/*! 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 38 Water is flowing in a pipe with ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070 m\(^2\), and the magnitude of the fluid velocity is 3.50 m/s. (a) What is the fluid speed at points in the pipe where the cross- sectional area is (a) 0.105 m\(^2\) and (b) 0.047 m\(^2\)? (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

Short Answer

Expert verified
(a) 2.33 m/s, (b) 5.21 m/s, (c) 882 m³

Step by step solution

01

Understand the Continuity Equation

The principle of conservation of mass in fluids is illustrated by the Continuity Equation. It states that the mass flow rate must remain constant from one cross-section to another in a closed system. The equation can be written as:\[ A_1V_1 = A_2V_2 \]where \( A \) is the area of the cross-section and \( V \) is the fluid velocity. This equation can help us find the velocity when the area changes.
02

Calculate Fluid Speed at Point (a) 0.105 m²

For point (a), we use the Continuity Equation:\[ A_1V_1 = A_2V_2 \]Given: \( A_1 = 0.070 \, \text{m}^2 \), \( V_1 = 3.50 \, \text{m/s} \), and \( A_2 = 0.105 \, \text{m}^2 \).Substitute the given values to find \( V_2 \):\[ (0.070)(3.50) = (0.105)(V_2) \]Calculate \( V_2 \):\[ V_2 = \frac{0.070 \times 3.50}{0.105} \approx 2.33 \, \text{m/s} \]
03

Calculate Fluid Speed at Point (b) 0.047 m²

For point (b), apply the Continuity Equation again:\[ A_1V_1 = A_3V_3 \]Given: \( A_3 = 0.047 \, \text{m}^2 \).Substitute the known values:\[ (0.070)(3.50) = (0.047)(V_3) \]Calculate \( V_3 \):\[ V_3 = \frac{0.070 \times 3.50}{0.047} \approx 5.21 \, \text{m/s} \]
04

Calculate Volume of Water Discharged in 1 Hour

To find the volume of water discharged, use the formula:\[ \text{Volume} = \text{Area} \times \text{Velocity} \times \text{Time} \]Choose the correct A and V for the outlet pipe. Assume the pipe ends at the original cross-section \( A_1 \):Use \( A_1 = 0.070 \, \text{m}^2 \), \( V = 3.50 \, \text{m/s} \), and \( \text{Time} = 1 \, \text{hour} = 3600 \, \text{seconds} \).\[ \text{Volume} = 0.070 \times 3.50 \times 3600 \approx 882 \, \text{m}^3 \]

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.

Conservation of Mass in Fluids
The Conservation of Mass in Fluids, beautifully illustrated by the Continuity Equation, is a fundamental concept in fluid dynamics. Imagine you're squeezing toothpaste from a tube, and despite the varying thickness of the tube, each segment contains the same amount of paste. Similarly, in any closed fluid system, the amount of fluid mass entering must equal the amount of fluid mass exiting. This is because mass cannot magically appear or disappear; it just gets redistributed along the flow path. When it comes to fluids traversing through pipes or channels, this principle is represented by the Continuity Equation:
  • \( A_1V_1 = A_2V_2 \)
Here, \( A \) stands for the cross-sectional area, and \( V \) for velocity. It means that if water flows through a narrower section of a pipe, it has to move faster to maintain the mass flow, like traffic speeding up to avoid jams. Understanding this equation allows us to predict changes in speed as pipe diameters fluctuate, making it a powerful tool in engineering and fluid dynamics education.
Fluid Speed Calculation
When tasked with Fluid Speed Calculation, the Continuity Equation becomes our trusty guide. Let's break this down with a practical example. Suppose we have a pipe with an initial cross-sectional area \( A_1 = 0.070 \, \text{m}^2 \) and an initial fluid velocity \( V_1 = 3.50 \, \text{m/s} \). If the section expands, say to \( 0.105 \, \text{m}^2 \), the water must slow down, and we compute the new speed like this:
Plugging into the equation:
  • \( 0.070 \times 3.50 = 0.105 \times V_2 \)
  • Solve for \( V_2 \), yielding \( V_2 \approx 2.33 \, \text{m/s} \)
Likewise, if the pipe narrows to \( 0.047 \, \text{m}^2 \), the water speeds up:
  • \( 0.070 \times 3.50 = 0.047 \times V_3 \)
  • From this, \( V_3 \approx 5.21 \, \text{m/s} \)
It’s like adjusting the nozzle on a garden hose; you decrease the opening size and the water jets out faster. Calculating fluid speed with changing areas informs us how different pipe sections affect flow speeds.
Volume Discharge Calculation
Volume Discharge Calculation is about determining how much fluid exits a system over a set period. For instance, if we wish to calculate how much water is discharged from the end of a pipe in an hour, we utilize the relationship:
  • Volume = Area \( \times \) Velocity \( \times \) Time
Assuming the water flows through a section with area \( 0.070 \, \text{m}^2 \) at a velocity of \( 3.50 \, \text{m/s} \), in an hour (3600 seconds), the volume can be computed as:
\[ \text{Volume} = 0.070 \times 3.50 \times 3600 \approx 882 \, \text{m}^3 \]
Think of it like filling a container; the area corresponds to the opening size, the velocity to the flow speed, and the time to how long you let it flow. This calculation is indispensable in determining capacities and can affect everything from irrigation system design to municipal water supply processes.

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 closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure (1.01 \(\times\) 10\(^5\) Pa) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 Pa. (a) What is the gauge pressure at the bottom of the water? (b) By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.

There is a maximum depth at which a diver can breathe through a snorkel tube (\(\textbf{Fig. E12.17}\) ) because as the depth increases, so does the pressure difference, which tends to collapse the diver's lungs. Since the snorkel connects the air in the lungs to the atmosphere at the surface, the pressure inside the lungs is atmospheric pressure. What is the external internal pressure difference when the diver's lungs are at a depth of 6.1 m (about 20 ft)? Assume that the diver is in freshwater. (A scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.)

In seawater, a life preserver with a volume of 0.0400 m\(^3\) will support a 75.0-kg person (average density 980 kg/m\(^3\)), with 20% of the person's volume above the water surface when the life preserver is fully submerged. What is the density of the material composing the life preserver?

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 11.20 N. Find the total volume and the density of the sample.

A barrel contains a 0.120-m layer of oil floating on water that is 0.250 m deep. The density of the oil is 600 kg/m\(^3\). (a) What is the gauge pressure at the oil\(-\)water interface? (b) What is the gauge pressure at the bottom of the barrel?

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.