FLUID POWER

TODAY, THE FOLLOWING WILL BE COVERED:

+ NATURE OF THIS COURSE;

+ SYLLABUS FOR COURSE;

+ BASIC FOUNDATIONS OF FLUID SYSTEMS: TERMS, UNITS, RELATIONSHIPS.

Bold denotes IMPORTANT TERMS !

INSTRUCTIONAL OBJECTIVES:

A. Develop an understanding of some basic principles:

     I. Pascal's Law

        1. Pressure

            a. head b. atmospheric c. gage

        2. Force

        3. Area

        4. Density and Specific gravity

II. Mechanical and Hydraulic leverage

III. Basic Hydraulic Circuit

    1. Hydraulic jack

    2. Linear hydraulic circuit

B. Dimensional calculations - English and Metric systems

     I. Units for:

1. Length

2. Area

        4. Pressure

        5. Force

II. Example Problems for:

        1. Head

        2. Hydraulic Jack (lever)

        3. Linear Circuit (sizing cylinder).

Homework Assignment: See Assignments Web Page.

C.  Standard Color Coding for Hydraulic Systems
 

BASIC TERMS, DESCRIPTIONS, UNITS, AND THEIR RELATIONSHIP.

TERM            DESCRIPTION          ENGLISH METRIC RELATIONSHIP

HEAD             Height of a fluid column         ft.                   m                1 ft. =.3048m

STROKE        Length of rod travel               in.                  mm              1 in.= 25.4mm

FORCE          Thrust of a cylinder                lbs.                N                 1 lb. =4.4482N

AREA              Surface of a piston                in.²               mm²

PRESSURE    Force/Area                           lbs/in²           N/m²

                                                                                         or N/mm²

          (NOTE: 1 N/m² is also called a Pascal denoted by Pa.)

                       (1 PSI = 6895 Pa.)

                       (1 N/mm² is also called a Mega Pascal denoted by MPa.)

                       (1 PSI = .006895 MPa.)

                       BAR (Average Barometric Pressure) 14.5 PSI = 10⁵ Pa.

VOLUME        Area X Length(height):  in.³
                                                                   or GAL.   (NOTE: 1 GAL. = 231 in.³)
                                                                   or cubic foot ( 1 cubic foot = 1728 cubic inches)
                                                                   or mm³
                                                                   or cm³
                                                                   or m³
                                                                   or Liter

                      (NOTE: 1000 liters = 1 m³ )

                      ( 1 liter = 1000 cm³ )

                      ( 1 liter = 10⁶ mm³ )

FACTORS AND PREFIXES

FACTOR           PREFIX              SYMBOL          FACTOR          PREFIX             SYMBOL

10¹²           tera                        T                       10^-2             centi                     c

10⁹                     gigi                         G                      10^-3             milli                       m

10⁶                     mega                      M                      10^-6           micro                    u

10³                     kilo                         k                        10^-9           nano                    n

10²                     hecto                     h                        10^-12           pico                      p

10                      deca                      da                       10^-15           femto                    f

10^-1                   deci                        d                         10^-18          atto                       a
 
 

EXAMPLE PROBLEMS:

1. Find the head pressure at the base of the storage tank shown if
    a) the tank is filled with fresh water;
    b) if the 20ft. tank is filled with oil having a weight density of 56 lbs./ft.³

SOLUTION: P= gx h

a. Weight Density for the fluid = 56 lbs/ft³

b. 56 lbs/ft³ x 20 ft. = 1248 lbs / ft.²  x 1 ft.² / 144 in.² = 7.78 psi
 
 

2. Find the maximum that can be lifted by the hydraulic jack.

FORCE @ HANDLE (FA) = 20 pounds;  SOLVE FORFORCE AT ROD (FB)

CYLINDER 1 = 1/2 INCH BORE (PUMP CYLINDER)

CYLINDER 2 = 4 INCH BORE (WORK CYLINDER)

NOTE:  BORE IS THE INSIDE DIAMETER OF THE CYLINDER.

 

SOLUTION: WORK @ A = WORK @ B OR 20 LBS. X 12 " = FB X 2"

240 LB-IN = 2 X FB, THEREFORE FB=120 LB.

ALTERNATE SOLUTION:   Sum the moments about point c.

                                                   FB(2 in.) = 240 lb-in

                                           FB = 120 lb.

AREA OF CYL 1 = .7854 X .5 IN.² = .19635 IN.²

AREA OF CYL 2 = .7854 X 4 IN.² = 12.57 IN.²

P=F/A_{CYL 1} ; P=120 LB./.19635 IN.² = 611.15 PSI

F_CYL2 =P X A_CYL2 = 611.15 X 12.57 = 7682.2 LBS.
 
