Calculating average horsepower of an elevator help

In summary, the conversation is about a problem in an introductory mechanical engineering course regarding the power requirement for an elevator that travels from the first floor to the fifth floor in 7 seconds. The problem involves calculating the change in kinetic energy and potential energy, as well as using equations for work, force, and power. The correct solution is to calculate the work using the weight and distance, and then divide by the time to get the power requirement, which is approximately 31.187 horsepower.
  • #1
3
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Hello. I hope someone here can give me some advice for solving this problem for an introductory mechanical engineering course. Below is the problem statement verbatim, and them my attempt at solving it. Advice and guidance is greatly appreciated!

Homework Statement



Here is the problem statement:

"An elevator has a capacity of 2000 lbf including its own weight. It
can travel from the first floor to the fifth floor in 7 seconds. The
vertical distance between each floor is 15 ft. Estimate the average
power requirement in horsepower for this elevator. The elevator
starts with an initial velocity of 0, and reaches the fifth floor (just
before stopping) with a final velocity that is equal to the distance
traveled divided by the time required to get to the fifth floor."

Homework Equations



Work = lbf X distance
Work = change in kinetic energy + change in potential energy
Force = lbm * 32.2
Units = English/American
1 Slug = lbm/32.2 ft/s
change in KE= 1/2mVf^2 - 1/2mVi^2
change in PE = force X change in height

The Attempt at a Solution



Since the elevator is traveling from the 1st floor to the 5th floor, and each floor is 15 ft, I think the distance traveled is 60 feet.

D = 15ft

Since the problem is asking for average horsepower, I'm assuming that I should calculate WORK, which is the change in potential energy + change in kinetic energy, and divide by the time of 7 seconds.

Work = (deltaKE + deltaPE)/t
t = 7 sends

SO, here's what I have:

To calculate the change in kinetic energy, I did this:

First, find the mass in SLUGS since I'm using American units:

F=MA
2000lbf = lbm*32.2
lbm=2000lbf/32.2
lbm= ~62.112

1slug=1lbm/32.2

so:

62.112/32.2=1.929 slugs

Then i need to calculate the initial and final velocities:

Vi=0 ft/sec
Vf=60ft/7sec= ~8.571 ft/sec

now I can plug this info. into the KE equation:

[1/2m(Vf)^2]-[1/2m(Vi)^2] = deltaKE

=

1/2(1.929slugs)(8.571)^2 - 1/2(1.929slugs)(0)^2 = 70.854lbf*ft

NOW, since I'm looking for total work, i need to add that to deltaPE, which is:

2000lbf*60ft = 120,000 lbf*ft

so: Work = deltaKE + deltaPE = 70.854lbf*ft + 120,000 lbf*ft = 120,070.854 lbf*ft

NOW, to calculate POWER, I use:

Power = Work/Time

so

120,070.854 lbf*ft / 7 seconds = 17,152.979 ft*lbf

and to calculate Horsepower I use:

550lbf*ft = 1 HP

so

17,152.979lbf*ft / 550 lbf*ft = 31.187 HP

is that correct?!
 
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  • #2
Welcome to PF!
I agree with your answer. I didn't bother with the kinetic energy, on the theory that it probably coasted the last few feet and converted its KE to PE. I just went with the weight x distance divided by time. My first time using feet and pounds in this way since 1967, I think!
 
  • #3
Thank you so much! I was wondering why the KE was SO LOW, and that's exactly what made me think I was approaching this problem the wrong way. I just assumed I was supposed to include KE because the problem included initial and final velocities. Thank you again!
 
  • #4
Most welcome!
 
  • #5




Hello! Your calculations and approach seem to be correct. However, a few things to note:

1. When calculating the initial and final velocities, make sure to use consistent units. In this case, since the distance is given in feet, the velocity should also be in feet per second (not feet per 7 seconds). So the final velocity would be 60 ft/7 sec = 8.571 ft/sec.

2. It's also important to note that the elevator is traveling from the first floor to the fifth floor, which means it is actually traveling a total distance of 60 ft (not 15 ft). So the change in potential energy would be 2000 lbf * 60 ft = 120,000 lbf*ft.

Other than those small corrections, your calculations and approach seem to be correct. Keep up the good work!
 

1. What is the formula for calculating average horsepower of an elevator?

The formula for calculating average horsepower of an elevator is as follows:
Average horsepower = (Total weight of elevator x distance traveled x acceleration due to gravity) / (Time taken to travel the distance)

2. How do I determine the weight of the elevator?

The weight of the elevator can be determined by adding the weight of all the components, including the cab, counterweight, and passengers, and the weight of the elevator ropes, pulleys, and other machinery.

3. What is the significance of distance and time in the calculation?

The distance and time variables are important in the calculation because they represent the work done by the elevator. The distance traveled is the vertical distance between the starting and ending points, and the time taken to travel the distance represents the rate at which work is being done.

4. How does the acceleration due to gravity affect the calculation?

The acceleration due to gravity plays a crucial role in the calculation as it determines the amount of force needed to lift the elevator and its contents against gravity. The higher the acceleration due to gravity, the more power is required to lift the elevator, resulting in a higher average horsepower.

5. Are there any other factors that can affect the average horsepower of an elevator?

Yes, there are other factors that can affect the average horsepower of an elevator, such as the efficiency of the elevator system, the speed of the elevator, and the number of trips made by the elevator in a given time period.

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