Elevator’s Power Requirement Use an order-of-magnitude calculation to estimate the capacity of an electric motor that will power a freight elevator in a four-story building. Express your estimate in the units of horsepower. The elevator car weighs 500 lb,and it should be able to carry an additional

Question

Elevator’s Power Requirement

Use an order-of-magnitude calculation to estimate the capacity of an electric motor that will power a freight elevator in a four-story building. Express your estimate in the units of horsepower. The elevator car weighs 500 lb, and it should be able to carry an additional 2500 lb of freight.

Approach
To make this calculation of the elevator’s power capacity, we will neglect air resistance, friction in the elevator’s drive mechanism, and all other sources of inefficiency. We also need to make reasonable estimates for information that is not explicitly known at this point in the elevator’s design but that is necessary to determine the motor’s power rating. We estimate that it will take the elevator 20 s to travel from ground level to the building’s top floor and that the total elevation change is 50 ft. The amount of work that the motor must perform is the product of the total weight (3000 lb) and the change in height. We can then apply Equation (7.5)

[latex]P_{avg}=\frac{W}{\Delta t} [/latex] (7.5)

to determine the average power produced by the motor.

Accepted Answer

The work performed in raising the fully loaded elevator’s car is

[latex]W = (3000 lb) (50 ft) \longleftarrow \left[W=F\Delta d ight] [/latex]

[latex]= 1.5 imes 10^5 ft .lb[/latex]

The average power is

[latex]P_{avg}=\frac{1.5 imes 10^5 ft .lb}{20s} \longleftarrow \left[P_{avg}=\frac{W}{\Delta t} ight] [/latex]

[latex]=7500\frac{ft.lb}{s} [/latex]

Finally, we will convert this quantity to the dimensions of horsepower by using the factor listed in Table 7.2:

Table 7.2
Conversion Factors Between Various Units for Power in the USCS and SI

(ft . lb)/s	W	hp
1	1.356	[latex]1.818×10^{-3}[/latex]
0.7376	1	[latex]1.341×10^{-3}[/latex]
550	745.7	1

[latex]P_{avg}=7500\left(\frac{ft.lb}{s} ight) \left(1.818 imes 10^{-3}\frac{hp}{(ft.lb)/s} ight) [/latex]

[latex]= 13.64\left(\frac{\cancel{ft.lb}}{\cancel{s}} ight) \left(\frac{hp}{\cancel{(ft.lb)/s}} ight) [/latex]

[latex]=13.64 hp[/latex]

Discussion
As a designer would later account for such other factors as the motor’s
inefficiency, the possibility of the elevator being overloaded, and a factor of safety, the calculation would become more accurate. From a preliminary standpoint, however, a motor rated for a few tens of horsepower, rather than one with a few horsepower or several hundred horsepower, would seem sufficient for the task at hand.

[latex]P_{avg}=13.64 hp[/latex]

Question 7.4: Elevator’s Power Requirement Use an order-of-magnitude calcu...

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