Electrical Motor HP for Hydraulic Jacking System

I take it you have an existing hydraulic drive motor which you wish to replace with an electric drive. In order to calculate the TORQUE applyed by the drive displacement of the motor in cm3/rev or in3/rev and the applied pressure is needed. From the torque applied and the rpm a HP figure can be calculated. Be carefull here as an electric motor does not like to be stalled while a hydraulic motor will accept stalling at near full torque. All the other numbers you have have no meaning as the applied torque (measured in pressure) take into account all of the effieciencies. Good luck! MIKE L.

Dear Mikel;

Thanks for your reply, however, please note that the given load at the final pinion is weight ( LBS ) not pressure and the torque can be calculated :

W x Radius = 650,000 x 1.555542 = 1011102.3 ft - lbs

I will come back to discuss the calculation with you if you are interesting, but sorry for now as I have to leave my office.

Regards.

M.Salem

Please go through my calculations and advise if I have missed something:

1- Torque on final pinion = r x weight (force)

= 1.555542 x 650000 = 1011102.3 ft - Lbs

2- Final pinion velocity (v) = 1 fpm = 1/2x3.14xr (rpm) = 0.1023666 rpm

3- Totat ratio = Motor speed / Final pinion speed

= 900 (rpm) / 0.1023666 (rpm) = 8792

4- Torque on the motor = Torque on final pinion / total ratio = 5252 x HP /N

= 1011102.3 / 8792 = 115 ft-Lbs

5- Motor HP = 115 x N/ 5252

= 115 x 900 / 5252 = 19.7 HP

If we take into account all effiecencies then:

Motor HP = 19.7 / (0.8 x 0.9 x 0.893 x 0.94 x 0.83 )

Motor HP = 19.7 / 0.5 = 39.4 = 40 HP

However if we toke safety factor of 1.15

then Motor HP should be 40 x 1.15 = 46 HP = 50 HP

Do you think that the above calculation is correct?

Appreciate your feedback - thanks -M-Salem

Put a 100Hp motor and VS drive on it and walk away from it.

Cost difference is so minimal why worry?

It would be nice to know the total mass and rate of lift required to do a sanity check.

I read your question to say that you are replacing a hydraulic jacking system with an electric rack and pinion system. Therefore, we don't need to think about pressure and and flow or hydraulic losses.

Your figures for friction between rack and pinion, and friction between rack and slides are extraordinarily high if interpreted as friction coefficients. I gather, then, that you are using these as efficiency figures: in other words, you are saying that 6% of the force that is applied to the pinion is lost to friction in the rack/pinion mesh. In that case, you are saying that the efficiency of the machine, from pinion through vertical motion is 79%. That's probably pretty close, given the size of the structure, the likelihood of misalignment, uneven loading on the platform, side load from anticipated winds, etc.

Your calculations look reasonable, and the figures are correct, however the "philosophy" may not be sufficiently conservative. In a perfect world, there would be no efficiency loss in the slides, because there would be no normal force on them (other than side load on the rack caused by the meshing forces). Given that the situation is not so perfect, your efficiency number could vary pretty widely, I'd think. Therefore, my inclination would be to use a larger safety factor. On the other hand, if you have historical data which indicates that the slide efficiency never goes below 84%, then your factor is probably OK. (Although I guess this is a new installation of the rack and pinion system?)

I'd talk this over with someone local also, who would know or could be shown the details of the installation. But from halfway around the world, it looks like you are very close to correct. I think is is reasonable to ignore acceleration loads, which would only occur for 1/32 of a second or so.

The only caution that I have, to go for bigger safety factor and eventually bigger size of motor, is that the new system does not have VFD for smooth speed / torque control, it is a direct connection between electrical non synchronous motors through pinions that climb up racks fixed on Rig's 3 legs (totally 18 motors), thus we might damage the teeth.

Frictions between rack and pinion and slides are given by the supplier without any idea about the real forces, and I multiplied them to other efficiencies as I dont know how to calculate the real loads of these frictions in order to add them to the total load.

The supplier is suggesting 40 HP motor which I have dought about it.

...and I multiplied them to other efficiencies as...

As others have mentioned, the motor efficiency does not belong in the calculation. The motor efficiency will effect how much you pay for electricity to operate the system, but does not effect the motor size, which is already rated in shaft hp.

