# Engineering Formulas

Useful engineering formulas for the design of fluid power systems.

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.

## 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 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

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

### 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 |