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A straight horizontal pipe with a diameter of \(1.0 \mathrm{~cm}\) and a length of \(50 \mathrm{~m}\) carries oil with a coefficient of viscosity of \(0.12 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\). At the output of the pipe, the flow rate is \(8.6 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{s}\) and the pressure is \(1.0 \mathrm{~atm}\). Find the gauge pressure at the pipe input.

Short Answer

Expert verified
The gauge pressure at the pipe input is approximately \(1.29 \times 10^{6}~\mathrm{Pa}\).

Step by step solution

01

Identify known quantities

From the problem, we can identify that the diameter \(d = 1.0 \mathrm{~cm} = 0.01 \mathrm{~m}\), the length \(L = 50 \mathrm{~m}\), the viscosity of the oil \(\eta = 0.12 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\), the flow rate \(Q = 8.6 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{s}\), and the pressure at the output \(P_{\mathrm{out}} = 1.0 \mathrm{~atm} = 1.013 \times 10^{5} \mathrm{~Pa}\). We need to find the gauge pressure at the pipe input \(P_{\mathrm{in}}\).
02

Use Poiseuille's equation

According to Poiseuille's law, the volume flow rate \(Q\) is given by: \[Q = \frac {\Delta P \pi d^{4}}{128 \eta L} \] Where \(\Delta P = P_{\mathrm{in}} - P_{\mathrm{out}}\) is the pressure difference between the two ends of the pipe. Rearranging this equation, we can find \( \Delta P \) and then \( P_{\mathrm{in}} \) as follows \[ \Delta P = \frac {Q \cdot 128 \eta L}{\pi d^{4}} \] and then \( P_{\mathrm{in}} = \Delta P + P_{\mathrm{out}} \]
03

Plug in the given values

Substituting the given values into the rearranged equation, we get \[ \Delta P = \frac {8.6 \times 10^{-5} \cdot 128 \cdot 0.12 \cdot 50}{\pi \cdot (0.01)^{4}} \] Solving for \(\Delta P\), and then adding \(P_{\mathrm{out}}\) to get \(P_{\mathrm{in}}\) gives the final answer.

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Key Concepts

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

Viscosity
Viscosity is a measure of a fluid's resistance to flow. It describes how much friction the fluid's molecules experience as they move past each other. Higher viscosity means greater resistance and, thus, slower flow under the same pressure condition. In our scenario, oil's viscosity is given as \(0.12 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\). This means the internal friction within the oil is moderate and relevant when considering how it flows through the pipe. The viscosity of fluids plays a crucial role in determining the nature of the flow and in calculating the flow rate using Poiseuille's Law.

It is essential to understand that viscosity is a temperature-dependent parameter. As the temperature increases, for most liquids, viscosity decreases and vice versa. This temperature relation is not needed in the exercise but knowing this might help to grasp real-world applicability to problems involving temperature changes, for example in engine oils or cooking.
Laminar Flow
Laminar flow refers to a flow regime characterized by smooth and ordered motion of fluid particles, often visualized as layers sliding past each other without mixing. In the context of Poiseuille's Law, it is assumed that the flow is laminar, which is a condition for the law to hold true. Laminar flow typically occurs at lower velocities and in smaller pipes, where the viscous forces are more significant than the inertial forces that cause turbulence.

Laminar flow is essential for the validity of the equation used in the exercise. If the flow were turbulent, we would need to use different methods to calculate the pressure and flow rate. In practice, the flow in a pipe can be assumed to be laminar if the Reynolds number, a dimensionless value determined by factors such as fluid velocity, viscosity, and pipe diameter, is below a critical value usually taken as 2000.
Pressure Gradient
The pressure gradient is the rate at which pressure changes in the direction of flow, often driving the fluid to move from higher to lower pressure regions. In the pipe flow problem, the pressure gradient is the force per unit area, causing the oil to flow from the pipe input to output. It results from the pressure difference between the pipe ends, symbolized as \(\Delta P\) and is a key parameter in Poiseuille's Law.

The pressure gradient is dependent on the pipe's dimensions and the viscosity of the fluid. A steeper pressure gradient indicates a higher force driving the flow for a given pipe and fluid combination. Interestingly, this pressure difference is used in Poiseuille's equation to calculate the flow rate, showing the direct relationship between pressure and flow.
Flow Rate
Flow rate, often denoted by \(Q\), is a term used to describe the volume of fluid that passes through a given cross-sectional area per unit time. It's a critical measure of how quickly a fluid is moving through a channel, like the oil in our pipe example. The flow rate is not constant and can change with pressure, pipe diameter, length, and fluid viscosity.

In our exercise, we are provided with the flow rate at the output of the pipe and must work backward to determine the pressure at the input. Understanding the flow rate is necessary for many applications, such as hydraulic systems, blood flow in medical diagnoses, and sizing pipes in engineering projects. The flow rate equation derived from Poiseuille's Law shows how these variables interrelate and can be manipulated to control the flow rate in practical situations.

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Most popular questions from this chapter

Suppose two worlds, each having mass \(M\) and radius \(R\), coalesce into a single world. Due to gravitational contraction, the combined world has a radius of only \(\frac{3}{4} R\). What is the average density of the combined world as a multiple of \(\rho_{0}\), the average density of the original two worlds?

The total cross-sectional area of the load-bearing calcified portion of the two forearm bones (radius and ulna) is approximately \(2.4 \mathrm{~cm}^{2}\). During a car crash, the forearm is slammed against the dashboard. The arm comes to rest from an initial speed of \(80 \mathrm{~km} / \mathrm{h}\) in \(5.0 \mathrm{~ms}\). If the arm has an effective mass of \(3.0 \mathrm{~kg}\) and bone material can withstand a maximum compressional stress of \(16 \times 10^{7} \mathrm{~Pa}\), is the arm likely to withstand the crash?

A sample of an unknown material appears to weigh \(300 \mathrm{~N}\) in air and \(200 \mathrm{~N}\) when immersed in alcohol of specific gravity \(0.700\). What are (a) the volume and (b) the density of the material?

In about 1657 , Otto von Guericke, inventor of the air pump, evacuated a sphere made of two brass hemispheres (Fig. P9.89). Two teams of eight horses each could pull the hemispheres apart only on some trials and then "with greatest difficulty," with the resulting sound likened to a cannon firing. Find the force \(F\) required to pull the thin-walled evacuated hemispheres apart in terms of \(R\), the radius of the hemispheres, \(P\) the pressure inside the hemispheres, and atmospheric pressure \(P_{0}\).

In a water pistol, a piston drives water through a larger tube of radius \(1.00 \mathrm{~cm}\) into a smaller tube of radius \(1.00 \mathrm{~mm}\) as in Figure \(\mathrm{P} 9.51\). (a) If the pistol is fired horizontally at a height of \(1.50 \mathrm{~m}\), use ballistics to determine the time it takes water to travel from the nozzle to the ground. (Neglect air resistance and assume atmospheric pressure is \(1.00 \mathrm{~atm}\).) (b) If the range of the stream is to be \(8.00 \mathrm{~m}\), with what speed must the stream leave the nozzle? (c) Given the areas of the nozzle and cylinder, use the equation of continuity to calculate the speed at which the plunger must be moved. (d) What is the pressure at the nozzle? (e) Use Bernoulli's equation to find the pressure needed in the larger cylinder. Can gravity terms be neglected? (f) Calculate the force that must be exerted on the trigger to achieve the desired range. (The force that must be exerted is due to pressure over and above atmospheric pressure.)

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