New loader valve not working properly

[J_J]

Was the original purpose of hose 510 from the loader valve to add flow to the drive motor circuits?

It tees up with hose 513 and looks like is adds flow to the input of the motor valves.

Is hose 510 now coming from the stacked valve using the single work port?

The HP necessary to run a pump is derived from the pressure desired and the GPM which is determined by the pump displacement and the rpm of the pump.

If you run out of HP, then you have to reduce the pressure or the GPM.

You are trying to transfer power and the hyd motors will develop the HP at about 85% efficiency.

12 GPM at 2800 psi = 23 HP

17 GPM at 2000 psi = 23 HP

20 GPM at 1700 psi = 23 HP

The 12 GPM pump will provide the most power from the above figures at the available HP, but at a reduced speed.

http://www.mgbryan.com/shop_image/product/Hatz-2G40-SS.pdf

Your max torque is around 2200 rpm, but if you run the engine at say 1800 rpm, the HP is about halved.

I believe a .9 cu in pump will use all of the 23 HP and pump 12 GPM, and have a pressure of 2800 psi.

So the hyd motors working at 85 % efficiency will develop about 19.55 HP

 

[jd4310man]

Yes hose 510 is still in place, I did not change that. OK, 12gpm seams to be the magic number for the hp rating. .900 cu in pump is a bit smaller than than the combined displacement of the two original pumps. The original track drive pump was .730 cu in, and the attachment pump was .509 cu in. So would it be safe to say that I could use a single pump of 1.200 cu in displacement? Is there less power required to run two pumps compared to one larger one? Looks like I have no choice but get a smaller pump anyway.

 

[J_J]

Your 23 HP engine, turning at 3000 rpm will pump 16 GPM's using a 1.2 cu in pump with the relief set to 2100 psi.

You can split that flow ever how you want.

But again, do you want power or speed?

Using two pumps, the displacement of the two pumps just add together.

Example:

Two pumps with .5 cu in each at 3000 rpm = 1 cu in = 12 GPM.

A .5 cu in pump turning at 3000 rpm will pump 6 GPM each X two = 12 GPM.

 

[jd4310man]

Does the flow divider make any difference in regards to power requirements? If I gave an 18gpm pump and I divide the flow with the priority flow divider, and only 12gpm is going to track drive circuit and the rest is going to the loader valves, would it provide the same affect as only having the 12gpm pump? Im not using the loader as much as the drive circuit.

 

[J_J]

If you do not use the loader valve , the fluid is just passing through the valve to tank and the flow is wasted.

However, if you activate that one SA spool lever, you add that flow to the drive circuit.

18 GPM, at 1800 psi = 23 HP

My thoughts on a situation like this would be to figure out the speed and torque you want for the drive.

Then use the displacement of the hyd motors to see how many GPM you need and how much pressure to operate the motors.

You know the cu in of the hyd motors, so based on the GPM which equals speed and the operating pressure, you figure the pump size you need in cu in, and then the HP required to drive it.

You have to compromise now using the 23 HP motor as to pump size and pressure.

The hyd motors can only develop the HP and speed if the pressure and the GPM's is there.

A set pump flow is also used by the loader valve for short periods based on the divider valve.

You could also eliminate the divider valve and put all the valves in series.

However if the first valve is using all the flow, there is none left for the other functions.

What you do in that case is to not use full lever, and allow a small amount of fluid to go to the loader valve, and you could eliminate the stacked valve.

The flow might look like this. Pump to motor valve, then to loader valve, then to tank.

All the valves should have the GPM rating of the pump. A little off will not matter much.

 

[J_J]

Example for a hyd wheel motor computing HP, speed, torque.

Disp. 11.0 cu. in. / rev.
Motor Type Geroler
Pressure:
3000 PSI cont.
4000 PSI int.
Torque:
5098 in-lbs. cont.
6776 in-lbs. int.
Speed:
381 RPM cont.
477 RPM int.
Flow:
20 GPM cont.
25 GPM int.

Need a 20 GPM pump turning at 3000 rpm = 1.5 cu in

So if you run that flow through a 11 cu in wheel motor, up to 3000 psi,

That pump will require 41 HP to run it.

That one motor will turn about 420 rpm and develop HP about 15 % HP less than the engine HP, and have torque of 5098 in-lbs .

That is just for one motor.

If using two hyd motors one for each track, then that machine will need an engine of about 82 HP.

 

[jd4310man]

The engineers at CF Struck corp designed this machine, Magnatrac 5000 with a 20hp Kohler gas engine, a drive pump of .730 cu in, an attachment pump of .500 cu in. The two drive motors are 10.6 cu in. each. Regular drive is 2.4 mph and overdrive is 3.5 mph. This thing is slow! I can live with that but lack of power I cannot. Right now with the 1.78 cu in pump, it will move the tracks and work the loader but it doesn't have quite the power I need. I dont think putting all the valves in series without the use of the flow divider would work very well. if one side of the divider isnt being used, it isnt taking any power to run it. I like the idea of having dedicated flow to each circuit so there will be power where and when I need it.

 

[J_J]

A .730 cu in pump, and a .5 cu in pump for a total of 1.23 cu in pump running at 3600 rpm, will pump 18 GPM.

19 GPM at 1500 psi requires about 20 HP from the orig Kohler engine.



New pump, 1.78 cu in, at 3000 rpm will pump 23 GPM

23 GPM at 1450 psi will require about 23 HP

If the full flow splits between the two 10 cu in hyd motors, and if each motor gets 11.5 GPM, at 1450 psi, the torque is only about 2309 in lbs per motor.

Can you verify if those hyd motors are 1500 psi motors?

 

[jd4310man]

The motors are rated at 3500 max psi. The motor valve is set at 3000 psi. Any lower pressure and the motors would stall easy. I never touched the pressure relief on the track drive valve. Attached is the current specs from Struck.

 

[J_J]

Something is not right in Struck advertising.

Using the MH8000 as an example, they say a 31 hp engine can produce 23.5 GPM at 3000 psi.

However the formula says different.Horsepower Required to Drive a Hydraulic Pump Calculator

Insert pressure of 3000 psi, then 23.5 GPM for the MH8000 and you will see, that 48.39 HP is required.

A big difference