1- flds 385 chapter 1 intro to hydraulics
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EATON Industrial Hydraulics Manual
Polytechnic
Manufacturing and Automation
Ted Nelson A.Sc.T.Rm T409
403-284-8242
ted.nelson@sait.ca
FLDS 385
Introduction to Hydraulics
Chapter 1
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Section1. Introduction to Hydraulics
Objectives:
1.1 Evaluate methods of providing mechanical power.
1.2 Describe the advantages and disadvantages of hydraulic power.
1.3 Examine the principles of hydraulic systems.
1.4 Explain simple circuit design.
1.5 Draw a simple circuit using appropriate schematic representation.
1.6 Build a simple hydraulic circuit.
1.7 Discuss and follow safety practices current in the hydraulic industry.
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Introduction to Hydraulics
The study of hydraulics deals with the use and
characteristics of liquids
Early recorded history shows devices such as pumps andwater wheels used in very ancient times
Modern hydraulic systems are based on a principlediscovered by Blaise Pascal
Hydraulic systems can transmit power, multiply force and modify
motions with the use of a confined fluid
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Introduction to Hydraulics
Pascals Law (simply stated):
Pressure applied on a confined fluid is transmitted in all directions,
and acts with an equal force on equal areas, at right angles to the area
1. The bottle is filledwith liquid, which isnot compressible
2. A 10 lb. force is appliedto thestopper with a surface area ofone square inch
3. This results in 10 lb of forceon every square inch ofsurface area in the container
4. If the bottom has an area of20 sq. in. and each squareinch is pushed on by 10lbs.of force, the entire bottom
of the container receives200 lbs push
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Methods of Providing Mechanical Power
Simple Hydraulic Press:
Joseph Bramah, a British mechanic, invented the hydraulic press by
applying Pascals law to his design Below: the stopper has an applied force of 10 lbs on its area of 1 in2,
the large piston has an area of 10 in2, therefore it can support a totalweight or force of 100 lbs
10 lbs
1. An input force of 10 lbs ona one square inch area
2. Develops a pressure of 10pounds per square inch (psi)throughout the container
3. This pressure will supporta 100 lb weight if this is a10 sq in piston
4. The forces are proportional to the piston areas10 lbs
1 sq in
100 lbs
10 sq in=
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Methods of Providing Mechanical Power
Simple Hydraulic Press: contd
The forces or weights will balance using a simple press
The forces are proportional to the piston areas
If the output piston area would have been 200 in2, the output forcewould be 2000 lbs (with the same 10 lbs/ in2 input force)
(input pressure) x (area of output piston) = (output force)10 lbs/ in2 x 200 in2 = 2000 lbs
This is the operating principle of the hydraulic jack, as well as the
hydraulic press
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Methods of Providing Mechanical Power
Simple Mechanical Lever:
The simple press and a mechanical lever are similar
Below: the left side of the fulcrum must equal the right side of thefulcrum
An applied force of 10 lbs on a length of 10 in, will balance 100 lbs ata length of 1 in
1. An inputforce
of10 lbs here
3. if this arm is10 times as long as 4. this arm
2. will balance100 lbs here
B. SIMPLE MECHANICAL LEVER
100 lbs
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Pressure
Pressure Defined:
In order to determine the total force exerted on a surface, it is
necessary to know the pressure or force on a unit of area
Pressure is usually expressed in pounds per square inch (psi)
It is also expressed in terms of Bar and kilopascals (kPa)
Force = Pressure x Area
Conversions from Eaton Textbook Appendix C page 557
1 kiloPascal = 100 Bars
1 Bar x 10-5 = 1 Newton/Sq. meter 1 Newton/Sq. meter = 1 Pascal
1 kilogram = 9.81 newtons
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Conservation of Energy
Conservation of Energy Defined:
A fundamental law of physics states that energy can neither be created
nor destroyed. The multiplication of force is not a matter of getting something for
nothing, the large piston is moved only by the liquid displaced by thesmall piston
What is gained in force is lost in distance
10 lbs
1. Moving the small piston 10 indisplaces 10 cu in of liquid(1 sq in x 10 in = 10 cu in)
2. 10 cu in of liquid will move
the larger piston only one inch(10 in sq x 1 in = 10 cu in)
10 in
1 in
3. The energy transfer here equals10 lbs x 10 in or 100 in lbs
4. The energy transfer here also 100 in lbs(1 in x 100 lbs = 100 in lbs)
is
100 lbs
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Hydraulic Power Transmission
Hydraulic Power Transmission Defined:
Hydraulic Power Transmission is a means of transmitting power by
pushing on a confined liquid.
