Lowering the max power without affecting the min

Discussion in 'Homework Help' started by Teszla, Jul 12, 2013.

  1. Teszla

    Thread Starter Member

    Jun 7, 2013
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    I get that when maximum power is used, there is a resistance of R = V^2/P = 230^2/80 = 661.25 Ω. This must therefore be a internal resistance that is active even when the varying resistance is inactive (i.e. when the pedal is not pushed).

    If the task was only to half the maximal power we could just connect another 661.25 Ω-resistance in series, but how are we supposed to do so that the lowest speed will not be affected?
     
  2. crutschow

    Expert

    Mar 14, 2008
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    You need a resistor is series with the pedal and a resistor in parallel with the pedal such that the combination resistance when the pedal at it's maximum resistance is the maximum resistance you want and is the minimum value you want when the pedal is at its minimum resistance.

    I'll leave the calculations of that to you. ;)
     
  3. Teszla

    Thread Starter Member

    Jun 7, 2013
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    First of all, how can you know that's how they should be connected? I mean what is the theory behind it?
     
  4. Teszla

    Thread Starter Member

    Jun 7, 2013
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    Secondly, I got these calculations:

    V = 230 V<br />
\\ P_{max}_{1}=80 W<br />
\\ R_{m}=\frac{V^{2}}{P}=\frac{230V^{2}}{80W}=661.25 \Omega<br />
 \\ \\ P_{max2}=40W \ll = \gg<br />
 \\  \ll = \gg R_{1}=\frac{V^2}{P_{max2}}-R_{m}=\frac{230V^{2}}{40W}-661.25\Omega = (1322.5-661.25)\Omega = 661.25\Omega \\ P_{min}=\frac{V^{2}}{R_{m}+R_{pmax}}=\frac{230V^{2}}{(661.25+550)\Omega}\approx 43.67W\\<br />
...
    Rm = Machine inner resistance
    R1 = Added resistance in series

    Please tell me if I'm on the right track...
     
    Last edited: Jul 14, 2013
  5. LDC3

    Active Member

    Apr 27, 2013
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    First, when the pedal is not pushed, the machine is not running. When the pedal is pushed slightly, the machine is running slowly. When the pedal is pushed all the way, the machine runs the fastest.

    When I first looked at Pmin and saw that you had 43.67W, I was surprised that it is greater than Pmax2. Then I realized that it just needed to be less than Pmax1. But it poses an interesting question, how can 1/2 the fastest speed be slower than the slowest speed (unless the question is in error)?
     
  6. crutschow

    Expert

    Mar 14, 2008
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    I agree with LDC3. I don't see how the objective can be achieved. To limit the maximum power to 40W you need a 661.25Ω resistor in series with the motor when the pedal is at zero , but you need no more than 550Ω in series to maintain the lowest speed with the pedal at it's maximum. Those are contradictory requirements.

    The problem, I think, is that the motor is a non-linear load. Not sure what term "The motor can be regarded as purely active" means. :confused:
     
  7. tubeguy

    Well-Known Member

    Nov 3, 2012
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    What I get from the given problem is that the resistance at max power (highest speed) would be 0 ohms, not ~ 660 ohms. Am I missing something :confused:
     
  8. LDC3

    Active Member

    Apr 27, 2013
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    Actually, I see no problem with the machine having the minimum speed being greater than half the fastest speed. The problem is that when the control is set to give this slower speed, the machine will probably stall.
    What I think the response should be is that the control provides the resistance to still give the minimum speed and 1/2 the difference between the minimum speed and the maximum speed. The control resistance is modified from '550Ω to 0Ω' to '550Ω to 275Ω'.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    The problem has a number of issues.

    First, we don't know what resistance corresponds to the motor just starting to turn. It merely states that she is happy with the speed that results when the pedal is pushed far enough just result in the motor turning. That may be at 549Ω or it may be at 500Ω, or 300Ω. About all we can do is assume that it starts turning with the barest touch of the pedal.

