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Chapter 4: Problem 15

A viscous liquid is flowing through a channel. The velocity profile is given as \(u=2\left(e^{0.3 y}-1\right) \mathrm{m} / \mathrm{s}\), where \(y\) is in meters. If the channel is \(1.5 \mathrm{~m}\) wide, find the volumetric discharge from the channel.

Short Answer

Expert verified
The volumetric discharge (Q) can be found by integrating the given velocity profile from 0 to channel width, it will be the result of \( Q = 2 \left[ \frac{1}{0.3} e^{0.3 * 1.5} - 1.5 - 0 \right] \) calculation, expressed in \(m^3/s\).

Step by step solution

01

Identify the integral limits

The flow rate is calculated by integrating the velocity from one side of the channel to the other. Therefore, the limits of the integration will be from 0 to the channel width, which is 1.5 meters. Thus, the limits are from 0 to 1.5.
02

Perform the integration

Next, perform the integration of \(u = 2(e^{0.3y} - 1)\) from 0 to 1.5. This gives: \[ Q = \int_0^{1.5} 2(e^{0.3y} - 1) dy \], which simplifies to: \[ Q = 2 \left[ \frac{1}{0.3} e^{0.3y} - y \right]_0^{1.5} \].
03

Calculate the value of the integral

By substituting the limits into the integral: \[ Q = 2 \left[ \frac{1}{0.3} e^{0.3 * 1.5} - 1.5 - 0 \right] \]. Solve this to find the value of Q.
04

Finding the final result

Solving the equation will give the volumetric discharge of the channel in \(m^3/s\).

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Key Concepts

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

Velocity Profile
A velocity profile describes how fluid speed varies across the cross-section of the channel. In this example, the velocity profile of the fluid is described by the equation \( u = 2(e^{0.3 y} - 1) \), where \( u \) is the velocity in meters per second and \( y \) is the vertical distance from one side of the channel measured in meters. This equation implies that the velocity is not uniform across the channel; instead, it varies exponentially with respect to \( y \).
  • The term \( e^{0.3y} \) indicates an exponential relationship, indicating growth in velocity as \( y \) increases.
  • The velocity is expected to be lowest at \( y = 0 \), since \( e^{0 \cdot 0.3} - 1 \) resets to zero.
  • As \( y \) increases towards the width of the channel, the velocity increases rapidly due to exponential growth.
Understanding the velocity profile is crucial in predicting how a viscous fluid will behave as it flows through a channel. It helps engineers optimize systems for precise control of fluid movements.
Volumetric Discharge
Volumetric discharge, often referred to as flow rate, is a measure of the volume of fluid that passes through a cross-section of a channel per unit of time. It is an important parameter in fluid dynamics because it helps determine the capacity and efficiency of a channel to transport fluid. In this case, we are interested in finding the flow rate of a viscous liquid flowing through a channel of width 1.5 meters.
  • To compute the volumetric discharge (\( Q \)), we need to integrate the velocity profile across the channel's width.
  • Since the velocity varies with \( y \), the integration should be taken from \( y = 0 \) to \( y = 1.5 \) meters.
  • The integration should account for all the variations in velocity across the width, providing an average volume of fluid transferred per second.
Calculating volumetric discharge helps assess situations like drainage systems or any application where controlling liquid movement is crucial. Understanding volumetric discharge ensures that channels are neither overwhelmed nor underutilized.
Integration in Fluid Dynamics
Integration is a mathematical technique used in fluid dynamics to calculate various properties, such as flow rates, when the variable changes across the length, width, or depth of a fluid's path. Here, integration is used to determine fluid flow through a channel with a non-uniform velocity field:
  • By integrating the velocity equation, \( u = 2(e^{0.3y} - 1) \), over the channel's width, we get the total volume of fluid flowing per second.
  • The definite integral \( \int_0^{1.5} 2(e^{0.3y} - 1) \, dy \) is calculated to find the volumetric discharge \( Q \).
  • Each part of the function inside the integral reflects a different aspect of the fluid's movement; using integration, these are summed to find the overall flow.
  • The resulting expression \( Q = 2 \left[ \frac{1}{0.3} e^{0.3y} - y \right]_0^{1.5} \) computes the volumetric discharge across the chosen channel width.
Integration simplifies the complex calculations involved in fluid dynamics, transforming them into a process we can solve to yield useful, applicable information about fluid flows.
Viscous Flow
Viscous flow refers to the movement of a fluid that has significant internal friction due to viscosity, which affects the flow characteristics. Viscosity is the measure of a fluid's resistance to deformation and flow. In this scenario:
  • The fluid's viscosity causes the velocity to vary across the channel, as depicted by the velocity profile equation.
  • A higher viscosity generally results in slower flow rates and greater velocity variation across the channel.
  • Viscous effects must be considered in engineering applications such as pipeline design, lubrication systems, and other fluid transport systems.
Understanding viscous flow is essential for designing systems that efficiently transport fluids. By comprehending the effects of viscosity, engineers can ensure that channels and tubes are designed appropriately to meet desired specifications and handle the intended fluid dynamics.

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Most popular questions from this chapter

The radius of the circular duct varies as \(r=\left(0.05 e^{-3 x}\right) \mathrm{m},\) where \(x\) is in meters. The flow of a fluid at \(A\) is \(Q=0.004 \mathrm{~m}^{3} / \mathrm{s}\) at \(t=0,\) and it is increasing at \(d Q / d t=0.002 \mathrm{~m}^{3} / \mathrm{s}^{2} .\) If a fluid particle is originally located at \(x=0\) when \(t=0\), determine the time for this particle to arrive at \(x=100 \mathrm{~mm}\).

The toy balloon is filled with air and is released. At the instant shown, the air is escaping at an average velocity of \(4 \mathrm{~m} / \mathrm{s}\), measured relative to the balloon, while the balloon is accelerating. For an analysis, why is it best to consider the control volume to be moving? Select this control volume so that it contains the air in the balloon. Indicate the open control surface, and show the positive direction of its area. Also, indicate the direction of the velocity through this surface. Identify the local and convective changes that occur. Assume the air to be incompressible.

The flat strip is sprayed with paint using the six nozzles, which are attached to the 20 -mm-diameter pipe. The strip is \(50 \mathrm{~mm}\) wide and the paint is to be \(1 \mathrm{~mm}\) thick. If the average speed of the paint through the pipe is \(1.5 \mathrm{~m} / \mathrm{s}\) determine the required speed \(V\) of the strip as it passes under the nozzles as a function of the diameter of the pipe. Plot this function of speed (vertical axis) versus diameter for \(10 \mathrm{~mm} \leq D \leq 30 \mathrm{~mm} .\) Give values for increments of \(\Delta D=5 \mathrm{~mm}\).

Water flows along a rectangular channel having a width of \(0.75 \mathrm{~m}\). If the average velocity is \(2 \mathrm{~m} / \mathrm{s}\), determine the volumetric discharge.

The jet engine is moving forward with a constant speed of \(800 \mathrm{~km} / \mathrm{h}\). Fuel from a tank enters the engine and is mixed with the intake air, burned, and exhausted with an average relative velocity of \(1200 \mathrm{~km} / \mathrm{h}\). Outline the control volume as the jet engine and the air and fuel within it. For an analysis, why is it best to consider this control volume to be moving? Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the magnitudes of the relative velocities and their directions through these surfaces. Identify the local and convective changes that occur. Assume the fuel is incompressible and the air is compressible.
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