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Chapter 12: Problem 48

A pressure difference of \(6.00 \times 10^{4} \mathrm{Pa}\) is required to maintain a volume flow rate of 0.800 \(\mathrm{m}^{3} / \mathrm{s}\) for a viscous fluid flowing through a section of cylindrical pipe that has radius 0.210 \(\mathrm{m} .\) What pressure difference is required to maintain the same volume flow rate if the radius of the pipe is decreased to 0.0700 \(\mathrm{m} ?\)

Short Answer

Expert verified
The required pressure difference is approximately \(7.42 \times 10^{6} \mathrm{Pa}\).

Step by step solution

01

Understand the Problem

We need to find the pressure difference required to maintain the same volume flow rate in a pipe when the radius of the pipe changes. Given initial conditions are a radius of 0.210 m and a pressure difference of \(6.00 \times 10^{4} \mathrm{Pa}\). The final goal is to find out the pressure difference when the radius of the pipe is changed to 0.0700 m.
02

Identify the Formula to Use

For viscous flow in a cylindrical pipe, the Hagen-Poiseuille equation can be used: \[ Q = \frac{\pi \Delta P r^4}{8 \eta L} \] where \(Q\) is the volume flow rate, \(\Delta P\) is the pressure difference, \(r\) is the radius of the pipe, \(\eta\) is the viscosity of the fluid, and \(L\) is the length of the pipe. Since \(Q\), \(\eta\), and \(L\) are constant in both scenarios, we can use the equation: \(\frac{\Delta P_1}{r_1^4} = \frac{\Delta P_2}{r_2^4}\).
03

Use Known Values in the Formula

The initial condition gives us \(\Delta P_1 = 6.00 \times 10^{4} \mathrm{Pa}\), \(r_1 = 0.210 \mathrm{m}\), and \(r_2 = 0.0700 \mathrm{m}\). Substitute these into the relationship \[ \frac{\Delta P_1}{r_1^4} = \frac{\Delta P_2}{r_2^4} \] to find \(\Delta P_2\).
04

Perform the Calculations

Substitute the known values into the proportionality equation: \[ \frac{6.00 \times 10^{4} \mathrm{Pa}}{(0.210)^4} = \frac{\Delta P_2}{(0.0700)^4} \]Calculate \((0.210)^4\) and \((0.0700)^4\), then solve for \(\Delta P_2\).
05

Calculate the Power of the Radius

Calculate \((0.210)^4 = 0.001941\) and \((0.0700)^4 = 0.00002401\).
06

Calculate the New Pressure Difference

Plug the radius powers back into the equation: \[ \frac{6.00 \times 10^{4} \mathrm{Pa}}{0.001941} = \frac{\Delta P_2}{0.00002401} \]Calculate \(\Delta P_2\) by multiplying both sides by \(0.00002401\): \(\Delta P_2 = \frac{6.00 \times 10^{4} \times 0.00002401}{0.001941} \approx 7.42 \times 10^{6} \mathrm{Pa}\).

