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Chapter 12: Problem 36

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 \(0.070 \mathrm{m}^{2},\) and the magnitude of the fluid velocity is 3.50 \(\mathrm{m} / \mathrm{s}\) (a) What is the fluid speed at points in the pipe where the cross-sectional area is (a) 0.105 \(\mathrm{m}^{2}\) and \((\mathrm{b}) 0.047 \mathrm{m}^{2} ?\) (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 m鲁 in 1 hour.

Step by step solution

01

Understanding the Problem

We need to calculate the fluid velocity at different cross-sectional areas of the pipe, knowing the velocity and area at one point. We'll use the equation of continuity, a key concept in fluid dynamics that states: **The product of cross-sectional area (A) and velocity (v) is constant for incompressible fluid flow**.
02

Applying the Equation of Continuity

The equation of continuity can be expressed as: \( A_1 v_1 = A_2 v_2 \).Given:- \( A_1 = 0.070 \text{ m}^2 \), \( v_1 = 3.50 \text{ m/s} \)To find \( v_2 \) when \( A_2 = 0.105 \text{ m}^2 \), rearrange the equation to solve for \( v_2 \): \( v_2 = \frac{A_1 v_1}{A_2} \).
03

Calculating Fluid Speed at 0.105 m虏

Substitute the known values into the equation:\[ v_2 = \frac{0.070 \, \text{m}^2 \times 3.50 \, \text{m/s}}{0.105 \, \text{m}^2} \]Compute \( v_2 \):\[ v_2 = \frac{0.245 \text{ m}^3/ ext{s}}{0.105 \, \text{m}^2} = 2.33 \text{ m/s} \]
04

Calculating Fluid Speed at 0.047 m虏

Now applying the continuity equation for \( A_2 = 0.047 \text{ m}^2 \):\( v_2 = \frac{A_1 v_1}{A_2} = \frac{0.070 \, \text{m}^2 \times 3.50 \,\text{m/s}}{0.047 \, \text{m}^2} \)Calculate \( v_2 \):\[ v_2 = \frac{0.245 \text{ m}^3/ ext{s}}{0.047 \, \text{m}^2} \approx 5.21 \text{ m/s} \]
05

Calculating Volume Discharge in 1 Hour

The volume flow rate at any point in a pipe is given by \( Q = Av \).Using initial conditions:\( Q = A_1 v_1 = 0.070 \text{ m}^2 \times 3.50 \text{ m/s} = 0.245 \text{ m}^3/\text{s} \).Thus, the volume discharged in 1 hour (3600 seconds) is:\[ \text{Volume} = Q \times \text{time} = 0.245 \text{ m}^3/\text{s} \times 3600 \text{ s} = 882 \text{ m}^3 \]

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Key Concepts

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

Equation of Continuity
The equation of continuity is a vital principle in fluid dynamics. It is based on the conservation of mass in fluid flow, stating that the mass flow rate must remain constant. For an incompressible fluid (one whose density does not change significantly), this translates into the equation: \( A_1 v_1 = A_2 v_2 \). This means the product of the cross-sectional area \( A \) and the velocity \( v \) of fluid remains constant throughout the flow. This concept helps us understand how fluids behave when they move through pipes of varying diameters, allowing us to calculate unknown variables like fluid velocity at different points.
  • It assumes no fluid is added or removed from the system.
  • This principle is foundational to solving problems related to fluid flow in pipes.
Volume Flow Rate
Volume flow rate pertains to the quantity of fluid passing through a point in a system over a given time period. It is denoted as \( Q \) and calculated using the formula \( Q = Av \). Here, \( A \) is the cross-sectional area and \( v \) is the fluid velocity.
This measure is crucial for determining how much fluid is transferred between two sections of a pipe in a given period, like in the calculation of volume discharge. For example, knowing the volume flow rate allows engineers to estimate capacities and design efficient piping systems.
  • The volume flow rate is directly affected by changes in pipe diameter and fluid velocity.
  • It helps in determining how long it will take for a fluid to fill a container or tank.
Cross-Sectional Area
The cross-sectional area of a pipe is the surface area through which the fluid flows. It plays a critical role in determining the velocity of fluid at any point along the pipe. If the cross-sectional area increases, the velocity of the fluid decreases, provided the volume flow rate remains constant, and vice versa. This inverse relationship is critical in understanding fluid dynamics in variable-diameter pipes.
Engineers often design systems with varying cross-sectional areas to control fluid speeds and pressures effectively.
  • The cross-sectional area is typically expressed in square meters (m虏).
  • It can be calculated using geometrical formulas based on the pipe's shape (e.g., for a circular pipe, \( A = \pi r^2 \)).
Incompressible Fluid Flow
In fluid dynamics, incompressible fluid flow refers to situations where the fluid density remains constant throughout. Most liquids, like water, are typically treated as incompressible under normal conditions as their density does not change significantly with pressure or temperature variations. This simplifies many calculations, such as the application of the equation of continuity, since the mass flow rate can be considered constant.
Understanding incompressible flow is instrumental in predicting and managing how fluids move through systems, especially those that do not involve extreme conditions that might lead to significant compressibility.
  • This assumption is often valid for water and other stable liquids.
  • It simplifies calculations in many practical engineering scenarios.

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Most popular questions from this chapter

At a certain point in a horizontal pipeline, the water's speed is 2.50 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(1.80 \times 10^{4}\) Pa. Find the gauge pressure at a second point in the line if the cross-sectional area at the second point is twice that at the first.

Viscous blood is flowing through an artery partially clogged by cholesterol. A surgeon wants to remove enough of the cholesterol to double the flow rate of blood through this artery. If the original diameter of the artery is \(D\) , what should be the new diameter (in terms of \(D\) ) to accomplish this for the same pressure gradient?

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water completely fills the pipe. (a) At one point in the pipe the radius is 0.150 \(\mathrm{m}\) . What is the speed of the water at this point if water is flowing into this pipe at a steady rate of 1.20 \(\mathrm{m}^{3} / \mathrm{s} ?\) (b) At a second point in the pipe the water speed is 3.80 \(\mathrm{m} / \mathrm{s} .\) What is the radius of the pipe at this point?

A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find (a) the speed of efflux of the water and (b) the volume discharged per second.

You win the lottery and decide to impress your friends by exhibiting a million-dollar cube of gold. At the time, gold is selling for \(\$ 426.60\) per troy ounce, and 1.0000 troy ounce equals 31.1035 g. How tall would your million-dollar cube be?
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