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Chapter 14: Problem 41

Water flows through a fire hose of diameter \(6.85 \mathrm{cm}\) at a rate of \(0.0120 \mathrm{m}^{3} / \mathrm{s} .\) The fire hose ends in a nozzle of inner diameter \(2.20 \mathrm{cm} .\) What is the speed with which the water exits the norsle?

Short Answer

Expert verified
By substituting the previously calculated values into the final formula for the speed of water exiting the nozzle, we can find the answer.

Step by step solution

01

Understand the equation of continuity

The equation of continuity states that for an incompressible fluid flowing in a continuous streamline, the volume flow rate (the product of the cross-sectional area and speed of the fluid) remains constant. In mathematical terms, this is given by \(A_1v_1 = A_2v_2\), where \(A_1\) and \(A_2\) are the cross-sectional areas, and \(v_1\) and \(v_2\) are the speeds of the fluid in different sections of the tube.
02

Calculate the cross-sectional area of the hose

The cross-sectional area \(A_1\) of the hose is given by \((\pi /4)d_1^2\), where \(d_1\) refers to the diameter of the hose. Substituting \(d_1 = 6.85 \mathrm{cm} = 0.0685 \mathrm{m}\) (remember to convert cm to m), we get \(A_1 = (3.142 /4)*(0.0685)^2 \mathrm{m^2}\).
03

Calculate the speed of water in the hose

The speed \(v_1\) of the water in the hose can be found from the volume flow rate equation, \(A_1v_1 = Q\), where \(Q\) is the volume flow rate given as \(0.0120 \mathrm{m^3/s}\). Thus, \(v_1 = Q / A_1\). Calculate this value.
04

Calculate the cross-sectional area of the nozzle

Repeat step 2 to calculate the cross-sectional area \(A_2\) of the nozzle using its diameter \(d_2 = 2.20 \mathrm{cm} = 0.0220 \mathrm{m}\). Get \(A_2 = (3.142/ 4)*(0.0220)^2 \mathrm{m^2}\).
05

Calculate the speed of water exiting the nozzle

Use the equation of continuity \(A_1v_1 = A_2v_2\) to find the speed \(v_2\) of the water exiting the nozzle. We get \(v_2 = (A_1v_1) / A_2\). Substitute the values calculated in previous steps to find \(v_2\).

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Key Concepts

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

Equation of Continuity
The equation of continuity is a fundamental principle in fluid dynamics. It ensures that the mass of an incompressible fluid remains constant as it flows through a pipe or channel. Essentially, it means that what flows in must flow out, without any loss or gain in the amount of liquid.

The equation is expressed as:
  • \(A_1v_1 = A_2v_2\)
Here, \(A_1\) and \(A_2\) are the cross-sectional areas, while \(v_1\) and \(v_2\) are the velocities of the fluid at different points. This relationship highlights that when the area decreases, the velocity must increase to maintain the same flow rate, and vice versa. This effect is commonly observed in everyday life, such as the speeding up of water as it exits a nozzle compared to its entry in a wider hose.
Volume Flow Rate
Volume flow rate refers to the amount of fluid that passes through a cross-section per unit of time. It is a crucial measure in fluid dynamics and represents the efficiency or performance of a fluid delivery system.

The formula for volume flow rate \(Q\) is:
  • \(Q = A v\)
Here, \(A\) is the cross-sectional area and \(v\) is the velocity of the fluid. This relationship implies that knowing either the flow rate or the velocity allows us to find the other if the cross-sectional area is known. In real-world applications like firefighting, correct calculations of flow rate ensure optimal water pressure to combat fires effectively.
Cross-Sectional Area
Cross-sectional area is essential when analyzing fluid flow as it directly influences the flow velocity and volume. Calculating this area helps determine the behavior of a fluid as it moves through different sections of a pipe or hose.

For a circular cross-section, the area \(A\) is determined by:
  • \(A = \frac{\pi}{4}d^2\)
Where \(d\) is the diameter of the pipe. In exercises such as the fire hose problem, knowing the cross-sectional area of both the hose and nozzle allows us to apply the equation of continuity accurately and determine how fast the water exits the nozzle. Understanding these calculations ensures that systems are designed to meet the necessary flow requirements.

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Most popular questions from this chapter

A Pilot tube can be used to determine the velocity of air flow by measuring the difference between the total pressure and the static pressure (Fig. \(\mathrm{P} 14.49) .\) If the fluid in the tube is mercury, density \(\rho_{\mathrm{Hg}}=13600 \mathrm{kg} / \mathrm{m}^{3},\) and \(\Delta h=5.00 \mathrm{cm},\) find the speed of air flow. (Assume that the air is stagnant at point \(A,\) and take \(\rho_{\mathrm{air}}=1.25 \mathrm{kg} / \mathrm{m}^{3} .\) )

A cube of wood having an edge dimension of \(20.0 \mathrm{cm}\) and a density of \(650 \mathrm{kg} / \mathrm{m}^{3}\) floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) How much lead weight has to be placed on top of the cube so that its top is just level with the water?

The United States possesses the eight largest warships in the world-aircraft carriers of the Nimilz class-and is building two more. Suppose one of the ships bobs up to float \(11.0 \mathrm{cm}\) higher in the water when 50 fighters take off from it in 25 min, at a location where the free-fall acceleration is \(9.78 \mathrm{m} / \mathrm{s}^{2} .\) Bristling with bombs and missiles, the planes have average mass 29 000 kg. Find the horizontal area enclosed by the waterline of the S1-billion ship. By comparison, its flight deck has area 18 000 \(\mathrm{m}^{2}\). Below decks are passageways hundreds of meters long, so narrow that two large men cannot pass each other.

A bathysphere used for decp-sca exploration has a radius of \(1.50 \mathrm{m}\) and a mass of \(1.20 \times 10^{4} \mathrm{kg} .\) To dive, this submarine takes on mass in the form of seawater. Determine the amount of mass the submarine must take on if it is to descend at a constant speed of \(1.20 \mathrm{m} / \mathrm{s},\) when the resistive force on it is \(1100 \mathrm{N}\) in the upward direction. The density of seawater is \(1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\)

A thin spherical shell of mass \(4.00 \mathrm{kg}\) and diameter \(0.200 \mathrm{m}\) is filled with helium (density \(=0.180 \mathrm{kg} / \mathrm{m}^{3}\) ). It is then released from rest on the bottom of a pool of water that is \(4.00 \mathrm{m}\) deep. (a) Neglecting frictional effects, show that the shell rises with constant acceleration and determine the value of that acceleration. (b) How long will it take for the top of the shell to reach the water surface?
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