Problem 56 - College Physics

Question

Water is pumped through a pipe of diameter \(15.0 \mathrm{~cm}\) from the Colorado River up to Grand Canyon Village, on the rim of the canyon. The river is at \(564 \mathrm{~m}\) elevation and the village is at \(2096 \mathrm{~m}\). (a) At what minimum pressure must the water be pumped to arrive at the village? (b) If \(4500 \mathrm{~m}^{3}\) are pumped per day, what is the speed of the water in the pipe? (c) What additional pressure is necessary to deliver this flow? Note: You may assume the free-fall acceleration and the density of air are constant over the given range of elevations.

Step-by-Step Solution

Answer

a) The minimum pressure required to pump the water to the village is 15013600 Pa. b) The speed of water flowing in the pipe is 2.94 m/s. c) The additional pressure to deliver this water flow is 4323 Pa.

Step 1: Calculation of Pressure Required to Pump the Water to Village

To find the minimum pressure, the height difference between the village and the river, the density of water, and the gravitational acceleration are necessary. The formula to use is \( P = \Delta{H} \times \rho \times g \), where \( \Delta{H} = 2096m - 564m = 1532m \) is the height difference, \( \rho = 1000 kg/m^3 \) is the density of water and \( g = 9.8 m/s^2 \) is the gravitational acceleration. Substituting these values into the formula, we find \( P = 1532m \times 1000 kg/m^3 \times 9.8 m/s^2 = 15013600 Pa \).

Step 2: Calculation of the Speed of Water

The speed of the water is found using the formula for flow rate, \( Q = Av \). We need to find \( v \), so we rearrange the formula to find \( v = Q/A \). Here, \( Q = 4500 m^3/day \), which needs to be converted to seconds for the unit to be consistent - \( Q = 4500 m^3/day \times (1day/86400s) = 0.0521 m^3/s \). The area \( A = \pi D^2/4 \), where \( D = 15 cm = 0.15 m \), so \( A = 3.14 \times (0.15)^2/4 = 0.0177 m^2 \). Substituting these values, we find \( v = 0.0521 m^3/s / 0.0177 m^2 = 2.94 m/s \).

Step 3: Calculation of Additional Pressure to Deliver the Flow

Using the Bernoulli equation, the additional pressure required is found using the formula \( \Delta{P} = 0.5 \times \rho \times v^2 \), where \( v \) is the speed of water calculated in the previous step. Substituting the known values, we find \( \Delta{P} = 0.5 \times 1000 kg/m^3 \times (2.94 m/s)^2 = 4323 Pa \).