Are you saying that this system has 18 motors, each lifting 650,000 lb? If so, then you should have a local professional engineer advise you. (Actually this applies even if there is only one 40 hp motor: 650,000 lb is a pretty large load, and if there is any possibility of injury if something jams or comes loose, it is worth double checking to make sure everything is correctly designed.)

Thanks Kern I will take out the motor efficiency from the calculation.

The design got ABS approval, however, the motor capacity is still under discussion.

Your comments are appreciated.

Your calulations appear correct until the effeciency is calculated. .8 is intial motor eff., (if this is the hydraulic motor about right, but does not calculate into electric drive eff.), .9 primary gear box OK, .893 final gear drive OK, but the next effeciency is this friction?. As for a 100 hp drive with a VSD yes it will work but this seems extremely oversized to do a 20HP or 40HP job. Up to you of course.

I dont' believe you can just mutliply the friction factors in to the efficiencies like that. I'm not exactly sure how these things work but I'm assuming its a giant vertial rack along the rig's legs and the pinions are connected to the rig itself which "climb up" the racks. I'm not sure what the slides are (perhaps a picture would help). I am guessing but I think the friction between the rack & pinion & slides is the frictional coefficient. In this case you'll need to determing the side loading (or normal forces of these components) which is probably related to the geometry of the rack and pinion. The side loading must then be caculated and multiplied by the coefficients to be added on to the total weight of the rig.

Since we know that Power = Work / time = force * distance / time

I can say that in a frictionless world:

297727kg * 9.81m/s^2 * .3048m / 60s = 14837W / 746HP/W = 19.88HP

So if you can figure out what those frictonal loads are this problem shouldn't be too hard from there. Don't mind the SI units but thats what I was taught and I finb the units much easier to deal with.

I hope my comments helped out a little. Good luck,

Stephen T

Thanks Stephent your comments are valid and it adds value.

I know now that I made a mistake by adding the friction and the motor efficiency to the calculation of HP.

I don't think that I can define the slide and friction real loads, however, they can be considered by using higher safety factor or we can consider them as losses from the total torque that can be delivered to the final pinion ( 6% + 14% ).

Your calculation method is another way to check the final results.

Thanks again and all the best. M-Salem

My quick check of your calculations and another very simple check indicate Hp is OK. (take 650,000# vertical-load x 1'/min-650000 ft-lb/min divide by 33,000 ft-lb/min per Hp, and that quotient is 20 Hp. Now multiply the efficiency factors and divide the '20 Hp 100% efficiency Hp by the product of the efficiencies, and 40Hp seems OK. However, electric motors will provide much more torque as they approach stall than when at rated speed--but they also quickly overheat.

The suggestion of 'install a 100 Hp motor and VSD' is a VERY GOOD recomendation. Be sure you have provided ample electrical power supply.

Question? What has caused the decision to change driver? Hydraulic drives have infinite speed control with full torque available at all speeds, and provide smooth starts. Only a real good VSD and oversized motor will emulate that smooth break away torque-speed characteristic. 'Inch-ing' height will be much more problematic with the electric drive in any case.

I would suggest that a good VSD these days with flux vector control can easily handle 1000:1 speed range with rated torque available at all speeds. 150% torque OL at all speeds is also very realistic.

Thanks Keith for your comment.

The decision to change the hydraulic drives has been taken because it is 30 years old system, and it does not have the automatic mode to jack up the rig leveled.

The new electrical system uses Siemens S7 300 PLC but without VFD to control 18 motors of 40 HP, 6 for each leg of the Rig.

Unfortunately, the owner didn't consult the engineering department before buying the system, and after the installation they ended up with 6 trips of the motors at the first jacking trail ( although overload setting has been increased from 35 A till 42 A for each motor of the total 18 motors of the 3 legs.)

Finally they asked us to check the calculation, to find a solution.

I think that I still need to know how to add the friction into the calculation as weight not as efficiency?

To be precise, motor efficiency should not be included in the HP calculations, since the motor NP HP is shaft power.

However, the AC supply would need the extra power to burn.

Thank you, all guys (ken,GW,Mikel, Stephen, and Keith) for your kind replies to my questions.

Your comments were very helpful to me.

Best Regards.

M-Salem