PT AL-629 ( can it be done ? )

[Bob999]

If that were the case, as I understand it, then each motor in the two series circuits would be producing only the power equivalent of 750 PSI -- 750 PSI to the front wheel and 750 PSI to the rear wheel. That means these 12.5ci motors would only be producing a maximum of 1386 in lbs of torque, done at a 2gpm flow, yielding 34 RPM. That's only 115.5 ft lbs of torque -- surely that is not the case.

According to the HP formulas at Surplus Center, at 750 PSI and 8 gpm, each of these wheel motors would only be putting out 2.845 HP. Multilply that times 4 and the 25-HP PT would only be putting 11.84 HP to the ground -- running full throttle. I don't think so...

12 HP to the ground doesn't sound too far off the mark to me. When the machine is used as a mower a lot of power goes to the PTO circuit for the mower. When all the losses are factored in I wouldn't expect a lot more power to the ground.

However, I take it that we are both guessing--I know I am--I still think it would be to your benefit to actually measure the pressures before investing big money in the project.

 

[KentT]

12 HP to the ground doesn't sound too far off the mark to me. When the machine is used as a mower a lot of power goes to the PTO circuit for the mower. When all the losses are factored in I wouldn't expect a lot more power to the ground.

However, I take it that we are both guessing--I know I am--I still think it would be to your benefit to actually measure the pressures before investing big money in the project.

That 12HP would be WITHOUT the PTO or steering/lift circuits robbing any power from the engine. That would be the max available HP at the wheels, regardless of the other constraints.

Pressure is not really generated by the pump, per se -- pressure is generated by the resistance to the wheel motors turning as a volume of fluid goes through them, in addition to the small percentage created by hoses, connections, etc. The pump is pumping a volume of oil, which happens to be correlated to pressure because it is to easier to pump against lower pressures. The more resistance to movement of the wheel motors the more the pressure -- to the point that the pump can no longer push that volume of oil through the system.

The only way I can see to measure the pressure would be to put the PT up against an immovable object such as a tree, and running wide open, measure the pressure at different treadle settings. That would be pretty difficult to do...

 

[Bob999]

The only way I can see to measure the pressure would be to put the PT up against an immovable object such as a tree, and running wide open, measure the pressure at different treadle settings.

I think about it differently. There is an outlet port on the variable displacement pump (maybe two--one for forward and one for reverse) and there is a hose connected to the port(s) that carries the fluid. To measure the maximum pressure developed by the pump you need to terminate that hose in a pressure guage--because we both agree that maximum pressure is generated at zero flow.

 

[KentT]

I think about it differently. There is an outlet port on the variable displacement pump (maybe two--one for forward and one for reverse) and there is a hose connected to the port(s) that carries the fluid. To measure the maximum pressure developed by the pump you need to terminate that hose in a pressure guage--because we both agree that maximum pressure is generated at zero flow.

There are two ports on the tram pump. Which is the inlet and which is the outlet simply depends on which way you're pushing the treadle, unless I'm mistaken.

But, I don't know if you can get meaningful info with your test. In the real world the flow is never quite limited to zero when the swash plate of the pump is open, because of leakage in the wheel motors. If you terminate the line completely into a gauge you should either (a) stall the engine when you push down on the treadle since the oil has no place to go, and all of it cannot "leak" through the pump, or (b) it should kick in any internal bypass/pressure relief valve in the pump. Based on my experience with the PT and cooler oil in the system, the first will likely occur -- the engine will die when you push down on the treadle -- after the pressure gauge momentarily spikes...

Perhaps I'm wrong, but I don't think that would be a good measurement. It would confirm that the pump is pumping -- yes, but I already know that. It won't tell me how much pressure that pump will produce under a constant load nor at maximum flow rate because I can't go "full treadle" to see. Right now the wheel motors can "leak" and relieve some of the pressure build-up. With the line terminated in a gauge, they can't, so I don't think you'd ever get to a meaningful treadle position (i.e. flow) to take a reading.

