/*! 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 67 Water is circulating through a c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Water is circulating through a closed system of pipes in a two-floor apartment. On the first floor, the water has a gauge pressure of \(3.4 \times 10^{5} \mathrm{Pa}\) and a speed of \(2.1 \mathrm{m} / \mathrm{s}\). However, on the second floor, which is \(4.0 \mathrm{m}\) higher, the speed of the water is \(3.7 \mathrm{m} / \mathrm{s}\). The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor?

Short Answer

Expert verified
The gauge pressure on the second floor is \(2.954 \times 10^5 \, \mathrm{Pa}\).

Step by step solution

01

Identify Given Values

First, we identify the given values from the problem: Gauge pressure on the first floor is \( P_1 = 3.4 \times 10^5 \, \mathrm{Pa} \), the speed of water on the first floor \( v_1 = 2.1 \, \mathrm{m/s} \), the height difference \( h = 4.0 \, \mathrm{m} \), and the speed of water on the second floor \( v_2 = 3.7 \, \mathrm{m/s} \).
02

Use Bernoulli's Equation

We use Bernoulli's equation which states: \[P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2\]Since \( h_1 = 0 \) and \( h_2 = 4.0 \, \mathrm{m} \), the equation simplifies to:\[P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) - \rho g h\]
03

Substitute Values into Equation

Substitute the known values into the simplified Bernoulli's equation:- Take the density of water \( \rho = 1000 \, \mathrm{kg/m^3} \) and \( g = 9.81 \, \mathrm{m/s^2} \).- Calculate:\[P_2 = 3.4 \times 10^5 + \frac{1}{2} \times 1000 \times (2.1^2 - 3.7^2) - 1000 \times 9.81 \times 4.0\]
04

Calculate Pressure Components

Compute each of the pressure components separately:- Kinetic energy change:\[\frac{1}{2} \times 1000 \times (2.1^2 - 3.7^2) = \frac{1}{2} \times 1000 \times (4.41 - 13.69) = -4640 \, \mathrm{Pa}\]- Potential energy change:\[1000 \times 9.81 \times 4.0 = 39240 \, \mathrm{Pa}\]Combine the components to find \( P_2 \).
05

Compute Final Gauge Pressure

Combine these values into the equation:\[P_2 = 3.4 \times 10^5 - 4640 - 39240\]Calculate to find:\[P_2 = 2.954 \times 10^5 \, \mathrm{Pa}\]
06

Result Interpretation

The gauge pressure on the second floor is calculated to be \( 2.954 \times 10^5 \, \mathrm{Pa} \), indicating a decrease in pressure due to increased height and kinetic energy.

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.

Bernoulli's Equation
Bernoulli's Equation is a fundamental principle in fluid dynamics that describes the conservation of energy in a fluid flow. It's crucial for understanding how pressure, velocity, and elevation interact in a moving fluid. Bernoulli's Equation can be expressed as:
\[P + \frac{1}{2} \rho v^2 + \rho gh = \, \text{constant} \]where:
  • \( P \) is the fluid pressure,
  • \( \rho \) is the fluid density,
  • \( v \) is the fluid speed, and
  • \( g \) is the acceleration due to gravity.
In our exercise, Bernoulli's Equation helps determine how gauge pressure changes between different floors in a building. It is noteworthy that the equation considers three energy terms: pressure energy, kinetic energy (due to fluid speed), and potential energy (due to elevation).
The equation tells us that if a fluid's speed increases, its pressure or potential energy must decrease to conserve energy. This principle is why, as water moves to a higher floor and speeds up, the pressure decreases, assuming the system is closed and energy is the focus.
Gauge Pressure
Gauge Pressure is a type of pressure measurement that differs from the absolute pressure because it measures the pressure relative to atmospheric pressure. This is beneficial in practical applications where atmospheric pressure can be considered negligible compared to what's being measured.
Think of gauge pressure as the difference between the absolute pressure and the atmospheric pressure, so:
\[ \, P_{gauge} = P - P_{atm} \, \]In the context of our exercise, the gauge pressure of water at different floors helps assess how changes in elevation and fluid speed affect the overall pressure system.
  • On the first floor, the initial gauge pressure is greater.
  • As the water rises to the second floor, despite an increase in speed, the gauge pressure decreases due to the potential energy increase from elevation gain.
This is a practical example of gauge pressure showing variations without needing atmospheric conditions, simplifying most calculations in closed systems like household piping.
Hydraulic Systems
Hydraulic Systems use fluid power to perform work and are built around the principles of fluid mechanics. These systems rely on the incompressible nature of liquids, transmitting force using confined fluid.
  • The fundamental operation is based on Pascal's Law, which states that a change in pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid.
  • Typical components include pipes, pumps, valves, and cylinders.
In the exercise, the water system in the apartment can be seen as a simple hydraulic system.
Changes in pressure between different floors reflect the mechanics of typical hydraulic systems, where pressure differences are used to move fluid through the system to supply various levels and components.
Understanding hydraulic systems also involves recognizing energy changes, like those calculated using Bernoulli's Equation,
to appreciate the force exerted by fluids in motion and elevation changes. These systems are everywhere from residential plumbing to complex industrial machinery. By exploring how pressure and velocity affect these systems, students learn key principles of fluid movement and force transmission.

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 log splitter uses a pump with hydraulic oil to push a piston, which is attached to a chisel. The pump can generate a pressure of \(2.0 \times 10^{7}\) Pa in the hydraulic oil, and the piston has a radius of \(0.050\) \(\mathrm{m} .\) In a stroke lasting \(25 \mathrm{s}\), the piston moves \(0.60\) \(\mathrm{m}\). What is the power needed to operate the log splitter's pump?

An airplane wing is designed so that the speed of the air across the top of the wing is \(251\) \(\mathrm{m} / \mathrm{s}\) when the speed of the air below the wing is \(225\) \(\mathrm{m} / \mathrm{s}\). The density of the air is \(1.29\) \(\mathrm{kg} / \mathrm{m}^{3} .\) What is the lifting force on a wing of area \(24.0\) \(\mathrm{m}^{2} ?\)

(a) The volume flow rate in an artery supplying the brain is \(3.6 \times 10^{-6} \mathrm{m}^{3} / \mathrm{s} .\) If the radius of the artery is \(5.2\) \(\mathrm{mm},\) determine the average blood speed. (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of \(3 .\) Assume that the volume flow rate is the same as that in part (a).

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of \(5.88 \times 10^{-2} \mathrm{kg} / \mathrm{s} .\) The density of the gasoline is \(735 \mathrm{kg} / \mathrm{m}^{3},\) and the radius of the fuel line is \(3.18 \times 10^{-3} \mathrm{m} .\) What is the speed at which the gasoline moves through the fuel line?

A Venturi meter is a device that is used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed \(v_{2}\) through a horizontal section of pipe whose cross-sectional area is \(A_{2}=\) \(0.0700 \mathrm{m}^{2} .\) The gas has a density of \(\rho=1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The Venturi meter has a cross-sectional area of \(A_{1}=0.0500 \mathrm{m}^{2}\) and has been substituted for a section of the larger pipe. The pressure difference between the two sections is \(P_{2}-P_{1}=120\) Pa. Find (a) the speed \(v_{2}\) of the gas in the larger, original pipe and (b) the volume flow rate \(Q\) of the gas.

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.