Chapter 2.8, Problem 34P

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, and each chair can seat three people. 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, determine the power required to operate this ski lift. Also estimate the power required to accelerate this ski lift in 5 s to its operating speed when it is first turned on.

To determine

The power required to operate the ski lift.

The power required to accelerate the ski lift in 5 s to its operating speed.

Answer to Problem 34P

The power required to operate the ski lift is $\underset{\_}{\text{68}\text{.1}\text{kW}}$.

The power required to accelerate the ski lift in 5 s to its operating speed is $\underset{\_}{\text{43}\text{.7}\text{kW}}$.

Explanation of Solution

Calculate the load of the lift at any given time.

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

Here, number of chairs at any given time is N and weight of each chair is w.

Calculate the work required to raise the mass by 200 m.

$W_g=mg\left(z_2-z_1\right)$ (II)

Here, acceleration due to gravity is g and difference between the elevation of ski lift is $z_2-z_1$.

Write the equation of change in time.

$\Delta t=\frac{D}{V}$ (III)

Here, distance is D and velocity is V.

Calculate the power needed.

${\dot{W}}_g=\frac{W_g}{t_2-t_1}$ (IV)

Calculate the acceleration of the ski lift.

$\begin{array}{c}a=\frac{\Delta V}{\Delta t}\\ =\frac{V_2-V_1}{t_2-t_1}\end{array}$ (V)

Here, change in velocity is $\Delta V$, initial and final time is $t_1$ and $t_2$ respectively.

During the acceleration, calculate the power needed.

${\dot{W}}_a=\frac{1}{2}\frac{m\left(V_2^2-V_1^2\right)}{t_2-t_1}$ (VI)

Here, the initial and final velocity of a ski lift are $V_1$ and $V_2$ respectively.

Calculate the vertical distance travelled during the acceleration.

$\begin{array}{l}{z}_2-z_1=h\\ h=\frac{1}{2}a{t}^2\sin \alpha \end{array}$ (VII)

Here, time to accelerate the ski lift to its operating speed is t and angle to raise the ski lift is $\alpha$.

Calculate the power needed to acceleate the ski lift in 5 s.

${\dot{W}}_g=\frac{mg\left(h\right)}{t_2-t_1}$ (VIII)

Calculate the total power required to accelerate the ski lift.

${\dot{W}}_{\text{total}}={\dot{W}}_a+{\dot{W}}_g$ (IX)

Conclusion:

Since the lift is 1 km long and chairs are spaced 20 m apart, calculate the number of chairs at any given time.

$\begin{array}{c}N=\frac{1\text{km}}{20\text{m}}\\ =\frac{1\text{km}\times \text{1000}\left(\frac{\text{m}}{1\text{km}}\right)}{20\text{m}}\\ =50\text{chairs}\end{array}$

Substitute 50 chairs for N and 250 kg/chair for w in Equation (I).

$\begin{array}{c}m=\left(\text{50}\text{chairs}\right)\left(250\text{kg/chair}\right)\\ =12,500\text{kg}\end{array}$

Substitute 12,500 kg for m, $9.81{\text{m/s}}^{\text{2}}$ for g, and 200 m for $z_2-z_1$ in Equation (II).

$\begin{array}{c}{W}_g=\left(12,500\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(200\text{m}\right)\\ =\left(12,500\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(200\text{m}\right)\left(\frac{1\text{kJ}}{1000\text{kg}\cdot {\text{m}}^2{\text{/s}}^{\text{2}}}\right)\\ =24,525\text{kJ}\end{array}$

Substitute 1 km for D and 10 km/h for V in Equation (III).

$\begin{array}{c}\Delta t=\frac{1\text{km}}{10\text{km/h}}\\ =0.1\text{h}\times 3600\frac{\text{s}}{1\text{h}}\\ =360\text{s}\end{array}$

Substitute 360 s for $\Delta t$ and 24,525 kJ for $W_g$ in Equation (IV).

$\begin{array}{c}{\dot{W}}_g=\frac{24,525\text{kJ}}{360\text{s}}\\ =68.1\text{kW}\end{array}$

Thus, the power required to operate the ski lift is $\underset{\_}{\text{68}\text{.1}\text{kW}}$.

Convert the unit of velocity from km/h to m/s.

$\begin{array}{c}V=10\frac{\text{km}}{\text{h}}\\ =10\frac{\text{km}}{\text{h}}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{h}}{3600\text{s}}\\ =2.778\text{m/s}\end{array}$

Substitute 0 for $V_1$, 2.778 m/s for $V_2$, 0 for $t_1$, and 5 s for $t_2$ in Equation (V).

$\begin{array}{c}a=\frac{2.778\text{m/s}-0}{5\text{s}-0}\\ =0.556{\text{m/s}}^{\text{2}}\end{array}$

Substitute 0 for $t_1$, 5 s for $t_2$, 0 for $V_1$, 12,500 kg for m, and 2.778 m/s for $V_2$ in Equation (VI).

$\begin{array}{c}{\dot{W}}_a=\frac{1}{2}\frac{\left(12,500\text{kg}\right)\left({\left(2.\text{778}\text{m/s}\right)}^2-\left(0\right)\right)}{5\text{s}-0}\\ =\frac{1}{2}\frac{\left(12,500\text{kg}\right){\left(2.\text{778}\text{m/s}\right)}^2}{5\text{s}}\left(\frac{1\text{kJ/kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =9.6\text{kW}\end{array}$

Calculate the value of $\sin \alpha$.

$\begin{array}{c}\sin \alpha =\frac{200\text{m}}{1\text{km}}\\ =\frac{200\text{m}}{1\text{km}\times \text{1000}\frac{\text{m}}{1\text{km}}}\\ =0.2\end{array}$

Substitute $0.556{\text{m/s}}^{\text{2}}$ for a, 5 s for t, and 0.2 for $\sin \alpha$ in Equation (VII).

$\begin{array}{c}h=\frac{1}{2}\left(0.556{\text{m/s}}^{\text{2}}\right){\left(5\text{s}\right)}^{2}0.2\\ =1.39\text{m}\end{array}$

Substitute 1.39 m for h, 5 s for $t_2$, 0 for $t_1$, $9.81{\text{m/s}}^{\text{2}}$ for g, and 12,500 kg for m in Equation (VIII).

$\begin{array}{c}{\dot{W}}_g=\frac{\left(\text{12,500}\text{kg}\right)9.81{\text{m/s}}^{\text{2}}\left(1.39\text{m}\right)}{5\text{s}-0}\\ =\frac{\left(\text{12,500}\text{kg}\right)9.81{\text{m/s}}^{\text{2}}\left(1.39\text{m}\right)}{5\text{s}}\left(\frac{1\text{kJ}}{1000\text{kg}\cdot {\text{m}}^2{\text{/s}}^{\text{2}}}\right)\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =34.08\text{kW}\\ \simeq \text{34}\text{.1}\text{kW}\end{array}$

Substitute 34.1 kW for ${\dot{W}}_g$ and 9.6 kW for ${\dot{W}}_a$ in Equation (IX).

$\begin{array}{c}{\dot{W}}_{\text{total}}=9.6\text{kW}+34.1\text{kW}\\ =43.7\text{kW}\end{array}$

Thus, the power required to accelerate the ski lift in 5 s to its operating speed is $\underset{\_}{\text{43}\text{.7}\text{kW}}$.