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jOmega (Electrical) (OP)
6 Nov 03 19:33

Josef --

The following is offered in response to your comments in the thread...DC Motor brush life....Thread237-77159

    "///Motor load is often considered with a constant shaft load (HP).
    Since HP~Watts=ExI, then the only way to preserve the relationship
    for E decreasing is by I increasing.

--- on the surface, you are absolutely correct and theory stands behind you in so far as defining constant watts. But where you fall down is in your assumption that constant HP operation of dc machines abides by that principle. The truth is, they do not, except in a very special case (as will be discussed herein)...

In 99.99% of dc motor applications, they motors operate,
  • CONSTANT TORQUE .. below base speed...

  • and
  • CONSTANT HP .. above base speed.

    • base speed = rated rpm at rated voltage, field, & arm. amps
    • base speed range is operation at or below base speed

Ergo, at rated armature voltage and below, DC motors are operated Constant Torque.... NOT Constant HP.

In order to operate constant HP in the base speed range the HP must be less than motor rated HP or, the amps must be allowed to increase above rated.

A couple of examples.

Consider a 10 HP, 240v, 38 amps, (81.8% eff), 1150 rpm, dc motor.

That motor develops 10 HP ONLY at 240v & 38 Amps

At 1/2 base speed (575 rpm) 120v armature, and 38 amps, the motor develops 5 HP ...

According to your hypothesis, you would have to allow the current to rise to 200% rated (76 amps) to develop 10 HP at 1/2 armature voltage (120v).

Now Joseph, you and I both know that ain't gonna happen.

There is one exception to this, however; that is in the case of winders and un-winders that operate constant tension (i.e.,constant HP) in the base speed range.  

This is a special case.

Using the same motor from the above example....we know that the motor can develop a maximum of 10 HP rated.

This winder is designed as a direct drive, center core, with a 6:1 buildup. Since you cannot obtain more than 10 HP from the motor,
  • the full core HP cannot be 10 HP (at 1/6 base speed ... 191.67 rpm) )... and
  • the empty core HP cannot be 10 HP at base speed (1150 rpm)

  • It must be something less.... How much less ???

    Consider that at full core, the motor will be operating at
  • 191.67 rpm,
  • 40 volts on the armature, and
  • 38 amps (rated motor current)....
  • then the maximum HP that the motor can developed (without exceeding it's stated design ratings) is 10/6 or 1.6667 HP ....

    That means that the motor starts at 1.6667 HP at empty core, at rated rpm.... and does not exceed that power rating over 6:1 buildup so as to maintain constant tension while winding ...

    The winder starts off at rated speed at rated voltage at 1/6 rated amps and at 1/6 rated HP. As the core builds up, the armature voltage reduces, and the current increases in direct proportion to maintain constant HP.
    At full roll, the armature voltage is 40 volts, 191.667 rpm, 38 amps, 1.667 HP ...

    That, Joseph is an example of constant HP operation in the base speed range of the motor.

    Such winders are typically controlled by dancers or some other type sensor that monitors tension and adjusts the winder torque accordingly to obtain constant hp winding.

    In this example... at empty core, the winder is operating at 1150 rpm and as each successive wrap is put on, the speed is reduced as the armature voltage is reduced and the current increases to keep the HP constant.

    But, Joseph, in order to obtain constant tension winding, the HP is not allowed to increase (or decrease)..... as the winder speed (armature voltage) is reduced with buildup).  

    In this example, the winder motor starts at empty core, 1150 rpm, 1/6 HP...... and maintains 1/6 HP over the buildup as the rpm and armature voltage are decreasedwhile armature current increases.

    At no time (other than perhaps acceleration) does the armature current exceed rated n/p value.

    This Joseph.... is an example of rated hp operation in the "base speed" range of the motor. This is a specialized application of a dc motor... and is not, as you alluded, a typical mode of dc motor operation in the base speed range.

    The typical.... and most common.... is Constant Torque in the base speed range.... not Constant HP.

    Constant HP is typical for operation ABOVE base speed; i.e., the field weakend range.......   which the original post never ascribed to as an application attribute for the question posed.

    In the field weakend range above base speed:

        HP = (Reduction in Torque  x  Increase in Speed) / 5250

    So, as flux is reduced, torque is reduced, speed is increased, and HP remains constant

    But Joseph, don't take my word for it. I've posted some references you can avail yourself of to obtain a better understanding of dc machines and how they operate.



    Siemens - Basics of DC Drives
    (from which the above graphic is borrowed)

    Siemens - Speed/Torque Relationships of Shunt Connected Motors 32 of paper  (page 7 of PDF file)

    Rockwell Automation (Allen-Bradley)
    "A Comparison of the Characteristics of AC and DC Motors
  • see pg. 11 - Aplication Issues of Speed Ranges and Fig. 19


    Summing it up Joseph:
    Constant HP operation of dc shunt motors occurs in the Field Weakend range above base speed of the motor.
    Below base speed, dc shunt motors are operated constant torque.... except in very special cases.

    Ergo, your premise that volts decrease and amps increase to obtain constant HP operation of dc machines ... is rejected except in the special case for the winder application. DC machine theory tells us that constant HP operation is obtained while holding arm volts and amps constant while reducing the shunt field flux.


