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What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 15.0 \(\mathrm{m} ?\) (Assume that the mains have a much larger diameter than the fire hose.)

Short Answer

Expert verified
The gauge pressure needed is 147,150 Pa.

Step by step solution

01

Understanding the Problem

We need to calculate the gauge pressure required in the city water mains for water to rise to a height of 15.0 meters. The pressure must overcome the gravitational force on the water column.
02

Apply Bernoulli's Equation

Use Bernoulli's equation to relate the pressures and heights:\[ P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \].Assume the water velocity in mains \( v_1 \) and the height \( h_1 \) are negligible compared to the hose nozzle. At the nozzle, the pressure \(P_2\) is atmospheric \(P_0\), velocity \(v_2\) is required for height 15 m, and height \(h_2\) is 15 m.
03

Simplify Bernoulli's Equation

Since \(P_2 = P_0\) (atmospheric pressure) and \(v_1 = 0\), Bernoulli simplifies to:\[ P_1 = P_0 + \rho g h_2 \].Here, \(\rho\) is the density of water \(1000 \text{ kg/m}^3\), \(g\) is the gravitational acceleration \(9.81 \text{ m/s}^2\), and \(h_2 = 15 \text{ m}\).
04

Solve for Gauge Pressure

The gauge pressure \(P_{\text{gauge}}\) is the difference between \(P_1\) and atmospheric pressure \(P_0\):\[ P_{\text{gauge}} = P_1 - P_0 = \rho g h_2 = 1000 \times 9.81 \times 15 \].Calculating gives:\[ P_{\text{gauge}} = 147150 \text{ Pascal (Pa)} \].
05

State the Final Answer

The required gauge pressure in the city water mains must be 147,150 Pa to ensure the fire hose can reach a height of 15 meters.

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Key Concepts

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

Bernoulli's Equation
When aiming to understand the flow of fluids, Bernoulli's Equation is a key concept. It is often termed the energy conservation law for fluid dynamics and helps in analyzing fluid flows.Here's the equation in a simplified form:\[ P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant} \]Where:
  • \( P \) is the pressure of the fluid at a certain point.
  • \( \rho \) is the fluid's density.
  • \( v \) represents the velocity of the fluid.
  • \( g \) is the gravitational acceleration (\( 9.81 \text{ m/s}^2 \) for Earth).
  • \( h \) is the height above a reference point.
In the context of a fire hose problem, we assume that the energy is flowing from higher pressure in the water mains to the atmosphere at the hose's end. Therefore, Bernoulli's equation helps us determine what pressure is lost due to height and velocity differences, allowing us to calculate how strong the initial pressure needs to be to push water to a certain height.
Gravitational Force in Fluid Mechanics
The role of gravity is essential when dealing with fluid movement and pressures. Gravity forces fluids to flow from higher to lower elevations, impacting pressures and velocities. In fluid mechanics, when dealing with height differences, the gravitational force is represented as \( \rho gh \), contributing to the total energy or pressure (\( P \)).
  • \( \rho \): Density of the fluid, which for water is typically \( 1000 \text{ kg/m}^3 \).
  • \( g \): Gravitational acceleration, usually \( 9.81 \text{ m/s}^2 \) on Earth.
  • \( h \): Height of the water column, which affects potential energy.
For a stream to reach a particular height, in this case, 15 meters, the upward pressure must equal the gravitational force attempting to pull it down. Understanding this helps us evaluate how much pressure is needed initially to overcome gravity, particularly when applied in calculating the gauge pressure in water mains.
Water Stream Height Calculation
When calculating the height to which water reaches, we need to consider both the initial pressure and subsequent forces at play, chiefly gravity. This is where our previous concepts integrate.From Bernoulli’s equation, we simplify terms to focus solely on height and initial pressure. Since we assume no energy is lost to friction or air (ideal conditions), the water leaves the hose at zero velocity as it reaches its peak height.Using the derived equation:\[ P_{\text{gauge}} = \rho g h \]Here, water needs enough pressure (\( P_{\text{gauge}} \)) added to the atmospheric pressure to maintain a vertical water flow to a desired height—a crucial application when designing systems like city water mains. Calculating gauge pressure involves simply multiplying the fluid density, gravitational acceleration, and desired height. In our case, it calculated to be 147,150 Pascal (Pa), demonstrating the practical application for the stream to reach 15 meters.

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Most popular questions from this chapter

A cubical block of density \(\rho_{\mathrm{B}}\) and with sides of length \(L\) floats in a liquid of greater density \(\rho_{\mathrm{L}}\) (a) What fraction of the block's volume is above the surface of the liquid? (b) The liquid is denser than water (density \(\rho_{\mathrm{W}} )\) and does not with it If water is poured on the surface of the liquid, how deep must the water layer be so that the water surface just rises to the top of the block? Express your answer in terms of \(L, \rho_{\mathrm{B}}, \rho_{\mathrm{L}},\) and \(\rho_{\mathrm{w}}\) . (c) Find the depth of the water layer in part (b) if the liquid is mercury, the block is made of iron, and the side length is \(10.0 \mathrm{cm} .\)

(a) What is the average density of the sun? (b) What is the average density of a neutron star hat has the same mass as the sun but a radius of only 20.0 \(\mathrm{km}\) ?

An object of average density \(\rho\) floats at the surface of a fluid of density \(\rho_{\text { fluid }}\). (a) How must the two densities be related? (b) In view of the answer to part (a), how can steel ships float in water? (c) In terms of \(\rho\) and \(\rho_{\text { fluid }}\) what fraction of the object is submerged and what fraction is above the fluid? Check that your answers give the correct limiting behavior as \(\rho \rightarrow \rho_{\text { fluid }}\) and as \(\rho \rightarrow 0 .\) (d) While on board your yacht, your cousin Throckmorton cuts a rectangular piece (dimensions \(5.0 \times 4.0 \times 3.0 \mathrm{cm} )\) out of a life preserver and throws it into the ocean. The piece has a mass of 42 g. As it floats in the ocean, what percentage of its volume is above the surface?

A rock is suspended by a light string. When the rock is in air, the tension in the string is 39.2 \(\mathrm{N}\) . When the rock is totally immersed in water, the tension is 28.4 \(\mathrm{N}\) . When the rock is totally immersed in an unknown liquid, the tension is 18.6 \(\mathrm{N}\) . What is the density of the unknown liquid?

At one point in a pipeline the water's speed is 3.00 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(5.00 \times 10^{4} \mathrm{Pa}\) . Find the gauge pressure at a second point in the line, 11.0 \(\mathrm{m}\) lower than the first, if the pipe diameter at the second point is twice that at the first.

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