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Hydraulic pump unit

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  • rohamm
    replied
    I don't.

    Leave a comment:


  • dian
    replied
    i think you got hp wrong.

    Leave a comment:


  • metalmagpie
    replied
    It would be useful to connect to an engine hoist cylinder, for example, or to power a hydraulic winch that won't be heavily loaded. Just some ideas ..

    If you want more information to allow you to understand hydraulics, read my crib notes:

    Fluid Power Formulas

    Hydraulic Pump Calculations

    Horsepower Required to Drive Pump

    GPM X PSI X .0007 (this is a 'rule-of-thumb' calculation)

    How many horsepower are needed to drive a 10 gpm pump at 1750 psi?

    GPM = 10
    PSI = 1750
    GPM X PSI X .0007 = 10 X 1750 X .0007 = 12.25 horsepower


    Pump Output Flow (in Gallons Per Minute)

    RPM X Pump Displacement / 231

    How much oil is produced by a 2.21 cubic inch pump operating at 1120 rpm?

    RPM = 1120
    Pump Displacement = 2.21 cubic inches
    RPM X Pump Displacement / 231 = 1120 X 2.21 / 231 = 10.72 gpm


    Pump Displacement Needed for GPM of Output Flow

    231 X GPM / RPM

    What displacement is needed to produce 7 gpm at 1740 rpm?

    GPM = 7
    RPM = 1740
    231 X GPM / RPM = 231 X 7 / 1740 = 0.93 cubic inches per revolution

    Cubic Inch Displacement in a Gear Pump

    Disassemble the pump.
    Remove the 4 bolts that hold the pump together.
    Then you can remove the end plate.
    Next remove the gear chamber (just the outside part) then the bearing
    carrier, the gear itself, and finally the front plate.

    Use calipers to measure (in inches) the length of both gear chambers L, the
    bore diameter of one chamber D, and the width of one gear W.
    L - D
    C.I.R. Displacement = 6 x W x (2D - L) x -----
    2
    Hydraulic Cylinder Calculations

    Cylinder Blind End Area (in square inches)

    PI X (Cylinder Radius) ^2

    What is the area of a 6" diameter cylinder?

    Diameter = 6"
    Radius is 1/2 of diameter = 3"
    Radius ^2 = 3" X 3" = 9"
    PI X (Cylinder Radius )^2 = 3.14 X (3)^2 = 3.14 X 9 = 28.26 square inches

    Cylinder Rod End Area (in square inches)

    Blind End Area - Rod Area

    What is the rod end area of a 6" diam. cylinder which has a 3" diam. rod?

    Cylinder Blind End Area = 28.26 square inches
    Rod Diameter = 2.25"
    Radius is 1/2 of rod diameter = 1.125"
    Radius^2 = 1.125" X 1.125" = 1.27"
    PI X Radius^2 = 3.14 X 1.27 = 3.98 square inches

    Blind End Area - Rod Area = 28.26 - 7.07 = 24.28 square inches


    Cylinder Output Force (in Pounds)

    Pressure (in PSI) X Cylinder Area

    What is the push force of a 6" diameter cylinder operating at 2,500 PSI?

    Cylinder Blind End Area = 28.26 square inches
    Pressure = 2,500 psi
    Pressure X Cylinder Area = 2,500 X 28.26 = 70,650 pounds (35 tons)

    What is the pull force of a 6" diameter cylinder with a 2.25" diameter rod
    operating at 2,500 PSI?

    Cylinder Rod End Area = 24.28 square inches
    Pressure = 2,500 psi
    Pressure X Cylinder Area = 2,500 X 24.28 = 60,710 pounds (30 tons)



    Fluid Pressure in PSI Required to Lift Load (in PSI)

    Pounds of Force Needed / Cylinder Area

    What pressure is needed to develop 50,000 pounds of push force from a 6"
    diameter cylinder?

    Pounds of Force = 50,000 pounds
    Cylinder Blind End Area = 28.26 square inches
    Pounds of Force Needed / Cylinder Area = 50,000 / 28.26 = 1,769.29 PSI

    What pressure is needed to develop 50,000 pounds of pull force from a 6"
    diameter cylinder which has a 3: diameter rod?

