Work of a Ski Lift

I wanted to post this example because I used to forget that when the ski lift accelerates to its operating speed it also rises the chairs. Hence, the total power required at start-up equals the sum of the power needed for acceleration as well as the power needed for the upward lift. It was such a silly mistake to make and, though harmless in problem-solving at school, can cause a lot of damage in real life situations. It is extremely important to consider the problem thoroughly before attempting to solve it.

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Problem 2-32 (Thermodynamics - An Engineering Approach 7th Edition)

A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart. The lift is operating at a steady speed of 10 km/h. Neglecting friction and air drag and assuming that the average mass of each loaded chair is 250 kg. Estimate the power required to accelerate this ski lift in 5 s to its operating speed when it is first turned on.

Power is needed for (1) acceleration and (2) lifting the loaded chairs. This two parts can be calculated separately and then added together.

(1) Power for acceleration:

The final speed of the lift is

$V=\left( 10\ km/h \right)\left( \frac{1\ h \times 1000\ m}{60\ sec \times 1\ km}\right)=2.887\ m/s$.

Then the power needed is

$P_a=\frac{1}{2}m\left(V^2-V_0^2\right)/\Delta t=\frac{1}{2}\left(50\times 250\ kg\right)\left(2.778\ m/s\right)^2=9.6\ kW$.

(2) Power for lift

Assume that the acceleration is constant (i.e. power supply is constant), its value will be

$a=\frac{\Delta V}{\Delta t}=\frac{2.778\ m/s}{5\ s}=0.556\ m/s^2$.

Then the vertical lift during acceleration will be

$\left(\frac{1}{2}at^2\right)\times\left(\frac{200}{1000}\right)=1.36\ m$.

Hence, the power needed to increase the potential energy of the lift is

$P_g=\frac{mg\Delta h}{\Delta t}=\left(50 \times 250\ kg \right)\left(9.89\ m/s^2\right) \left(1.36\ m \right)/ \left(5\ s \right)=3.41\ kW$.

Then the total Power required is

$P_{total}=P_a+P_g=9.6+34.1=43.7\ kW$.