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A pipe is horizontal and carries oil that has a viscosity of \(0.14\) \(\mathrm{Pa} \cdot \mathrm{s}\) The volume flow rate of the oil is \(5.3 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s} .\) The length of the pipe is \(37\) \(\mathrm{m},\) and its radius is \(0.60 \mathrm{cm} .\) At the output end of the pipe the pressure is atmospheric pressure. What is the absolute pressure at the input end?

Short Answer

Expert verified
The absolute pressure at the input end is approximately \(2.319 \times 10^6\) Pa.

Step by step solution

01

Understand the Problem and Required Formula

We need to find the absolute pressure at the input end of a pipe. The pipe carries oil with given viscosity, flow rate, length, and radius. This situation involves Poiseuille's Law: \[ Q = \frac{\pi r^4 (P_1 - P_2)}{8 \eta L} \]where \(Q\) is the volume flow rate, \(r\) is the radius, \(\eta\) the viscosity, \(L\) the length of the pipe, and \(P_1\) and \(P_2\) are the pressures at the start and end of the pipe respectively.
02

Convert Units

First, convert the radius from centimeters to meters:\[ r = 0.60 \text{ cm} = 0.0060 \text{ m} \]
03

Rearrange the Formula to find Input Pressure

Rearrange the Poiseuille's Law formula to solve for \(P_1\):\[ P_1 = \frac{8 \eta L Q}{\pi r^4} + P_2 \]
04

Substitute Known Values

Substitute all known variables into the rearranged equation:- \(Q = 5.3 \times 10^{-5} \text{ m}^3/\text{s} \)- \(\eta = 0.14 \text{ Pa} \cdot \text{s} \)- \(L = 37 \text{ m} \)- \(r = 0.006 \text{ m} \)- \(P_2 = 101325 \text{ Pa} \) (atmospheric pressure)Into the formula:\[ P_1 = \frac{8 \times 0.14 \times 37 \times 5.3 \times 10^{-5}}{\pi \times (0.006)^4} + 101325 \]
05

Calculate Pressure Difference

Calculate the first part of the equation which is the pressure difference:\[ \Delta P = \frac{8 \times 0.14 \times 37 \times 5.3 \times 10^{-5}}{\pi \times (0.006)^4} \approx 2.217 \times 10^6 \text{ Pa} \]
06

Compute Absolute Input Pressure

Now add the atmospheric pressure:\[ P_1 = 2.217 \times 10^6 + 101325 \approx 2.319 \times 10^6 \text{ Pa} \]Thus, the absolute pressure at the input end is approximately \(2.319 \times 10^6\) Pa.

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is the study of how fluids move and interact with their environment. In our case, we are primarily concerned with how oil flows through a pipe. Key factors in fluid dynamics include:
  • Velocity of the fluid as it travels through a medium.
  • The influence of external forces, such as pressure and gravity.
  • The properties of the fluid, such as density and viscosity.
Understanding fluid dynamics is crucial for analyzing the behavior of fluids in different scenarios, such as the flow of oil through a pipeline. In scenarios governed by Poiseuille's Law, fluid moves through a circular pipe in laminar flow, which means it flows in parallel layers without disruption between them. This law is essential for predicting the rate of flow and understanding how changes in pressure or pipe dimensions affect the movement of the fluid.
Viscosity
Viscosity is a measure of a fluid's resistance to deformation or flow. Think of it as the "thickness" or internal friction of a fluid. In the context of oil flowing through a pipeline, viscosity is a critical factor that influences how easily the oil can move from one end of the pipe to the other. Here are some key points about viscosity:
  • High viscosity means the fluid is thicker and flows less easily.
  • Low viscosity indicates a thinner fluid that flows more readily.
  • Viscosity can change with temperature; typically, it decreases as temperature increases.
In Poiseuille's Law, viscosity (\(\eta\)) acts as a resistance to the flow, playing an important role in determining the flow rate. The oil in the exercise has a viscosity of \(0.14 \, \mathrm{Pa} \cdot \mathrm{s}\), indicating its level of resistance to flow. Understanding viscosity helps explain why certain fluids move more freely than others in pipes.
Pressure Difference
The pressure difference (\(\Delta P\)) across a pipe is the driving force that causes fluid to flow. In our exercise, we need to calculate the difference in pressure between the input and output ends of the pipe to determine how the oil flows. Here's what you need to know:
  • When there's a higher pressure at one end than the other, fluid moves from high to low pressure.
  • The greater the pressure difference, the higher the flow rate, assuming other factors remain constant.
  • In Poiseuille's Law, the pressure difference is calculated using a relationship that involves the volume flow rate, pipe dimensions, and fluid viscosity.
Accurately calculating the pressure difference is essential to solving problems in fluid dynamics, such as determining the absolute pressure at the pipe's input end. This makes it easier to analyze how changes in pressure affect flow in various applications.
Volume Flow Rate
Volume flow rate (\(Q\)) tells us how much fluid passes through a cross-section of the pipe in a given time, typically measured in \(\mathrm{m}^3/\mathrm{s}\). It is a pivotal concept in Poiseuille's Law as it encapsulates the flow efficiency within a pipe. Key aspects of volume flow rate include:
  • High flow rate means a large volume of fluid is passing through a point quickly.
  • It’s affected by pipe radius: larger radius allows more fluid to pass through.
  • Volume flow rate also depends on the pressure difference and fluid viscosity.
In the exercise, the volume flow rate of the oil is \(5.3 \times 10^{-5} \, \mathrm{m}^3/\mathrm{s}\). By understanding \(Q\) and its interactions with other variables, students can predict and manipulate how fluids flow through pipes and similar conduits for various engineering and science applications.