    Pounds of Force = 50,000 pounds
    Cylinder Rod End Area = 21.19 square inches
    Pounds of Force Needed / Cylinder Area = 50,000 / 21.19 = 2,359.60 PSI


    Cylinder Speed (in inches per second)

    (231 X GPM) / (60 X Net Cylinder Area)

    How fast will a 6" diameter cylinder with a 3" diameter rod extend with 15
    gpm input?

    GPM = 6
    Net Cylinder Area = 28.26 square inches
    (231 X GPM) / (60 X Net Cylinder Area) = (231 X 15) / (60 x 28.26) = 2.04
    inches per second

    How fast will it retract?

    Net Cylinder Area = 21.19 square inches
    (231 X GPM) / (60 X Net Cylinder Area) = (231 X 15) / (60 x 21.19) = 2.73
    inches per second


    GPM of Flow Needed for Cylinder Speed

    Cylinder Area X Stroke Length in Inches / 231 X 60 / Time in seconds for one
    stroke

    How many GPM are needed to extend a 6" diameter cylinder 8 inches in 10 seconds?

    Cylinder Area = 28.26 square inches
    Stroke Length = 8 inches
    Time for 1 stroke = 10 seconds
    Area X Length / 231 X 60 / Time = 28.26 X 8 / 231 X 60 / 10 = 5.88 gpm


    If the cylinder has a 3" diameter rod, how many gpm is needed to retract 8
    inches in 10 seconds?

    Cylinder Area = 21.19 square inches
    Stroke Length = 8 inches
    Time for 1 stroke = 10 seconds
    Area X Length / 231 X 60 / Time = 21.19 X 8 / 231 X 60 / 10 = 4.40 gpm

    Cylinder Blind End Output (GPM)

    Blind End Area / Rod End Area X GPM In

    How many GPM come out the blind end of a 6" diameter cylinder with a 3" diameter
    rod when there is 15 gallons per minute put in the rod end?

    Cylinder Blind End Area =28.26 square inches
    Cylinder Rod End Area = 21.19 square inches
    GPM Input = 15 gpm
    Blind End Area / Rod End Area X GPM In = 28.26 / 21.19 * 15 = 20 gpm

    Hydraulic Motor Calculations

    GPM of Flow Needed for Fluid Motor Speed

    Motor Displacement X Motor RPM / 231

    How many GPM are needed to drive a 2.51 cubic inch motor at 1200 rpm?

    Motor Displacement = 2.51 cubic inches per revolution
    Motor RPM = 1200
    Motor Displacement X Motor RPM / 231 = 2.51 X 1200 / 231 = 13.04 gpm

    Fluid Motor Speed from GPM Input

    231 X GPM / Fluid Motor Displacement

    How fast will a 0.95 cubic inch motor turn with 8 gpm input?

    GPM = 8
    Motor Displacement = 0.95 cubic inches per revolution
    231 X GPM / Fluid Motor Displacement = 231 X 8 / 0.95 = 1,945 rpm


    Fluid Motor Torque from Pressure and Displacement

    PSI X Motor Displacement / (2 X PI)

    How much torque does a 2.25 cubic inch motor develop at 2,200 psi?

    Pressure = 2,200 psi
    Displacement = 2.25 cubic inches per revolution
    PSI X Motor Displacement / (2 x PI) = 2,200 X 2.25 / 6.28 = 788.22 inch
    pounds


    Fluid Motor Torque from Horsepower and RPM

    Horsepower X 63025 / RPM

    How much torque is developed by a motor at 15 horsepower and 1500 rpm?

    Horsepower = 15
    RPM = 1500
    Horsepower X 63025 / RPM = 15 X 63025 / 1500 = 630.25 inch pound

    Fluid Motor Torque from GPM, PSI and RPM

    GPM X PSI X 36.77 / RPM

    How much torque does a motor develop at 1,250 psi, 1750 rpm, with 9 gpm input?

    GPM = 9
    PSI = 1,250
    RPM = 1750
    GPM X PSI X 36.7 / RPM = 9 X 1,250 X 36.7 / 1750 = 235.93 inch pounds second

    Fluid & Piping Calculations

    Velocity of Fluid through Piping

    0.3208 X GPM / Internal Area

    What is the velocity of 10 gpm going through a 1/2" diameter schedule 40 pipe?

    GPM = 10
    Internal Area = .304 (see note below)
    0.3208 X GPM / Internal Area = .3208 X 10 X .304 = 10.55 feet per second

    Note: The outside diameter of pipe remains the same regardless of the thickness
    of the pipe. A heavy duty pipe has a thicker wall than a standard duty pipe, so
    the internal diameter of the heavy duty pipe is smaller than the internal
    diameter of a standard duty pipe. The wall thickness and internal diameter of
    pipes can be found on readily available charts.