I know you'd definitely kill the engine on a fixed displacement pump -- I've dealt with "hydraulic locks" before. I'm just not sure how this variable displacement pump would respond, nor how meaningful the test result might be....

 

[Bob999]

If that were the case, as I understand it, then each motor in the two series circuits would be producing only the power equivalent of 750 PSI -- 750 PSI to the front wheel and 750 PSI to the rear wheel. That means these 12.5ci motors would only be producing a maximum of 1386 in lbs of torque, done at a 2gpm flow, yielding 34 RPM. That's only 115.5 ft lbs of torque -- surely that is not the case.

According to the HP formulas at Surplus Center, at 750 PSI and 8 gpm, each of these wheel motors would only be putting out 2.845 HP. Multilply that times 4 and the 25-HP PT would only be putting 11.84 HP to the ground -- running full throttle. I don't think so...

I need to go back to your power calculation--it seems to me that it may be low by a factor of 2.

Wouldn't the flow through each motor be 4 gallons rather than two (this assumes the total pump output is 8 gallons/minute and that the flow is divided into two circuits)? If the flow is double in each circuit then the total HP would also double.

 

[J_J]

There are two ports on the tram pump. Which is the inlet and which is the outlet simply depends on which way you're pushing the treadle, unless I'm mistaken.

But, I don't know if you can get meaningful info with your test. In the real world the flow is never quite limited to zero when the swash plate of the pump is open, because of leakage in the wheel motors. If you terminate the line completely into a gauge you should either (a) stall the engine when you push down on the treadle since the oil has no place to go, and all of it cannot "leak" through the pump, or (b) it should kick in any internal bypass/pressure relief valve in the pump. Based on my experience with the PT and cooler oil in the system, the first will likely occur -- the engine will die when you push down on the treadle -- after the pressure gauge momentarily spikes...

Perhaps I'm wrong, but I don't think that would be a good measurement. It would confirm that the pump is pumping -- yes, but I already know that. It won't tell me how much pressure that pump will produce under a constant load nor at maximum flow rate because I can't go "full treadle" to see. Right now the wheel motors can "leak" and relieve some of the pressure build-up. With the line terminated in a gauge, they can't, so I don't think you'd ever get to a meaningful treadle position (i.e. flow) to take a reading.

I know you'd definitely kill the engine on a fixed displacement pump -- I've dealt with "hydraulic locks" before. I'm just not sure how this variable displacement pump would respond, nor how meaningful the test result might be....

Just tee in a pressure gage at the pump outlet, then you can read the pressure at any time. You could also extend the pressure gage line up to the front console.

On my 1445, I have installed a tee with a quick attach fitting. I put one in the PTO circuit, and the lift circuit. On the test gage, I put the other half of the quick attach. I just plug in the gage into the circuit I want to test, or just leave it plugged in.

 

[KentT]

I need to go back to your power calculation--it seems to me that it may be low by a factor of 2.

Wouldn't the flow through each motor be 4 gallons rather than two (this assumes the total pump output is 8 gallons/minute and that the flow is divided into two circuits)? If the flow is double in each circuit then the total HP would also double.

I mentioned 2 gpm in my example because the performance charts for the White 12.5ci RS motor shows that its torque at 750 PSI is highest at 2 gpm -- i.e. that would be the "best case" scenario of the amount of torque it could produce at 750 PSI. GPM has nothing to do with PSI or torque necessarily -- GPM determines RPM, not torque. Pressure (PSI) determines torque. The combination of torque (pressure) and flow (gpm) determines horsepower. For the HP calculation in my earlier post I used 8 gpm flow at 750 PSI -- the maximum power that the wheel motor could provide at that PSI if the 16 GPM output of the pump is correct (16 gpm/2 circuits = 8 gpm at each wheel motor).

