/*! 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 31 Compute the pressure required fo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 31

Compute the pressure required for a water supply system that will raise water \(50.0 \mathrm{~m}\) vertically.

Short Answer

Expert verified
The required pressure is 490,000 Pascals (490 kPa).

Step by step solution

01

Understand the Problem

To determine the pressure required to raise water vertically by a height of 50.0 m, we need to use the concept of hydrostatic pressure. This pressure is typically calculated using the formula that relates height, density, gravitational acceleration, and pressure.
02

Recall the Hydrostatic Pressure Formula

The hydrostatic pressure formula is given by \( P = \rho g h \), where \( P \) is the pressure, \( \rho \) (rho) is the density of the fluid (water), \( g \) is the acceleration due to gravity, and \( h \) is the height of the water column.
03

Identify Known Values

The density of water \( \rho \) is approximately \( 1000 \, \text{kg/m}^3 \). The acceleration due to gravity \( g \) is approximately \( 9.8 \, \text{m/s}^2 \). The height \( h \) is given as 50 m.
04

Substitute Values into Formula

Substitute the known values into the hydrostatic pressure formula: \[P = (1000 \, \text{kg/m}^3)(9.8 \, \text{m/s}^2)(50 \, \text{m})\]
05

Calculate the Pressure

Perform the multiplication to find the pressure:\[P = 490,000 \, \text{Pa}\]This shows that the pressure required is 490,000 Pascals or 490 kPa.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding the Water Supply System
A water supply system is a complex network designed to deliver water from its source to consumers, ensuring sufficient pressure and flow. To accomplish this, engineers must consider various factors, including the elevation from the water source to the delivery point. This elevation, or height, directly affects the pressure required to move water. When water supply systems need to raise water to a higher level, for instance, to reach upper floor buildings, hydrostatic pressure becomes a crucial factor. To ensure adequate water flow, systems often integrate pumps or utilize gravity-fed reservoirs. These strategies help maintain the desired pressure at various outlets. Being able to calculate the correct pressure needed is vital to prevent pipes from bursting and to ensure efficient water distribution.
The Process of Pressure Calculation
The calculation of pressure in fluid systems is essential for both practical and safety reasons. In our exercise, this involves using the hydrostatic pressure formula. Hydrostatic pressure specifically deals with the pressure exerted by a stationary fluid due to the force of gravity acting on it.Key components of the hydrostatic pressure formula are:
  • Density of the fluid, represented by \( \rho \).
  • Gravitational acceleration, symbolized by \( g \).
  • Height of the fluid column, denoted by \( h \).
The formula appears as \( P = \rho g h \), where \( P \) is the pressure. By understanding this relationship, one can calculate the necessary pressure to achieve particular engineering tasks, such as lifting water to heights or preventing leaks in fluid containers.
The Role of Fluid Density
Fluid density is a measure of how much mass of fluid occupies a given volume. It is an intrinsic property of fluids and plays a crucial role in determining the hydrostatic pressure. For water, the density is typically around \( 1000 \, \text{kg/m}^3 \), making it a relatively dense fluid in practical scenarios.The higher the fluid density, the greater the weight of a given volume and, consequently, the greater the pressure it exerts at a specific depth or height. This is why substances with higher densities exert more force, impacting how much pressure is created within a system. Therefore, understanding the density of the fluid is vital to calculating pressure accurately, as changes in density can lead to significant differences in required pressure.
Gravity's Influence on Fluid Dynamics
Gravity is the natural force pulling objects toward the center of the Earth and plays a central role in fluid dynamics. When calculating hydrostatic pressure, gravity is quantified as acceleration due to gravity, denoted by \( g \). On Earth, this acceleration is approximately \( 9.8 \, \text{m/s}^2 \).The force of gravity influences many aspects of how fluids move and are distributed. In water supply systems, gravity can either assist or resist the flow of water, depending on how the system is designed. For example, gravity naturally drives water flow in downward sloping pipes, whereas lifting water requires additional pressure to counteract gravity's pull.By understanding the effects of gravity, engineers can design systems that maximize energy efficiency, ensuring water is supplied where it is needed with minimal additional energy use.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A metal cube, \(2.00 \mathrm{~cm}\) on each side, has a density of \(6600 \mathrm{~kg} / \mathrm{m}^{3}\). Find its apparent mass when it is totally submerged in water.

