/*! 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 69 A firehose must be able to shoot... [FREE SOLUTION] | 91Ó°ÊÓ

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A firehose must be able to shoot water to the top of a building 28.0 m tall when aimed straight up. Water enters this hose at a steady rate of 0.500 m\(^3\)/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) Diameter: 0.165 m; (b) New height: 1.75 m.

Step by step solution

01

Identify Known Values and Formulas

We need to find the maximum diameter of the nozzle for the water to reach 28.0 m, using the initial velocity of the water and the constant acceleration due to gravity. From the kinematic equation for projectile motion, we have \( v^2 = u^2 - 2gh \), where \( v = 0 \) m/s at the maximum height \(h\), \( u \) is the initial velocity, and \( g = 9.81 \ m/s^2 \). Also, use the continuity equation \( Q = A \cdot u \), where \( Q \) is the flow rate and \( A \) is the cross-sectional area of the nozzle.
02

Calculate Initial Velocity

Using the kinematic equation \( 0 = u^2 - 2 \, g \, h \), solve for \( u \): \[ u = \sqrt{2gh} = \sqrt{2 \times 9.81 \, m/s^2 \times 28.0 \, m} \]This equals \( u \approx 23.43 \, m/s \). This is the initial velocity needed for the water to reach 28.0 meters.
03

Calculate Cross-Sectional Area of Nozzle

From the continuity equation \( Q = A \cdot u \), we find:\[ A = \frac{Q}{u} = \frac{0.500 \, m^3/s}{23.43 \, m/s} \approx 0.02133 \, m^2 \]. This is the required cross-sectional area of the nozzle.
04

Calculate Maximum Diameter

The area of the nozzle's circular opening is \( A = \pi (d/2)^2 \). Solve for \( d \):\[ d = 2 \sqrt{A/\pi} = 2 \sqrt{0.02133 \, m^2/\pi} \approx 0.165 \, m \]. This is the maximum diameter of the nozzle.
05

Determine New Height with Modified Nozzle

For the nozzle twice the diameter, the area is quadrupled due to the relationship \( A \propto d^2 \). Thus, new velocity \( u' \) is:From \( Q = A' \cdot u' \), where \( A' = 4A \), we find:\[ u' = \frac{Q}{4A} = \frac{23.43 \, m/s}{4} \approx 5.86 \, m/s \].Using the kinematic equation again, \( h' \) is:\[ h' = \frac{(u')^2}{2g} = \frac{(5.86 \, m/s)^2}{2 \times 9.81 \, m/s^2} \approx 1.75 \, m \].

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Key Concepts

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

Continuity Equation
The Continuity Equation is a fundamental principle in fluid dynamics. It ensures that mass is conserved as it flows through a system. For fluids like water flowing through a pipe or nozzle, this concept is crucial. The equation can be expressed as: \( Q = A \cdot u \), where \( Q \) is the volumetric flow rate, \( A \) is the cross-sectional area of the pipe or nozzle, and \( u \) is the fluid velocity.

  • Volumetric Flow Rate \( (Q) \): This measures how much fluid passes a point per unit time, usually in cubic meters per second \(( m^3/s )\).
  • Cross-sectional Area \(( A )\): This is the size of the opening through which the fluid flows, measured in square meters \(( m^2 )\).
  • Fluid Velocity \(( u )\): This is the speed at which the fluid moves through the area.
By ensuring the product of area and velocity remains constant, the continuity equation helps us determine unknown values within the fluid flow system. In this exercise, it allows us to find the nozzle diameter when given the flow rate and desired velocity.
Kinematic Equations
Kinematic equations describe the motion of objects without considering the forces that cause the motion. They are very useful in solving problems involving projectile motion, like shooting water from a hose.

In this context, we primarily use the kinematic equation: \( v^2 = u^2 - 2gh \), where
  • \( v \) is the final velocity (0 m/s at maximum height).
  • \( u \) is the initial velocity.
  • \( g \) is the acceleration due to gravity (\(9.81 \, m/s^2\)).
  • \( h \) is the height to which the water is projected.
For the water to reach a given height, we first determine the required initial velocity \( u \) using this equation. This initial velocity is critical in determining the compatibility of nozzle size with the desired height of water projection.
Nozzle Diameter Calculation
Calculating the nozzle diameter involves using information from both the continuity equation and kinematic equations. First, by using the kinematic equation, we find the initial velocity \( u \) needed for water to reach the target height.

After determining \( u \), we go back to the continuity equation \( Q = A \cdot u \) to find the cross-sectional area \( A \). With \( A \) calculated, we can determine the diameter \( d \) of the nozzle opening using the formula for the area of a circle:

\[ A = \pi (d/2)^2 \]
By solving for \( d \), we conclude the maximum diameter that the nozzle can have to achieve the desired projectile height.

It's important to remember that doubling the diameter quadruples the area (since it's squared in the area formula). Hence, the larger diameter nozzle means a reduced exit velocity and consequently, a reduced maximum height for the water.

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