 
 

3. Determine the size (bore) of the cylinder for the system
    shown below.

SYSTEM PRESSURE = 8 Mpa;   CLAMP FORCE = 36KN

SOLUTION: A=F/P = 36000 N / (8 N/mm²) = 4500 mm²

A=.7854 X (D²)

D= (A/.7854)^.5 = (4500/.7854)^.5 = 75.69 mm

+ FLOWRATE, VELOCITY, AREA RELATIONSHIPS FOR LINEAR ACTUATORS;

+ TERMS, UNITS, RELATIONSHIPS;

+ REGENERATIVE CIRCUIT.
 
 

Normal Linear Circuit.  Under normal operation, the piston velocity (Vp) during the extension mode for a hydraulic cylinder is dependent on the pump flow rate (Q) entering the cap side of the piston and the cap area (Acap) as shown in the general  equation below:

            Velocity = Flow Rate / Area

Velocity of the piston during the extension phase of motion, can be determined using the formula as shown below:

            Velocity of piston during extension in normal linear mode equation: Vp_ext  =    Qpump
                                                                                                                    A_cap

Similarly, the velocity during retraction is calculated as follows:
 

             Velocity of piston during retraction in normal mode equation:     Vp_ret  =   Qpump
                                                                                                                                     A_an

The time required for a hydraulic cylinder to extend is dependent on the Stroke (S) or linear displacement of the actuator and piston velocity.  The general equation
is shown below:

            Time =  Stroke / Velocity

To determine time during extension,  the velocity during extension much be substituted into the equation.  Similarly, to determine the time during retraction, the retraction velocity must be used.

Regenerative Circuit.   Regenerative circuits are used when it is desirable to rapidly advance an actuator into position to reduce cycle time.  When configured as
a regenerative system, cylinders can be advanced more rapidly than in normal operation with the pump flow rate alone.  In order to accomplish regeneration, the fluid leaving the rod end of the cylinder is routed back to the cap side of the cylinder to combine with the pump flow rate from the pump as depicted in Figure 1 below:

Therefore, Q_Total = Q _Pump + Q _{Regenerative.}

Solving for Q _Pump:   Q _Pump = Q_Total - Q_Regenerative

Remember that Flow Rate (Q) = Piston Velocity (Vp) x Area.

Substituting the appropriate areas, the equation for Qpump becomes:

        Q_pump = Q _Total - Q _Regenerative

        Q_pump =[Vp x A_cap] - [Vp x (A_cap - A_rod)]

    Factoring out Vp, the equation becomes:

        Q_pump =  Vp (A_cap - (A_cap - A_rod)

                    =  Vp (_Acap - A_cap + A_rod)

                    =  Vp (A_rod)

    Therefore:  To find the velocity of piston IN REGENERATIVE MODE,  the equation below can be used.

        Velocity of piston during extension in regenerative mode equation:   Vp_Regeneration  =  Qpump
                                                                                                                       A_rod
 

Note:  Regeneration circuits are typically only used on cylinders having a single rod and only in the extending direction!

Next we will consider force during extension in the regenerative circuit.   Since both the capside and rodside are pressurized at the same time,
the pressure in the system is constant and equal on both the capside and rodside of the cylinder as shown in Figure 2 below:

Resulting force available on the external rod is the net force or difference between the force on the capside and the rodside cylinder.  Note that even
though both sides are pressurized, the force on the capside will be greater that the force on the rodside due to the cap area being greater than the annular area.
Force in a regenerative circuit during extension of a single rod, double acting cylinder is show in Figure 3.

Note:  If (and only if) the ratio of CAP area to ANNULAR area is exactly 2:1, equal speed and force will result for a
          regenerative circuit operating during extension AND a normal linear circuit operating during retraction.

_{A typical complete regenerative circuit in extension mode is shown in Figure 4 and retraction mode in Figure 5.}

 
 

WEEK 3

THE FOLLOWING WILL BE COVERED:

+ Fluid Properties

+ WORK, POWER AND HORSPOWER;

+ Review FLOWRATE, VELOCITY, AREA RELATIONSHIPS FOR LINEAR ACTUATORS;

+ EXAMPLE PROBLEMS

Blue denotes important terms

DENSITY and SPECIFIC GRAVITY

MASS :  Property of a body of fluid that is a measure of inertia or resistance to change in motion.
              The term can also be used as a measure of the "quantity of fluid"

Note:  In the Imperial System,  the SLUG is used for MASS DENSITY (r).