Find the pressure due to the fluid at a depth of \(76 \mathrm{~cm}\) in still \((a)\) water \(\left(\rho_{w}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and (b) mercury ( \(\left.\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)\). (a) \(P=\rho_{w} g h=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=7450 \mathrm{~N} / \mathrm{m}^{2}=7.5 \mathrm{kPa}\) (b) \(P=\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \approx 1.0 \mathrm{~atm}\)

A glass stopper has a mass of \(2.50 \mathrm{~g}\) when measured in air, \(1.50 \mathrm{~g}\) in water, and \(0.70 \mathrm{~g}\) in sulfuric acid. What is the density of the acid? What is its specific gravity? The \(F_{B}\) on the stopper in water is \((0.00250-0.00150)(9.81) \mathrm{N}\). This is the weight of the displaced water. Since \(\rho=m / V\), or \(\rho g=F_{W} / V\), $$\begin{aligned} \text { Volume of stopper } &=\text { Volume of displaced water }=\frac{\text { weight }}{\rho \mathrm{g}} \\ V &=\frac{(0.00100)(9.81) \mathrm{N}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=1.00 \times 10^{-6} \mathrm{~m}^{3} \end{aligned}$$ The buoyant force in acid is $$\left[(2.50-0.70) \times 10^{-3}\right](9.81) \mathrm{N}=(0.00180)(9.81) \mathrm{N}$$ But this is equal to the weight of displaced acid, \(m g\). Since \(\rho=m / V\), and since \(m=0.00180 \mathrm{~kg}\) and \(V=1.00 \times 10^{-6} \mathrm{~m}^{3}\) $$\rho \text { of acid }=\frac{0.00180 \mathrm{~kg}}{1.00 \times 10^{-6} \mathrm{~m}^{3}}=1.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$$ Then, for the acid, $$\text { sp } \mathrm{gr}=\frac{\rho \text { of acid }}{\rho \text { of water }}=\frac{1800}{1000}=1.8$$ Alternative Method $$\begin{array}{l} \text { Weight of displaced water }=\left[(2.50-1.50) \times 10^{-3}\right](9.81) \mathrm{N} \\ \text { Weight of displaced acid }=\left[(2.50-0.70) \times 10^{-3}\right](9.81) \mathrm{N} \end{array}$$ so sp gr of acid \(=\frac{\text { Weight of displaced acid }}{\text { Weight of equal volume of displaced water }}=\frac{1.80}{1.00}=1.8\) Then, since sp gr of acid \(=(\rho\) of acid \() /(\rho\) of water \()\), $$ \rho \text { of acid }=(\text { sp gr of acid })(\rho \text { of water })=(1.8)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)=1.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} $$

A metal object "weighs" \(26.0 \mathrm{~g}\) in air and \(21.48 \mathrm{~g}\) when totally immersed in water. What is the volume of the object? Its mass density?

A mass (or load) acting downward on a piston confines a fluid of density \(\rho\) in a closed container, as shown in Fig. 13-2. The combined weight of the piston and load on the right is \(200 \mathrm{~N}\), and the cross-sectional area of the piston is \(A=8.0 \mathrm{~cm}^{2}\). Find the total pressure at point- \(B\) if the fluid is mercury and \(h=25 \mathrm{~cm}\left(\rho_{\mathrm{Hg}}=13600 \mathrm{~kg} / \mathrm{m}^{3}\right) .\) What would an ordinary pressure gauge read at \(B ?\) Recall what Pascal's principle tells us about the pressure applied to the fluid by the piston and atmosphere: This added pressure is applied at all points within the fluid. Therefore, the total pressure at \(B\) is composed of three parts: Pressure of the atmosphere \(=1.0 \times 10^{5} \mathrm{~Pa}\) Pressure due to the piston and load \(=\frac{F_{W}}{A}=\frac{200 \mathrm{~N}}{8.0 \times 10^{-4} \mathrm{~m}^{2}}=2.5 \times 10^{5} \mathrm{~Pa}\) Pressure due to the height \(h\) of fluid \(=h p g=0.33 \times 10^{5} \mathrm{~Pa}\) In this case, the pressure of the fluid itself is relatively small. We have Total pressure at \(B=3.8 \times 10^{5} \mathrm{~Pa}\) The gauge pressure does not include atmospheric pressure. Therefore, Gauge pressure at \(B=2.8 \times 10^{5} \mathrm{~Pa}\)
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Thermodynamics

Read Explanation

Electricity

Read Explanation

Scientific Method Physics

Read Explanation

Energy Physics

Read Explanation

Physical Quantities And Units

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.