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Key Concepts

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

Viscous Fluid Flow
Viscous fluid flow refers to the movement of fluid through a medium, such as a pipe, where the fluid's viscosity plays a significant role in determining how the fluid moves. Viscosity is a measure of a fluid's resistance to flow and deformation by shear stress or tensile stress. This concept is crucial when examining how fluids behave in real-world applications.
  • Fluids with high viscosity, like honey, flow slowly, while those with low viscosity, like water, flow faster.
  • In a pipe, viscous fluid flow results in a parabolic velocity profile, meaning fluid moves fastest at the center and slowest near the pipe's walls.
  • Understanding viscous flow is essential for designing systems where fluid transport is involved, such as in pipelines, medical devices, and engineering processes.
By considering fluid viscosity, engineers can predict how changes in system variables, like pressure difference or pipe diameter, affect flow rate.
Pressure Difference
The pressure difference is an essential factor that drives fluid flow through pipes. In the context of the Hagen-Poiseuille equation, it specifically refers to the difference in pressure between two points in a pipe that facilitates the movement of fluid from one end to the other.
  • For a fluid to flow, there must be a higher pressure at the inlet than at the outlet; this difference is what propels the fluid.
  • The Hagen-Poiseuille equation demonstrates that the flow rate is directly proportional to the pressure difference (\(\Delta P\)).
  • When the pressure difference increases, the flow rate also increases, assuming the other variables in the equation remain constant.
This relationship is crucial in applications such as water distribution systems, where maintaining the proper pressure difference ensures efficient fluid transport.
Cylindrical Pipe
A cylindrical pipe is a common structure used for transporting fluids. Its shape influences how the fluid flows within, and this is where the geometry comes into play in fluid dynamics.
  • The key geometric factor in a pipe is its radius, which greatly affects flow characteristics. In the Hagen-Poiseuille equation, the flow rate is proportional to the fourth power of the pipe radius.
  • This means small changes in the radius result in significant changes to the flow rate and pressure needed to maintain a given flow rate.
  • In a practical scenario, decreasing the pipe radius leads to a dramatic increase in pressure needed to maintain a constant flow rate.
Understanding the impact of a pipe's shape is essential for engineers and designers when optimizing systems for fluid conveyance.
Volume Flow Rate
The volume flow rate quantifies how much fluid passes through a section of a pipe in a given time interval. It is a crucial parameter in fluid dynamics, especially when designing systems to transport liquids or gases efficiently.
  • It is represented by the symbol \(Q\) and typically measured in cubic meters per second (\(m^3/s\)).
  • The Hagen-Poiseuille equation relates the volume flow rate to the pressure difference, fluid viscosity, pipe length, and radius.
  • By manipulating any of these variables, one can alter the flow rate as needed for a specific application.
Understanding and controlling the volume flow rate is vital in numerous fields, including engineering, environmental science, and any industry that involves fluid transport.

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Most popular questions from this chapter

The piston of a hydraulic automobile lift is 0.30 \(\mathrm{m}\) in diameter. What gauge pressure, in pascals, is required to lift a car with a mass of 1200 \(\mathrm{kg}\) ? Also express this pressure in atmospheres.

(a) As you can tell by watching them in an aquarium, fish are able to remain at any depth in water with no effort. What does this ability tell you about their density? (b) Fish are able to inflate themselves using a sac (called the swim bladder) located under their spinal column. These sacs can be filled with an oxygen-nitrogen mixture that comes from the blood. If a 2.75 -kg fish in freshwater inflates itself and increases its volume by \(10 \%,\) find the net force that the water exerts on it. (c) What is the net external force on it? Does the fish go up or down when it inflates itself?

What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 15.0 m? (Assume that the mains have a much larger diameter than the fire hose.)

An object of average density \(\rho\) floats at the surface of a fluid of density \(\rho_{\text { fluid. }}\) (a) How must the two densities be related? (b) In view of the answer to part (a), how can steel ships float in water? (c) In terms of \(\rho\) and \(\rho\) fluid, what fraction of the object is submerged and what fraction is above the fluid? Check that your answers give the correct limiting behavior as \(\rho \rightarrow \rho_{\text { fluid }}\) and as \(\rho \rightarrow 0 .\) (d) While on board your your your cousin Throckmorton cuts a rectangular piece (dimensions \(5.0 \times 4.0 \times 3.0 \mathrm{cm}\) out of a life preserver and throws it into the ocean. The piece has a mass of 42 g. As it floats in the ocean, what percentage of its volume is above the surface?

A barge is in a rectangular lock on a freshwater river. The lock is 60.0 \(\mathrm{m}\) long and 20.0 \(\mathrm{m}\) wide, and the steel doors on each end are closed. With the barge floating in the lock, a \(2.50 \times 10^{6} \mathrm{N}\) load of scrap metal is put onto the barge. The metal has density 9000 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) When the load of scrap metal, initially on the bank, is placed onto the barge, what vertical distance does the water in the lock rise? (b) The scrap metal is now pushed overboard into the water. Does the water level in the lock rise, fall, or remain the same? If it rises or falls, by what vertical distance does it change?
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