The hydraulic system is not a source of power
The power source is the prime mover (electric motor, or an engine) that drives a
pump
A hydraulic system provides versatility and flexibility which is anadvantage over other methods of transmitting power
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Hydraulic Power Transmission
The input component of the system is called a pump
Most pumps have multiple pistons, vanes or gears as their pumping elements
The output components are called actuators
Actuators can be linear (a cylinder) or rotary (a hydraulic motor)
Load
Pump
ElectricMotor
1 The pump pushes the hydraulicfluid into the lines
.
2. Lines carry the fluid to actuators which arepushed internally to produce a mechanicaloutput which moves the load
Piston
& Rod
To Reservoir
3. Some actuators operate in a straight line(linear actuators). They are called rams orcylinders. They are used to lift weight, exertforce, clamp, ect.
A. LinearActuator
Fig 1-4 (A) Linear and rotary actuatorCOPYRIGHT (2001) EATON CORPORATIONC
Fig 1-4 (B) Linear and rotary actuatorCOPYRIGHT (2001) EATON CORPORATIONC
PUMP
ElectricMotor
Pump
4. Rotary actuators or motors give the systemrotating output. They can be connected topulleys, gears, rack and pinions, conveyors, ect.
RotaryDriveShaft
Motor
B. RotaryActuator
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Advantages of Hydraulic Power
Variable Speed:
Most electric motors run at a constant speed
It is also desirable to operate an engine at a constant speed
An Actuator (linear or rotary) can be driven anywhere from amaximum speed to a reduced speed by varying a flow control valve
Figure 1-5 (B) Hydraulic drive speed is variable
COPYRIGHT (2001) EATON CORPORATIONC
Load
Pump
ElectricMotor
B. Reduced Speed
6. the actuator receives only 5 gallons andonly travels half as far in one minute
4. If the pump delivers10 gpm,
5. but a valve restricts the flowto 5 gallons per minute
FlowControlValve
5 gpm
7. Excess 5 gpm is diverted over therelief valve
Relief
Valve
Load
Pump
ElectricMotor
A. Maximum Speed
1. If the pump constantly delivers 10 gallonsper minute
2. and the volume is 10 gallons
3. the piston will move this far on one minute
Figure 1-5 (A) Hydraulic drive speed is variable
COPYRIGHT (2001) EATON CORPORATIONC
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Advantages of Hydraulic Power
Reversible:
Few Prime Movers are reversible
If they are reversible they must be stopped before reversing them
Hydraulic actuators can be reversed instantly while in full motionwithout damage to the hydraulic system
A cross-port relief valve is used on a hyd motor to reduce pressure spikes
A four way directional control valve provides the reversing control
Load
Pump
ElectricMotor
1. In this position of thedirectional valve
2. pump delivery is directed to thecap end of the cylinder
3. The piston extends
4. Exhaust oil is pushedout of the rod endand back to tank
Figure 1-6 (A) Hydraulic drives are reversible
COPYRIGHT (2001) EATON COR PORATIONC
Figure 1-6 (B) Hydraulic drives are reversible
COPYRIGHT (2001) EATON CORPORATIONC
Load
Pump
ElectricMotor
5. In another position, oilis directed to the rodend of the cylinder
6. The piston rod retracts