    Here's where you started:

      jbartos (Electrical) Nov 1, 2003
      Suggestion: If the motor terminal voltage decreases, the motor current through the brushes increases, since HP is a constant. Therefore, the brushes would have to be adjusted to the higher commutation currents through the brushes.

    which you  posted in response to Clyde38's original post :
      Clyde38 (Electrical) Oct 31, 2003
      Why do some dc motor manufacturers say you shouldn't run a motor at a voltage lower that rated?  These are small motors, less that 2" in diameter.
      12 volt motor at 6 volts.
      They state that they need to install different brushes.



    Clyde never mentioned operating the motors in the "constant HP" range - standard definition i.e., above base speed.....

    You posted in response comments about "motor terminal voltage decreasing, motor current thru the brushes increases since HP is constant."

    Your comments Joseph, were both incorrect
    • dc machines operate in constant HP with arm volts and current fixed (held constant) ... while field flux is reduced,
    and not relevant to Clyde's original post; i.e., he never said anything about operating above base speed or operating in Constant HP mode.


    edison123 (Electrical)
    6 Nov 03 21:26

    one question, what will happen to torque and hp if a dc motor originally designed for 1000 RPM is rewound for 500 RPM ? same original torque and 50% of original HP ?
    cbarn24050 (Industrial)
    7 Nov 03 2:55
    Hi edison, if a motor dc motor is rewound for a lower speed, at the same arm volts/current/field flux it will not have the same torque output.
    edison123 (Electrical)
    7 Nov 03 6:48
    Hi cbarn,

    Assuming the same field flux, the armature rewound (arm voltage and current immaterial since ultimately the Amp turns are going to be the same in both the cases)to half the speed would result in 50% of original output. The torque will remain the same using the equation,
    Power = 2*Pi*N*T/60 (metric) since both power and speed are reduced in the same proportion.

    cbarn24050 (Industrial)
    7 Nov 03 7:42
    Hi edison, if you are going to change the arm voltage and current then you have a completly different setup.
    edison123 (Electrical)
    7 Nov 03 8:00

    As I said, arm voltage and current doesn't matter in the torque equation I posted. For example I can convert (and I have) 230 V to 460 V. The armature turns will double and current will halve so that net AT is the same for both 230 and 460 V for the same HP.
    cbarn24050 (Industrial)
    7 Nov 03 8:59
    that wasn't the question you posed on 6 nov.
    Helpful Member!  jOmega (Electrical) (OP)
    7 Nov 03 17:21

    Advise changes made to motor to cut RATED speed in half.



    edison123 (Electrical)
    7 Nov 03 20:46

    My first question and other posts dealt with the same topic "what happens to the torque and HP " when the motor is rewound for half the RPM. Since you stated " if a dc motor is rewound for a lower speed at the same arm volts/current/field flux it will not have the same torque output", I pointed out with the new lower speed, the armature voltage can also be changed without affecting the torque.


    The armature, interpoles and series field were redesigned and rewound for half the base speed (1200 RPM to 600 RPM) and twice the armature voltage (230 V to 460 V). The shunt field was not touched. The motor was derated from 150 HP to 75 HP. The motor is running quite well from a base speed of 600 RPM to the field-weakened speed of 1800 RPM.

    I was keen to know whether the motor can develop higher torque than the original at the lower speed (though I know from the equations, it cannot). I wanted to know if anyone else any practical experience in this matter.
    jbartos (Electrical)
    7 Nov 03 21:37
    Suggestion to jOmega (Electrical) Nov 6, 2003 marked ///\\\
    Josef --
    The following is offered in response to your comments in the thread...DC Motor brush life....Thread 237-77159
          "///Motor load is often considered with a constant shaft load (HP). Since HP~Watts=ExI, then the only way to preserve the relationship for E decreasing is by I increasing.\\\"

    --- on the surface, you are absolutely correct and theory stands behind you in so far as defining constant watts. But where you fall down is in your assumption that constant HP operation of dc machines abides by that principle.
    ///I never stated this. This appears to be part of your false assumption.\\\
     The truth is, they do not, except in a very special case (as will be discussed herein)...
    ///DC motors are more frequently scrutinized for Torque Speed characteristics. Therefore, the literature is somewhat scarce for motor HP output versus input current and voltage characteristics. It is better to consider DC Generators for these relationships since they are better documented in the literature.
    1. Self excited shunt generator will have decreasing terminal voltage with increasing terminal current. The generator is rated at certain kW. There is so much power only it can be transferred over the iron and airgap. Therefore, there is a limit of the DC generator kW that can be delivered. The shunt generator output voltage versus current curve generically demonstrates it.
    2. Compound self-excited generator terminal voltage versus current also exhibits noticeable decrease in the terminal voltage when current exceeds its rated value. Therefore, again it is seen that the approximately Max kW power available from this generator has been reached and the generator cannot deliver more than kW=Max ~ constant.

    Obviously, the DC generators are assumed to be driven at an approximately constant speed while the motor speed usually varies. However, the common essence of the DC rotating machine stays. The limiting factors of the iron and airgap do not permit the motor to deliver more power than its Pout=Max ~ constant.

    The exact answer to that referenced posting would require exact mathematical relationships. However, this eng-tips Forum shows very little of mathematics besides links, which I frequently post in my postings.\\\
    cbarn24050 (Industrial)
    8 Nov 03 23:32
    Hi edison, yes that is true but nobody has suggested that arm voltage and output torque are related.

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