Forging new generations of engineers

Hydraulics

Hydraulics

An area of engineering science that deals with

liquid flow

and

pressure

Hydraulic Fluids • Liquid pumped through a hydraulic system • Petroleum-based or synthetic oil • Serve four major functions: 1. Power transmission 2. Lubrication of moving parts 3. Sealing of spaces between moving parts 4. Heat removal • Relatively Incompressible!

Two Types or Conditions of Hydraulic Systems Hydrostatic Hydrodynamic

Hydrostatics a “No Flow” Scenario • “Static” means “stationary” or “non flowing” in a hydraulic system • Hydraulic systems are considered static when there is no flow • Pascal’s Law (for hydrostatics): – a pressure applied to a confined hydrostatic fluid is transmitted with equal intensity throughout the fluid – Same pressure all throughout!

Hydrodynamics – a “Flow” scenario • “Dynamic” means “moving” or “flowing” in a hydraulic system • Hydraulic systems are considered dynamic when there is flow • Pascal’s Law does not apply! – Pressure does not have equal intensity in a flowing dynamic system – Pressure drops along the length of a hydraulic line in flowing systems

Flow and Pressure • Flow, Q – volume flow rate – amount of fluid moving through system per unit time • Pressure, P – force per unit area of fluid moving through a system Pressure  Force Area P  F A

Mechanical Advantage • Ideal mechanical advantage (IMA) – Assumes no frictional losses – Calculated as ratio of output force to input force

IMA



F

– always less than ideal

F output

– difficult to calculate

Application of Pascal’s Law in a Simple Hydrostatic System How much force must you exert on piston A to lift a load on piston B of 500 lbs? What is the ideal mechanical advantage of this system?

Problem Solving Step 1: Determine the pressure in the system using information about piston B Known Unknown A = 500 in 2 P=?

F = 500 lb Equation No algebra needed P  F A Substitution & Solution P  500lb 500in 2  lb 1 in 2  1psi STEP 2: Use the pressure calculated in STEP 1 and information about piston A to calculate force Known Unknown A = 1.0 in 2 F=?

P  1 psi  1 lb in 2 Equation and algebra: P  F A (A)P  F A (A) Substitution

&

Solution F  PA     1 lb in 2     1.0in

2 F   1.0lb

 PA

Problem Solving Step 3: Determine the ideal mechanical advantage (IMA) of the system using information from STEPS 1 & 2 Known Unknown F(input) = 1 lb IMA=?

F(output) = 500 lb Equation No algebra needed IMA  F(output) IMA  500lb 1.0lb

 500

A Hydraulic System

Tank/Reservoir • Storage device which is open and not pressurized Filter

Pumps • Positive displacement pump (Gear Pump) : a specific amount of fluid passes through the pump for each rotation • Centrifugal pump (Vane Pump) : no specific amount of fluid flow per rotation; flow depends on speed of blades

Accumulators • Storage device which is closed and is under pressure

Valves • Check Valve • Directional Control

Linear Actuators Use hydraulic power to move linearly Single Acting Double Acting

Rotary Actuators Use hydraulic power to rotate Single-Vane Double-Vane

Applications • Robotics • Oil systems in vehicles (e.g. brakes) • Presses • Heavy equipment • Wood splitter • Aircraft control systems

In-line Pressure Gauge

The Hydraulic Trainer

Inline-Tee Flow Control Valve Check Valve Actuators Return line from reservoir Pressure line Pump Motor Directional Control Valve