ence between inlet and outlet pressure) in PSI.
G -
Specific Gravity (Taken from Properties of Liquids)
EXAMPLE
GIVEN:
Flow - 20 GPM of Water
Inlet pressure - 100 PSIG
Outlet pressure - 95 PSIG
FIND THE C
v
REQUIRED.
SOLUTION
Q = 20 GPM
Inlet pressure = 100 PSI
Outlet pressure = 95
P = 5 PSI
Media = Water
Specific Gravity of Water = 1.0
Cv = Q G = 20 1.0
P 5
Cv = 20 x 1 = 8.9
2.24
NOTE
1 GALLON OF WATER EQUALS 8.336 LBS.
1 LB. OF WATER EQUALS .1198 GALLONS PRODUCT APPLICATIONS ENGINEERING DATA - 4
GAS FLOW C
v
EQUATION
SUBSONIC FLOW
DEFINITION
Flow is subsonic when the
P (differential pressure) is less than 1/2 of the inlet pres-
sure.
C
v
= Q G
P
2
P
LEGEND
C
v
-
Flow coefficient
Q -
Flow in SCFM
P - Differential Pressure (Difference between inlet and outlet pressure) in PSI.
G -
Specific gravity of Media (Taken from Properties of Gases)
P
1
-
Inlet pressure in PSIA (PSIG + 14.7)
P
2
-
Outlet pressure in PSIA (PSIG + 14.7)
EXAMPLE
GIVEN: Flow - 100 SCFM of N
2
Inlet Pressure - 100 PSIG
Outlet Pressure - 75 PSIG
FIND THE C
v
REQUIRED.
SOLUTION
Q = 100 SCFM N
2
Inlet Pressure = 100 PSIG
P
1
= 100 PSIG + 14.7 = 114.7 PSIA
Outlet Pressure = 75 PSIG
P
2
= 75 PSIG + 14.7 = 89.7 PSIA
P = P
1
- P
2
= 114.7 PSIA - 89.7 PSIA
P = 25 PSI
Media = N
2
Specific Gravity of N
2
= 0.067
C
v
= Q G
P
2

P
C
v
= 100 0.967
89.7 x 25
C
v
= 100 x 0.983 = 98.33
C
v
= 2.07
2242 47.4 PRODUCT APPLICATIONS ENGINEERING DATA - 5
GAS FLOW C
v
EQUATION
SONIC FLOW
DEFINITION
Flow is sonic when the
P (Differential Pressure) is equal to or greater than 1/2 of the
inlet pressure.
C
v
= Q G
P
1
/ 2
LEGEND
C
v
-
Flow coefficient.
Q

-
Flow in SCFM.
P - Differential Pressure (Difference between inlet and outlet pressure) in PSI.
G -
Specific Gravity of Media. (Taken from Properties of Gases)
P
1
-
Inlet Pressure in PSIA. (PSIG + 14.7)
P
2
-
Outlet Pressure in PSIA. (PSIG + 14.7)
EXAMPLE
GIVEN: Flow = 100 SCFM of N
2
Inlet Pressure = 100 PSIG
Outlet Pressure = 25 PSIG
FIND THE C
v
REQUIRED.
SOLUTION
Q = 100 SCFM of N
2
Inlet Pressure = 100 PSIG
P
1
= 100 PSIG + 14.7 = 114.7 PSIA
Outlet Pressure = 25 PSIG
P
2
= 25 PSIG + 14.7 = 39.7 PSIA
P = P
1
- P
2
= 114.7 - 39.7 = 75 PSI
Media - N
2
Specific Gravity of N
2
= 0.967
C
v
= Q _G = 100 0.967 = 100 x 0.9533
P
1
/ 2 114.7 / 2 57.35
C
v
= 1.7