HYDRAULICS & PNEUMATICS

Presented by: Dr. Abootorabi

Basic ConceptsFluidSIM-H

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Energy

Potential energy:

W=m.g.h

Press with elevated reservoir:

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EnergyPressure energy:

W=p.ΔV

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EnergyMotion energy:

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Thermal energy:

Power

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Power

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Efficiency

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Calculation of input and output power

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Throttle points

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Physical Properties of Hydraulic FluidsSpecific Weight

Also known as unit weight, is the weight per unit

volume of a material.

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Volume

=

Specific weight =Weight

W

VN.m-3 = m

Vkg.m-3

VolumeDensity =

Mass

Specific weight – Density relationship:

W = mg = g [N. m-3]

Specific gravity is a dimensionless unit defined as the

specific weight of the fluid divided by the specific weight of

water.

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water

SGoil = oil

9800 N/m3

=oil

water

Density of the oil [kg/m3]

Density of water [kg/m3]

1000 kg/m3

Specific weight of the oil [N/m3]

Specific Gravity (SG)

Pressure Head It represents the height of a fluid column that produces

the static pressure.

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g

pH =

In other words, due to its weight, a 1-ft column of water develops at its base a pressure of 0.433 psi. The 1-ft height of water is commonly called a pressure head.

0.433 psi

1 ft

Pressure - force relationship: p = F / A

Pressure - head relationship: p = H

The Continuity Equation for Hydraulic Systems

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Use of Volume Flow Rate Q

Q1(m3/s) = A1(m2) 1(m/s) = A2 2 = Q2

Flow rates are frequently specified in units of liters per second (Lps) or liters per minute (Lpm). 1m3 = 1000 L

Q is the volume flow rate ( volume of fluid passing a given station per unit time).

Hence, for hydraulic systems, the volume flow rate is also constant in a pipe line. The continuity equation for hydraulic systems can be rewritten as follows:

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(/4) D12

(/4) D221

2

=A2

A1

=

This equation shows the smaller the pipe size, the greater the velocity.

1

2

=D1

( D2 )2

Where D1 and D2 are the pipe diameters at stations 1 and 2, respectively.The final result is:

Continuity equation forhydraulic system

The Continuity Equation for Hydraulic Systems

Hydraulic PowerHydraulic Cylinder

Rod

F Load

Barrel

p

Q

Piston

• Hydraulic power is the power delivered by a hydraulic fluid to a load-driving device such as hydraulic cylinder.

• Let’s analyze the hydraulic cylinder (above figure) by developing equations that will allow us to answer the following 3 questions:

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Hydraulic PowerHydraulic Cylinder

1. How do we determine how large a piston diameter is

required for the cylinder?

2. What is the pump flow rate required to drive the cylinder

through its stroke in a specific time?

3. How much hydraulic horsepower does the fluid deliver to

the cylinder?

QUESTIONS

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Pressure p acts on the area of the piston to produce

the force required to overcome the load:

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FF loadloadA =

p

Load is known from the application

Pressure is established based on the pump design

ANSWER 1 – Piston Size

Rod

F Load

Barrel

p

Q

Piston

Piston velocity

Q [m3/s] = A [m2] × [m/s]

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ANSWER 2 – Pump Flow Rate

Piston velocityPiston area

The larger the piston area and velocity, the greater

must be the pump flow rate.

Volume displacement VD of the hydraulic cylinder = A X S

A

S

Q = VD / t

= (A X S) / t

= A X

Pump flow rate in a specific time

Time

ANSWER 3 – Hydraulic power

Power =Energy

= Force X Distance= F X S= p A X S

p A S

t=

= p A = p Q

Hydraulic power (W) = p [N/m2] X Q [m3/s]

Hydraulic power (kW) = (p [bar] X Q

[lit/min])/600

or Pa

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Mechanical Power The mechanical output power delivered by a hydraulic

motor:

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Power (W) = T (N.m) X (rad/sec)

If RPM is given , must change to rad/sec by X 2/60

Introduction to FluidSIM-H Demo

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Introduction to Simulating and Creating Circuits

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Introduction to Simulating and Creating Circuits

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The end.

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