     Since m = F / a;  the units = lbs / ft/sec² = lb-sec²/ ft.  Therefore 1 slug = 1 lb-sec² / ft.

     Typically we are more concerned with MASS DENSITY or SLUGS / cubic feet

          Example:  Suppose a hydraulic fluid has a weight density of 58 lbs / ft³

              m = F / a  =  58 lbs / 32.2 ft/sec²   =  1.8 lb-sec² / ft  OR 1.8 SLUGS / ft³

For the SI system,  the unit of kilogram (kg) is used to denote MASS DENSITY.

The Newton (N) is the standard unit of Force and is defined as 1 kg - m / sec²

          Example:   A boulder has a mass of 5.75 kg,  what is the force in N due to the earth's gravity?

          Answer:   F = m a   =  5.75 kg  x 9.81 m / sec²  =  56.4075 kg - m / sec² = 56.4075 N

Note:  On the earth's surface, acceleration is assumed to be constant, and weight density or specific weight
          is typically used.   MASS DENSITY is crucial when designing hydraulic systems for space craft.

WEIGHT DENSITY:

          Weight density is defined as specific weight per volume.  In the Imperial system typical units would be lbs / ft³
                                                                                                In the SI system typical units are N / m³

SPECIFIC GRAVITY:

           Specific gravity is a ratio of the density of a fluid compared to the density of fresh water.   Specific gravity
           can be expressed as a ratio of MASS DENSITY OR WEIGHT DENSITY.

           NOTE:  The constant for fresh water (weight density) used as a standard is 62.4 lbs / ft³ in the Imperial system

                       For the SI System,  the MASS DENSITY of fresh water is 1000 kg / m³

                       Therefore the weight density (g) = 1000 kg / m³ x 9.81 = 9810 N / m³

           Example Specific Gravity Problem:  Find the specific gravity of sea water if the SPECIFIC WEIGHT  is 10.1 kN / m³

           Answer:  10.1 kN/m3 = 10,100 N/m3   Sg  = g_{sea water} / g _{fresh water} = 10,100 / 9810 = 1.02956
 

Week 3

    I. Power, Torque and Mechanical Horsepower

            Power = Work / time;     Torque = Force x distance of moment arm;

            HP = Power / Basic Unit of Horsepower      (e.g. 746 watts,  550 Foot-lbs./sec, 746 N-m/sec.)

     II. Horsepower in Hydraulic Systems

    1. Fluid Horsepower

          Generic Formula: FHP =  Pressure x Flowrate x C
                                    (C = constant for converting to appropriate units)

  English System:   FHP = (P  X Q) / 1714
                           where:  P = Pressure in PSI
                                      Q = Flowrate in GPM

              Metric System:    FHP = (P x Q) / 44.76
                           where:  P = Pressure in mega Pascals  (1 MPa = 1 N/mm²)
                                       Q= Flowrate in LPM  (liters per minute)

                   Note for computations, brake and torque horsepower use essentially the same formula.
                   Brake howepower reflects the braking intensity or load placed on the prime mover (e.g. electric motor).
                   The formula will be covered below.

        3. Torque Horsepower

                In hydraulic systems, indicates the horsepower available at a rotary
                actuator:

                Generic formula:  THP = T x N x C  (constant for converting appropriate units)

   English System:  THP = (T x N) / 63025
                           Where:  T = torque in lb-in.
                                        N = rpm

   Metric System:    THP = (T x N) / 7121
                           Where:  T = torque in N-m
                                        N = rpm
        4. Efficiency
                   Since no systems are 100% efficient,  the formulae must me adjusted to account for losses.  Typically,
                   an efficiency of 80 to 85 % is assumed for hydraulic fluid horsepower.

                          Therefore the formula becomes:   FHP = (P x Q) / (1714 x e)     for English system.
                          Similarly, efficiency must be included for the metric formula as well as for THP formulas when
                          efficiencies are know.   More attention will be given to efficiency when pumps and actuators are covered.
 
 
 

     Formulae and Example Problems  - English and Metric systems for:

1. Flow, velocity and area
    Flow rate (Q) = (V)elocity X (A)rea
    Q = V x A;  units must be compatible.
        Example problem:  A cylinder has a stroke (linear movement) of
        of .25 meters and must extend in 5 seconds.  Cylinder bore is 100 mm.
        What is the required flow rate for the system?

