Chapter 6, Problem 68P

To determine

The maximum total power needed.

Answer to Problem 68P

Solution:

Maximum total power needed to lift, to move $47000$ people per hour up to average height $200\text{m}$ and average mass of man $=70\text{kg}$

  $\text{=1}{\text{.79×10}}^{\text{6}}\text{ Watt or 2401}\text{.10 hp}$

Explanation of Solution

Given:

Lifts can move $=47000\left(\text{people per hour}\right)$

Average height lift by lift $=200\text{m}$

Concept used:

Power $=\frac{\text{work }}{\text{time}}$

Work done by force

  ${\overrightarrow{F}}_p$ : pulling force

  $\overrightarrow{S}$: Displacement vector

Newton’s second law:

  $\sum {\text{F}}_{\text{x}}=0$

  $\sum {\text{F}}_{\text{y}}=0$

Because lift moves with slowly with constant speed.

Calculation:

  

Work done by pulling force

  $\left({\text{F}}_{\text{p}}\right)\text{ to move}$

  $\text{200}\text{m}$ up work

  $\begin{array}{l}W={\overrightarrow{F}}_p\cdot \overrightarrow{S}\\ =F_P S\cos 0 \\ =F_{p}S\\ \left(F_p=mg\right)\\ W=mgs\end{array}$

Let average mass of a man $\text{=70 kg}$

  $\begin{array}{c}mgs=70\times 9.8\times 200\\ =137200\text{ J}\end{array}$

Work done by lift to a man $=137200\text{ J}$

Work done by lift to $47000\text{ people}=137200\times 47000\text{ J}$

Time taken by lift to do above work $=1\text{ h}$

  $=60\times 60\text{ s}$

Power of lift $=\frac{\text{work done by lift}}{\text{time}}$

  $=\frac{137200\times 47000}{60\times 60}\text{watt}$

  $=1791222.22\text{ watt}$

  $\left(746\text{ watt}=1\text{ hp}\right)$

  $=\frac{1791222.22\text{ watt}}{746\text{ watt}}\times 1\text{ hp}$

  $\text{= 2401}\text{.10 hp}$

Maximum total power of lift to move $47000$ per hour at average height $200\text{ m}$

  $\text{=}\text{2401}\text{.10 hp}$

Conclusion: Maximum total power of lift to move $47000$ per hour at average height $200\text{m},$

(Average mass of man $=70\text{ kg}$ ) is equal to $2401.10\text{ hp}$ .