1- FLDS 385 Chapter 1 Intro to Hydraulics

7/29/2019 1- FLDS 385 Chapter 1 Intro to Hydraulics

1/70

EATON Industrial Hydraulics Manual

Polytechnic

Manufacturing and Automation

Ted Nelson A.Sc.T.Rm T409

403-284-8242

[email protected]

FLDS 385

Introduction to Hydraulics

Chapter 1


2/70

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.


3/70


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


4/70


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


5/70

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=


6/70


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


7/70


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


8/70

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


9/70

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


10/70

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


11/70

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


12/70

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



13/70


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


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


14/70


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


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


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


15/70


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


16/70


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


17/70

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


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


18/70

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


19/70

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


20/70

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


21/70

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



22/70


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



23/70


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


24/70

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


25/70


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


26/70


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


27/70


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


28/70


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


29/70


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


30/70


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:



31/70


Pressure created from a restriction (or orifice):

Below: No restriction causes No Pressure



32/70



Below: A small restriction causes an increase in PressureThe restriction is not enough to limit the quantityof flow out of the faucet



33/70



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



34/70


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


35/70

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


36/70

Parallel Flow Paths


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


37/70

Parallel Flow Paths


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


38/70

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


39/70

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


40/70

Orifices


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


41/70

Orifices


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


42/70

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


43/70

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


44/70

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


45/70

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


Area


46/70

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


47/70

Formulas

Force, Pressure and Area relationships:

F = P x A

P = F AA = F P

A = 4 x d2

Speed


48/70

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


49/70

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


50/70

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


51/70

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


52/70

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


53/70

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


54/70

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


55/70

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


56/70

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


57/70

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


58/70

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


59/70

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



60/70

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



61/70

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


62/70

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


63/70

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


64/70

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




65/70

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


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



66/70

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



67/70

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



68/70

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


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




69/70

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


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



70/70

Any Questions?

Documents

1- FLDS 385 Chapter 1 Intro to Hydraulics