Drives with different motor slip:

You need to make clear:

1. are the motors on the same pulley shaft or different pulleys

2. belt elastic modulus if drives are on different pulleys

3. if on different pulleys, how far apart are the pulleys from one-another and what are the tolerances of the pulley diameters or better the absolute diameters of the pulleys

4. type of fluid coupling and rating or slip curve wrt motor size

5. fill level of coupling vs its rated nameplate and fill level

Motors have a manufacturing tolerance of around 7% if taken from distributor stock. The tolerance can be reduced by machting the motor rotor material components. Stock drives are guaranteed to lpull within 7% +/- of nameplate ie a 100 hp drive , at 1450 RPM, will pull between 93 and 107 hp at 1450 RPM.

One percent difference on a 1% slip motor is not meaningful. When the manufacturer states an electrical slip, it is a nominal value for the motor design.

If the motors are on the same shaft, and the variation of drive slip is 1%, then one drive will pull 0.5% more than nominal and one drive will pull 0.5% less than nominal. Add the fluid coupling slip at 3% will reduce the difference by a factor of 4. Neither of these conditions do I believe to be the case. I anticipate you have mis-stated the problem you are trying to solve.

Lawrence Nordell

Conveyor Dynamics, Inc.

www.conveyor-dynamics.com ■

When two motors are mounted on the same shaft they will load share as follows:

1. the two motors must, in total, pull the required load demanded of the conveyor -- ie the sum of their power, in kW, equals the necessary power of the conveyor.

2. each motor will pull according to the electric slip curve --

i.e. kW vs rpm.

Solve these two conditions by an example:

1. give: conveyor demands 1000 kW to move load by both motors

2. given motors are rated 670 kW each and @ zero kW=1500 RPM

3. motor #1 newly rewound and has an electrical slip of 1480 rpm at motor 670 kW nameplate

4. motor #2 old and feeble has a tired rotor with an electric slip of 1450 rpm

5. motors 1 and 2 are on same shaft -- what do they each pull in kW and what is their RPM???

6. Equation 1: RPM1 = RPM2 = rpm

Equation 2: kW1 + kW2 = 1000 kW

Equation 3: kW1 = 670/(1500-1480) * (1500 - rpm)

Equation 4: kW2 = 670/(1500-1450) * (1500 - rpm)

Equation 5: 1000 kW = constant (1500)- constant (rpm)

where: constant = 670/(1500-1480)+670/(1500-1450)

= 33.5 + 13.4 = 46.9

Solving from the above we get:

KW1 = 714.286 kW

kW2 = 285.714 kW

rpm = 1478.678 RPM

We are consultants in the field and can give good technical assistance in all manners of belt conveyor design and control.

Lawrence Nordell

Conveyor Dynamics, Inc.

website: www.conveyor-dynamics.com

email: nordell@conveyor-dynamics.com ■

In addition to required (matching) value of motor kW, rpm & type; the crucial item of comparison is torque-speed graph of both the drives (i.e. motor at steady speed operation). Please compare torque-speed graph of both the motors around steady speed operation zone.

If these are reasonably matching; you may use them. After cautious trial by gradually increasing mtph, the marginal difference in load sharing can be adjusted by minor adjustments in fluid quantity / flow within operating circuit (for simple fluid coupling it is level of fluid within fluid coupling). Please do not opt for more than minor adjustment in fluid, as it could result in to other problems for fluid coupling.

Please note that it is torque-speed graph of the drive, which matters.

Regards,

Ishwar G Mulani

Author of Book : Engineering Science and Application Design for Belt Conveyors.

Email: parimul@pn2.vsnl.net.in ■

I assume that both motors are driving the common shaft at the same time. That is, you do noy have a main and standby motor situation.

If this is the case, you would be better off in the long run to keep the motors matched by replacing both of them with your new motor of choice. The information provided by contributors Nordell and Mulani highlight the potential dangers of a mismatch. ■

Last Word in Solving the Requested Problem:

Solving for the fluid coupling fill relationship, given your two motors with stated electric slips of 1490 and 1480 RPM is as follows:

There are an infinite number of solutions due to the variations that can be selected between couplings. Let us assume that both coupling can produce 100% torque at 2% slip beyond the motor slip. From the above procedure the following can be deduced:

motor #1 @ 1490 RPM = 100% torque, let mtr+cplg slip to 1450 which is same as motor2 @ 100% torque when total slip is 1450.

Solving this state, the net pulley RPM= 1462.69 and both drives = 500kW

ie 670*(1500-1462.69)/(1500-1450) = 500 kW.

Thus, the fluid coupling of motor #1 must have 2.67% slip curve while motor #2 has the initial 2% slip.

As Mr. Mulani suggests, this is done by increase a small amount of fluid coupling slip (0.67%= removing a small amount of oil) from the steeper motor #1 slip curve while maintaining the factory fluid coupling fill (2%)recommendation on the other. The above analytic result help in estimating about how much oil should be removed and from which coupling.

Cheers,

Lawrence Nordell

Conveyor Dynamics, Inc. ■