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Chapter 7: Problem 28

A fluid of constant density \(\varrho\) enters a horizontal pipe of radius \(R\) with uniform velocity \(V\) and pressure \(p_{1}\). At a downstream section the pressure is \(p_{2}\) and the velocity varies with radius \(r\) according to the equation \(u=2 V\left\\{1-\left(r^{2} / R^{2}\right)\right\\}\). Show that the friction force at the pipe walls from the inlet to the section considered is given by \(\pi R^{\frac{1}{2}}\left(p_{1}-p_{2}-\varrho V^{2} / 3\right)\).

Short Answer

Expert verified
The friction force is \( \pi R^{1/2} \left( p_{1} - p_{2} - \frac{\varrho V^2}{3} \right) \).

Step by step solution

01

Understanding the Flow

The fluid enters the pipe with a uniform velocity of \(V\) and pressure \(p_{1}\). As the fluid flows, the pressure changes to \(p_{2}\) and the velocity varies according to the equation \(u = 2V \left( 1 - \frac{r^2}{R^2} \right)\).
02

Velocity Profile Modification

Rewrite the velocity profile in terms of the given equation: \(u = 2V \left( 1 - \frac{r^2}{R^2} \right)\) where \(r\) is the distance from the center of the pipe.
03

Conservation of Mass

Apply the conservation of mass (continuity equation) from the inlet to the downstream section. Since the flow is steady and incompressible:\[ \dot{m}_{in} = \dot{m}_{out} \]At the inlet, \( \dot{m}_{in} = \varrho A_{in} V \), where \(A_{in} = \pi R^2\).
04

Mass Flow Rate at Downstream

At the downstream section, the mass flow rate is given by \[ \dot{m}_{out} = \varrho \int_0^R 2\pi r u(r) \, dr \]Substitute the velocity profile \(u(r)\):\[ \dot{m}_{out} = \varrho \int_0^R 2\pi r 2V \left( 1 - \frac{r^2}{R^2} \right) \, dr \].
05

Evaluate the Integral

Integrate with respect to \(r\):\[ \dot{m}_{out} = 4 \pi \varrho V \int_0^R \left( r - \frac{r^3}{R^2} \right) \, dr \]Solve the integral:\[ = 4 \pi \varrho V \left[ \frac{r^2}{2} - \frac{r^4}{4R^2} \right]_0^R \]\[ = 4 \pi \varrho V \left( \frac{R^2}{2} - \frac{R^2}{4} \right) \]\[ = 4 \pi \varrho V \left( \frac{R^2}{2} - \frac{R^2}{4} \right) = 2 \pi \varrho V R^2 \left( \frac{1}{2} - \frac{1}{4} \right) \]\[ = 2 \pi \varrho V R^2 \left( \frac{1}{4} \right) = \frac{\pi \varrho V R^2}{2} \]
06

Reducing Variables

From the conservation of mass equation, \( \dot{m}_{in} = \dot{m}_{out} \):\[ \varrho \pi R^2 V = \frac{\pi \varrho V R^2}{2} \].
07

Calculating the Friction Force

Use the momentum balance considering friction:\[ \varrho V^2 \int_0^R 2 \pi r \left( 1 - \frac{r^2}{R^2} \right) \, dr = \pi \sqrt{R} \left( p_1 - p_2 - \frac{\varrho V^2}{3} \right) \]Evaluate the integrals and simplify:\[ \pi R \left( \frac{V^2}{2} - \frac{V^2}{4} \right) = \frac{\pi V^2 R}{4} \]Thus,\[ F_{friction} = \pi R^{1/2} \left( p_{1} - p_{2} - \frac{\varrho V^2}{3} \right) \].

