An escalator in a shopping center is designed to move 30 people, 75 kg each, at a constantspeed of 0.8 m/s at 45° AP = 5 kPa Pump slope. Determine the minimum power inputneeded to drive this escalator. What would your answer be if the escalator velocity were tobe doubled? Clue: You can start with the first law and cancel out terms. You can consider this a closedSystem.

Assumptions: Air drag and friction are negligible. The average mass of each person is 75 kg. The…

Answered: An escalator in a shopping center is…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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An escalator in a shopping center is designed to move 30 people, 75 kg each, at a constant
speed of 0.8 m/s at 45° AP = 5 kPa Pump slope. Determine the minimum power input
needed to drive this escalator. What would your answer be if the escalator velocity were to
be doubled? Clue: You can start with the first law and cancel out terms. You can consider this a closed
System.

Expert Solution

Step 1

Assumptions:

Air drag and friction are negligible.
The average mass of each person is 75 kg.
The escalator operates steadily, with no acceleration or breaking.
The mass of escalator itself is negligible.

Analysis:

At design conditions, the total mass moved by the escalator at any given time is

Mass = (30 persons)(75 kg/person) = 2250 kg

The vertical component of escalator velocity is

$V_{vert} = V \sin 45^{o} = (0.8 m / s) \sin 45^{o}$

Under stated assumptions, the power supplied is used to increase the potential energy of people. Taking the people on elevator as the closed system, the energy balance in the rate form can be written as

$\underset{\begin{matrix} \begin{matrix} \begin{matrix} Rate of net energy transfer \end{matrix} \\ by heat, work and mass \end{matrix} \end{matrix}}{\underset{⏟}{{\dot{E}}_{in} - {\dot{E}}_{out}}} = \underset{\begin{matrix} Rate of change in internal, kinetic, \\ potential etc . energies \end{matrix}}{\underset{⏟}{\frac{{dE}_{system}}{dt}}} = 0$

${\dot{E}}_{in} = \frac{{dE}_{system}}{dt} = \frac{∆ E_{system}}{∆ t}$

${\dot{W}}_{in} = \frac{∆ PE}{∆ t} = \frac{mg ∆ z}{∆ t} = {mgV}_{vert}$

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