Question by alfredmorkos: What power would be required if the escalator has to carry twice the number of people?
An office complex needs an escalator that can carry people, at a constant speed of 0.6 m/s at 35o slope. Calculate the maximum number of persons, each weighing 80 kg, if the maximum power input to the elevator is 10.8 kW. What power would be required if the escalator has to carry twice the number of people?
Best answer:
Answer by Munkstump
Escalators are designed to carry a load slightly higher than the maximum amount of people that can fit on its steps. You would not be able to get “twice as many” people on there, if the steps are already at full capacity. So it is an invalid question.
Anyway, if someone was foolish enough to design and install an elevator that could only move half the weight capacity required, then obviously you would have to double the power (size) of the motor. If that somehow increases the speed (it doesn’t always. Bigger motors most often spin more slowly than smaller ones), then you would also need to re-gear it to the appropriate setting.
Keep in mind, you would be able to double your load capacity with the same motor, if you could settle on traveling 1/2 the speed.
What do you think? Answer below!
If this is a Physics question, then you can assume everything is lossless, and no safety factors are required.
The sine of 35 degrees is about 0.57, so moving those people on the slope at 0.6 m/S is like moving them straight up at 0.57 * 0.6 m/S.
From there, you can calculate the power needed to move one person of 80 kg at this rate. I remember a horsepower was 550 ft-lb/sec, but I’m sure there’s a set value you can look up for kg-m/sec to kilowatts.
And to carry twice as many people, twice the power input.
If this is an Engineering problem, you will need to make additional real-world considerations about efficiency and design margin.