Results 1 to 7 of 7

Thread: 2-Motors Driven Conveyor

  1. #1
    Join Date
    Feb 03
    Posts
    2

    2-motors driven a conveyor

    Pls advice for amateur...

    Ratio:~20
    Motor(1)-->Fluid coupling--->Bevel Gear--|
    Ratio:~20 |-->Conveyor
    Motor(2)-->Fluid coupling--->Bevel Gear--|

    Both motors are same rating,type&manufacturer.
    Could different motor brand with almost same rating
    be replaced to existing motor? Rated speed is different
    approx 1 %from existing motor.

  2. #2
    Lawrence K. Nordell

    Lawrence K. Nordell

    President and CEO

    Conveyor Dynamics Inc. [eDir]

    Conveyor Dynamics Inc.

    Professional Experience 57 Years / 11 Month Lawrence K. Nordell has 57 Years and 11 Month professional experience

    Discussions 2608 Lawrence K. Nordell acceded to 2608 discussions, Articles 0 Lawrence K. Nordell wrote 0 articles, Publications 0 Lawrence K. Nordell Nordell released 0 publications

    Searching nothing specified

    Know-How Design (1521) Lawrence K. Nordell used this tag 1521 times, Pipe Conveyor (238) Lawrence K. Nordell used this tag 238 times, Chutes (119) Lawrence K. Nordell used this tag 119 times

    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

  3. #3
    Join Date
    Feb 03
    Posts
    2

    2 motrs drive conveyor on same pulley

    Thank you for your reply and sorry for my
    unclear problem...

    1. Both motors are on same pulley.
    2. Rated of both existing motors : 670hp,1490rpm
    Application: Spreader conveyor in mining
    3. Slip of fluid coupling could be adjusted.

    Due to one of existing motors must be overhauled,
    I would like to replace different motor manufacturer
    rated 670hp,1480rpm to the exisitng. Pls recomend
    me;

    -Is it possible?,suggestion or reason

  4. #4
    Lawrence K. Nordell

    Lawrence K. Nordell

    President and CEO

    Conveyor Dynamics Inc. [eDir]

    Conveyor Dynamics Inc.

    Professional Experience 57 Years / 11 Month Lawrence K. Nordell has 57 Years and 11 Month professional experience

    Discussions 2608 Lawrence K. Nordell acceded to 2608 discussions, Articles 0 Lawrence K. Nordell wrote 0 articles, Publications 0 Lawrence K. Nordell Nordell released 0 publications

    Searching nothing specified

    Know-How Design (1521) Lawrence K. Nordell used this tag 1521 times, Pipe Conveyor (238) Lawrence K. Nordell used this tag 238 times, Chutes (119) Lawrence K. Nordell used this tag 119 times

    Follow-up on Two Mtrs Load Sharing

    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

  5. Conveyor driven by 2 motors

    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

  6. 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.
    Dave Miller
    ADM Consulting
    10668 Newbury Ave., N.W.,
    Uniontown, Ohio 44685 USA
    Tel: 001 330 265 5881
    FAX: 001 330 494 1704
    E-mail: admconsulting@cs.com

  7. #7
    Lawrence K. Nordell

    Lawrence K. Nordell

    President and CEO

    Conveyor Dynamics Inc. [eDir]

    Conveyor Dynamics Inc.

    Professional Experience 57 Years / 11 Month Lawrence K. Nordell has 57 Years and 11 Month professional experience

    Discussions 2608 Lawrence K. Nordell acceded to 2608 discussions, Articles 0 Lawrence K. Nordell wrote 0 articles, Publications 0 Lawrence K. Nordell Nordell released 0 publications

    Searching nothing specified

    Know-How Design (1521) Lawrence K. Nordell used this tag 1521 times, Pipe Conveyor (238) Lawrence K. Nordell used this tag 238 times, Chutes (119) Lawrence K. Nordell used this tag 119 times

    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.

Similar Threads

  1. Unbalanced Motors
    By Sandeep Nair in forum Screening & Feeding
    Replies: 1
    Last Post: 5th August 2005, 22:49
  2. Selection of Motors
    By Divya Pandya in forum Screening & Feeding
    Replies: 5
    Last Post: 30th May 2005, 14:29
  3. Electric Driven Mobile Conveyor Belt
    By wadee in forum New Projects, Tenders, Inquiries
    Replies: 1
    Last Post: 19th October 2003, 13:28
  4. - Electric Driven Mobile Convoyer Belt
    By wadee in forum Mechanical Conveying
    Replies: 0
    Last Post: 17th October 2003, 11:14
  5. Drum Motors
    By Author in forum New Projects, Tenders, Inquiries
    Replies: 0
    Last Post: 24th June 2002, 12:18

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Single Sign On provided by vBSSO