    The biggest issue is that the problem calls for the motor to be limited to half power, not half speed. But the speed is, in general, not proportional to power, but the voltage across the motor (assuming it is a DC permanent magnet motor, which may or may not be what was intended) is.
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Stina should stick to sewing and leave the electrical modifications alone.

    My Grandma had a foot pedal human driven sewing machine - no motor at all.:rolleyes:
     
  11. LDC3

    Active Member

    Apr 27, 2013
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    My grandma had one of those too. I think it was a Singer.
     
  12. WBahn

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    Same here. It was a Singer. When I was really young I would work the peddle with my hands (the sewing machine was folden down so the only load was the flywheel) and try to get it going as fast as I could. It was tricky to keep your movements in sync with it at speed. When I was a bit older and staying at her place from time to time she taught me to sew using it.
     
  13. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Grandma's workout so she could kick your butt no matter how old you were. :D
     
  14. t_n_k

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    Strange how one memory leads to another.
    My oldest aunt also owned a similar Singer machine. To this day I have a vivid recollection of two of my female cousins doing some sewing when I happened to be visiting with them. Efforts came to an abrupt halt with one of them holding out a bloodied thumb with a Singer sewing needle embedded through the nail and out the other side.
    Awesome.
     
  15. t_n_k

    AAC Fanatic!

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    What's with the 661.25Ω?

    The motor is rated for 0.348A.

    With a linear emf to speed factor the motor voltage at half rated speed would be 115V. Thus to limit the motor to half speed at rated current draw requies only 115/0.348 or 330.625Ω.
     
  16. WBahn

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    Mar 31, 2012
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    Two things. The problem didn't ask for half speed, it asked for half power. Also, the current draw will not be independent of speed, if for no other reason than that it requires more power to turn faster even with just parasitic frictions.
     
  17. t_n_k

    AAC Fanatic!

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    There was no mention of this being a real motor in a genuinely real situation - the OP's question indicated the motor was somehow miraculously "purely active".

    Half the rated voltage at rated current would be half power.

    I could have assumed the hypothetical motor was driving a constant torque load.
     
  18. WBahn

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    The real issue, of course, is that we shouldn't have to be making all of these assumptions to begin with.

    I'm still waiting to get some kind of an explanation of what "purely active" means.
     
  19. Teszla

    Thread Starter Member

    Jun 7, 2013
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    I written it the way the task was given, however I noticed that I wrote: "The motor can be regarded as purely active, and its speed is controlled by a series resistor in the pedal." It actually says "The motor can be regarded as purely active, and its rotational speed is controlled by a series resistor in the pedal." I don't know if this makes it clearer though?

    I think "purely active" means something it should be considered as "true power"?
     
  20. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    @ Teszla,

    Sorry, that didn't clarify anything. Enough people "in the know" have looked at the question and commented on the inadequacies in its drafting.

    I gave it a parting "shot", with my thinking based on an idealized universal motor or idealized DC motor model. I had to assume [with no justification] certain conditions about the motor loading with speed variation. Namely, I used a constant torque loading. On that basis I did some further number crunching. Using the 80W / 230V values as rated conditions, I denote the speed to be 100% in that case.

    I then found the following [steady state] values whilst keeping the load torque fixed as the same value applied for the 100% speed reference at 80W output. For the given series R values ...

    1. R=550Ω, Speed = 16.6% [of rated speed], P_motor=13.4W
    2. R=330Ω, Speed = 50% [of rated speed], P_motor=40W
    3. R=150Ω, Speed = 77% [of rated speed], P_motor=62W
    4. R=50Ω, Speed = 92% [of rated speed], P_motor=74W
    Would I suggest you trust these values? They are probably worth very little, particularly as the original question offers no insights into how the machine loading varies with motor speed. You are welcome to use any or all of them to devise a possible resistance based speed control scheme. Without giving too much away, two additional resistor values come to mind - one of which is 367Ω.

    I believe the consensus is that it's a flawed question. Unless something has been lost in translation.
     
    Last edited: Jul 15, 2013
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