8. The relief valve protects the system bymomentarily diverting flow to tank duringreversing, and when the piston is stalled orstops at the end of stroke
7. Exhaust oil from thecap end is directedto tank
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Advantages of Hydraulic Power
Overload Protection:
A Pressure Relief Valve protects the hydraulic system from overloaddamage due to excessive pressures
When the load exceeds the valve setting, the fluid is directed back tothe tank (reservoir)
The Relief Valve also provides a means of setting a machine for a
specified amount of torque or force (ie: a press or clamping operation)
Load
Pump
ElectricMotor
1. In this position of thedirectional valve
2. pump delivery is directed to thecap end of the cylinder
3. The piston extends
4. Exhaust oil is pushedout of the rod endand back to tank
Figure 1-6 (A) Hydraulic drives are reversible
COPYRIGHT (2001) EATON COR PORATIONC
Figure 1-6 (B) Hydraulic drives are reversible
COPYRIGHT (2001) EATON CORPORATIONC
Load
Pump
ElectricMotor
5. In another position, oilis directed to the rodend of the cylinder
6. The piston rod retracts
8. The relief valve protects the system bymomentarily diverting flow to tank duringreversing, and when the piston is stalled orstops at the end of stroke
7. Exhaust oil from thecap end is directedto tank
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Advantages of Hydraulic Power
Small Packages:
Hydraulic components provide high power output with very smallweight and size
Due to their high speeds and high pressure capabilities
Can Be Stalled:
Stalling an electric motor will cause damage or blow a fuse A stalled engine will have to be restarted to continue an operation
A Hydraulic actuator can be overloaded and stall without damage, andwill start up immediately when the load is reduced
During the stall, the relief valve dumps the fluid back to the tank The only loss is the energy lost in wasted horsepower
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Advantages of Hydraulic Power
Hydraulic Oil:
Any liquid is essentially noncompressible and will transmit powerinstantaneously in a hydraulic system
The most common liquid used in hydraulic systems is petroleum oil
Oil is virtually incompressible
It will compress about 0.5% at 1000 psi (70 bar, 7000 kPa) This is a negligible amount in most systems
A hydraulic oils most desirable property is its lubricating ability
Moving parts in a system must be lubricated
With a hydraulic system you dont have to worry about lubricatingcomponents because it is done automatically
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Pressure in a Column of Fluid
The weight of a volume of oil varies slightly as the viscosity changes
Most hydraulic oils weigh from 55 to 58 lbs per cubic foot
The weight of the oil is an important consideration on the pump inlet
Figure 1-7 Weight of oil creates pressure
COPYRIGHT (2001) EATON CORPORATIONC
4 psi
2 psi
0.4 psi
1 cubic foot of oil weighs approximately 58 lbs
1. If this weight is divided equally over the 144 sq in theforce on the bottom of each square inch is .4 lbsThus the pressure at the bottom is .4 psi
2. If the fluid column height is 10 ft then the pressure atthe base of the column will be 4 psi. It is the height of
the column, not its volume that determines thepressure.
