/*! 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 37 Compute the pressure required fo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
490,500 Pascals are needed to raise water 50 m vertically.

Step by step solution

01

Understand the Problem

We need to compute the pressure required to lift water to a height of 50 meters. This involves calculating the pressure at the base necessary to overcome this height.
02

Apply the Hydrostatic Pressure Formula

The pressure needed to lift water to a specific height is determined by the hydrostatic pressure formula: \[ P = \rho g h \] where \( P \) is the pressure, \( \rho \) is the density of the fluid (water in this case), \( g \) is the acceleration due to gravity, and \( h \) is the height.
03

Define Known Values

For water, the density \( \rho \) is typically \( 1000 \mathrm{~kg/m^3} \). The acceleration due to gravity \( g \) is approximately \( 9.81 \mathrm{~m/s^2} \). The height \( h \) is given as \( 50.0 \mathrm{~m} \).
04

Plug Values into the Formula

Substitute the known values into the hydrostatic pressure equation:\[ P = 1000 \times 9.81 \times 50.0 \]
05

Calculate the Pressure

Perform the multiplication:\[ P = 1000 \times 9.81 \times 50.0 = 490500 \mathrm{~Pa} \]Thus, the required pressure is 490,500 Pascals.

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 91Ó°ÊÓ!

Key Concepts

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

Fluid Mechanics
Fluid mechanics is the branch of physics that deals with the study of fluids (liquids and gases) and the forces on them. Key aspects include understanding how fluids move and behave under various conditions. In the context of a water supply system, fluid mechanics helps explain how pressure can be applied to move water through pipes and up to different elevations.
When dealing with fluid systems, concepts such as fluid flow, pressure variations, and fluid dynamics come into play. By comprehending fluid mechanics, we can better design systems that efficiently transport water, prevent leakage, and minimize energy consumption.
Two categories primarily define fluid mechanics:
  • Hydrostatics: The study of fluids at rest, focusing on pressure in a fluid at a given depth.
  • Hydrodynamics: The study of fluids in motion, which involves understanding how the flow changes due to different forces.
Pressure Calculation
The pressure calculation in fluid systems is vital to determine the force exerted by a fluid at any given point. It forms the core understanding of how a water supply system can lift water to different heights.
Pressure is defined as the force per unit area applied on a surface in a direction perpendicular to that surface. In fluids, it's important to understand how depth affects pressure. The deeper you go in a fluid, the greater the weight of the fluid above, thus the greater the pressure.
For water systems, we use the hydrostatic pressure formula: \[ P = \rho g h \]Here:
  • \( P \): Pressure (in pascals)
  • \( \rho \): Density of the fluid (in \( \mathrm{kg/m^3} \))
  • \( g \): Gravitational acceleration (in \( \mathrm{m/s^2} \))
  • \( h \): Height (in meters), indicating the vertical distance the fluid is moved.
Density of Water
The density of water is a crucial aspect when calculating the required pressure to move water vertically. Density refers to the mass per unit volume of a substance. For water, the density is typically considered to be \( 1000 \mathrm{~kg/m^3} \) under standard conditions.
Understanding water's density helps in assessing how much mass is being affected by gravity and, consequently, how much pressure is needed to move it. Higher density means more mass per unit volume, requiring more force to lift the same volume of fluid.
This factor is a constant in most calculations dealing with water, making computations often straightforward for standard conditions. However, temperature or impurities can cause water's density to vary slightly, affecting certain precise applications.
Acceleration Due to Gravity
Acceleration due to gravity, denoted as \( g \), is a constant that represents the force imparted by the Earth which causes objects to fall towards its center. At or near the surface of the Earth, this value is approximately \( 9.81 \mathrm{~m/s^2} \).
In the context of fluid mechanics, gravity plays a significant role in determining how fluids behave, especially regarding pressure. The hydrostatic pressure formula, \( P = \rho g h \), directly incorporates gravity because it defines how the weight of a column of fluid contributes to pressure at a given depth.
Understanding gravity is crucial for calculating the amount of force needed to lift water in supply systems, accounting for gravity's pull on every kilogram of water. This knowledge helps engineers design efficient systems that ensure water reaches the desired height without excessive 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

Atmospheric pressure is about \(1.0 \times 10^{5} \mathrm{~Pa}\). How large a force does the still air in a room exert on the inside of a window pane that is \(40 \mathrm{~cm} \times 80 \mathrm{~cm}\) ? The atmosphere exerts a force normal to any surface placed in it. Consequently, the force on the window pane is perpendicular to the pane and is given by \(F=P A=\left(1.0 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.40 \times 0.80 \mathrm{~m}^{2}\right)=3.2 \times 10^{4} \mathrm{~N}\) Of course, a nearly equal force due to the atmosphere on the outside keeps the window from breaking.

Find the density \(\rho\) of a fluid at a depth \(h\) in terms of its density \(\rho_{0}\) at the surface. If a mass \(m\) of fluid has volume \(V_{0}\) at the surface, then it will have volume \(V_{0}-\Delta V\) at a depth \(h\). The density at depth \(h\) is then which gives $$ \begin{array}{c} \rho=\frac{m}{V_{0}-\Delta V} \quad \text { while } \quad \rho_{0}=\frac{m}{V_{0}} \\ \frac{\rho}{\rho_{0}}=\frac{V_{0}}{V_{0}-\Delta V}=\frac{1}{1-\left(\Delta V / V_{0}\right)} \end{array} $$ However, from Chapter 12, the bulk modulus is \(B=P /\left(\Delta V / V_{0}\right)\) and so \(\Delta V / V_{0}=P / B\). Making this substitution, we obtain $$ \frac{\rho}{\rho_{0}}=\frac{1}{1-P / B} $$ If we assume that \(\rho\) is close to \(\rho 0\), then the pressure at depth \(h\) is approximately \(\rho_{0} g h\), and so $$ \frac{\rho}{\rho_{0}}=\frac{1}{1-\left(\rho_{0} g h / B\right)} $$

The area of a piston of a force pump is \(8.0 \mathrm{~cm}^{2}\). What force must be applied to the piston to raise oil \(\left(\rho=0.78 \mathrm{~g} / \mathrm{cm}^{2}\right)\) to a height of \(6.0 \mathrm{~m}\) ? Assume the upper end of the oil is open to the atmosphere.

A cube of wood floating in water supports a 200-g mass resting on the center of its top face. When the mass is removed, the cube rises \(2.00 \mathrm{~cm}\). Determine the volume of the cube.

A beaker contains oil of density \(0.80 \mathrm{~g} / \mathrm{cm}^{3}\). A \(1.6-\mathrm{cm}\) cube of aluminum \(\left(\rho=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)\) hanging vertically on a thread is submerged in the oil. Find the tension in the thread.

See all solutions

Recommended explanations on Physics Textbooks

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.