Hi Raymond
It’s a very convoluted and error prone calculation, but if you really do want to try to work out the performance of your proposed hydraulic system then it goes like this:
Image you had a motor of 225 cc/rev, when you give this motor a flow of 45.4 L/min [45,400 cc/min] you would expect a [theoretical] speed of 45,400/225 = 201.7 rpm. You will find, however that the speed is less than that because of internal leakage. The ratio of the actual speed to the theoretical speed is called the “volumetric efficiency” and you will see from your datasheet that this is a function of torque (a higher pressure causes a greater internal leakage) and also a function of speed (flow).
Let’s imagine that the motor was driving a load such that the torque was 158 Nm. The theoretical [differential] pressure induced at the ports of the motor would be 225/(20*pi) Nm per bar, i.e., 3.58 Nm/bar. Theoretically, you would need 44.1 bar to drive that motor, but when you look it up on the datasheet, you would find that you actually needed 55 bar. The ratio of the actual torque to the theoretical torque is called the “hydro-mechanical efficiency” and you will see that this is also a function of torque and speed (pressure and flow).
So this is our situation: it takes a pressure of 55 bar to push 45.4 L/min of oil into the motor when the shaft has to overcome a torque of 158 Nm and, when the motor is loaded like this, our flow only gives us a speed of 191 rpm. Our input power (hydraulic) is 45.4*55/600 = 4.16 kW and our output power (mechanical) is 191*2*pi/60*158/1000 = 3.16 kW. So it looks like the “overall efficiency” is 3.16/4.16 = 76% (which is just like Hydtools Ted said it would be).
When driving this Char-Lynn unit like a pump we have no option other than to assume that the efficiencies as a pump are the same as the efficiencies as a motor (as long as the operating conditions are the same). Our volumetric efficiency was 191/201.7 = 94.7% at 191 rpm and our hydro-mechanical efficiency was 44.1/55 = 80.2% at 55 bar. When running as a motor the internal losses in the unit meant that the speed was less than we expected for the flow input and the pressure differential was more than we expected for the torque output. If we run it as a pump then it’s all the other way round: the losses now mean the flow output will be less than we expect for the speed and the torque input will be more than we expect for the pressure differential.
Driving the “pump” at 191 rpm against a circuit resistance which causes the outlet pressure to be 55 bar would give us a flow of 94.7% of the theoretical flow figure: 191*225/1000*0.947 = 40.7 L/min and would require an input torque of 55*3.58/0.802 = 245.5 Nm. To get a hydraulic power output of 40.7*55/600 = 3.73 kW we had to put in a mechanical power input of 191*2*pi/60*245.5/1000 = 4.91 kW. The overall efficiency is 3.73/4.91 = 76%. It’s not surprising that this overall efficiency figure appears to the same for the pump duty and the motor duty: I used the same efficiency figures for both equations.
[Note that the big assumptions I’ve made here are that both the volumetric efficiency and the hydro-mechanical efficiency remain constant if shaft speed stays at 191 rpm and the motor input pressure and the pump outlet pressure are both at 55 bar – ignoring the fact that the motor inlet flow is slightly higher than the pump outlet flow.]
This is something like what your application works out to be: your power input is much more limited (sorry – got to work in SI units): 1.5 HP = 1.118 kW so we will have to work at significantly lower pressures and speeds than used in the example above. We have yet to learn what sort of outlet pressure your pump is going to have to work against, so imagine you put enough cans in the crusher to cause the pump outlet pressure to be 55 bar – what will be the pump’s hydro-mechanical efficiency? Let’s guess that the speed is around 23 rpm (because that’s a number on the table in the datasheet), the motor torque output at 23 rpm and 55 bar is 169 Nm but it should have been 55*3.58 = 196.9 Nm so your hydro-mechanical efficiency is 85.8%. If you want to run this as a pump you will have to put in 55*3.58/0.858 = 229.5 Nm.
Your available input power is 1.118 kW (1118 Watts) and your electric motor shaft speed is 1725 rpm (180.6 rad/s) so your available electric motor output torque is 1118/180.6 = 6.19 Nm. Let’s say you lose 3% of this in your gearing/pulley/chain arrangement so you only have 6.00 Nm left to driv ethe pump and your gear ratio will have to be 229.5/6.00 = 38.25:1. This will give you a resultant pump shaft speed of 45.1 rpm.
When the motor is running at 55 bar the tabulated figure we have for its speed is 23 rpm for an input flow of 7.6 L/min. But the theoretical input speed is 7.6*1000/225 = 33.8 rpm so the volumetric efficiency is 23/33.8 = 68%. If we assume this volumetric efficiency figure doesn’t change with speed (that’s not really true but it will actually improve at higher speeds) then we can say that a pump shaft speed of 45.1 rpm will give an actual outlet flow of 225/1000*45.1*0.68 = 6.9 L/min.
Here we are then; if your can crusher presented a resistance of 55 bar then you would get a flow of 6.9 L/min from your “pump” which is running at 45.1 rpm. Your mechanical input power would be 1.118 kW, your hydraulic output power would be 6.9*55/600 = 0.632 kW and your overall efficiency would be 56%. Your 2” hydraulic cylinder would extend at 56.7 mm/sec and the force would be 11.1 kN (assuming no losses in the cylinder, pipework, back pressures etc.). Then converting that back to US units: ~2¼ in/sec cylinder extension speed and ~2490 lbf compression, ~800 psi system pressure, ~1.84 (US)GPM pump flow and motor input power equals 1.5 HP.
Of course if your load were more you would overload your electric motor, if you wanted to run at a higher pressure you would need a greater reduction ratio and you would get less flow. As the pressure goes up the volumetric efficiency (at low speeds) gets even worse and eventually will reach zero (the pump shaft turns and there is no outlet flow – it all goes into leakage before you get up to the required pressure).
Actually now I’ve done some sums it doesn’t look too bad – except that’s one rather large input torque you need to put on your motor shaft and the effort involved in gearing down by nearly 40:1 isn’t a small task either. And your pump still needs a 1 bar boost pressure on its inlet (that’s a header tank 40 ft above the pump inlet and a fat enough, 40 ft long suction line so that your ~ 2 gpm doesn’t cause the pump inlet pressure to drop below 1 bar). Are you sure you don’t want to buy a simple gear pump?
DOL
