Help in hydraulic motor sizing for a firewood processor needed

[polemidis]

I need to make a decision for my firewood processor and I have hard time doing so

I have a 13.2gpm circuit (OC). (hopefully @1800psi with my 16hp gas engine and the 0.85in pump). But I assume I have a 12.5gpm system taking into consideration inefficiencies. (Should I lower that??)

The circuits that will run simultaneously are:
a)The chainsaw motor
b)The outfeed conveyor of the splits
c)The feed cylinder for the chainbar (but it draws only 0.5gpm or maybe less (1" bore cylinder) and I am not worried for this)

My intentions were to run the a) at 11.5gpm on a 0.58 motor. and the b) at 0.5gpm on a 2.99in motor, but I am worried that with such small flow the conveyor motor will have lots of loss due to slippage. So I start to questioning myself if there is another way to do it. (and also if my worry is rational or not important at all!!)
If my math is correct I have:
a) ｱ4500rpm on the motor (5000rpm is its max rating)
b) I have no clue how many ft/s this motor will give me as I leave open the reduction ratio, neither how much pressure is going to be needed to lift the splits 8ft high. The splitter cycling time is 14s at least so a belt moving at only 1.5ft/s is convenient.

So my question is: should I plan for more gpm on the conveyor motor in order to reduce the loss in inefficiency of that gear motor (and do the appropriate changes in the motor size etc) , and also change the a) to a)0.45in motor that runs on less flow of 9.7gpm??

I definitely want to keep almost 5000rpm on the saw motor. This is the most hungry part of the system, so any hp that I am saving elsewhere its a plus. Any other suggestions?
I know I have to do the math on the conveyor but Its hard for me. I prefer to play with the reduction gears

I am ready to order the last parts before that baby starts breathing and I am questioning myself. Does what I question make sense??

 

[IceStationZebra]

You have to assume that the pump is ~95% efficient when figuring supply power. But the when looking at the output side you have to include the motor efficiency as well, and you may want to err on the low side to make sure you get the work done you expect (0.93 x 0.93 = .865 or 86.5%). This will change due to wear, oil viscosity, oil temperature and pressure so don't get too hung up on the exact percentage.

Are you sure the pump displacement is right, or are you gearing it up? I only get 6.6gpm from a .85cc pump. To get 12.5-13.2gpm would need approx 1.6cc pump.

HP = (psi x gpm x efficiency)/1714 = (psi x pump displacment (in3) x RPM x efficiency)/(1714 * 231) = (1800*.85*1800*.95)/(1714*231) = 6.6hp

ISZ

 

[polemidis]

For RPM, put 3600 in your calculations

Other than that the simplified question is if I am wrong on these scenario calculations:
The facts:
Gas engine 16hp
Pump 0.85 @3600 ===>13.2gpm theoretical, so maybe 12gpm for real after hoses, losses, etc?
1800 psi @ 13.2 needs 15.8hp
Chainbar feed 1" cylinder takes 0.5gpm and insignificant pressure, maybe 300psi

Scenario a)
Conveyor out feed motor 3ci, takes 0.5gpm, turns at 40rpm needs maybe 1500psi to lift the splits, will require 0.4hp
Saw motor 0.58in takes all the remaining flow 11gpm turns at 4380rpm and have 1700psi available , 13ft.lb torque will require 11hp,

Scenario b)
Conveyor outfeed motor 12ci takes 2gpm, turns at 40rpm, takes maybe 375psi to lift the same weight of splits (375psi so it will be the same as the a) scenario) will require 0.4hp
Saw motor 0.45in takes all the remaining flow 9.5gpm, turns at 4850rpm (faster), and have 1980psi available, 11.8ft.lb torque will require 11hp

So total is about 11.4 hp, with the rest up to 16hp in losses. (16hp *85%pump*85%motor=11.5hp)

Which scenario is better?
I now prefer the b) because the chainsaw will run faster, the conveyor motor will run more efficient at higher flow, and because it leaves me with more margin to adjust the system (like to provide more flow on the saw motor if needed or change the gear ratio on the conveyor)

What do you think guys?