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Hydraulic Force: Formulas & Calculator

Hydraulic Cylinder: Volume Displacement

Hydraulic Oil Pump: HP Required

Horsepower for driving a pump:

For every 1 hp of drive, the equivalent of 1 gpm @ 1500 psi can be produced.

Horsepower for idling a pump:

To idle a pump when it is unloaded will require about 5% of it's full rated power

Wattage for heating hydraulic oil:

Each watt will raise the temperature of 1 gallon of oil by 1° F. per hour.

Flow velocity in hydraulic lines:

Pump suction lines 2 to 4 feet per second, pressure lines up to 500 psi - 10 to 15 ft./sec., pressure lines 500 to 3000 psi - 15 to 20 ft./sec.; all oil lines in air-over-oil systems; 4 ft./sec.

Variable |
Word Formula w/ Units |
Simplified Formula |

Fluid Pressure - P |
(PSI) = Force (Pounds) / Area ( Sq. In.) | P = F / A |

Fluid Flow Rate - Q |
GPM= Flow (Gallons) / Unit Time (Minutes) | Q = V / T |

Fluid Power in Horsepower - HP |
Horsepower = Pressure (PSIG) × Flow (GPM)/ 1714 | HP = PQ / 1714 |

Variable |
Word Formula w/ Units |
Simplified Formula |

Cylinder Area - A |
( Sq. In.) = π × Radius (inch)2 | A = π × R2 |

(Sq. In.) = π × Diameter (inch)2 / 4 | A = π × D2 / 4 | |

Cylinder Force - F |
(Pounds) = Pressure (psi) × Area (sq. in.) | F = P × A |

Cylinder Speed - v |
(Feet / sec.) = (231 × Flow Rate (gpm)) / (12 × 60 × Area) | v = (0.3208 × gpm) / A |

Cylinder Volume Capacity - V |
Volume = π × Radius2 × Stroke (In.) / 231 |
V = π × R2 × L / 231 (L = length of stroke) |

Cylinder Flow Rate - Q |
Volume = 12 × 60 × Velocity (Ft./Sec.) × Net Area(In.)2 / 231 | Q = 3.11688 × v × A |

Fluid Motor Torque - T |
Torque (in. lbs.) = Pressure (psi) × disp. (in.3 / rev.) / 6.2822 | T = P × d / 6.2822 |

Torque = HP × 63025 / RPM | T = HP × 63025 / n | |

Torque = Flow Rate (GPM) × Pressure × 36.77 / RPM | T = 36.77 × Q × P / n | |

Fluid Motor Speed - n |
Speed (RPM) = (231 × GPM) / Disp. (in.)3 | n = (231 × GPM) / d |

Fluid Motor Horsepower - HP |
HP = Torque (in. lbs.) × rpm / 63025 | HP = T × n / 63025 |

Variable |
Word Formula w/ Units |
Simplified Formula |

Pump Output Flow - GPM |
GPM = (Speed (rpm) × disp. (cu. in.)) / 231 | GPM = (n ×d) / 231 |

Pump Input Horsepower - HP |
HP = GPM × Pressure (psi) / 1714 × Efficiency | HP = (Q ×P) / 1714 × E |

Pump Efficiency - E |
Overall Efficiency = Output HP / Input HP | EOverall = HPOut / HPIn X 100 |

Overall Efficiency = Volumetric Eff. × Mechanical Eff. | EOverall = EffVol. × EffMech. | |

Pump Volumetric Efficiency - E |
Volumetric Efficiency = Actual Flow Rate Output (GPM) / Theoretical Flow Rate Output (GPM) × 100 | EffVol. = QAct. / QTheo. X 100 |

Pump Mechanical Efficiency - E |
Mechanical Efficiency = Theoretical Torque to Drive / Actual Torque to Drive × 100 | EffMech = TTheo. / TAct. × 100 |

Pump Displacement - CIPR |
Displacement (In.3 / rev.) = Flow Rate (GPM) × 231 / Pump RPM | CIPR = GPM × 231 / RPM |

Pump Torque - T |
Torque = Horsepower × 63025 / RPM | T = 63025 × HP / RPM |

Torque = Pressure (PSIG) × Pump Displacement (CIPR) / 2π | T = P × CIPR / 6.28 |

Throughout the year, we will hold various training events to inform and educate attendees on Hydrotech products, divisions, and principles of hydraulic technologies.

### CLICK HERE to view upcoming training events and register today

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MK Technology Group QuickDesigner Configurator

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- Basic Fluid Power Formulas
- Actuator Formulas
- Pump Formulas
- General Fluid Power Guidelines

- Valve Sizing for Cylinder Actuation Direct Formula
- Calculators
- Converters

- Electrical Units
- Electrical Formulas

**09-25-2017 | Level 2 Hydraulic Training - Hydraulic System Maintenance**The goal of this 2nd level industrial hydraulic course is to develop maintenance personnel with advanced hydraulic skills.

**01-30-2017 | Level 1 Hydraulic Training - Principles of Hydraulics**This is a 5 day training program that teaches maintenance personnel the principles of industrial hydraulics.