I have 25hp Chinese diesel tractor with a small hydraulic pump ( I think it's 11GPM) I have a new 25GPM 3500PSI hydraulic pump. It will require about 12HP from the PTO. I don't plan to use the added pump while driving, only for stationary work. The question is: would it be OK to use the hydraulic fluid from the main system? Tee off from the low pressure supply from the case, to supply the new pump, and return from whatever drive unit the new pump supplies, back to the case.? If I don't need a separate hyd. fluid tank, that would be great for the amount space saved!
Add a PTO pump
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oldnslo
Super Member
I am a little confused on some of your numbers.
1) 25 GPM @ 3500 PSI is more than 12 HP. Generic rule of thumb is takes 1HP to produce 1 GPM at 1500 PSI.
2) how is an 11 GPM inlet supposed to supply a 25 GPM pump?
3) I doubt you have the oil volume or cooling capacity to operate this pump.
also suspect tractor maybe 11 liter per minute flow vs 11 GPM.
1) 25 GPM @ 3500 PSI is more than 12 HP. Generic rule of thumb is takes 1HP to produce 1 GPM at 1500 PSI.
2) how is an 11 GPM inlet supposed to supply a 25 GPM pump?
3) I doubt you have the oil volume or cooling capacity to operate this pump.
also suspect tractor maybe 11 liter per minute flow vs 11 GPM.
PILOON
Super Star Member
I'd say a lot would depend on the type of usage.
A log splitter would not cycle much so that might be OK but driving a snow blower or bush hog would heat up the oil quite fast.
Basically it is all about heat dissipation.
Hey, perhaps a cooler on the return might do the trick!
A log splitter would not cycle much so that might be OK but driving a snow blower or bush hog would heat up the oil quite fast.
Basically it is all about heat dissipation.
Hey, perhaps a cooler on the return might do the trick!
Thanks! As i said, the prime use would be not while driving. Mostly for the logsplitter, which is way too slow on the exsiting hyd. pump. But I was going to build a driver to set Groundscrews. A new method to build structures from the ground up, all gaivanized steel. Here in a Northern California fire zone, new or replacement decks, and Solar Array mounting systems need to be fire resistant.
The existing 11 GPM pump is fine for the FEL and brush hook attachment, but real slow on my logspltter. The question is about taking the hyd. fluid from the tractor case directly to a 25 GPM pump inlet, to whatever device is attached, and then back to the tractor's case reservoir. Advice to add a line cooler before fluid returns, seems to be good advice. Do you have a source for the rule of thumb Hp/flow/PSI rate calc?
oldnslo
Super Member
Actual formula to figure hydraulic pump input power is
HP = (GPM times PSI) divided by (1714 times hydraulic efficiency)
I typically just divide by 1500 since that is approximately 87 percent efficiency
I also suspect do a search for hydraulic HP for hydraulic pump and you will find that rule of thumb
You can calculate existing pump flow by measuring cylinder extend distance vs time. You will have to estimate cylinder bore diameter if you do not what it actually is.
example:
2 inch bore cylinder extends 12 inches in 4 seconds or 3 inches per second
formula: GPM = velocity times cylinder area
convert GPM to cubic inches multiply GPM by 231 then divide by 60 to cubic inches per second
using cubic inches per second and velocity of inches per second keeps all the results in inches per second format
2 inch bore cylinder area is 3.14 square inches time 3 inch per second velocity is 9.42 cubic inches per second
9.42 times 60 = 565.2 cubic inches per minute divide by 231 = 2.45 GPM per cylinder
HP = (GPM times PSI) divided by (1714 times hydraulic efficiency)
I typically just divide by 1500 since that is approximately 87 percent efficiency
I also suspect do a search for hydraulic HP for hydraulic pump and you will find that rule of thumb
You can calculate existing pump flow by measuring cylinder extend distance vs time. You will have to estimate cylinder bore diameter if you do not what it actually is.
example:
2 inch bore cylinder extends 12 inches in 4 seconds or 3 inches per second
formula: GPM = velocity times cylinder area
convert GPM to cubic inches multiply GPM by 231 then divide by 60 to cubic inches per second
using cubic inches per second and velocity of inches per second keeps all the results in inches per second format
2 inch bore cylinder area is 3.14 square inches time 3 inch per second velocity is 9.42 cubic inches per second
9.42 times 60 = 565.2 cubic inches per minute divide by 231 = 2.45 GPM per cylinder
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On your advice I went online and found this page with a variety of pretty easy calculators: Learn with Bailey: Calculators
My plan is to double the PTO shaft RPM at the pump input shaft to increase flow rate. ( 1000rpm@PTO > 2000rpm@pump) It looks like the lowest HP requirement for this pump is just about 25HP.
My plan is to double the PTO shaft RPM at the pump input shaft to increase flow rate. ( 1000rpm@PTO > 2000rpm@pump) It looks like the lowest HP requirement for this pump is just about 25HP.
Sawyer Rob
Super Member
If you double the speed, it will take even more HP...
SR
SR
oldnslo
Super Member
Sailon
Is your new pump a hi-lo pump designed for log splitter service? If yes then you could have enough PTO power to operate it but I seriously doubt that your tractor has enough reservoir capacity or a large enough opening for the new pumps inlet
Is your new pump a hi-lo pump designed for log splitter service? If yes then you could have enough PTO power to operate it but I seriously doubt that your tractor has enough reservoir capacity or a large enough opening for the new pumps inlet