Engineering Formulas
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
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
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
Cylinder Rod End Area (in square inches)
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
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
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
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
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
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
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.
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
Pneumatic Valve Sizing
Notes:
-
All these pneumatic formulas assume 68 degrees F at sea level
-
All strokes and diameters are in inches
-
All times are in seconds
-
All pressures are PSI
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
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)
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
Conversions
| To Convert | Into | Multiply By |
|---|---|---|
| Bar | PSI | 14.5 |
| cc | Cu. In. | 0.06102 |
| °C | °F | (°C x 1.8) + 32 |
| Kg | lbs. | 2.205 |
| KW | HP | 1.341 |
| Liters | Gallons | 0.2642 |
| mm | Inches | 0.03937 |
| Nm | lb.-ft | 0.7375 |
| Cu. In. | cc | 16.39 |
| °F | °C | (°F – 32) / 1.8 |
| Gallons | Liters | 3.785 |
| HP | KW | 0.7457 |
| Inch | mm | 25.4 |
| lbs. | Kg | 0.4535 |
| lb.-ft. | Nm | 1.356 |
| PSI | Bar | 0.06896 |
| In. of HG | PSI | 0.4912 |
| In. of H20 | PSI | 0.03613 |