12 in
12 in
12 in
Area = 144 in2
Pressure = = 0.4 psiForce
Area =58 lbs
144 in2
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Oil Reservoir Location
Reservoir above the Pump Inlet:
When the reservoir oil level is above the pump inlet, a positivepressure is available to force the oil into the pump
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Oil Reservoir Location
Reservoir below the Pump Inlet:
If the reservoir oil level is below the pump, a vacuum equivalent to0.4 psi per foot is needed to lift the oil to the pump inlet
Oil is actually forced by atmospheric pressure into the vacuum createdat the pump inlet when the pump is in operation
C
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AtmosphericPressure
Inlet
Outlet1. On the intake stroke, the pump piston movesout expanding the pumping chamber
2. A partial vacuum or void is created here
3. Atmospheric pressure pushes the fluid into the pumping chamberto fill the void. Fluid is pushed, not pulled into a pump
Atmospheric Pressure Charges the Pump
A pumps inlet is charged with oil due to a difference in pressurebetween the reservoir and the pump inlet
On the intake stroke, the piston
creates a partial vacuum, and the
atmospheric pressure in the
reservoir pushes the oil into the
pumping chamber vacuum
In a rotary pump, successive
pumping chambers increase in size
as they pass the inlet, creating thesame vacuum condition
P bl ith V t th P I l t
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Problems with Vacuum at the Pump Inlet
Cavitation:
If it were possible to pull a complete vacuum at the pump inlet,there would be available some 14.7 psi to push the oil in
Pump manufacturers recommend a vacuum of less than 12.2 psi (5 ofmercury) absolute at the pump inlet
This leaves about 2.5 psi pressure difference to push oil into the pump
If there is too much of a pressure difference the liquid will vaporize in
this vacuum, which in turn will put gas bubbles into the oil
P bl ith V t th P I l t
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Problems with Vacuum at the Pump Inlet
Cavitation:
When these gas bubbles get to the pump the higher pressure in thepump causes them to collapse (implode) against pump componentswith considerable force and cause damage, which impairs the pumpsoperation and reduces its life
Driving a pump at too high a speed increases fluid velocity whichincreases friction in the line, which in turn causes a lower pressurecondition and will increase the possibility of cavitation
P bl ith V t th P I l t
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Problems with Vacuum at the Pump Inlet
Aeration:
If the inlet fittings are not tight, air can get into the inlet line and intothe pump
This air-oil mixture also causes trouble and noise, but not as much
damage as cavitation
This air is not dissolved in the oil and continues throughout the systemas compressible bubbles causing erratic valve and actuator operation
P iti Di l t P
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Positive Displacement Pumps
Most pumps used in hydraulic systems are positive displacement
This means that the pumps output is constant regardless of thepressure
The outlet is positively sealed from the inlet, so whatever goes into thepump is forced out of the pump through the outlet
The sole purpose of a pump is to create flow
Pressure is caused by a resistance to flow
Pressure can be lost only by a leakage path that diverts all the flow
from the pump
P iti Di l t P
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Positive Displacement Pumps
If a 10 GPM pump is used to push oil under a 10 in2 piston and raisean 8000 lb load, the pressure must be 800 psi while the load is being
moved or supported by the hydraulic oil
P = F APump
Electric Motor
Load8000 lbs
1. The load is 8000 lbs2. The area is 10 sq in
3. The pressure equals the force areaequals 8000 lbs 10 in sq = 800 psi
psi0
250
500
7501000
1250
1500
1750
2000
A.
Figure 1-10 (A) Pressure loss requires full loss of pump outletCOPYRIGHT (2001) EATON CORPORATIONC
P iti Di l t P
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Positive Displacement Pumps
Even if a hole in the piston allows 9.5 GPM to leak at 800 psi,pressure will still be maintained.
Only 0.5 GPM is available to move the load very slowly The pressure required to move the load still remains at 800 psi
P iti Di l t P
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Positive Displacement Pumps
Now imagine that the 9.5 GPM leak is in the pump instead of thecylinder
There is still 0.5 GPM available to move the load and there is stillpressure at 800 psi
Thus, a pump can be badly worn, losing nearly all of its efficiency,and pressure can still be maintained
Pressure alone is no indicator of a pumps condition
Its necessary to measure the flow at a given pressure to determinewhether a pump is in good or bad condition
P iti Di l t P
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Positive Displacement Pumps
What will happen?Scenario
There is a hole through the piston Both the Rod end and Cap end
ports are blocked off
Will the load drop quickly?
Will the load drop slowly?
Will it stay in place?
??????????????????
LOAD8000 lbs
Positi e Displacement P mps
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Positive Displacement Pumps
What will happen?
The load will stay in place!
