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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