Chapter 2.8, Problem 49P

An escalator in a shopping center is designed to move 50 people, 75 kg each, at a constant speed of 0.6 m/s at 45° slope. Determine the minimum power input needed to drive this escalator. What would your answer be if the escalator velocity were to be doubled?

To determine

The minimum power input needed to drive the escalator.

If the escalator velocity is doubled, what would be the answer?

Answer to Problem 49P

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

If the escalator velocity is doubled, the minimum power input needed to drive the escalator would be $\underset{\_}{31.2\text{kW}}$.

Explanation of Solution

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

$m=\left(n\right)\left(w\right)$ (I)

Here, the number of persons is n and weight per person is w.

Calculate the vertical component of escalator velocity.

$V_{\text{vert}}=V\sin \theta$

Here, constant speed of escalator is V and the angle of the escalator from ground is $\theta$.

Write the energy balance equation for a control volume when people on the elevator is taken as the system.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=d{E}_{\text{system}}/dt$ (II)

Here, the rate of energy transfer entering and leaving the system is ${\dot{E}}_{\text{in}}$ and ${\dot{E}}_{\text{out}}$, rate of change in internal, kinetic, and potential energy per unit time is $d{E}_{\text{system}}/dt$.

Since no energy is leaving the room, ${\dot{E}}_{\text{out}}=0$.

Substitute 0 for ${\dot{E}}_{\text{out}}$ in Equation (II).

$\begin{array}{l}{\dot{E}}_{\text{in}}-0=d{E}_{\text{system}}/dt\\ {\dot{E}}_{\text{in}}=d{E}_{\text{system}}/dt\\ {\dot{E}}_{\text{in}}=\frac{\Delta E_{\text{sys}}}{\Delta t}\end{array}$ (III)

Rewrite Equation (III) when rate of work is done on the system.

$\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{\Delta PE}{\Delta t}\\ =\frac{mg\Delta z}{\Delta t}\\ =mg{V}_{\text{vert}}\end{array}$ (IV)

Here, the total mass moved by th escalator at any given time is m, the gravitational acceleration is g, and change in the elevation of the center of gravity of a system is $\Delta z$, and change in time is $\Delta t$.

Substitute $V\sin \theta$ for $V_{\text{vert}}$ in Equation (IV).

${\dot{W}}_{\text{in}}=mgV\sin \theta$ (V)

If the escalator velocity were to be doubled, calculate the minimum power input needed to drive the escalator.

${\dot{W}}_{\text{in}}=mg\left(2V\right)\sin \theta$ (VI)

Conclusion:

Substitute 50 persons for n and 75 kg/person for w in Equation (I).

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

Substitute 3,750 kg for m, $9.81{\text{m/s}}^{\text{2}}$ for g, 0.6 m/s for V, and $45°$ for $\theta$ in Equation (V).

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

     $=15.6\text{kW}$

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

Substitute 3,750 kg for m, $9.81{\text{m/s}}^{\text{2}}$ for g, 0.6 m/s for V, and $45°$ for $\theta$ in Equation (VI).

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

                $=31.2\text{kW}$

Thus, if the escalator velocity is doubled, the minimum power input needed to drive the escalator would be $\underset{\_}{31.2\text{kW}}$.