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  • Hydraulic pump unit

    Picked this up at a recent estate sale.

    We plugged it in and the motor ran
    Got it for peanuts as the guy seemed to want stuff gone.

    It was filthy nasty so I gave it a clean up when
    I got it back to the home shop.

    I grabbed it thinking maybe it would have
    enough oomph to use as part of my up coming
    hydraulic shop press build.

    Its a Dayton Model 5W033.
    A quick search showed 1000 psi pump.
    The tank/ reservoir is about the size
    of a one gallon fuel can ...for reference,

    I don't have any cylinders other than a few
    small bottle jacks, so one would have to
    be obtained perhaps.

    I don't think this pump would have
    enough grunt to be used for the
    shop press project.

    At the moment I can't think of
    another use for this unit.

    Anybody nearby have a project that this
    unit would fit in?

    If so, it's up for grabs. Click image for larger version

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    Currently (no pun intended), it's wired for 110 volts.
    Mike

    Mike Green

  • #2
    I was thinking maybe a log splitter. But if you don't think it has enough power for a press then I doubt it would have enough power for a log splitter.

    JL.....

    Comment


    • #3
      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.
      Mike Green

      Comment


      • #4
        Would be interesting to find out just
        what this whole little unit would have been
        built for in the first place ?
        Mike
        Mike Green

        Comment


        • #5
          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
          People say I'm getting crankier as I get older. That's not it. I just find I enjoy annoying people a lot more now. Especially younger people!!!

          Comment


          • #6
            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.

            Comment


            • #7
              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. ?
              Mike Green

              Comment


              • #8
                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

                Comment


                • #9
                  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.

                  Comment


                  • #10
                    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.

                    Comment


                    • #11
                      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.

                      Comment


                      • #12
                        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!

                        Comment


                        • #13
                          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
                          Mike Green

                          Comment


                          • #14
                            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.

                            Comment


                            • #15
                              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




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