Hydraulic pump unit

[MGREEN]

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.

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

[JoeLee]

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.....

[MGREEN]

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.

[MGREEN]

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

[v860rich]

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

[tom_d]

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.

[MGREEN]

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. ?

[tom_d]

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-Help/Hydraulics/Calculators/Cylinder-Force-Speed/

lots of reading material here too:

https://northernhydraulics.net/catalog/support.html#calculator

[alanganes]

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.

[Mike Burch]

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.

[SVS]

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.

[Bob Engelhardt]

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!

[MGREEN]

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

[DR]

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.

[metalmagpie]

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