concepts over hydraulic

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Basic Hydraulic Concepts

Myounggyu NohDept. Mechatronics EngineeringChungnam National University

Spring Semester, 2004



1

Objectives of the Lecture

• Difference between hydrostatics and hydrodynamics

• Pascal’s Law and its signicance

• Advantages and disadvantages of hydraulic uid power

• Pressure-force-area relationship

• Denitions of work, power, and horsepower

• Basic concept of a uid power system

• Denitions of input, hydraulic and output horsepower

• Continuity equation

Basic Concepts



2

Hydrodynamics versus Hydrostatics

Two ways of trasmitting uid energy

1. Hydrodynamic: the kinetic or the impact of the moving uid is converted tomechanical energy.

2. Hydrostatic: energy is transferred through a conned uid by the pressure that iscreated by the application of a force to that conned uid.

Basic Concepts



3

Pascal’s Law

Pressure exerted on a conned liquid at rest is transmitted equally in all directions, is thesame at any point in the liquid, and acts at right angles to the surfaces of the containerF

p

A

1. Pascal’s law is valid irrespective of the shape of the vessel2. Any change in the exerted pressure is seen almost instantly throughout the liquid.3. Fluid in a system can be as rigid as steel for the transmission of power

Basic Concepts



4

Advantages of Hydraulic Power Systems

1. Provide high levels of readily regulated torque and force2. Offer innitely variable linear or rotary speed over a wide range3. Instantly reversible4. Can be stalled without damage5. High power output is possible6. High accuracy and extreme stiffness

7. Possible to be automated without electronics8. Fully adaptable to electrical or electronic controls9. Provide cushioning to reduce the effects of impact or shoch loads

10. The uid itself provides lubrication

Basic Concepts



5

Disadvantages of Hydraulic Power Systems

1. Hazards exist due to high-pressure2. Fluids are ammable3. Leakage is possible4. Adequate ltration is necessary

Hydraulics are ideal for applications requiring high forces for heavy loads, long strokelinear motion, rigidity and accuracy with heavy loads, high-stall torque, etc.

Basic Concepts



6

Press–Force–Area Relationship

Pressure is dened as

Pressure = ForceAreaor

p =F A

Several conversion factors for pressure

• 1 psi = 6895 Pa = 6.895 kPa

• 1 bar = 14.5 psi = 100 kPa

Basic Concepts



7

Multiplication of Force

A A A B

F A F B

F B = F A ×AB

AA, F B = F A ×

D 2B

D 2A

Basic Concepts



8

Hydraulic Jack

Basic Concepts



9

Work and Power

Work is dened as a force acting over a distance.

W = F × d

With reasonable accuracy (if losses are negligible), W in = W out .

F A×

dA = F B×

dB

Denition of Power:Power =

WorkUnit Time

p =W

t=

F × d

t= F × v

Conversion Factor: 1 HP = 550 ft · lb/sec = 33,000 ft · lb/min = 0.746 kW.

Basic Concepts



10

James Watt’s Experiment

Basic Concepts



11

Fluid Power Systems

PressureControl

FlowControl

DirectionalControl

Actuator LoadHydraulicPumpMover

Prime

Power Input Control Power Output

Basic Concepts



12

Input Power

Input power: power transmitted from the prime mover to the hydraulic pump. A function

of the torque and the rotational speed of the input device.

IHP =T (lb· in) × 2πN (rev/min)(ft/12in)(60 sec/min) × (550 ft · lb/sec/HP)

IHP =T × N 63, 025

In SI units,P IN =

T × N 9550

where P is in kW, T in N· m, and N in rpm.

Basic Concepts



13

Hydraulic Power

Hydraulic horsepower: power output from a pump.

Hydraulic Horsepower =Pressure (psi) × Flow Rate (gpm)

1714

HHP =p × Q1714

In SI unitsP HY D =

p × Q60, 000

where p in kPa, Q in liters per minute (Lpm), and P HY D in kW.

Basic Concepts



14

Fluid Flow

Fluid power systems are generally considered to be neither uid sources nor uid sinks

as far as uid ow through piping is concerned.To determine the ow rate at any given point,

Flow Rate(Q ) = Pipe Area(A) × Fluid Velocity(v)

Flow rate in USCS is gallons per minute (gpm)Conversion factor: 1gallon = 231 in3 .

Basic Concepts



15

Continuity Equation

Continuity equation states that the uid ow rate remains constant through a given

portion of a cicuit.Q 1 = Q 2 = Q 3 = . . .

A 1 v1 = A 2 v2 = A 3 v3 = . . .

1 2 3

Basic Concepts

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