If the load were lowered it wouldintroduce more rod into the cylinder
and none of the fluid can escape
The load may drop a very small amountdue to the oil compressing
LOAD8000 lbs
How Pressure is Created
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How Pressure is Created
Pressure is created whenever the flow of a fluid is resistedThe resistance may come from:
1) A load on an actuator or
2) A restriction (or orifice) in the piping
Below: Pressure is being created from the load on the actuator:
How Pressure is Created
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How Pressure is Created
Pressure created from a restriction (or orifice):
Below: No restriction causes No Pressure
How Pressure is Created
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How Pressure is Created
Pressure created from a restriction (or orifice):
Below: A small restriction causes an increase in PressureThe restriction is not enough to limit the quantityof flow out of the faucet
How Pressure is Created
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How Pressure is Created
Pressure created from a restriction (or orifice):
Below: An increased restriction causes an increase inpressure up to the relief valve setting
The restriction limits the quantityof flow out of the faucet, the excess
flow goes over the relief valve and back to tank
How Pressure is Created
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How Pressure is Created
Summary of Pressure created from a restriction (or orifice):
As the faucet is gradually closed it resists flow and causes the pressureto build up on the upstream side
It takes more and more pressure to push the 10 GPM through therestriction
Without the relief valve there would be no limit to the pressure buildup, until something breaks or stalls the prime mover (electric motor)
Note:
It is always advised to have a relief valve or some other pressure
limiting device in a hydraulic system with a positive displacementpump
Parallel Flow Paths
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Parallel Flow Paths
Path of least resistance:
Liquids will always take the path of least resistance
When two or more parallel flow paths have different resistances, thepressure in the system will increase only to the amount required totake the easiest path (100 psi, path A) and no flow will go through
path B or C
Parallel Flow Paths
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Parallel Flow Paths
Path of least resistance:
If path A is blocked the next easiest path in this parallel system is path
B, therefore the system pressure will be 200psi, and there will be noflow through path C
Parallel Flow Paths
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Parallel Flow Paths
Path of least resistance:
Similarly, when a pumps flow goes to two actuators, the actuator
which needs lower pressure will move first
It is difficult to balance loads exactly, therefore cylinders are oftenphysically connected to each other if they must move together
Series Flow Paths
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Series Flow PathsAdditive flow path:
Pressures are additive when resistance to flow is connected in series
Pressure in the line, at any given point, will be the pressure to open avalve plus the back pressure from the valve downstream
The pressure at the pump is the sum of the pressures required to openeach valve
Orifices
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Orifices
Pressure Drop through an Orifice:
An orifice is a restricted passage in a hydraulic line or component,
used to control flow or create a pressure difference (pressure drop)
If there is no flow, there is no difference in pressure across the orifice
Orifices
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Orifices
Pressure Drop through an Orifice:
Increasing the flow through an orifice will cause an increased pressure
drop across the orifice
A high flow causes a large 500 psi pressure drop
Orifices
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Orifices
Pressure Drop through an Orifice:
Reducing the flow through an orifice will cause a reduced pressure
drop across the orifice
A medium flow causes a smaller 200 psi pressure drop
Pressure
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Pressure
Pressure Indicates Work Load:
Pressure equals the force of a load divided by the piston area
Formula:
P = F AWhere: P = pressure (pounds per square inch (lbs/in2) or (kPa))
F = force (pounds (lbs) or (newtons))
A = area (square inches (in2) or (square centimeters))
Pressure
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Pressure
Pressure Indicates Work Load:
An increase or decrease in the load will result in an increase or
decrease in the operating pressure
Pressure is proportional to the load and the pressure gauge readingindicates the work load at any given moment
Pressure gauge readings (psig) normally ignore atmospheric pressure
A standard gauge reads zero at atmospheric pressure
An absolute gauge reads 14.7 psi (101 kPa) at sea level
Absolute pressure is usually designated as psia (pounds per squareinch absolute)
Force
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Force