  SOLUTION:  A = .7854 x d²  = .7854 x (.1m)² = .07854 m^{2
                             V = Stroke / time of extension
                                 = .25 m / 5 seconds = .05 m / sec.}
               Q = V x A = .05 m / sec.  x  .07854 m²
               Q = .003927 m³ / sec.
                                 =.003927 m³ / sec. x 60 sec. / min. x 1000 liters/ m^{3
                             Q = 3.927 LPM (liters per minute)}

2.  Power and mechanical horsepower.
     Example problem:  A forklift raises a 2000 lb. pallet 5 feet in 5 seconds.
      What is the power output?
  Solution:  Power = Work / time = (2000 lbs x 5 feet) / 5 seconds = 2000 ft-lbs/sec
       How much horsepower is required?
  Solution:  HP = Power / (Basic Unit of Horsepower)
                           HP = (2000 ft-lbs / sec)  / (550 ft-lbs/sec)
                           HP = 3.64

3. Fluid Horsepower.
   Example problem:  A hydraulic system operates at 10 MPa pressure with
    pump flow rate of 20 LPM.  What horsepower is required for the electric
    motor driving the pump assuming 85% efficiency?
  Solution:  FHP = (P X Q) / (44.76 x e)
        Note the formula requires flow in LPM therefore units are compatible.
                      FHP = (10 x 20) / (44.76 x .85)
                      FHP = 4.47

Homework Assignment: See Assignments Web Page.

THIS WEEK WILL BE DEDICATED TO PROJECT WORK.

+  ORGANIZING WORKSHEETS IN EXCEL

+  EXCEL EXAMPLES:  DEALING WITH CIRCULAR REFERENCE

+   Introduction to systems and preparation for Lab 4 (putting together what we have learned to date).

TOPICS:

    + Review SPECIFIC GRAVITY
    + Review WEIGHT DENSITY
    + Review MASS DENSITY
    + Review HEAD PRESSURE
    + VISCOSITY
    + VISCOSITY INDEX
    + BULK MODULUS
NOTE:  EXAMPLE PROBLEMS WILL BE USED TO EXPLAIN THESE TOPICS.

RELATIONSHIP OF SPECIFIC GRAVITY, WEIGHT DENSITY, MASS DENSITY.

Sg = Weight Density of "X" / Weight Density of Water

    =  Mass Density of "X" / Mass Density of Water

Properties of fresh water will be used as a standard.

VISCOSITY
        ABSOLUTE
        KINEMATIC
        SAYBOLT UNIVERSAL SECONDS

VISCOSITY INDEX FOR RATING THE STABILITY OF A FLUID

BULK MODULUS AND THE COMPRESSIBILITY OF A FLUID

LAB:  SYSTEM SPECIFICATIONS FOR AN APPLEJUICE PROCESSING SYSTEM.
           (NOTE:  This lab will incorporate virtually everything that has been covered thus far in this class)
           The purpose of the lab is provide an application for system design, AND to serve as a review for test one.

  TEST ONE

 TOPICS
    + BULK MODULUS
           Example:  Hydraulic feed for a milling machine.

    + VOLUMETRIC DISPLACEMENT (VANE PUMP)
           Deriving the formula:
      Vd = Pi/2 (Dc+Dr) (e) (L)

                    Example problem for a variable displacement pump.

   + VOLUMETRIC DISPLACEMENT (PISTON PUMP)
            Deriving the formula:
      Vd = D x A x Y x (tan O )

                    Example problem for a variable displacement pump.

    + EFFICIENCIES AND RELATIONSHIPS

WEEK  8

TOPIC:  PUMPS THEORY AND EXAMPLES (Video tapes)

     Review UNIT 3 on Pumps from the following site:
  http://64.78.42.182/free-ed/MechTech/hydraulics01/default.asp

            Topics:
               Conservation of Energy
               Continuity Equation
               Bernoulli's Principle
               Pressure Losses Through Conductors

Conservation of Energy (Review)
            Energy cannot be created nor destroyed
            Three froms of energy related to incompressable fluid flow
                   Pressure Energy (PE)
                   Kinetic Energy   (KE)
                   Elevation (Potential) Energy (EE)
 

Continuity Equation

          Volume Flow Rate in must equal Volume Flow rate out.

            Q_in = Q_out
            Q = A x V
            A₁ x V₁ = A₂ x V₂
            V₂ = (A₁ x V₁) / A₂

Bernoulli's Principle
         When the velocity of a fluid increases, the pressure exerted by that fluid decreases.

        Curve of a Baseball Example

      For an animation of why curve balls curve, click here

   To view an animated air foil example, click here.

