/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 Water is pumped through a pipe o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 56

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.

Short Answer

Expert verified
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 by step solution

01

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 \( P = 1532m \times 1000 kg/m^3 \times 9.8 m/s^2 = 15013600 Pa \).
02

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 \( v = 0.0521 m^3/s / 0.0177 m^2 = 2.94 m/s \).
03

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 \( \Delta{P} = 0.5 \times 1000 kg/m^3 \times (2.94 m/s)^2 = 4323 Pa \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A high-speed lifting mechanism supports an \(800-\mathrm{kg}\) object with a steel cable that is \(25.0 \mathrm{~m}\) long and \(4.00 \mathrm{~cm}^{2}\) in cross-sectional area. (a) Determine the elongation of the cable. (b) By what additional amount does the cable increase in length if the object is accelerated upward at a rate of \(3.0 \mathrm{~m} / \mathrm{s}^{2} ?\) (c) What is the greatest mass that can be accelerated upward at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) if the stress in the cable is not to exceed the elastic limit of the cable, which is \(2.2 \times 10^{8} \mathrm{~Pa}\) ?

Four acrobats of mass \(75.0 \mathrm{~kg}, 68.0 \mathrm{~kg}, 62.0 \mathrm{~kg}\), and \(55.0 \mathrm{~kg}\) form a human tower, with each acrobat standing on the shoulders of another acrobat. The \(75.0-\mathrm{kg}\) acrobat is at the bottom of the tower. (a) What is the normal force acting on the \(75-\mathrm{kg}\) acrobat? (b) If the area of each of the 75.0-kg acrobat's shoes is \(425 \mathrm{~cm}^{2}\), what average pressure (not including atmospheric pressure) does the column of acrobats exert on the floor? (c) Will the pressure be the same if a different acrobat is on the bottom?

A wooden block of volume \(5.24 \times 10^{-4} \mathrm{~m}^{3}\) floats in water, and a small steel object of mass \(m\) is placed on top of the block. When \(m=0.310 \mathrm{~kg}\), the system is in equilibrium, and the top of the wooden block is at the level of the water. (a) What is the density of the wood? (b) What happens to the block when the steel object is replaced by a second steel object with a mass less than \(0.310 \mathrm{~kg}\) ? What happens to the block when the steel object is replaced by yet another steel object with a mass greater than \(0.310 \mathrm{~kg}\) ?

A \(1.00-\mathrm{kg}\) beaker containing \(2.00 \mathrm{~kg}\) of oil (density \(=916 \mathrm{~kg} / \mathrm{m}^{3}\) ) rests on a scale. A \(2.00-\mathrm{kg}\) block of iron is suspended from a spring scale and is completely submerged in the oil (Fig. P9.43). Find the equilibrium readings of both scales.

Whole blood has a surface tension of \(0.058 \mathrm{~N} / \mathrm{m}\) and a density of \(1050 \mathrm{~kg} / \mathrm{m}^{3}\). To what height can whole blood rise in a capillary blood vessel that has a radius of \(2.0 \times 10^{-6} \mathrm{~m}\) if the contact angle is zero?
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Work Energy and Power

Read Explanation

Magnetism

Read Explanation

Radiation

Read Explanation

Scientific Method Physics

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.