Force is Proportional to Pressure and Area:
Output force of a cylinder can be computed as follows
Formula:
F = P x AWhere: P = pressure (pounds per square inch (lbs/in2) or (kPa))
F = force (pounds (lbs) or (newtons))
A = area (square inches (in2) or (square centimeters))
Force
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Force
Force is Proportional to Pressure and Area:
Below the hydraulic press has an operating pressure of 2000 psi
The pressure is applied to a ram area of 20 in2, Find output force:
F = P x A
Substitute F = 2000 lbs/in2 x 20 in2
Therefore F = 40,000 lbs
psi
psi
0
0
250
250
500
500
750
750
1000
1000
1250
1250
1500
1500
1750
1750
2000
2000
Relief valveset at 2000 psi
1. This valve limits the maximumpressure in the system to2000 psi. This controls the
maximum force of the press
2. The force is 2000 psi x 20sq in = 40,000 lbs or 20 tonsof pressing force
Movingplaten
Fixedplaten
Figure 1-15 Force equals pressure multiplied by area
COPYRIGHT (2001) EATON CORPORATIONC
Area
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Area
Computing Piston Area:
Formula:
A = x d2 4Where: A = area (square inches (in2) or (square centimeters (cm2))
d = diameter of the piston (inches (in) or centimeters(cm))
= pi (3.1416)
4 = 0.7854
Area of a piston can also be calculated with the formula A = r2but pistons are measured by diameters so you would have to makesure to use the radius of the piston for any calculations
Formulas
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Formulas
Force, Pressure and Area relationships:
F = P x A
P = F AA = F P
A = 4 x d2
Speed
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Speed
Speed of an Actuator:
Speed of a piston or motor depends on its size and the rate of oil flow
Below, both cylinders have the same volume, yet the piston in cyl Bwill travel twice as fast as cyl A because the flow from the pump isdoubled
If one of the cyl diameters were smaller its speed would be faster
As long as the pump delivery remained constant
psi0
250
500
7501000
1250
1500
1750
2 0 0 0
psi0
250
500
7501000
1250
1500
1750
2 0 0 0
Load
1 gpmpump
The cylinder has a 2 ft stroke
1. The one gpm pump will cause the cylinderpiston to move 2 ft. in one minute
6010
20
3040
50
60 Seconds
A.
psi0
250
500
7501000
1250
1500
1750
2000
Load
2 gpmpump
The cylinder has a 2 ft stroke
2. The 2 gpm pump will cause the cylinderpiston to move 2 ft in 30 seconds
6010
20
3040
50
30 Seconds
3. The rate of fluid delivery and its areadetermine the speed of the cylinder
psi0
250
500
7501000
1250
1500
1750
2000
B.
Speed
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Speed
Speed of an Actuator:
Formula:
speed (s) = flow (Q)
area (a)
flow (Q) = speed (s) x area (a)
area (a) = flow (Q)
speed (s)
Where: Q = flow (in3
/ minute or cm3
/ minute)a = area (in2 or cm2)
s = speed (in/ minute or cm/ minute)
Actuator Summary
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Actuator Summary
From the previous formulas we can conclude:
1) The force or torque of an actuator is directly proportional to thepressure and independent of the flow
2) An actuators speed or rate of travel will depend upon the amountof fluid flow without regard to pressure
Velocity
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Velocity
Velocity in Pipes:
The velocity at which hydraulic fluid flows is an important design
consideration because of the friction developed by velocity
Recommended Velocity Ranges:
Pump Inlet Line = 24 feet per second (0.61.2 meters/second)
Working Lines = 720 feet per second (2.16.1 meters/second)
Velocity
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Velocity
Velocity in Pipes:
Doubling the inside diameter of a line will quadruple the cross-
sectional area This would mean that the velocity would only be one fourth as fast in
the large line
If you use half the size of pipe, the velocity will quadruple
2 in dia
1 in dia
1. The diameter of the large pipe is
twice that of the smaller one
2. It would take four of the smaller pipes to
equal the flow area of the larger one
Velocity
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VelocityVelocity in Pipes:
The velocity in the 2 dia line is 20 fps and the flow is laminar(smooth), as it flows into the smaller 1 dia pipe the flow istransitional (part laminar and turbulent), the velocity in the 1 dia pipeis 80 fps and the flow is turbulent
Turbulent flow increases friction in the pipe and resists flow, whichresults in an increased pressure drop through the line
A very low velocity is recommended for the pump inlet line becausevery little pressure drop can be tolerated there, the inlet lines areusually one size larger than the outlet lines on a pump
2 in dia 1 in dia area
Turbulent FlowTransitional flowLaminar flow
Pipe Sizes
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pe S es
Determining Pipe Size Requirements:
Formula: (If GPM (l/m) and desired velocity are known)
Areapipe cross-section= GPM x 0.3208Velocity (ft/sec)
Velocity(ft/sec) = GPM x 0.3208Areapipe cross-section
Typically increase the size of a pipe by 20% of thecalculated size to account for inaccuracies, wear etc.