Here's the specs at 750 PSI and their conversion to HP:
.5 gpm = 1250 in lbs torque, 6 RPM (.12 HP)
1 gpm = 1360 in lbs torque, 15 RPM (.32 HP)
2 gpm = 1386 in lbs torque, 34 RPM (.75 HP)
4 gpm = 1349 in lbs torque, 72 RPM (1.54 HP)
6 gpm = 1322 in lbs torque, 110 RPM (2.31 HP)
8 gpm = 1228 in lbs torque, 146 RPM (2.84 HP)


1228 in lbs X 146 RPM divided by 63025 = 2.84 HP at each wheel motor

 

[KentT]

Just tee in a pressure gage at the pump outlet, then you can read the pressure at any time. You could also extend the pressure gage line up to the front console.

On my 1445, I have installed a tee with a quick attach fitting. I put one in the PTO circuit, and the lift circuit. On the test gage, I put the other half of the quick attach. I just plug in the gage into the circuit I want to test, or just leave it plugged in.

Have you tested the variable displacement tram pump with your test gauge? If so, what kind of readings did you get? I'm guessing that the PSI moves all over the place, unless you're doing something like tramming down a level road at a steady treadle. In normal use working, I'd presume as the load on the wheel motors goes up (climbing even a slight slope, pushing or pulling something, making a turn, etc.) the PSI increases, then as the load goes down the PSI decreases...

IMO, the only way to test the max PSI output would be to essentially stall the engine when at full throttle -- and see the spike reading just before the engine died -- or the relief valve opens up.

 

[Bob999]

I mentioned 2 gpm in my example because the performance charts for the White 12.5ci RS motor shows that its torque at 750 PSI is highest at 2 gpm -- i.e. that would be the "best case" scenario of the amount of torque it could produce at 750 PSI. GPM has nothing to do with PSI or torque necessarily -- GPM determines RPM, not torque. Pressure (PSI) determines torque. The combination of torque (pressure) and flow (gpm) determines horsepower. For the HP calculation in my earlier post I used 8 gpm flow at 750 PSI -- the maximum power that the wheel motor could provide.

Here's the specs at 750 PSI and their conversion to HP:
.5 gpm = 1250 in lbs torque, 6 RPM (.12 HP)
1 gpm = 1360 in lbs torque, 15 RPM (.32 HP)
2 gpm = 1386 in lbs torque, 34 RPM (.75 HP)
4 gpm = 1349 in lbs torque, 72 RPM (1.54 HP)
6 gpm = 1322 in lbs torque, 110 RPM (2.31 HP)
8 gpm = 1228 in lbs torque, 146 RPM (2.84 HP)


1228 in lbs X 146 RPM divided by 63025 = 2.84 HP at each wheel motor

I went to two different calculators on the web and used 1500 PSI and 8 gallons per minute. That equated to 7 HP--assuming 100% efficiency of the motor which is obviously unrealistic. If you reduce the pressure to 750 PSI the HP drops to 3.5 --again assuming 100% effeciency. Alternatively doubling the flow will double the HP. What is missing in all of this is any anchoring in data--what is the output of the variable displacement pump.

I really don't think it is realistic to try to back into this by making assumptions--either you need specs for the variable displacement pump or you need good real world measurements to make good decisions about replacing wheel motors.

 

[KentT]

I went to two different calculators on the web and used 1500 PSI and 8 gallons per minute. That equated to 7 HP--assuming 100% efficiency of the motor which is obviously unrealistic.

I really don't think it is realistic to try to back into this by making assumptions--either you need specs for the variable displacement pump or you need good real world measurements to make good decisions about replacing wheel motors.

But, the two wheel motors in a series circuit where the pump only puts out 1500 PSI (your propositition) would only be using 750 PSI each -- 750 PSI at the front motor, and 750 PSI at the rear motor... that's where you're misunderstanding lies. For a wheelmotor in a series circuit to produce the power that corresponds to 1500 PSI, the pump would have to output 3000 PSI.

The output of a wheel motor is determined by the delta PSI ... the PSI at the inlet less the PSI at the outlet. The front wheel motor has 3000-1500=1500 while the rear wheel motor has 1500-0=1500. (Those are "100% efficiency numbers" assuming no resistance from hoses, fittings, valves, etc.)

To be semantically correct, the pump does not output PSI, it outputs GPM. The PSI is the resistance to flow at the wheelmotors, 1/2 at the rear and 1/2 at the front.