Fluid Power Formulas
The following formulas are readily available in many engineering textbooks, fluid power design guides, and hydraulic handbooks. Every effort has been made to insure the accuracy of the formulas and the examples shown. However, it is possible that a typographical error or two has slipped in. Please double check any results that don't seem right.
This page is copyright 2003 by Customized Creations. You may link to it, but please do not copy it onto your web site. Industrial Marketing does have permission to include this page on their web site.
Click on a link below to go to the formula.
Hydraulic Pump Calculations Horsepower Required to Drive Pump GPM X PSI X .0007 (this is a 'rule-of-thumb'
calculation) How many horsepower are needed to drive a 10 gpm pump at 1750 psi? GPM
= 10
PSI = 1750
GPM X PSI X .0007 = 10 X 1750 X .0007 = 12.25 horsepower
top of page
Pump Output Flow (in Gallons Per Minute) RPM
X Pump Displacement / 231 How much oil will be produced by a 2.21 cubic inch pump operating
at 1120 rpm? RPM = 1120
Pump Displacement = 2.21 cubic inches
RPM X Pump Displacement / 231 = 1120 X 2.21 / 231 = 10.72 gpm
top of page
Pump Displacement Needed for GPM of Output Flow 231
X GPM / RPM What displacement is needed to produce 7 gpm at 1740 rpm? GPM = 7
RPM = 1740
231 X GPM / RPM = 231 X 7 / 1740 = 0.93 cubic inches per revolution
Hydraulic Cylinder Calculations Cylinder Blind End Area (in square inches) PI
X (Cylinder Radius) ^2 What is the area of a 6" diameter cylinder? Diameter = 6"
Radius is 1/2 of diameter = 3"
Radius ^2 = 3" X 3" = 9"
PI X (Cylinder Radius )^2 = 3.14 X (3)^2 = 3.14 X 9 = 28.26 square inches
top of page
Cylinder Rod End Area (in square inches) Blind
End Area - Rod Area What is the rod end area of a 6" diameter cylinder which has a 3" diameter
rod? Cylinder Blind End Area = 28.26 square inches
Rod Diameter = 3"
Radius is 1/2 of rod diameter = 1.5"
Radius ^2 = 1.5" X 1.5" = 2.25"
PI X Radius ^2 = 3.14 X 2.25 = 7.07 square inches Blind End Area - Rod
Area = 28.26 - 7.07 = 21.19 square inches
top of page
Cylinder Output Force (in Pounds) Pressure
(in PSI) X Cylinder Area What is the push force of a 6" diameter cylinder operating
at 2,500 PSI? Cylinder Blind End Area = 28.26 square inches
Pressure = 2,500 psi
Pressure X Cylinder Area = 2,500 X 28.26 = 70,650 pounds
What is the pull force of a 6" diameter cylinder with a 3" diameter rod
operating at 2,500 PSI? Cylinder Rod End Area = 21.19 square inches
Pressure = 2,500 psi
Pressure X Cylinder Area = 2,500 X 21.19 = 52,975 pounds
top of page
Fluid Pressure in PSI Required to Lift Load (in PSI) Pounds
of Force Needed / Cylinder Area What pressure is needed to develop 50,000 pounds of
push force from a 6" diameter cylinder? Pounds of Force = 50,000 pounds
Cylinder Blind End Area = 28.26 square inches
Pounds of Force Needed / Cylinder Area = 50,000 / 28.26 = 1,769.29 PSI
What pressure is needed to develop 50,000 pounds of pull force from a 6" diameter
cylinder which has a 3: diameter rod? Pounds of Force = 50,000 pounds
Cylinder Rod End Area = 21.19 square inches
Pounds of Force Needed / Cylinder Area = 50,000 / 21.19 = 2,359.60 PSI
Cylinder Speed (in inches per second) (231
X GPM) / (60 X Net Cylinder Area) How fast will a 6" diameter cylinder with a 3" diameter
rod extend with 15 gpm input? GPM = 6
Net Cylinder Area = 28.26 square inches
(231 X GPM) / (60 X Net Cylinder Area) = (231 X 15) / (60 x 28.26) = 2.04 inches per second
How fast will it retract? Net Cylinder Area = 21.19 square inches
(231 X GPM) / (60 X Net Cylinder Area) = (231 X 15) / (60 x 21.19) = 2.73 inches per second
top of page
GPM of Flow Needed for Cylinder Speed Cylinder
Area X Stroke Length in Inches / 231 X 60 / Time in seconds for one stroke How many
GPM are needed to extend a 6" diameter cylinder 8 inches in 10 seconds? Cylinder Area = 28.26 square inches
Stroke Length = 8 inches
Time for 1 stroke = 10 seconds
Area X Length / 231 X 60 / Time = 28.26 X 8 / 231 X 60 / 10 = 5.88 gpm
If the cylinder has a 3" diameter rod, how many gpm is needed to retract 8 inches
in 10 seconds? Cylinder Area = 21.19 square inches
Stroke Length = 8 inches
Time for 1 stroke = 10 seconds
Area X Length / 231 X 60 / Time = 21.19 X 8 / 231 X 60 / 10 = 4.40 gpm
top of page
Cylinder Blind End Output (GPM) Blind
End Area / Rod End Area X GPM In How many GPM come out the blind end of a 6" diameter cylinder with a 3" diameter
rod when there is 15 gallons per minute put in the rod end? Cylinder Blind End Area =28.26 square inches
Cylinder Rod End Area = 21.19 square inches
