Calculate Efficiency of 10.0 kW Motor for 8.5 kg Elevator

  • Thread starter petuniac
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In summary, an 8.5 kg elevator is being pulled at a constant velocity of 1.0 m/s by a 10.0 kW motor. To calculate the efficiency of the motor, the equation Pout/Pin was used. Pout was found to be 833.85 W while Pin was given as 10000 W. This resulted in an efficiency of 8.33%. However, the correct answer is 83.3%, indicating an error in the original problem.
  • #1
petuniac
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Homework Statement



An 8.5 kg elevator is pulled at a constant velocity of 1.0 m/s by a 10.0 kW motor. Calculate the efficiency of the motor.


Homework Equations



W = change in E = Fd

e = Work out/Work in * 100 = energy out/energy in *100

Epotential = mgh
Ekinetic = 1/2mv^2

P = W/t

The Attempt at a Solution



Ekinetic = 1/2mv^2 = 8.5 J (Energy in)

I don't know how to get the Work from the motor as I don't have the time?
 
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  • #2
petuniac said:
I don't know how to get the Work from the motor as I don't have the time?

Power is also written as: Power = Force x Velocity. You have two known quantities viz. Power & velocity. Can you find the force?
 
  • #3
if you want to approach this using time and simply potential energy change, you can set an initial condition and a final condition:

ex)
initial condition:
height=0m
t=0

final condition:
height=10m
t=10s

Then play with (delta)E in a much simpler way.
 
  • #4
Sorry Reshma but I am still confused ... if I solve for F using the P = Fv equation what does this accomplish as I need to calculate the efficiency, and why would I use the 10 kW power given for the motor in that equation and the speed of the elevator?? Please help.
 
  • #5
petuniac, you should not be using 10kW and the elevator's speed in the same equation.

It sounds like 10kW is your Power in, the power supplied to the motor.

efficiency is also Pout/Pin.

So Pout is the amount of power used while lifting the elevator. For this calculation you can in fact use P=Fv. Your unknown here is NOT F, but rather P. If you're stuck on figuring out F, remember gravity's effect on the elevator's mass.
 
  • #6
Thank you Matheyrichs! You've made it much clearer to me!

I made an error in quoting the original question.. the elevator mass is 85 kg.

Ok, so for the solution I get...

Pout = Fv = mgh = (85*9.81*1) = 833.85 W

Pin = 10000 W

e = Pout/Pin = (833.85)/10000 *100 = 8.33%

The only issue is that the answer is 83.3% so I am off by an order of magnitude and can't seem to see why?
 
  • #7
are there any people in the elevator?? :tongue:

I'm not sure. 833/10,000 is certainly only 8.33%. I also did the problem a different way, substituting the initial start and end and got the same answer.

Check the problem again and make sure V_elevator and P_motor are actually 1 m/s and 10kW as well.

83% sounds like a reasonable efficiency, 10kW is ~13horsepower, and 85kg is ~187lbs.

I'd say that the most unreasonable piece of that is for an elevator to only weigh 85kg.
 
  • #8
Thanks again!
 

1. How is efficiency calculated for a motor?

Efficiency is calculated by dividing the output power of the motor by the input power, and then multiplying by 100 to get a percentage. In this case, the output power is the power needed to lift the elevator (10.0 kW) and the input power is the power consumed by the motor (10.0 kW).

2. What is the output power of a 10.0 kW motor?

The output power of a 10.0 kW motor is 10.0 kW, as stated in the question.

3. How much power is needed to lift an 8.5 kg elevator?

The power needed to lift an 8.5 kg elevator would depend on the acceleration and height of the elevator, as well as any friction or resistance in the system. However, for this calculation, we are assuming a constant force and no resistance, so the power needed would be equal to the force (in Newtons) multiplied by the velocity (in meters per second).

4. Can a 10.0 kW motor lift an 8.5 kg elevator?

Yes, a 10.0 kW motor should be able to lift an 8.5 kg elevator, depending on the height and acceleration required. However, this calculation does not take into account any external factors such as friction or resistance in the system.

5. How efficient is a 10.0 kW motor for an 8.5 kg elevator?

The efficiency of a 10.0 kW motor for an 8.5 kg elevator can be calculated by dividing the output power (10.0 kW) by the input power (10.0 kW) and multiplying by 100. This results in an efficiency of 100%, meaning that all of the input power is being used to lift the elevator.

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