triangel
Silver Member
I need someone to translate what is on this pump. I am planning to use this on a wood processor I am building and I would like to see if I can use this pump on it.
Hydraulic pump
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pmsmechanic
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Fordermenge = the output = 5.55 cubic cm/revolution = .33868 cubic inches/revolution. Max. Drehzahl = 2500 rpm. Max. Betreibsdruck = working or operating pressure 300 kg/cm2 = (kgf/cm2) = 4267 psi.
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triangel
Silver Member
Thank you for your reply, One more question, how many gallons per min flow at 2000 rpm. would this pump be fast enough for a splitter with a 4" x24 stroke cylinder
pmsmechanic
Elite Member
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- Southern Alberta, Canada
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- 4410 and F-935 John Deere, MF 245
5.55 cm3/revolution = 0.00146615 US gpm or 0.00122083 Imp. gpm. Since most hydraulics are in US gpm I'll go with that. 0.00146615 x 2000 rpm = 2.9323 US gpm. Now this is the part that I'm not sure of. There might be an efficiency factor in here that I'm not aware of. JJ or someone who knows more can correct me.
I do know a bit of German but I wanted to double check so all I did was Google those German words and I had my hunches confirmed. The conversions I did with an app on my phone.
I do know a bit of German but I wanted to double check so all I did was Google those German words and I had my hunches confirmed. The conversions I did with an app on my phone.
J_J
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Here is some data on pump efficiency.
Theoretical flow is calculated by multiplying the pump’s displacement per revolution by its driven speed.
Actual flow has to be measured using a flow meter.
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triangel
Silver Member
Thank you to all who replied, I guess this pump will have a lot of power but will not have any speed for this ram. need about 14-16 gpm.