Size Ratings of Lines
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gDetermining Pipe Size:
In standard pipes, the actual inside diameter is larger than the nominalsize quoted, so use a manufacturers chart to find accurate sizes
For steel and copper tubing, the quoted size is the outside diameter
The inside diameter will be the outside diameter minus 2 times thewall thickness
2 in dia
1. Tubing size is quoted asoutside diameter
2. To find the internal diameter,subtract two times the wall
thickness from the quoted size
InternalDiameter
Wall Thickness
Work and Power
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Work:
Work is done whenever a force or push is exerted through a distance
Formula:
Work = distance (ft) x force (lbs)
Where: Work units are foot-pounds (in metric it is joules)
Work Example:A 10 lb (44.5N) weight is lifted 10 ft (3 m)
Work = distance (ft) x force (lbs)
Substitute Work = 10 ft x 10 lbs OR Work = 3m x 44.5 N
Therefore Work = 100 ft-lbs Work = 135 J
Work and Power
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Power:
The rate of doing work is called power
To visualize power, think of climbing a flight of stairs, It is moredifficult to run up the stairs than to walk. When you run you aredoing the same work but it is done at a faster rate (more power)
Formula:
Power = distance x force OR Worktime time
Where: Power units are horsepower (hp) (in metric its a watt (W))
1 horsepower is equivalent to 33,000 lbs lifted 1 foot in 1 minute1 watt is equal to 1 newton applied over a distance of 1 meter in 1 sec
Work and Power
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Power:
Formula:
1 hp = 33,000 ft-lbs OR 550 ft-lbs
minute second
1 hp = 746 watts (electrical power)
1 hp = 42.4 BTU/minute (heat power)
We convert hydraulic power to horsepower (watt) so that the
mechanical, electrical and heat power equivalents will be known
HP in a Hydraulic System
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y y
Hydraulic Power:
In the hydraulic system, speed and distance are represented by flow,
and force is indicated by pressure, hydraulic power is expressed as: Formula:
Power = GPM x psi
Conversions:
1 gallon = 232 cubic inches (in3)
12 inches = 1 foot
Thus: 1 gallon per minute flow at 1 psi equals 0.000583 hp
HP in a Hydraulic System
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y yHydraulic Horsepower Out of the Pump:
The horsepower out of the pump is the exact power being used in asystem
Formula:
Hyd hpout = GPM x psi x 0.000583
OR
Hyd hpout = GPM x psi
1714
HP in a Hydraulic System
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y yMechanical Horsepower in to the Pump:
The horsepower required to drive the pump will be somewhat higherthan the pump puts out because the system is not 100% efficient
If we assume an average efficiency of 83%, then power inputrequirements can be estimated using the following formulas:
Formula:
Imperial
hpin = GPM x psi x 0.0007
Metric
kW = L/min x bar x 0.002
Horsepower and Torque
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p qGeneral torque-power formulas for rotating equipment:
It is desirable to convert back and forth from horsepower to torquewithout computing pressure and flow
Formula:
Imperial
torque = 63,025 x hp
RPM
hp = torque x RPM
63,025
Where:torque is in pound-inches
Horsepower and Torque
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p qGeneral torque-power formulas for rotating equipment:
Formula:
Metric
kW = torque x RPM
9,550
Where:
torque is in newton-meters
Designing a Simple Hydraulic System
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Job to Be Done:
All circuit design starts with the job to be done
