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Find power of an escalator

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A typical escalator in the London Underground rises at an angle of 30° to the horizontal. It lifts people through a vertical height of 15 m in 1 minute. Assuming all the passengers stand still whilst on the escalator, 75 people can step on at the bottom and off at the top in each minute. Take the average mass of a passenger to be 75 kg.

    (a) Find the power needed to lift the passengers when the escalator is fully laden. For this calculation assume that any kinetic energy given to the passengers by the escalator is negligible.

    (b) The frictional force in the escalator system is 1.4 * 104 N when the escalator is fully laden. Calculate the power needed to overcome the friction. Hence find the power input for the motor driving the fully laden escalator, given that the motor is only 70 % efficient.

    (c) When the passengers walk up the moving escalator, is more or less power required by the motor to maintain the escalator at the same speed? Explain your answer.

    Answers: (a) 1.4 * 104 W, (b) 7.0 * 103 W, 3.0 * 104 W

    2. The attempt at a solution
    (a) I. P = W / t = Fs / t = mgs / t
    mg = (75 * 75) * (10) [simplified g) = 56 250 kg
    Since the force (F = mg) is inclined at an angle 30°, therefore F = mg cos 30°, and the height h equals s / tan 30°.
    P = [56 250 * cos 30° * (15 / tan 30°)] / 60 seconds = 21 094 W -- wrong.
    II. P = mgs / t = (56 250 * 15) / 60 = 14 063 W -- looks correct. But why don't we use the 30°?

    (b) I. P = [(F - R) * s] / t
    14 063 = (15F - 210 000) / 60
    F = 70 252 N
    II. P = Rs / t = (14 000 * 15) / 60 = 3500 W

    In any case I am confused about (a) and not sure about (b) either. Any ideas please?
     
  2. jcsd
  3. Sep 16, 2016 #2

    gneill

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    Staff: Mentor

    mg should result in a force (Newtons), not a mass (kg).
    You already have the height: h = 15 m. Perhaps you were looking for the slope length of the escalator?
    upload_2016-9-16_8-32-5.png
    You should choose another trig function to relate h, s, and θ.
    Because no work is done against gravity by motion in the horizontal direction. Only changing height (vertical motion) does work. So you only need to consider the total height through which the mass is lifted. When you consider friction you'll be interested in the length of the slope because friction acts all along the slope as the mass moves along it.
    You'll have to explain what you attempted to do in (b); At the very least define your variables.

    The way I see it the work (and hence power) being done falls into two lumps. The first is the work done to lift the passengers against gravity, which you've already found in part (a). The second is the work done against friction. You can assume that the friction acts along the length of the slope. You'll want to add them together to get a total, then apply the efficiency criterion to that total to find your motor rating.
     
  4. Sep 16, 2016 #3
    Oh yes, that's correct: kg * m/s2 = N.

    Well, I thought that W = Fs for horizontal motion. So I need the adjacent side of the triangle F = mg cos 30° and the adjacent side for the distance travelled s = h / tan 30°.

    Alright, then (a) should be clear.

    Having found (a) I tried to find the driving force (F) which turned out to be F = 70 252 N. But I don't see any use in it.

    For II. I used the P = Resistance * s / t to find the power required to overcome resistance and it is equal to 3500 W.


    Work done against friction: isn't it P = Rs / t = (14 000 * 15) / 60 = 3500 W? Or P = Rs / t = (14 000 cos 30) * 15 / 60 = 3031 W. Then we combine 14 062.5 + 3031 = 17 093. Since 30 % of power is lost we need X * 0.7 = 17 093 and therefore total power needed = 24 419 W.
     
    Last edited: Sep 16, 2016
  5. Sep 16, 2016 #4

    gneill

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    Staff: Mentor

    The friction acts along the slope. So the distance involved in WORK = Fd is the length of the elevator. That's the hypotenuse s in the image that I posted back in post #2. So the power associated with the friction will be much higher than the 3500 W that you've found.

    What's the length of the elevator?
     
  6. Sep 16, 2016 #5
    Length is: we have h = 15 (opposite side), 30°. So sin 30° = 15 / X → X = 15 / sin 30° = 30 m. Therefore (14 000 * 30) / 60 = 7000 W

    14 + 7 = 21 / 0.7 = 30 kW
     
  7. Sep 16, 2016 #6

    gneill

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    Staff: Mentor

    Right. That looks much better. Although you should't use equal signs between items where you don't perform the same operations on both sides of the "=". It can lead to misinterpretation by others reading it. Instead write something like:

    P = (14 + 7) / 0.7 kW = 30 kW
     
  8. Sep 16, 2016 #7
    Thank you! And what could be the logic behind "(c) When the passengers walk up the moving escalator, is more or less power required by the motor to maintain the escalator at the same speed? Explain your answer." My guess would be that more power is required, since I imagined carrying on my back a box of books or a box of moving puppies of the same size. I guess the movement would make it more difficult to carry the box. Maybe because the passengers / puppies would create more opposing energy (directed downwards)?
     
  9. Sep 16, 2016 #8

    gneill

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    Staff: Mentor

    Well, it's true that the elevator needs to support the people as they lift themselves on what is essentially a stairway from their point of view. The question is, is the force that people exert on the treads greater while walking than standing? You should think about this and present an argument.

    You might also investigate in terms of the rate at which people are "processed" if the stream of people walk up the escalator. Individual "on board" times will be shortened (i.e. less than 1 minute by an amount that depends on the walking speed), but presumably the escalator remains filled to capacity (75 persons) at all times.

    I did a quick web search and came across a report where the energy consumption for an escalator was measured for both standing and walking loads. You might want to take a look at the relevant sections here. It's a pdf file.
     
  10. Sep 16, 2016 #9

    jbriggs444

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    Science Advisor

    There are various presumptions that could be made. Those presumptions and, perhaps, some justifiction for them ought to figure into the explanation for part c.

    Does the escalator handle a fixed number of people per minute (bottleneck elsewhere in the system) or is it filled to capacity at all times?
    When people are walking up as opposed to standing, is capacity effected?
    What units are you using to measure "capacity"?
     
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