    Hydraulic steel tubing also maintains the same outside diameter regardless of
    wall thickness.

    Hose sizes indicate the inside diameter of the plumbing. A 1/2" diameter hose
    has an internal diameter of 0.50 inches, regardless of the hose pressure rating.

    Suggested Piping Sizes

    Pump suction lines should be sized so the fluid velocity is between 2 and 4
    feet per second.

    Oil return lines should be sized so the fluid velocity is between 10 and 15
    feet per second.

    Medium pressure supply lines should be sized so the fluid velocity is
    between 15 and 20 feet per second.

    High pressure supply lines should be sized so the fluid velocity is below 30
    feet per second.



    Heat Calculations

    Heat Dissipation Capacity of Steel Reservoirs

    0.001 X Surface Area X Difference between oil and air temperature

    If the oil temperature is 140 degrees, and the air temperature is 75 degrees,
    how much heat will a reservoir with 20 square feet of surface area dissipate?

    Surface Area = 20 square feet
    Temperature Difference = 140 degrees - 75 degrees = 65 degrees
    0.001 X Surface Area X Temperature Difference = 0.001 X 20 X 65 = 1.3
    horsepower

    Note: 1 HP = 2,544 BTU per Hour



    Heating Hydraulic Fluid

    1 watt will raise the temperature of 1 gallon by 1 degree F per hour

    and

    Horsepower X 745.7 = watts

    and

    Watts / 1000 = kilowatts



    Pneumatic Valve Sizing

    Notes:

    All these pneumatic formulas assume 68 degrees F at sea level
    All strokes and diameters are in inches
    All times are in seconds
    All pressures are PSI

    Valve Sizing for Cylinder Actuation

    SCFM = 0.0273 x Cylinder Diameter x Cylinder Diameter x Cylinder Stroke / Stroke
    Time x ((Pressure-Pressure Drop)+14.7) / 14.7

    Cv Required = 1.024 x SCFM / (Square Root of (Pressure Drop x (Pressure-Pressure
    Drop+14.7)))

    Pressure 2 (PSIG) = Pressure-Pressure Drop

    Air Flow Q (in SCFM) if Cv is Known

    Valve Cv x (Square Root of (Pressure Drop x ((PSIG - Pressure Drop) + 14.7))) /
    1.024



    Cv if Air Flow Q (in SCFM) is Known

    1.024 x Air Flow / (Square Root of (Pressure Drop x ((PSIG-Pressure Drop) +
    14.7)))

    Air Flow Q (in SCFM) to Atmosphere

    SCFM to Atmosphere = Valve Cv x (Square Root of (((Primary Pressure x 0.46) +
    14.7) x (Primary Pressure x 0.54))) / 1.024

    Pressure Drop Max (PSIG) = Primary Pressure x 0.54

    Flow Coefficient for Smooth Wall Tubing

    Cv of Tubing =(42.3 x Tube I.D. x Tube I.D. x 0.7854 x (Square Root (Tube I.D. /
    0.02 x Length of Tube x 12)



    Conversions


    To Convert Into Multiply By
    Bar PSI 14.5
    cc Cu. In. 0.06102
    °C °F (°C x 1.8) + 32
    Kg lbs. 2.205
    KW HP 1.341
    Liters Gallons 0.2642
    mm Inches 0.03937
    Nm lb.-ft 0.7375
    Cu. In. cc 16.39
    °F °C (°F - 32) / 1.8
    Gallons Liters 3.785
    HP KW 0.7457
    Inch mm 25.4
    lbs. Kg 0.4535
    lb.-ft. Nm 1.356
    PSI Bar 0.06896
    In. of HG PSI 0.4912
    In. of H20 PSI 0.03613




    Leave a comment:


  • DR
    replied
    Originally posted by MGREEN View Post
    v860rich,
    thanks

    .................................................. ..............

    Perhaps another online search of the
    required info that you mention would
    help to answer whether or not this unit
    and an appropriate cylinder could put
    out 20 tons. ?

    Yes, sure, this pump could generate 20 tons with a large enough cylinder. But the cylinder might need to be so large the pump's tank isn't enough to contain the needed amount of hydraulic fluid. Plus, it would probably be painfully slow to move the cylinder's ram because of low flow rate.

