) (m3/h)
by interpolation in the graphs. However,                                                         Fgm = 15, 83 ∆p(2P1 − ∆P) (m3/h)
Fgl = 1, 13 ∆p (2p1 − ∆p) (l/min)
more accurate results can be obtained for
Fgl = 0, 95 ∆p(2P1 − ∆P) (l/min)
the calculation of the required values by
using the following equations (on which         Example: ∆p = 0,4 bar; p1 = 3 bar
the flow graphs are based):                     gauge or 4,013 bar absolute.                     Example : ∆P = 7 bar,
P1 = 40 bar or 41,013 bar abs.
p1 = absolute inlet pressure (bar) =            Calculation:
gauge pressure plus atmospheric                                                             Calculation :
Fgm = 18, 9 0, 4(8, 026 − 0, 4) = 33 m3 /h
pressure of 1,013 bar
Fgmgm = 15, 83 (82, 026 − 7) 7) = 363 m3 /h
F = 15, 83 7 7(82, 026 − = 363 m /h
3

p2 = absolute outlet pressure (bar) =           Fgl = 1, 13 0, 4(8, 026 − 0, 4) = 1, 97 l/min
gauge pressure plus atmospheric
pressure of 1,013 bar                      Note: The gas equations only apply accu-             gl 0, 0, 7 7(82, 026 − = 21, 8 8 l/ min
FglF= = 9595 (82, 026 − 7) 7) = 21,l/ min
∆p = p1 - p2 = pressure drop across the         rately to a medium temperature of 20°C
valve (bar)                                (for the purpose of this catalogue the
t  = 0°C                                                                                         Note 1: The steam formulas apply to
standard cubic meter nm3 has been de-
saturated steam. For superheated steam
fined at 20°C and 1,013 bar absolute or
Note: In most systems it is desirable to                                                         a correction factor is required.
760 mm mercury).
keep the pressure drop to a minimum. If                                                          Consult your ASCO/JOUCOMATIC sup-
At a different temperature (= t2 °C) the
necessary - in case of liquids - the pres-                                                       plier.
sure drop may equal the total inlet (gauge)     by the following correction factor.
pressure. This also applies to air, gases                                                        Note 2: For vapours (e.g. Freon) various
and steam up to 1,013 bar inlet (gauge)                                                          other factors have to be considered.
pressure but for these media never us a         Ft =     293
∆p greater than 50% of the absolute inlet              273 + t 2
pressure because excessive pressure
drops will cause an irregular flow. If∆p is     Specific gravity of various liquids at
not specified and this information is needed    20°C (related to water at 4°C)
to size the valve, a rule of thumb is to take                                                    Specific gravity of various gases (at
10% of the inlet pressure as pressure           Alcohol, Ethyl                0,79
20°C and atm. pressure and related to
Bezene                        0,88
drop.                                                                                            air)
Carbon-tetrachloride          1,589
Castor Oil                    0,95
Liquids                                                                                          Acetylene                          0,91
Fuel Oil no. 1                0,83
Air                                1,000
Fgm =     ∆p      3
(m /h)                           Fuel Oil no. 2                0,84
Ammonia                            0,596
Fuel Oil no. 3                0,89
et                                                                                               Butane                             2,067
Fuel Oil no. 4                0,91
Fuel Oil no. 5                0,95               Carbon-dioxide                     1,53
Fgl = 0, 06 ∆p (l/min)                                                                           Chloride                           2,486
Fuel Oil no. 6                0,99
Ethane                             1,05
Gasoline (petrol)             0,75 to 0,78
Example: for ∆p = 1,7 bar will be found                                                          Ethyl-chloride                     2,26
Glycerine                     1,26
Fgm = 1,3 (m3/h) and Fgl = 0,08 (l/min)                                                          Helium                             0,138
Linseed Oil                   0,94
Methane                            0,554
Olive Oil                     0,98
Methyl-chloride                    1,785
Note: if the medium viscosity is higher         Turpentine                    0,862
Nitrogen                           0,971
than 300 SSU (approx. 9°E), the deter-          Water                         1,000
Oxygen                             1,105
mined Kv-value must be adjusted. Con-                                                            Propane                            1,56
sult your ASCO/JOUCOMATIC supplier.                                                  Kv1         Sulphur-dioxide                    2,264
The actual flow factor is: Kv 2 =
Ft

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V1215-7   12
ENGINEERING INFORMATION SECTION 12

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V1215-8

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