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Chapter 14: Problem 53

Water is pumped steadily out of a flooded basement at a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) through a uniform hose of radius \(1.0 \mathrm{~cm}\). The hose passes out through a window \(3.0 \mathrm{~m}\) above the waterline. What is the power of the pump?

Short Answer

Expert verified
The power of the pump is approximately 65.89 W.

Step by step solution

01

Determine the Area of the Hose

First, calculate the cross-sectional area of the hose using the radius. The formula for the area of a circle is \( A = \pi r^2 \). The radius is given as \( 1.0 \ cm \), which is equivalent to \( 0.01 \ m \). Thus the area \( A \) is calculated as:\[A = \pi (0.01 \, \text{m})^2 = \pi \times 0.0001 \, \text{m}^2 \approx 3.14 \times 10^{-4} \, \text{m}^2\]
02

Calculate the Volume Flow Rate

Calculate the volume flow rate, \( Q \), which is the product of the cross-sectional area \( A \) and the speed \( v \) of the water. The formula is \( Q = Av \).Given \( v = 5.0 \, \text{m/s} \) and \( A \approx 3.14 \times 10^{-4} \, \text{m}^2 \):\[Q = 3.14 \times 10^{-4} \, \text{m}^2 \times 5.0 \, \text{m/s} = 1.57 \times 10^{-3} \, \text{m}^3/\text{s}\]
03

Calculate Gravitational Potential Energy per Second

Calculate the gravitational potential energy per second gained by the water, which is given by \( P_{\text{potential}} = \rho g h Q \), where \( \rho = 1000 \, \text{kg/m}^3 \) (density of water), \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity), and \( h = 3.0 \, \text{m} \).Using the value of \( Q \) from Step 2:\[P_{\text{potential}} = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 3.0 \, \text{m} \times 1.57 \times 10^{-3} \, \text{m}^3/s \approx 46.26 \, \text{W}\]
04

Calculate Kinetic Energy per Second

Calculate the kinetic energy per second of the water, which is \( P_{\text{kinetic}} = \frac{1}{2} \rho v^2 Q \).Using \( \rho = 1000 \, \text{kg/m}^3 \), \( v = 5.0 \, \text{m/s} \), and \( Q = 1.57 \times 10^{-3} \, \text{m}^3/s \):\[P_{\text{kinetic}} = \frac{1}{2} \times 1000 \, \text{kg/m}^3 \times (5.0 \, \text{m/s})^2 \times 1.57 \times 10^{-3} \, \text{m}^3/s \approx 19.625 \, \text{W}\]
05

Calculate Total Power

Sum the power contributions from gravitational potential energy and kinetic energy. The total power is calculated as:\[P_{\text{total}} = P_{\text{potential}} + P_{\text{kinetic}} = 46.26 \, \text{W} + 19.625 \, \text{W} \approx 65.885 \, \text{W}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy
Gravitational potential energy is a form of energy that an object possesses due to its position in a gravitational field. For example, when water is pumped from a lower level to a higher level, it gains height and thus increases its gravitational potential energy.
In the context of the exercise, the water has to be lifted 3 meters vertically above the initial waterline. The potential energy gained by the water as it is raised can be calculated using the formula:
  • \( P_{\text{potential}} = \rho g h Q \)
Here:
  • \( \rho \) is the density of water (1000 kg/m\( ^3 \)),
  • \( g \) is the acceleration due to gravity (9.81 m/s\( ^2 \)),
  • \( h \) is the height (3.0 meters),
  • \( Q \) is the volume flow rate.
This formula allows us to calculate how much energy per second is needed just to lift the water to the higher level. In this particular example, the gravitational potential energy calculated is approximately 46.26 watts.
Kinetic Energy
Kinetic energy is the energy that an object has due to its motion. When water is pumped out at a certain speed, it has kinetic energy. The faster the water moves, the more kinetic energy it possesses.
In our exercise, water is being expelled at 5 meters per second through a hose. The kinetic energy per second for the water can be found using the formula:
  • \( P_{\text{kinetic}} = \frac{1}{2} \rho v^2 Q \)
Where:
  • \( \rho \) is again the density of water (1000 kg/m\(^3\)),
  • \( v \) is the velocity of the water (5.0 m/s),
  • \( Q \) is the volume flow rate.
This equation helps us understand the amount of energy per second required to propel the water out through the hose at the specified speed. For this problem, the kinetic energy comes out to be around 19.625 watts.
Volume Flow Rate
The volume flow rate is a measure of the volume of fluid that passes through a cross-section of a pipe or channel per unit time. It is an essential concept in fluid dynamics.

The volume flow rate can be calculated by multiplying the cross-sectional area of the hose by the velocity of the water flowing through it:

  • \( Q = A v \)
In this context:
  • \( A \) is the area of the hose, which can be found using \( A = \pi r^2 \), and
  • \( v \) is the speed of the water (5.0 m/s).
The volume flow rate essentially informs us how much water is being moved out of the basement per second. With our examples, we find that the volume flow rate is approximately 1.57 x 10\( ^{-3} \) m\(^3\)/s, which helps determine both the gravitational potential and kinetic energy contributions.

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Most popular questions from this chapter

An object hangs from a spring balance. The balance registers \(30 \mathrm{~N}\) in air, \(20 \mathrm{~N}\) when this object is immersed in water, and \(24 \mathrm{~N}\) when the object is immersed in another liquid of unknown density. What is the density of that other liquid?

To suck lemonade of density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) up a straw to a maximum height of \(4.0 \mathrm{~cm}\), what minimum gauge pressure (in atmospheres) must you produce in your lungs?

An iron casting containing a number of cavities weighs 6000 \(\mathrm{N}\) in air and \(4000 \mathrm{~N}\) in water. What is the total volume of all the cavities in the casting? The density of iron (that is, a sample with no cavities) is \(7.87 \mathrm{~g} / \mathrm{cm}^{3}\).

A water pipe having a \(2.5 \mathrm{~cm}\) inside diameter carries water into the basement of a house at a speed of \(0.90 \mathrm{~m} / \mathrm{s}\) and a pressure of \(170 \mathrm{kPa}\). If the pipe tapers to \(1.2 \mathrm{~cm}\) and rises to the second floor \(7.6 \mathrm{~m}\) above the input point, what are the (a) speed and (b) water pressure at the second floor?

Suppose that you release a small ball from rest at a depth of \(0.600 \mathrm{~m}\) below the surface in a pool of water. If the density of the ball is \(0.300\) that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)
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