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A pressure difference of \(1.8 \times 10^{3}\) Pa is needed to drive water \(\left(\eta=1.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) through a pipe whose radius is \(5.1 \times 10^{-3} \mathrm{m} .\) The volume flow rate of the water is \(2.8 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s} .\) What is the length of the pipe?

Short Answer

Expert verified
The length of the pipe is approximately 0.213 meters.

Step by step solution

01

Identify Known Variables

We are given:- Pressure difference: \( \Delta P = 1.8 \times 10^3 \text{ Pa} \).- Viscosity of water: \( \eta = 1.0 \times 10^{-3} \text{ Pa} \cdot \text{s} \).- Radius of the pipe: \( r = 5.1 \times 10^{-3} \text{ m} \).- Volume flow rate: \( Q = 2.8 \times 10^{-4} \text{ m}^3/\text{s} \).We need to find the length of the pipe \( L \).
02

Use the Hagen-Poiseuille Equation

The flow of liquid through a cylindrical pipe is described by the Hagen-Poiseuille equation: \[Q = \frac{\pi r^4 \Delta P}{8 \eta L}\]We will rearrange this formula to solve for the length \( L \).
03

Rearrange the Formula to Solve for Pipe Length

Rearranging the Hagen-Poiseuille equation to solve for \( L \):\[L = \frac{\pi r^4 \Delta P}{8 \eta Q}\]
04

Substitute Known Values into the Formula

Substitute the known values into the rearranged equation:\[L = \frac{\pi (5.1 \times 10^{-3})^4 (1.8 \times 10^3)}{8 \times (1.0 \times 10^{-3}) \times (2.8 \times 10^{-4})}\]
05

Calculate the Length of the Pipe

Calculate the expression:\[L = \frac{\pi \times 6.75 \times 10^{-13} \times 1.8 \times 10^3}{8 \times 1.0 \times 10^{-3} \times 2.8 \times 10^{-4}} \approx 0.2128\]Thus the length of the pipe is approximately 0.2128 meters.

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Key Concepts

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

Hagen-Poiseuille equation
The Hagen-Poiseuille equation is a fundamental principle in fluid dynamics that describes the flow of a viscous fluid through a cylindrical pipe. It is named after two engineers, Gotthilf Hagen and Jean Léonard Marie Poiseuille, who independently discovered it in the 19th century. This equation is essential for calculating how fluids behave when they move through narrow channels. The equation is given by:\[Q = \frac{\pi r^4 \Delta P}{8 \eta L}\]where:
  • \(Q\) is the volume flow rate.
  • \(r\) is the radius of the pipe.
  • \(\Delta P\) is the pressure difference between the two ends of the pipe.
  • \(\eta\) refers to the fluid's viscosity.
  • \(L\) is the length of the pipe.
This relationship shows that flow rate is directly proportional to the pressure difference and the fourth power of the radius, but inversely proportional to the fluid's viscosity and the pipe's length. The equation underlines the significant impact even a small change in radius can have, due to the fourth power term. It's commonly applied in systems where smooth, laminar flow is expected in a closed conduit.
viscosity of fluids
Viscosity is a measure of a fluid's resistance to flow and is an important property in fluid dynamics. It's akin to the internal friction between fluid layers when they slide past each other. High viscosity fluids, like honey, flow slowly, whereas low viscosity fluids, like water or air, flow more readily.
Viscosity is represented by the symbol \(\eta\) and has units of pascal-seconds (Pa·s) in the SI system. In the problem you encountered, water's viscosity was given as \(1.0 \times 10^{-3} \text{ Pa} \cdot \text{s}\), which is relatively low. This low viscosity explains why water can flow easily through a pipe.
To affect the volume flow rate in a pipe, viscosity plays a crucial role. Reducing a fluid's viscosity can increase flow rates, while increasing it does the opposite. Understanding viscosity helps engineers and scientists design systems that effectively manage the flow of various substances.
pressure difference
The pressure difference, often denoted as \(\Delta P\), is vital in determining the flow of fluids through a pipe. It is essentially the driving force of fluid motion, pertaining to the difference in pressure at two different points along the pipe.
In the context of the Hagen-Poiseuille equation, this pressure difference propels the fluid through the pipe. The greater the pressure difference, the higher the volume flow rate, assuming other factors like viscosity and pipe length stay constant.
In our example, the pressure difference was noted as \(1.8 \times 10^{3} \text{ Pa}\). This is a positive indication of the force pushing the water through the pipe. If you imagine the scenario, pressure at one end pushes the fluid; if there were no pressure difference, the fluid would cease to flow. Understanding pressure differences is crucial for designing systems where fluid transport efficiency is paramount.
volume flow rate
Volume flow rate, symbolized by \(Q\), defines the volume of fluid that moves through a section of a pipe per unit of time. It's measured in cubic meters per second (\(\text{m}^3/\text{s}\)) in the SI system.
This rate is an essential measure in fluid dynamics as it helps quantify how much fluid flows through a pipe within a specific period. In engineering applications, maintaining a desired flow rate is crucial for the efficient operation of fluid handling systems.
In your original problem, the given volume flow rate was \(2.8 \times 10^{-4} \text{ m}^3/\text{s}\). It illustrates how rapidly the fluid is moving through the pipe. This aspect of the problem ties together with the Hagen-Poiseuille equation, indicating how the fluid's characteristics and conditions influence flow rate. To manage and optimize systems with fluid flow, professionals need to understand how different factors, like pipe dimensions and pressure, impact the volume flow rate.

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Most popular questions from this chapter

The human lungs can function satisfactorily up to a limit where the pressure differ- ence between the outside and inside of the lungs is one-twentieth of an atmosphere. If a diver uses a snorkel for breathing, how far below the water can she swim? Assume the diver is in salt water whose density is 1025 \(\mathrm{kg} / \mathrm{m}^{3}\)

The Mariana trench is located in the floor of the Pacific Ocean at a depth of about 11 000 m below the surface of the water. The density of seawater is 1025 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius \(=0.10 \mathrm{m} ) ? \quad\) (b) For comparison, determine the weight of a jetliner whose mass is \(1.2 \times 10^{5} \mathrm{kg}\) .

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