Chapter 3, Problem 49P

To determine

The minimum power input needed to drive the escalator.

The minimum power input needed to drive the escalator when the escalator velocity is doubled.

Explanation of Solution

Given:

Mass of each people $\left(m_{\text{people}}\right)$ is $75\text{kg}$.

Number of people can move in the escalator $\left(n\right)$ is $50$.

Velocity of the escalator $\left(V\right)$ is $0.6\text{m}/\text{s}$.

Slope of the escalator from the ground $\left(\theta \right)$ is $45°$.

Acceleration due to gravity $\left(g\right)$ is $9.81\text{m}/{\text{s}}^2$.

Calculation:

Calculate the total mass moved by the escalator at any given time.

  $\begin{array}{c}m=n{m}_{\text{people}}\\ =\left(50\text{persons}\right)\left(75\text{kg}/\text{person}\right)\\ =3,750\text{kg}\end{array}$

Write the expression for the vertical component of escalator velocity as follows:

  $V_{\text{vert}}=V\sin \theta$

Calculate the minimum power input needed to drive the escalator $\left({\dot{W}}_{\text{in}}\right)$ using the energy balance equation.

  $\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=d{E}_{\text{system}}/dt\\ {\dot{E}}_{\text{in}}-0=d{E}_{\text{system}}/dt\end{array}$

  $\begin{array}{c}{\dot{E}}_{\text{in}}=d{E}_{\text{system}}/dt\\ =\frac{\Delta E_{\text{sys}}}{\Delta t}\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{\Delta PE}{\Delta t}\\ =\frac{mg\Delta z}{\Delta t}\\ =mg{V}_{\text{vert}}\\ =mgV\sin \theta \end{array}$

  $\begin{array}{c}=\left(3,750\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(0.6\text{m}/\text{s}\right)\sin 45°\\ =15607.61\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^3\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =15.6\text{kJ}/\text{s}\left(\frac{\text{1}\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =\text{15}\text{.6}\text{kW}\end{array}$

Thus, the minimum power input needed to drive the escalator is $\underset{\_}{15.6\text{kW}}$.

Calculate the minimum power input needed to drive the escalator $\left({\dot{W}}_{\text{in}}\right)$ when the escalator velocity is doubled.

  $\begin{array}{c}{\dot{W}}_{\text{in}}=mg\left(2V\right)\sin \theta \\ =2mgv\sin \theta \end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{in}}=2\left(3,750\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(0.6\text{m}/\text{s}\right)\sin 45°\\ =31,215.23\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^3\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =31.2\text{kJ}/\text{s}\left(\frac{\text{1}\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =\text{31}\text{.2}\text{kW}\end{array}$

Thus, the minimum power input needed to drive the escalator is $\underset{\_}{\text{31}\text{.2}\text{kW}}$.