Question

Water is pumped up from the Colorado River to supply Grand Canyon Village, located on the rim of the canyon. The river is at an elevation of 564 m, and the village is at an elevation of 2 096 m. Imagine that the water is pumped through a single long pipe 15.0 cm in diameter, driven by a single pump at the bottom end. (a) What is the minimum pressure at which the water must be pumped if it is to arrive at the village? (b) If$\mathbf{4500}\mathbf{}\mathbf{}{\mathit{m}}^{\mathbf{3}}$of water is pumped per day, what is the speed of the water in the pipe? Note: Assume the free-fall acceleration and the density of air are constant over this range of elevations. The pressures you calculate are too high for an ordinary pipe. The water is actually lifted in stages by several pumps through shorter pipes.

Short answer

(a) The minimum pressure at which the water must be pumped if it is to arrive at the village is$P_1=1.50\times {10}^{7}pa$

(b) The speed of the water in the pipe is$v=2.95m/s$

Step-by-step solution

Step 1:

The sum of the pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume has the same value at all points along a streamline for an ideal fluid. This result is summarized in Bernoulli’s equation:

$P+\frac{1}{2}\rho v^2+\rho gy=cons\tan t$

Step 2:

Part(b):

Given:

Diameteroflongpipe(d),$d=15cm$

Radiusof long pipe(r),$r=7.5cm=7.5\times {10}^{-2}m$

Applying Bernoulli’s equation, we get

$P_1+\rho g{h}_1=P_2+\rho g{h}_2$

$P_1$Pressure of elevation (1)

$P_2$Pressure of elevation (2)

$g$Acceleration due to gravity

$\rho$Density of water

$h_1$Height at elevation (1)

$h_2$Height at elevation (2)

$h_1=564m$

$h_2=2096m$

$$\begin{gathered} P_1-P_2=\rho g{h}_2-\rho g{h}_1 \\ {P}_1-P_2=\rho g\left(h_2-h_1\right) \end{gathered}$$

At minimum pressure, $P_2=0$

$$\begin{gathered} P_1={10}^3\times 9.8\left(2096-564\right) \\ {P}_1=1.50\times {10}^{7}pa \end{gathered}$$

Step 3:

Part(b):

Given:

Diameteroflongpipe(d),$d=15cm$

Radiusof long pipe(r),$r=7.5cm=7.5\times {10}^{-2}m$

Applying Bernoulli’s equation, we get

$P_1+\rho g{h}_1=P_2+\rho g{h}_2$

$P_1$=Pressure of elevation (1)

$P_2$=Pressure of elevation (2)

$g=$Acceleration due to gravity

$\rho =$Density of water

$h_1$Height at elevation (1)

$h_2=$Height at elevation (2)

$h_1=564m$

$h_2=2096m$

Volume of rate flow $4500m^3$per day:

We know volume rate flow is given by:

$$\begin{gathered} Q=\frac{V}{t} \\ Q=\frac{V}{t}=Av \end{gathered}$$

Where v is the velocity of flow of water

Area of pipe:

$$\begin{gathered} A=\pi r^2=3.14\times {\left(7.5\times {10}^{-2}\right)}^2=176.6\times {10}^{-4}m \\ v=\frac{V}{t}\left(\frac{1}{A}\right) \end{gathered}$$

$$\begin{gathered} v=\frac{4500}{24\times 60\times 60}\left(\frac{1}{176.6\times {10}^{-4}}\right) \\ v=2.95m/s \end{gathered}$$