    All is not lost though. Low pressure hydraulics can usually be used in pneumatic cylinders too. Cylinders that typically will be used with 100psi shop air can handle low pressure hydraulics. I had a toggle press that operated at rated capacity on 100psi air. By supplying 250 psi hydraulics I increased the press capacity 2.5x. You do have to use good judgement, don't put 1000psi to pneumatic equipment.

    Leave a comment:


  • MGREEN
    replied
    Thanks all for working out the possibilities!
    But, this unit isn’t going to work out for
    what I’d like to do.

    Anybody want it ? it’s up for grabs.
    Mike

    Leave a comment:


  • Bob Engelhardt
    replied
    Originally posted by Mike Burch View Post
    At 1000psi a 7 1/2" diameter piston (say an 8" cylinder) would give you 20 tons of push.
    The flow rate is said to be 1.2GPM.
    8" diam => 50 sq in; 1.2gl = 277 cu-in; 277/50 = 5-1/2" per minute! That would get real old real quick!

    Leave a comment:


  • SVS
    replied
    Originally posted by Mike Burch View Post
    At 1000psi a 7 1/2" diameter piston (say an 8" cylinder) would give you 20 tons of push.
    The flow rate is said to be 1.2GPM.
    Problem is 8” cylinders are rare and expensive beasts in the used/surplus market. 4” and maybe 5” are more common.

    Reservoir size matters for cooling (not a big deal with intermittent use on a press) but definitely needs to hold the volume difference between cylinder extended and retracted.

    Leave a comment:


  • Mike Burch
    replied
    At 1000psi a 7 1/2" diameter piston (say an 8" cylinder) would give you 20 tons of push.
    The flow rate is said to be 1.2GPM.

    Leave a comment:


  • alanganes
    replied
    There are lots of factory type applications for units like this. I have seen units very similar to this used for hydraulic crimping tools used for crimping large electrical connections and for small dedicated assembly presses and clamping applications.

    Leave a comment:


  • tom_d
    replied
    Originally posted by MGREEN View Post
    v860rich,
    thanks

    tom-d,
    Thanks for clarifying.
    it was just a quick glance on google.
    Like I said, I know so little about these
    systems, that I don't know how to think
    it through, nor what I need to digest.

    Perhaps another online search of the
    required info that you mention would
    help to answer whether or not this unit
    and an appropriate cylinder could put
    out 20 tons. ?
    Pass along as much information as you can glean off that unit. Can you post a picture of the motor id tag too? Many here are eager to help research, and help with solutions. It's what this forum is all about. Here are some links to calculators that can assist in determining what you will need for your project:

    tabs on left of page for specific formulae:
    https://www.surpluscenter.com/Tech-H...r-Force-Speed/
    lots of reading material here too:
    https://northernhydraulics.net/catal...tml#calculator

    Leave a comment:


  • MGREEN
    replied
    v860rich,
    thanks

    tom-d,
    Thanks for clarifying.
    it was just a quick glance on google.
    Like I said, I know so little about these
    systems, that I don't know how to think
    it through, nor what I need to digest.

    Perhaps another online search of the
    required info that you mention would
    help to answer whether or not this unit
    and an appropriate cylinder could put
    out 20 tons. ?

    Leave a comment:


  • tom_d
    replied
    Looks like it's designed to be foot operated. Where did you find the specs for that pump? 1000psi doesn't give all the required information. Is that its maximum pressure, or pressure at a specific rpm? What is its max rpm, pressure, and gallons per minute output at various speeds? Depending on the specs, and the diameter of the piston used in you project, this may well be perfect for your application.

    Leave a comment:


  • v860rich
    replied
    Force equals area X pressure.
    So if you have a cyl with a total of 4 sq inches of fluid contact and the unit puts out 1000 lbs of press, you will have 4000 lbs of force on the business end of the cyl.

    THANX RICH

    Leave a comment:


  • MGREEN
    replied
    Would be interesting to find out just
    what this whole little unit would have been
    built for in the first place ?
    Mike

    Leave a comment:


  • MGREEN
    replied
    I'm not too savvy with hydraulic systems, so
    I think that there are various factors that
    would need to be worked out.

    Cylinder size, piston size, pump psi etc.....

    Someone with knowledge in that area
    might chime in.

    Leave a comment:

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