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Battery driven winch

  1. Oct 18, 2016 #1
    1. The problem statement, all variables and given/known data

    A standard D cell can supply 10mA at 1.5 volts for about 300 hours. An alkaline D cell can do about the same. Assume - 2 - that the chemistry of the batteries will produce the same amount of energy and will maintain the same EMF regardless of the current drawn from the battery. However, the internal resistance of a standard D cell is about 1 ohm and the internal resistance of an alkaline cell is about 0.1 ohm, so the amount of energy dissipated internally is different. Let us suppose that you have a multi-speed winch that is 50% efficient and that your mass is 60kg.
    (a) If the winch is set to super-slow speed how high can one of these batteries lift you before it is exhausted?
    (b) What is the fastest rate (in cm/sec) that each battery can lift you?
    (c) At this fastest lift rate, what is the maximum height that you achieve before the battery dies? How long does it take?

    2. Relevant equations

    P=I*V=W/t
    U(gravity)=mgh

    3. The attempt at a solution

    a). I said power dissipated here = 1.5V*10mA = 0.015 W.
    50% efficiency means 0.0075 W are used by the winch. 0.0075 * (3600 s/hr * 300 hr) = 8100 J supplied by the battery. Weight = 60 kg * 10 m/s^2 = 600 N, so total distance traveled = 8100/600 = 13.5 meters.

    b.) For a standard D cell,
    - maximum current draw = 1.5V/(R_internal) = 1.5 A. Power dissipated = 1.5 V * 1.5 A = 2.25 W. At 50% efficiency 1.125 W are used by the winch.
    - Power = Energy / time. P=mg(delta(h)/delta(t))
    1.125 W = 60kg * 10m/s^2 * (h/t), so h/t = speed = 0.1875 cm/s.

    For an alkaline D cell,
    same as above but multiplied by a factor of 10 = 1.1875 cm/s.

    c.) Total energy supplied by the battery over its lifetime = 0.01A*1.5V*(300 hr * 3600 s/hr) = 16,200 J.
    Time = Energy/power. For a standard D cell, t = 16,200 J / (2.25 J/s) = 7,200 s.
    For an alkaline D cell, t = 16,200 / 22.5 = 720 s.

    Total distance traveled = speed * time = 0.1875 cm/s * 7200 s = 13.5 m total.
     
  2. jcsd
  3. Oct 19, 2016 #2

    CWatters

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    Perhaps check answer b. What is the voltage delivered to the winch if 1.5A is being drawn? It's not 1.5 V. Would a lower current deliver a higher voltage and more power?
     
  4. Oct 19, 2016 #3
    Doesn't the voltage delivered by the battery remain constant? As in, "the chemistry of the batteries will produce the same amount of energy and will maintain the same EMF regardless of the current drawn from the battery."

    Reference https://www.physicsforums.com/threads/battery-driven-winch.889785/#post-5597336
     
  5. Oct 19, 2016 #4

    gneill

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    Staff: Mentor

    Nope. The internal EMF remains the same but the battery's internal resistance lies between the source of the EMF and the battery terminals. In other words, a real battery is modeled as an ideal voltage source in series with a resistance.
     
  6. Oct 19, 2016 #5
    How could I figure out the voltage across the winch without knowing its resistance?
     
  7. Oct 19, 2016 #6

    gneill

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    The questions are posed in such a way that you're looking for extreme situations. One thing I'd advise reviewing is the Maximum Power Transfer Theorem :wink:
     
  8. Oct 19, 2016 #7

    CWatters

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    The winch will draw a current that depends on it's load. You can show that the output of the battery will be somewhere between 0A @ 1.5V and 1.5A @ 0V.

    Try picking some values of the current (say 0A, 0.5, 0.75, 1, 1.25 1.5A ) and calculate the output voltage and then output power. Then sketch two graphs of output power vs current and Output power vs Voltage. Where is the power a maximum?

    Once you "see the light" you may not have to plot the graphs next time.
     
  9. Oct 19, 2016 #8
    I tried the max power transfer theorem and revised parts (b) and (c) (I don't think part a changes since it's just an energy calculation):

    b. Max power: R_internal = R_load.
    For a standard D cell, R_load = 1 ohm. Power dissipated by the load = I^2 * R_load = R_load * ([V_source] / [R_int+R_load])^2. If V=1.5V and both R's = 1 ohm, Power = 0.5625 W.
    At 50% efficiency, P = 0.281 W = mg(h/t). Thus h/t = 0.047 cm/s.

    For an alkaline D cell, same as above but with R_int = R_load = 0.1 ohm, which leads to v = 0.47 cm/s.

    c. Dividing total energy by power dissipated in the load: 16200J / 0.5625 J/s = 28800s for standard D cell, and 2880 s for alkaline. Running at their respective speeds, both batteries give the same result for total height, 13.5 m. Does this sound right? Or do I need to divide energy by 2*power to account for power dissipation in the battery itself?
     
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