In this problem we will make an order-of-magnitude estimate of the cost to opera

Question

In this problem we will make an order-of-magnitude estimate of the cost to operate the elevators in the MEOB building for one month (assume 30 days). For this purpose we will assume:

1) On average, 5000 persons per day visit the MEOB building (including weekend days),

2) A person has a mass of 75 kg,

3) Each person takes one trip up the elevators entering at the lobby, i.e. level one, and one trip back down to the lobby.

4) An equal number of persons visit each of four levels above lobby level, (i.e. 5000/4 visit level two, 5000/4 visit level three, etc.)

5) The vertical distance between levels is 3.6 m,

6) The elevator mechanisms are counter-balanced with a dummy weight equal in weight to an empty elevator car; thus the motors do not lift the weight of the car

7) The electric motor/mechanical efficiency of the elevator system if 50%. Also, the motors do not act as generators when torque is applied during a descending load.

8) The cost of electricity is $0.12 per kW-hr.

With these assumptions, what is the cost to operate the ETC elevators for one month?

Explanation / Answer

work done in lifting 5000/4 persons to 1st floor = (5000/4)*75*9.81*3.6 = 3310875 J

work done in lifting 5000/4 persons to 2nd floor = (5000/4)*75*9.81*3.6*2 = 6621750 J

work done in lifting 5000/4 persons to 3rd floor = (5000/4)*75*9.81*3.6*3 = 9932625 J

work done in lifting 5000/4 persons to 4th floor = (5000/4)*75*9.81*3.6*4 = 13243500 J

so.. total work done = 3310875 + 6621750 + 9932625 + 13243500 = 33108750 Joules

so... electric energy consumed by motors in one day= 33108750 / 50% = 66217500 J

so... electric enecry consumed in 30 days = 66217500 * 30 = 1986525000 J = (1986525000 / (1000*3600) ) Kw-hr = 551.8125 Kw-hr

so... cost = 551.8125 * 0.12 = $ 66.2175

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