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Chapter 12: Problem 70

A firehose must be able to shoot water to the top of a building 28.0 \(\mathrm{m}\) tall when aimed straight up. Water enters this hose at a steady rate of 0.500 \(\mathrm{m}^{3} / \mathrm{s}\) and shoots out of a round nozzle. (a) What is the maximum diameter this nozzle can have? (b) If the only nozzle available has a diameter twice as great, what is the highest point the water can reach?

Short Answer

Expert verified
(a) Maximum diameter is 16.5 cm. (b) Highest point with larger nozzle is 1.75 m.

Step by step solution

01

Define Relevant Equations for Part (a)

To solve this, we will use the equations for projectile motion and fluid dynamics. First, for straight upward shooting of water, we have the projectile motion equation: \( v^2 = u^2 + 2as \), where initial velocity \( u = 0 \), final velocity \( v = 0 \) at the highest point, \( a = -9.8 \text{ m/s}^2 \) (acceleration due to gravity), and \( s = 28 \text{ m} \) (height of the building). We rearrange the equation to find \( u \):\[ u = \sqrt{2gs} \]
02

Calculate Initial Velocity

Calculate the initial velocity required to reach 28 meters:\[ u = \sqrt{2 \cdot 9.8 \cdot 28} \approx \sqrt{548.8} \approx 23.42 \text{ m/s} \]
03

Utilize Continuity Equation

According to the continuity equation, \( A \cdot v = Q \), where \( A \) is the cross-sectional area of the nozzle, \( v \) is the velocity, and \( Q \) is the volume flow rate. Rearrange to find \( A \):\[ A = \frac{Q}{v} = \frac{0.5}{23.42} \approx 0.02136 \text{ m}^2 \]
04

Determine the Diameter of the Nozzle Part (a)

The cross-sectional area \( A \) is related to the diameter \( d \) by \( A = \frac{\pi d^2}{4} \). Solve for \( d \):\[ d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{0.02136 \times 4}{\pi}} \approx 0.165 \text{ meters} \] or approximately 16.5 cm.
05

Re-evaluate with Larger Nozzle Diameter Part (b)

If the nozzle diameter is twice as large, then the diameter \( d' = 2 \times 0.165 = 0.33 \text{ meters} \). First, we determine the new cross-sectional area \( A' \):\[ A' = \frac{\pi (0.33)^2}{4} \approx 0.0855 \text{ m}^2 \]
06

Calculate New Initial Velocity for Larger Nozzle

Using the continuity equation \( A' \cdot v' = Q \), we solve for the new initial velocity \( v' \):\[ v' = \frac{Q}{A'} = \frac{0.5}{0.0855} \approx 5.85 \text{ m/s} \]
07

Compute Maximum Height for Larger Nozzle

With the new velocity, use the projectile motion formula \( v^2 = u^2 + 2as \) to find the new maximum height:\[ 0 = (5.85)^2 - 2 \times 9.8 \times s' \]Solve for \( s' \):\[ s' = \frac{(5.85)^2}{2 \times 9.8} \approx 1.75 \text{ meters} \]

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Key Concepts

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

Projectile Motion
Projectile motion is a common way to analyze the movement of objects, including fluids such as water, which are projected into the air. When the water is shot straight up from a firehose, it behaves like a projectile. Thus, the kinematic equations of motion are applicable.
The main equation used here is:
  • \( v^2 = u^2 + 2as \)
In this equation:- \( v \) is the final velocity at the highest point (0 m/s since the water stops rising).- \( u \) is the initial velocity you want to calculate.- \( a \) is the acceleration due to gravity (\(-9.8 \text{ m/s}^2\)).- \( s \) is the height the water reaches.
Understanding these elements helps you determine how fast the water must be projected to reach a specific height, like the 28-meter tall building in the problem.
Continuity Equation
In fluid dynamics, understanding how fluid flows and maintains its volume is crucial. The continuity equation aids in this by stating that the product of the cross-sectional area (\( A \)) and velocity (\( v \)) at one point is equal to the product at another point:
  • \( A \, v = Q \)

Here, \( Q \) represents the flow rate, or how much fluid passes a point per second (0.500 \( m^3/s \) in this scenario).
The equation is foundational for determining nozzle size as a change in nozzle diameter changes the velocity of the water projected. A smaller area means a higher velocity, while a larger area reduces it.
Velocity Calculation
Calculating velocity is essential when designing systems in fluid mechanics, as it affects how high and far fluids can travel. When water exits a nozzle, its velocity is determined by dividing the flow rate by the nozzle area using the continuity equation:
  • \( v = \frac{Q}{A} \)

For the nozzle in the problem, the velocity needed for the water to reach 28 meters was found to be approximately 23.42 m/s. This calculation was made using the initial diameter, ensuring that the water shoots high enough.
When the nozzle diameter doubles, the area where water exits increases, decreasing the exit velocity significantly to about 5.85 m/s. This reduction limits the maximum height the water can reach.
Fluid Mechanics
Fluid mechanics encompasses the study of fluids (liquids and gases). It explores how fluids move and the forces with which they interact.
This problem utilizes some fluid mechanics principles:
  • Projectile motion, as fluids can act like projectiles.
  • The continuity equation, focusing on fluid flow.
  • Variables such as velocity and area profoundly affect how fluids behave.
These principles help us solve practical problems like predicting how high water from a hose can shoot. By applying these equations, we account for various conditions such as nozzle size and flow rate. Understanding these factors is crucial for engineers and anyone working with fluid systems to ensure efficiency and meet desired outcomes.

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