Question

A chair-lift is designed to transport 1000 skiers per hour from the base A to the summit B. The average mass of a skier is 70 kg and the average speed of the lift is 75 m/min. Determine (a) the average power required, (b) the required capacity of the motor if the mechanical efficiency is 85 percent and if a 300 percent overload is to be allowed. 300 m 750 m

Answer 1

a. The average power required for the chair-lift is approximately 3428.5 watts or 3.43 kilowatts.

b. The required capacity of the motor, considering the
`$300 \%$` overload, is approximately 21.05 kilowatts.

To determine the average power required for the chair-lift, we can use the formula for power:

`P=(W)/(t)`

Where:

-
`$P$` is the power (in watts, W)

-
`$W$` is the work done (in joules, J)

-
`$t$` is the time (in seconds, s)

First, let's calculate the work done to transport 1000 skiers from the base A to the summit B, which is a vertical distance of 750 meters. We'll also account for the
`$300 \%$` overload, which means we need to calculate the work for transporting
`$300 \%$` more skiers in the same time.

(a) Work Done:

The work done to lift a mass against gravity is given by the formula:

`$$W=m g h$$`

Where:

-
`$\mathrm{m}$` is the mass (in kg)

-
`$g$` is the acceleration due to gravity (approximately
`$9.81 \mathrm{~m} / \mathrm{s}^2$` )

-
`$\mathrm{h}$` is the height (in meters)

Let's calculate the work done for one skier:

`$$W_{\text {skier }}=(70 \mathrm{~kg}) \cdot(9.81)$$`

Now, we need to account for the $300 \%$ overload, so the total work done for 1000 skiers is:

`$$W_{\text {total }}=1000 \text { skiers } \cdot(1+3) \cdot W_{\text {skier }}=4000 \cdot 514275 \mathrm{~J}=2,057,100 \mathrm{~J}$$`

Now, we'll calculate the time it takes to transport the skiers at a speed of
`$75 \mathrm{~m} / \mathrm{min}$`:

`t=(d)/(v)`

Where:

-
`$d$` is the distance
`$(750 \mathrm{~m})$`

-
`$v$` is the speed
`$(75 \mathrm{~m} / \mathrm{min})$`

Convert the speed to meters per second:

`v=\frac{75 \mathrm{~m} / \mathrm{min}}{60}=1.25 \mathrm{~m} / \mathrm{s}`

Now, calculate the time:

`t=\frac{750 \mathrm{~m}}{1.25 \mathrm{~m} / \mathrm{s}}=600 \mathrm{~s}`

Now, we can calculate the average power required:

`P=\frac{W_{\text {total }}}{t}=\frac{2,057,100 \mathrm{~J}}{600 \mathrm{~s}} \approx 3428.5 \text { watts }`

So, the average power required for the chair-lift is approximately 3428.5 watts or 3.43 kilowatts.

(b) Required Capacity of the Motor:

The mechanical efficiency is given as
`$85 \%$`. Mechanical efficiency is defined as the ratio of useful work output to the total work input. So, we can write:

Efficiency = Useful Work Output/Total Work Input

We can rearrange this equation to find the total work input:

Total Work Input = Useful Work Output / Efficiency

In this case, the useful work output is the work done to transport the skiers, which is
`$2,057,100 \mathrm{J}$`, and the efficiency is
`$85 \%(0.85)$`.

Total Work Input
`$=\frac{2,057,100 \mathrm{~J}}{0.85} \approx 2,418,941.18 \mathrm{~J}$`

Now, we need to account for the
`$300 \%$` overload, so the total work input for the motor is: Total Work Input
`$=2,418,941.18 \mathrm{~J} *(1+3)=9,675,764.72 \mathrm{~J}$`

The motor has to do this amount of work in 600 seconds (the time it takes to transport the skiers):

Power Required = Total Work Input,
`{t}=\frac{9,675,764.72 \mathrm{~J}}{600 \mathrm{~s}} \approx 16,126.27$` watts

To account for the overload, the motor should have a capacity that can handle this power, so we'll consider the power with the overload:

Required Motor Capacity:

`$=\frac{16,126.27 \text { watts }}{1000} * 1.3 \approx 21.05 \mathrm{~kW}$`

So, the required capacity of the motor, considering the
`$300 \%$` overload, is approximately 21.05 kilowatts.