Problem 17 - Applied Physics

Question

An escalator is needed to carry 75 passengers per minute a vertical distance of \(8.0 \mathrm{~m}\). Assume that the mass of each passenger is \(7 \overline{0} \mathrm{~kg}\). (a) What is the power (in \(\mathrm{kW}\) ) of the motor needed? (b) Express this power in horsepower. (c) What is the power (in \(\mathrm{kW}\) ) of the motor needed if \(35 \%\) of the power is lost to friction and other losses?

Step-by-Step Solution

Answer

(a) 6.86 kW; (b) 9.195 HP; (c) 10.554 kW.

Step 1: Calculate Total Weight

First, determine the total weight of the 75 passengers. The weight of one passenger is the mass multiplied by the gravitational force, \[ W = m \cdot g = 70 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 686 \text{ N}. \] Thus, for 75 passengers, the total weight is \[ 75 \times 686 = 51,450 \text{ N}. \]

Step 2: Calculate the Work Done

The work done in carrying the passengers is the product of the force (weight) and the distance: \[ \text{Work} = \text{Force} \times \text{Distance} = 51,450 \text{ N} \times 8.0 \text{ m} = 411,600 \text{ J}. \]

Step 3: Calculate Power in Watts

Power is the work done per unit time. Here, we want to calculate the power needed per minute: \[ P = \frac{\text{Work}}{\text{Time}} = \frac{411,600 \text{ J}}{60 \text{ s}} = 6,860 \text{ W}. \] This is the power needed in Watts.

Step 4: Convert Watts to Kilowatts

To find power in kilowatts, convert from Watts: \[ 6,860 \text{ W} = 6.86 \text{ kW}. \]

Step 5: Convert Kilowatts to Horsepower

To convert kilowatts to horsepower, use the conversion factor: \[ 1 \text{ kW} = 1.341\text{ HP}. \] Thus, \[ 6.86 \text{ kW} = 6.86 \times 1.341 = 9.195 \text{ HP}. \]

Step 6: Account for Efficiency Loss

Since 35% of power is lost to friction and other losses, only 65% of the power is effective. Therefore, the required input power for the motor is: \[ \frac{6.86}{0.65} = 10.554 \text{ kW}. \]

Key Concepts

Escalator Physics

Understanding the physics of an escalator involves acknowledging it as a machine that facilitates the transport of people vertically through a building. Escalators work continuously and must operate smoothly as they carry numerous passengers.
One important aspect is the mechanical load they carry, which includes the weight of passengers. In the given exercise, the escalator moves 75 people every minute over a vertical distance of 8 meters. Observing such parameters gives us the basic idea of the forces and motion at play.
Escalators are powered by motors, and their performance relies on converting electrical energy into mechanical energy necessary to perform work against the gravitational force. Calculations start with determining the total weight the escalator must lift to understand the demand on the motor thoroughly.

Mechanical Work

Mechanical work in the context of escalators relates to moving a mass over a distance under the action of a force, particularly against gravity. Here, work done is calculated as the product of force and distance.
The force in question is the total gravitational force on the passengers, which is their combined weight ( 51,450 N). The distance they move is 8 meters. Thus, the work done can be represented as: \[\text{Work} = 51,450 \text{ N} \times 8.0 \text{ m} = 411,600 \text{ J.}\]
This calculation is a direct application of the fundamental formula for work, \( ext{Work} = ext{Force} \times ext{Distance} \).

• Understanding this helps in estimating the energy an escalator consumes.
• It also provides insight into the potential demands for more efficient energy transfers.

Energy Conversion

Energy conversion is a pivotal aspect of escalator functions—changing electrical energy to mechanical energy. During this transition, some power is lost, often due to heat, sound, or other inefficiencies.
Power metrics are essential since escalators need a specific amount of energy over time to function correctly. Initially, the machine requires 6.86 kW of power to lift 75 passengers per minute. Conversions of these power measurements are then expressed in kilowatts, and further, into horsepower through the relation: \[ 1 \text{ kW} = 1.341 \text{ HP} \].
This conversion translates practical power outputs into terms more commonly used in industries and allows more intuitive understanding and comparisons of machinery.

Friction Losses

Friction losses in escalators occur as some of the energy geared for lifting passengers is dissipated due to friction and other inefficiencies.
These losses can be substantial, consuming up to 35% of the total energy. Thus, actual power requirements for operation are higher than theoretical calculations. If 35% is lost, only 65% remains effective for lifting:

• Escalators, similar to many machines, are subject to losses in energy efficiency.
• This requires additional power from the motor to maintain the same level of performance.

Calculating the adjusted power demand involves dividing the needed power by the efficiency fraction (here, 65%), showing escalators’ dependence on effective energy management.