  Demo::  Take a sheet of paper, and tear a strip (approximately 1 1/2 inches x 11 inches)
  Hold the paper strip under your bottom lip (lenghtwise facing from you).
  Blow across the top of the strip of paper....what happens?  Why?

  ROLLING CANS DEMO

FLOW THROUGH CONDUITS (PIPES, TUBING, HOSE)

         Where:
                     A = Area
                      r =  Radius
                     h = height or Elevation
                     P = Pressure
                     v = Velocity

        Venturi Demo

           Flow Example 1 :  http://library.thinkquest.org/27948/bernoulli.html

           Flow Example 2 (with Q,V,A plot:  http://home.earthlink.net/~mmc1919/venturi.html#animation

           Bernoulli on UTube

Darcy's Equation
Pressure losses through fluid systems

WEEK 12 CONDUCTORS AND SIZING

  Piping Selection Chart: http://www.onlinepipe.com/pipechart.html

  Hydraulic Tubing Selection Chart:  http://www.hydraulic-supply.com/html/productline/prodcat/hydraulic-tubing.htm

  Flexile Hose     Types :http://www.hydraulicspneumatics.com/200/TechZone/HydraulicHoseTu/Article/True/6417/TechZone-HydraulicHoseTu

  Flexible Hose Sizes and Pressure Ratings (SAE)

Fundamentals of Pneumatics (tutorial with video clips)
     http://www.nfpa.com/Training-Pneumatic/
 

KEY EQUATIONS:

(P1V1)/T1 = (P2V2)/T2

    CONSUMPTION RATES:  CR = (atmospheric pressure + gage pressure) / atmospheric pressure; Based on standard air,
            atmospheric pressure is assumed to be 14.7 pisa for the English System,  101 KPa for the Metric System.

          CR = (14.7 + gage pressure) / 14.7

            Qr(cyliner) = (A x S x N x CR) / K   Note: In the English System:  A= in sq. S = in. N=cpm ; CR=compression rate; K = 1728

            Qr (cylinder) = (A x S x N X CR)/ 1728

            Qr(motor) = (Vd x N x CR) / K  ;   Note  In the English system:  Vd in cubed / rev; N = rpm; CR = compresstion ration; K = 1728

           Qr(motor) = (Vd x N x CR) / 1728

    SIZING RECEIVERS: (an example problem is also shown in your text)

            Vr = atmospheric pressure x (t) x (Qr-Qc) / (Pmax - Pmin)
                     t   is the time the system is consuming air (usually the maximum flow condition);
            For the English system, using standard air and SCFM for Q the equation is as follows:

           Vr = [14.7 x (time in minutes) x (Qr - Qc)] / [ PSI(max) - PSI(min) ]

    PRESSURE LOSSES DUE TO FRICTION: ( English System - assuming schedule 40, commercial steel pipe)

           General equation  for  Delta P = c x L x Q² / CR x d⁵

           Specifically for commercial steel pipe, schedule 40, English system only.

         Delta P = (0.1025 x L x Q²)  / ( 3600 x CR x d ^{5. 31} )

                  Where 0.1025 = friction factor for commercial steel pipe (these coefficients are experimentally derived
                                             based on the type of conduit being used.
                                      L  = Total length of pipe + equivalent lenght for valves and fittings.
                                      Q = flow in SCFM
                                 3600 = Constant to convert units
                                    CR = Compression ratio
                                       d = pipe diameter in inches (use actual inside diameter for schedule 40 pipe).

POWER AND HORSE POWER (For adiabatic conditions which assumes no loss or gain of heat).

        To determine the horsepower and power required to drive an air compressor, the following formulae can be used:

        For the English System, Use:

       HP = [ (P_in x Q) / (65.4)] x [ (P_out / P_in)^0.286    -   1  ]

           Where:  Pin   = inlet atmospheric pressure ( psia)
                          Pout = outlet pressure (psia)
                          Q     = Flow rate (standard cubic feet per minute)
                Note:  This is a variation of the formula used in the Norvelle text.  The 0.0286 constant is for air under adiabatic conditions.
                            for an explanation of the adiabatic process click HERE

       For the metric System to determine theoretical power (kW) use:

       Power (kW)  = [ (Pin x Q) / 17.1 ] x [ ( Pout / Pin) ^0.286)     -     1   ]

                Where:  Pin  = inlet atmospheric pressure (kPa abs)
                              Pout = outlet pressure (kPa abs)
                               Q    = flow rate (standard meters cubed per minute)

WEEK 15 FINAL EXAM

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