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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?

Short Answer

Expert verified
The speed of water at the first point in the pipe is approximately \(16.9 \: m/s\). At the second point in the pipe, the radius is approximately \(0.318 \: m\).

Step by step solution

01

Foundational Equation

The equation of continuity is given by: \( A_1v_1=A_2v_2 \), where \( A_1,A_2 \) are the cross-sectional areas at points 1 and 2 respectively, and \( v_1, v_2 \) are the speeds of the fluid at these points. The flow rate \( Q \) is equal to \( Av \). Since the flow is steady, our task is finding \( v_1 \), and then \( A_2 \), from which we can find radius at point 2.
02

Solving Part (a)

The cross-sectional area \( A_1 \) at point 1 is given by \( \pi (R_1^2) \), where \( R_1=0.150 \: m \) is the radius at point 1. Thus, \( A_1=\pi(0.150^2 \: m^2)=0.071 \: m^2 \) (rounded to three decimal places). As \( Q=A_1v_1 \), we have \( v_1=Q/A_1=1.2 \: m^3/s/0.071 \: m^2=16.9 \: m/s \). Therefore, the speed of water at point 1 is approximately 16.9 m/s.
03

Solving Part (b)

Given the speed \( v_2=3.80 \: m/s \), at point 2, we find the area \( A_2 \) at point 2 from the equation \( Q=A_2v_2 \), i.e., \( A_2=Q/v_2=1.2 \: m^3/s/3.80 \: m/s=0.316 \: m^2 \) (rounded to three decimal places). Now, we calculate the radius \( R_2 \), such that \( A_2=\pi R_2^2 \), which gives \( R_2=\sqrt{A_2/\pi}=0.316 \: m^2/\pi=0.318 \: m \). Therefore, the radius of the pipe at point 2 is approximately 0.318 m.

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Key Concepts

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

Continuity Equation
If you’ve ever watched water flow through a garden hose, you might have noticed the flow of water remains constant even as the hose twists and turns. This constancy is a principle in fluid dynamics described by the continuity equation. Imagine slicing through a pipe with water flowing in it; this slice lets you see the cross-section of the flow. Now, what the continuity equation tells us is quite elegant: as long as the flow is steady, the amount of water passing through each cross-section per unit of time must be the same.

To put it mathematically, if we have two points along the pipe, the product of the area at the first point (\(A_1\)) and the speed of the water there (\(v_1\)) equals the product of the area at the second point (\(A_2\)) with the water speed there (\(v_2\)). So, \(A_1v_1 = A_2v_2\). Now, why should the water care about maintaining this equation? It s\blobfish{}o the mass of water coming in and going out of any section of the pipe in a given time is conserved; nothing is being lost or magically created along the way. This allows us to solve various problems where the pipe’s width changes, impacting the speed of the water flow.
Flow Rate
Let’s talk about a concept that is quite crucial in daily life, from turning on your shower to industrial piping systems: the flow rate. The flow rate is closely tied to the continuity equation we just discussed, but instead of focusing on the changes in speed and area, the flow rate gives us the volume of fluid that passes by a point in the pipe over a certain time.

Represented by the letter 'Q', flow rate is found by multiplying the area of the cross-section of the pipe (\(A\)) by the speed (\(v\)) of the fluid at that point: \(Q = Av\). This formula is practical: if you know how fast the fluid is moving and the size of the pipe, you can figure out how much volume is transported over any period. An increase in flow speed or area will result in more fluid passing through, which makes sense, right? When looking at water flowing into a narrowing pipe, its speed increases to maintain the flow rate, much like how you'd squeeze the end of a hose to make the water shoot out faster.
Circular Cross-Section Area
Visualize cutting a cylindrical carrot; you’re left with a perfect circle, right? That's the idea of a cross-sectional area for a pipe with a circular shape. This area is significant because it affects both the flow rate and the speed of the fluid through the continuity equation. To calculate the cross-sectional area of a pipe with a circular section, you just need the radius (\(R\)), with the formula \(A = \pi R^2\).

Simple Calculation in Action

Let me give you a glimpse into why this matters. If we have a pipe with a smaller cross-sectional area, it means the space for water to flow through is reduced. In everyday terms, it's like putting your thumb partially over the end of a hose - the water has less area to escape and thus speeds up. Conversely, if the area is larger, water can flow more leisurely, because there’s more 'room' for it to move. For any problems involving pipes with varying diameters, understanding the relationship between circular cross-sectional area and fluid dynamics is crucial.

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Most popular questions from this chapter

Water runs into a fountain, filling all the pipes, at a steady rate of \(0.750 \mathrm{~m}^{3} / \mathrm{s}\). (a) How fast will it shoot out of a hole \(4.50 \mathrm{~cm}\) in diameter? (b) At what speed will it shoot out if the diameter of the hole is three times as large?

A uniform plastic block floats in water with \(30.0 \%\) of its volume above the surface of the water. The block is placed in a second liquid and floats with \(20.0 \%\) of its volume above the surface of the liquid. What is the density of the second liquid?

A closed and elevated vertical cylindrical tank with diameter \(2.00 \mathrm{~m}\) contains water to a depth of \(0.800 \mathrm{~m}\). A worker accidently pokes a circular hole with diameter \(0.0200 \mathrm{~m}\) in the bottom of the tank. As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure of \(5.00 \times 10^{3} \mathrm{~Pa}\) at the surface of the water. Ignore any effects of viscosity. (a) Just after the hole is made, what is the speed of the water as it emerges from the hole? What is the ratio of this speed to the efflux speed if the top of the tank is open to the air? (b) How much time does it take for all the water to drain from the tank? What is the ratio of this time to the time it takes for the tank to drain if the top of the tank is open to the air?

An electrical short cuts off all power to a submersible diving vehicle when it is \(30 \mathrm{~m}\) below the surface of the ocean. The crew must push out a hatch of area \(0.75 \mathrm{~m}^{2}\) and weight \(300 \mathrm{~N}\) on the bottom to escape. If the pressure inside is 1.0 atm, what downward force must the crew exert on the hatch to open it?

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 (b) \(0.047 \mathrm{~m}^{2}\) ? (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

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