From this you can determine the actuator that will be used If you need to raise a load, a hydraulic cylinder would do the job
The stroke length must be at least equal to the distance the load mustbe moved
Load8000 lbs
Load8000 lbs
30 in
30 in
1. To raise an 8000 lb load 30 inches,a cylinder with at least a 30 inchstroke is required
Figure 1-19 Use a cylinder to raise a load
COPYRIGHT (2001) EATON CORPORATIONC
Designing a Simple Hydraulic System
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LOAD8000 lbs
1. If the piston area is 10 sq in
2. The pressure required to liftthe load equals the loaddivided by the piston area
8000 lbs
10 sq in= 800 psi
Figure 1-20 Choosing cylinder size
COPYRIGHT (2001) EATON CORPORATIONC
Size of Cylinder:
The piston area is determined by the force required to raise the loadand the desired operating pressure
Assuming a weight of 8000 lb is to be lifted 30 in and the maximumoperating pressure is limited to 1000 psi
Cylinder stroke = At least 30 inches
Piston area = 8000 lb 1000 lb/in2 = 8 in2
This gives no margin for error
By choosing a Piston area of 10 in2 we would
have the capability of lifting up to 10,000 lbs
The system pressure would only need to be
800 psi to lift the 8000 lbs
Designing a Simple Hydraulic System
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Size of Pump required:
The rate at which the load must travel will determine the pump size
The 10 in
2
piston will displace 10 in
3
for every inch it lifts the load Extending the rod 30 will require 300 in3 of fluid
If it is to move at 10 inches per second, it will require 100 in3 of fluidper second or 6000 in3 per minute
Pumps are rated in gallons per minute, so a conversion will be done
Divide 6000 in3 per minute by 231 in3 per gallon to get GPM
GPM = 6000 in3/minute = 26 GPM
231 in3/gallon
The pump required needs to be sized at 26 GPM
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Horsepower needed to drive the pump:
Assume maximum operating pressure, therefore 1000 psi
hpin = GPM x psi x 0.0007
hpin = 26 gpm x 1000 psi x 0.0007
hpin = 18.2 hp
It will take 18.2 hp to operate the prime mover (electric motor) thatdrives the hydraulic pump
Designing a Simple Hydraulic System
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Overload Prevention:
To protect the pump and other components from excessive pressure
due to overloads and stalling, a relief valve set to limit the maximumsystem pressure should be installed in the line between the pumpoutlet and the inlet port of the directional control valve
Load
Pump
ElectricMotor
2. In this position of thedirectional valve
3. pump delivery is directed tothe cap end of the cylinder
4. The piston rod extends
5. Exhaust oil ispushed out of
the rod end andback to tank
Directional Valve
1. The relief valve protects thesystem from over pressureby diverting the pump flow totank when the maximumpressure setting is reached
Figure 1-21 Valving to protect and control the system
COPYRIGHT (2001) EATON CORPORATIONC
Designing a Simple Hydraulic System
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Reservoir Size:
The reservoir should hold approximately two to three times the pump
capacity in gallons per minute Filters and adequate piping would complete the system
Load
Pump
ElectricMotor
2. In this position of thedirectional valve
3. pump delivery is directed tothe cap end of the cylinder
4. The piston rod extends
5. Exhaust oil ispushed out of
the rod end andback to tank
Directional Valve
1. The relief valve protects thesystem from over pressureby diverting the pump flow totank when the maximumpressure setting is reached
Figure 1-21 Valving to protect and control the system
COPYRIGHT (2001) EATON CORPORATIONC
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