Question

When referring to two separate chair lifts, and their different power output against the force of gravity, how would one calculate the power of each chair lift when given the same distance, but two different masses and times?

Answer 1

Power=Work/time=Force*distance/time
if they move uniformly, Force=ma=mg
so P=mg/t

Answer 2

That's what I used, thanks for the reassurance! Why, then, would they require more power than the amount used to overcome the force of gravity?

Answer 3

And would the faster chair lift be less efficient because of greater energy lost in heat?

Answer 4

If you wanted to actually move the chairlift upwards from a stand still, you would need more power.

\[P = {m(g+a) d \over t}\]

This amount of power would accelerate the chairlift upwards. Once the desired velocity was reached, the power could be reduced to the "gravity only" case and the chairlift would continue to move upwards at a constant velocity (wind resistance ignored of course).

Answer 5

Your coefficient of friction (according to Coulomb's model) doesn't change with speed.

Your wind resistance would indeed change with speed. This is why jet liners fly at about 85% throttle, because they are more efficient at the slower speed.

Answer 6

So it's less efficient due to air resistance, not friction?

Answer 7

Friction would be your primary heat generation source. Wind resistance in incompressible flow (Mach numbers much less than one) tend not to generate too much heat unless the fluid is highly viscous. Once we approach sonic speeds, we will see the development of a stagnation point, where we have huge releases of enthalpy.

Answer 8

Didn't follow that.. haha

Answer 9

Let me rephrase that.

Answer 10

If we define efficiency as\[\eta = {W_o \over W_i}\]we can say that\[W_o = W_{motor} - W_{friction} - W_{air~resistance}\]
\[W_{friction} = \mu N d\]\[W_{air~resistance} = C v^2d\]where C is some constant defining the nature of the fluid in which we move through and the profile of the chairlift the fluid is exposed to.

We can see that here velocity only comes into play when dealing with air resistance.

Answer 11

\[W_i = W_{motor}\]

Answer 12

Awesome! So how would you maximize the efficiency of the chairlift? By making it aerodynamic?

Answer 13

Thank so much for all the help

Answer 14

We need to focus on both causes of inefficiency.

Considering that most chairlift have flat tops, I would assumes that engineers aren't too concerned with the aerodynamic drag. This is due to several factors. Increased manufacturing cost and little benefit due to the low speeds of chairlifts.

Answer 15

So how would we maximize it? Haha

Answer 16

Low friction bearings and a high efficiency motor driving the chairlift would be your best staring points.

Answer 17

Thanks so much!!!