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Most popular questions from this chapter

A paperweight, when weighed in air, has a weight of \(W=6.9 \mathrm{N}\). When completely immersed in water, however, it has a weight of \(W_{\text {in water }}=4.3 \mathrm{N}\). Find the volume of the paperweight.

A cylindrical air duct in an air conditioning system has a length of \(5.5 \mathrm{m}\) and a radius of \(7.2 \times 10^{-2} \mathrm{m} .\) A fan forces air \(\left(\eta=1.8 \times 10^{-5}\right.\) \(\mathrm{Pa} \cdot \mathrm{s})\) through the duct, so that the air in a room (volume \(\left.=280 \mathrm{m}^{3}\right)\) is replenished every ten minutes. Determine the difference in pressure between the ends of the air duct.

Poiseuille's law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds number Re is less than about \(2000: \mathrm{Re}=2 \bar{v} \rho R / \eta\) Here \(\bar{v}, \rho,\) and \(\eta\) are, respectively, the average speed, density, and viscosity of the fluid, and \(R\) is the radius of the pipe. Calculate the highest average speed that blood \(\left(\rho=1060 \mathrm{kg} / \mathrm{m}^{3}, \eta=4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) could have and still remain in laminar flow when it flows through the aorta \(\left(R=8.0 \times 10^{-3} \mathrm{m}\right)\).

When an object moves through a fluid, the fluid exerts a viscous force \(\overrightarrow{\mathbf{F}}\) on the object that tends to slow it down. For a small sphere of radius \(R,\) moving slowly with a speed \(v,\) the magnitude of the viscous force is given by Stokes' law, \(F=6 \pi \eta R v,\) where \(\eta\) is the viscosity of the fluid. (a) What is the viscous force on a sphere of radius \(R=5.0 \times 10^{-4} \mathrm{m}\) that is falling through water \(\left(\eta=1.00 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) when the sphere has a speed of \(3.0 \mathrm{m} / \mathrm{s} ?\) (b) The speed of the falling sphere increases until the viscous force balances the weight of the sphere. Thereafter, no net force acts on the sphere, and it falls with a constant speed called the "terminal speed." If the sphere has a mass of \(1.0 \times 10^{-5} \mathrm{kg},\) what is its terminal speed?

A tube is sealed at both ends and contains a 0.0100-m-long portion of liquid. The length of the tube is large compared to \(0.0100 \mathrm{m}\). There is no air in the tube, and the vapor in the space above the liquid may be ignored. The tube is whirled around in a horizontal circle at a constant angular speed. The axis of rotation passes through one end of the tube, and during the motion, the liquid collects at the other end. The pressure experienced by the liquid is the same as it would experience at the bottom of the tube, if the tube were completely filled with liquid and allowed to hang vertically. Find the angular speed (in \(\mathrm{rad} / \mathrm{s}\) ) of the tube.

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