COURSE INTRODUCTION

  • Welcome to the IIT Basic Hydraulics Interactive Computer Based Training Course.
  • Purpose: To provide a broad understanding of important hydraulic concepts.
  • Goals upon completion:   - Understanding various basic physics laws as they apply to fluid power.   - Understanding schematics and system design.   - Studying the components of a hydraulic system and their functions and interactions.
  • Navigation: Select topics from the course menu.   - Recommended progression: Start with Fluid Power Physics and continue in order.
  • Interactive nature: Study at your own pace and review materials periodically after course completion.
  • Manual supplement for the course includes:   - Notes   - Additional applications   - Expanded formulas   - Quiz questions
  • Manual available on the Web Portal: www.fluidpowerzone.com
  • Note: Quiz answers found in the quiz sections on the CD. A Post Test is provided online.
  • Copyright © 2000 by Interactive Industrial Training

FLUID POWER PHYSICS

Introduction

  • Goals after completion: Better understanding of basic physics principles governing fluid power.
  • These principles provide a solid foundation for further learning in fluid power.

Energy

  • Definition: Fluid power as a method of transferring energy.   - Energy transferred from a prime mover (input source) to an actuator (output device).   - Transfer methods can vary in efficiency but can yield optimal work control when applied correctly.
  • Energy: The ability to do work.   - Work: Defined as force through distance.   - Example: Moving 1000 pounds a distance of 2 feet equals 2000 foot-pounds of work.   - Units: Work is measured in foot-pounds.
  • Power: The rate of doing work, defined as work over time.   - Example: Lifting 1000 pounds 2 feet in 2 seconds yields 1000 units of power, calculated as     1000imes22extseconds\frac{1000 imes 2}{2 ext{ seconds}}.   - Horsepower (hp) is often used to measure power.

NOTES

Work and Power Definitions

  • Work (in lbs) = Force (lbs) x Distance (in)
  • Power = Force x Distance / Time
  • Important Note: Systems are generally less than 10% efficient; efficiency factors must be considered in calculations.
  • Example Calculation: Input horsepower = 10extgpmimes1500extpsi1714=8.75exthp\frac{10 ext{ gpm} imes 1500 ext{ psi}}{1714} = 8.75 ext{ hp}; thus, rounded to 10 hp, considering efficiency.
  • Rule of Thumb: 1 gpm at 1500 psi = 1.0 input hp.

Horsepower Calculation

  • Hydraulic horsepower formula:   extHorsepower=extgpmimesextpsi1714ext{Horsepower} = \frac{ ext{gpm} imes ext{psi}}{1714}
  • Example of lifting weight:   - Lifting 10,000 pounds a distance of 1 foot in 2 seconds with a flow rate of 10 gpm at 1500 psi gives a theoretical requirement of 8.75 hp.

Heat

  • Law of Conservation of Energy: Energy cannot be created or destroyed, only transformed.
  • Hydraulics can convert unused energy into heat (e.g., through resistance in a relief valve).

Other Important Calculations

  1. Torque: Defined as twisting force, measured in foot-pounds (e.g., 10 foot-pounds from applying 10 pounds of force to a wrench). Applies to hydraulic motors rated by specific torque values.
  2. Quiz Questions:    1. Work accomplished by moving 500 lbs 2 feet is 1000 foot-pounds. (True/False?)    2. Power is defined as the rate of doing work. (True/False?)    3. Wasted energy in a hydraulic system is destroyed. (True/False?)   

Additional Notes on Horsepower

  • 1 hp = 33,000 ft • lbs/min = 746 W = 42.4 Btu/min.
Formulas
  • Energy Formulas:   - 1 kW = 1.3 hp   - 1 hp = 550 ft • lbs/s   - extHydraulichp=extgpmimesextpsi1714ext{Hydraulic hp} = \frac{ ext{gpm} imes ext{psi}}{1714}   - Torque (in • lbs) = extpsiimesextdisp(in3/extrev)6.28\frac{ ext{psi} imes ext{disp (in}^3/ ext{rev})}{6.28}   - extTorque(inlbs)=exthpimes63025extrpmext{Torque (in lbs)} = \frac{ ext{hp} imes 63025}{ ext{rpm}}   - extBtu(perhour)=extΔpsiimesextgpmimes1.5ext{Btu (per hour)} = ext{Δpsi} imes ext{gpm} imes 1.5
  • To determine the volume (in3) of a piston moving a distance, use:   - Volume = Piston Area (in2) x Stroke (in)
      - Cylinder Speed (ft/min) = extgpmimes19.25extEffectiveArea(in2)\frac{ ext{gpm} imes 19.25}{ ext{Effective Area (in2)}}   - Flow (theoretical) = extPumprpmimesextin3/extrevimes1231ext{Pump rpm} imes ext{in}^3/ ext{rev} imes \frac{1}{231}
Flow Principles
  1. Flow in a hydraulic system:    - Produced from a positive displacement pump (unlike a centrifugal pump).
Three Key Principles of Flow:
  • Principle 1: Flow makes it go.   - An actuator must be supplied with flow to operate.
  • Principle 2: Rate of flow determines speed.   - Usually measured in gallons per minute (gpm).
  • Principle 3: Changes in actuator volume displacement alter the flow rate and, consequently, speed.
Quiz Questions on Flow Concepts
  1. Changes to flow rate do not affect actuator speed. (True/False?)
  2. A larger diameter cylinder will affect its speed on extension/retraction. (True/False?)

Torque and Rotational Relationships

  • Torque is crucial in projecting hydraulic motors' performance and efficiency at given pressures.
  • Example: Torque (in • lbs) = 63025imesexthp/extrpm63025 imes ext{hp} / ext{rpm}

Summary of Key Concepts

  • Managing energy transfer, flow, and pressure control in hydraulic systems is foundational for understanding hydraulics.
  • Essential calculations underpin hydraulic performance, and understanding component interactions is critical for optimizing system design.