GPM Input = 15 gpm
Blind End Area / Rod End Area X GPM In = 28.26 / 21.19 * 15 = 20 gpm
Hydraulic Motor Calculations GPM of Flow Needed for Fluid Motor Speed Motor Displacement X Motor RPM / 231 How
many GPM are needed to drive a 2.51 cubic inch motor at 1200 rpm? Motor Displacement = 2.51 cubic inches per revolution
Motor RPM = 1200
Motor Displacement X Motor RPM / 231 = 2.51 X 1200 / 231 = 13.04 gpm
Fluid Motor Speed from GPM Input 231
X GPM / Fluid Motor Displacement How fast will a 0.95 cubic inch motor turn with 8
gpm input? GPM = 8
Motor Displacement = 0.95 cubic inches per revolution
231 X GPM / Fluid Motor Displacement = 231 X 8 / 0.95 = 1,945 rpm
Fluid Motor Torque from Pressure and Displacement PSI
X Motor Displacement / (2 X PI) How much torque does a 2.25 cubic inch motor develop
at 2,200 psi? Pressure = 2,200 psi
Displacement = 2.25 cubic inches per revolution
PSI X Motor Displacement / (2 x PI) = 2,200 X 2.25 / 6.28 = 788.22 inch pounds
Fluid Motor Torque from Horsepower and RPM Horsepower
X 63025 / RPM How much torque is developed by a motor at 15 horsepower and 1500 rpm? Horsepower = 15
RPM = 1500
Horsepower X 63025 / RPM = 15 X 63025 / 1500 = 630.25 inch pound
Fluid Motor Torque from GPM, PSI and RPM GPM
X PSI X 36.77 / RPM How much torque does a motor develop at 1,250 psi, 1750 rpm, with
9 gpm input? GPM = 9
PSI = 1,250
RPM = 1750
GPM X PSI X 36.7 / RPM = 9 X 1,250 X 36.7 / 1750 = 235.93 inch pounds second
Fluid & Piping
Calculations
Velocity of Fluid through Piping 0.3208 X GPM / Internal Area What
is the velocity of 10 gpm going through a 1/2" diameter schedule 40 pipe?
GPM = 10
Internal Area = .304 (see note below)
0.3208 X GPM / Internal Area = .3208 X 10 X .304 = 10.55 feet per second
Note: The outside diameter of pipe remains the same regardless of the thickness
of the pipe. A heavy duty pipe has a thicker wall than a standard duty pipe, so the
internal diameter of the heavy duty pipe is smaller than the internal diameter of
a standard duty pipe. The wall thickness and internal diameter of pipes can be found
on readily available charts. Hydraulic steel tubing also maintains the same outside
diameter regardless of wall thickness. Hose sizes indicate the inside diameter of
the plumbing. A 1/2" diameter hose has
an internal diameter of 0.50 inches, regardless of the hose pressure rating.
Suggested Piping Sizes
-
Pump suction lines should be sized so the fluid velocity is between 2 and 4 feet per second.
-
Oil return lines should be sized so the fluid velocity is between 10 and 15 feet per second.
-
Medium pressure supply lines should be sized so the fluid velocity is between 15 and 20 feet per second.
-
High pressure supply lines should be sized so the fluid velocity is below 30 feet per second.
top of page
Heat Calculations
Heat Dissipation Capacity of Steel Reservoirs 0.001 X Surface Area
X Difference between oil and air temperature If the oil temperature is 140 degrees,
and the air temperature is 75 degrees, how much heat will a reservoir with 20 square
feet of surface area dissipate?
Surface Area = 20 square feet
Temperature Difference = 140 degrees - 75 degrees = 65 degrees
0.001 X Surface Area X Temperature Difference = 0.001 X 20 X 65 = 1.3 horsepower
Note: 1 HP = 2,544 BTU per Hour
Heating Hydraulic Fluid 1 watt will raise the temperature of 1 gallon by 1 degree F per hour
and
Horsepower X 745.7 = watts
and
Watts / 1000 = kilowatts top of page
Pneumatic Valve Sizing
Notes:
Valve Sizing for Cylinder Actuation SCFM = 0.0273 x Cylinder
Diameter x Cylinder Diameter x Cylinder Stroke / Stroke Time x ((Pressure-Pressure
Drop)+14.7) / 14.7 Cv Required = 1.024 x SCFM / (Square Root of (Pressure Drop
x (Pressure-Pressure Drop+14.7))) Pressure 2 (PSIG) = Pressure-Pressure Drop
Air Flow Q (in SCFM) if Cv is Known Valve Cv x (Square Root of (Pressure Drop x ((PSIG - Pressure Drop) + 14.7))) / 1.024
Cv if Air Flow Q (in SCFM) is Known 1.024 x Air Flow / (Square Root of (Pressure Drop x ((PSIG-Pressure Drop) + 14.7)))
Air Flow Q (in SCFM) to Atmosphere SCFM
to Atmosphere = Valve Cv x (Square Root of (((Primary Pressure x 0.46) + 14.7) x (Primary
Pressure x 0.54))) / 1.024 Pressure Drop Max (PSIG) = Primary Pressure x 0.54
Flow Coefficient for Smooth Wall Tubing Cv of Tubing =(42.3 x Tube I.D. x Tube I.D. x 0.7854 x (Square Root (Tube I.D. / 0.02 x Length of Tube x 12)
- Conversions
-
Inches = Millimeters/25.4
PSIG = Pressure (Kgf/cm2) x 14.223
SCFM = NI/min x 0.035
Millimeters = Inches x 25.4
Pressure (Kgf/cm2) = PSIG x 0.0703
NI/min = SCFM x 28.571
- top of page
|