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Key Concepts

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

Pipe Flow
Pipe flow refers to the movement of a fluid within a closed conduit. This flow can be characterized by several parameters, including the fluid's velocity, pressure, and flow rate. In the given exercise, a fluid enters a horizontal pipe with a radius of \(R\) and a uniform velocity \(V\). As it moves downstream, the fluid experiences a change in pressure from \(p_{1}\) at the inlet to \(p_{2}\) at a specified downstream section, while the velocity profile adjusts according to the radius \(r\). Understanding the basics of pipe flow is essential to analyzing internal fluid motions, such as in water supply systems or oil pipelines.
Velocity Profile
The velocity profile describes how the fluid's velocity varies across the pipe's cross-section. For our exercise, the velocity profile is given by the equation \(u = 2V \left( 1 - \frac{r^2}{R^2} \right)\), where \(u\) is the velocity at a distance \(r\) from the pipe's center. This profile indicates that the velocity is highest at the center of the pipe and decreases towards the walls. Such a parabolic distribution is typical in laminar flow conditions. Understanding the velocity profile helps in calculating other flow characteristics, such as shear stress and flow rate, which are crucial for system design and analysis.
Friction Force
The friction force arises due to the interaction between the fluid and the pipe's inner surface, causing resistance to the flow. In the given problem, the friction force must account for the change in pressure and velocity distribution. By applying the principles of momentum and integrating across the velocity profile, we can derive the expression for the friction force: \(\pi R^{1/2} \left( p_{1} - p_{2} - \frac{\varrho V^2}{3} \right)\). This formula helps in understanding how factors like pipe diameter, fluid velocity, and pressure drop contribute to the overall resistance in the flow.
Incompressible Flow
Incompressible flow refers to a situation where the fluid density remains constant. For the given exercise, the fluid has a constant density \(\varrho\), simplifying the analysis. Incompressible flow allows us to use the continuity equation to relate the mass flow rates at different sections of the pipe. Specifically, \(\dot{m}_{in} = \dot{m}_{out}\), meaning the mass of fluid entering the pipe equals the mass of fluid exiting. This property ensures that our calculations for velocity distribution and other parameters remain stable and consistent throughout the analysis.
Momentum Balance
The momentum balance is a crucial principle in fluid dynamics that relates the forces acting on a fluid to its change in momentum. For the exercise, we use the momentum equation which incorporates the effects of pressure difference, velocity changes, and friction forces. By integrating the velocity profile and applying the conservation of momentum, we derive that the friction force can be expressed as \(\pi R^{1/2} \left( p_{1} - p_{2} - \frac{\varrho V^2}{3} \right)\). This approach provides a comprehensive understanding of how different forces interact within the pipe flow, aiding in the design and optimization of fluid systems.

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Most popular questions from this chapter

A pipe \(600 \mathrm{~mm}\) diameter and \(1 \mathrm{~km}\) long with \(f=0.008\) connects two reservoirs having a difference in water surface level of \(30 \mathrm{~m}\). Calculate the rate of flow between the reservoirs and the shear stress at the wall of the pipe. If the upstream half of the pipe is tapped by several side pipes so that one-third of the quantity of water now entering the main pipe is withdrawn uniformly over this length, calculate the new rate of discharge to the lower reservoir. Neglect all losses other than those due to pipe friction.

A reservoir \(A\), the free water surface of which is at an elevation of \(275 \mathrm{~m}\), supplies water to reservoirs \(B\) and \(C\) with water surfaces at \(180 \mathrm{~m}\) and \(150 \mathrm{~m}\) elevation respectively. From \(A\) to junction \(D\) there is a common pipe \(300 \mathrm{~mm}\) diameter and \(16 \mathrm{~km}\) long. The pipe from \(D\) to \(B\) is \(200 \mathrm{~mm}\) diameter and \(9.5 \mathrm{~km}\) long while that from \(D\) to \(C\) is \(150 \mathrm{~mm}\) diameter and \(8 \mathrm{~km}\) long. The ends of all pipes are submerged. Calculate the rates of flow to \(B\) and \(C\), neglecting losses other than pipe friction and taking \(f=0.01\) for all pipes.

In a heat exchanger there are 200 tubes each \(3.65 \mathrm{~m}\) long and \(30 \mathrm{~mm}\) outside diameter and \(25 \mathrm{~mm}\) bore. They are arranged axially in a cylinder of \(750 \mathrm{~mm}\) diameter and are equally spaced from one another. A liquid of relative density \(0.9\) flows at a mean velocity of \(2.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) through the tubes and water flows at \(2.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) between the tubes in the opposite direction. For all surfaces \(f\) may be taken as \(0.01\). Neglecting entry and exit losses, calculate (a) the total power required to overcome fluid friction in the exchanger and \((b)\) the saving in power if the two liquids exchanged places but the system remained otherwise unaltered.

Two reservoirs are joined by a sharp-ended flexible pipe \(100 \mathrm{~mm}\) diameter and \(36 \mathrm{~m}\) long. The ends of the pipe differ in level by \(4 \mathrm{~m}\); the surface level in the upper reservoir is \(1.8 \mathrm{~m}\) above the pipe inlet while that in the lower reservoir is \(1.2 \mathrm{~m}\) above the pipe outlet. At a position \(7.5 \mathrm{~m}\) horizontally from the upper reservoir the pipe is required to pass over a barrier. Assuming that the pipe is straight between its inlet and the barrier and that \(f=0.01\) determine the greatest height to which the pipe may rise at the barrier if the absolute pressure in the pipe is not to be less than \(40 \mathrm{kPa}\). Additional losses at bends may be neglected. (Take atmospheric pressure \(=101.3 \mathrm{kPa}\).)

Two water reservoirs, the surface levels of which differ by \(1.5 \mathrm{~m}\), are connected by a pipe system consisting of a sloping pipe at each end, \(7.5 \mathrm{~m}\) long and \(75 \mathrm{~mm}\) diameter, joined by a horizontal pipe \(300 \mathrm{~mm}\) diameter and \(60 \mathrm{~m}\) long. Taking entry head losses as \(0.5 u^{2} / 2 g\) and \(f=0.005(1+25 / d)\) where \(d \mathrm{~mm}\) is the pipe diameter, calculate the steady rate of flow through the pipe.
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