/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 At a particular axial station, v... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At a particular axial station, velocity and temperature profiles for laminar flow in a parallel plate channel have the form $$ \begin{aligned} &u(y)=0.75\left[1-\left(y / y_{o}\right)^{2}\right] \\ &T(y)=5.0+95.66\left(y / y_{o}\right)^{2}-47.83\left(y / y_{o}\right)^{4} \end{aligned} $$ with units of \(\mathrm{m} / \mathrm{s}\) and \({ }^{\circ} \mathrm{C}\), respectively. Determine corresponding values of the mean velocity, \(u_{m}\), and mean (or bulk) temperature, \(T_{m}\). Plot the velocity and temperature distributions. Do your values of \(u_{m}\) and \(T_{m}\) appear reasonable?

Short Answer

Expert verified
The mean velocity, \(u_m\), in a laminar flow in a parallel plate channel is \(u_m=0.5y_o\), and the mean temperature, \(T_m\), is \(T_m=27.324y_o\). Both of these values appear reasonable as they are consistent with the expected behavior of velocity and temperature profiles in such a flow.

Step by step solution

01

Find the mean velocity \(u_m\)

To find the mean velocity \(u_m\), we need to integrate the given velocity profile, \(u(y)\), over the channel width and normalize by the channel width. The mean velocity is calculated using the following formula: \[ u_m = \frac{1}{y_o} \int_0^{y_o} u(y) dy \] Given the velocity profile, \(u(y) = 0.75\left[1-\left(\frac{y}{y_o}\right)^2\right]\), let's compute \(u_m\):
02

Perform the integration

We start by integrating \(u(y)\) over the domain \(0\) to \(y_o\): \[\int_0^{y_o} 0.75\left[1-\left(\frac{y}{y_o}\right)^2\right] dy = 0.75\int_0^{y_o} \left[1-\left(\frac{y}{y_o}\right)^2\right] dy\] Now, apply substitution. Let \(z = \frac{y}{y_o}\), so \(dz = \frac{dy}{y_o} \implies dy = y_o dz\): \[0.75 y_o \int_0^1 \left[1-z^2\right] dz\] Next, integrate the equation: \[0.75 y_o \left[z - \frac{1}{3}z^3\right]_0^1\]
03

Compute the mean velocity \(u_m\)

Now, we plug in the limits of integration to get \(u_m\): \[u_m = 0.75 y_o \left[\left(1 - \frac{1}{3}\right) - (0)\right] = 0.75 y_o \left(\frac{2}{3}\right)\] Hence, the mean velocity \(u_m\) is: \[u_m = 0.5 y_o\]
04

Calculate the mean (or bulk) temperature, \(T_m\)

To find the mean temperature \(T_m\), we need to integrate the given temperature profile, \(T(y)\), over the channel width and normalize by the channel width. The mean temperature formula is: \[ T_m = \frac{1}{y_o} \int_0^{y_o} T(y) dy \] Given the temperature profile, \(T(y)=5.0+95.66\left(\frac{y}{y_o}\right)^{2}-47.83\left(\frac{y}{y_o}\right)^{4}\), let's compute \(T_m\):
05

Perform the integration

We start by integrating \(T(y)\) over the domain \(0\) to \(y_o\): \[\int_{0}^{y_o} \left[5.0+95.66\left(\frac{y}{y_o}\right)^{2}-47.83\left(\frac{y}{y_o}\right)^{4}\right] dy\] Now, apply substitution. Let \(z = \frac{y}{y_o}\), so \(dz = \frac{dy}{y_o} \implies dy = y_o dz\): \[y_o \int_0^1 \left[5+95.66z^2-47.83z^4\right] dz\] Next, integrate the equation: \[y_o\left[5z+31.89z^3-9.566z^5\right]_0^1\]
06

Compute the mean temperature \(T_m\)

Now, we plug in the limits of integration to get \(T_m\): \[T_m = y_o \left[\left(5+31.89-9.566\right) - (0)\right]\] Hence, the mean temperature \(T_m\) is: \[T_m = 27.324 y_o\]
07

Plot the velocity and temperature distributions

To plot the velocity and temperature distributions, we will create a graph with \(y\) on the horizontal axis and \(u(y)\) and \(T(y)\) on the vertical axis. The velocity profile should be parabolic in shape, decreasing to zero at the channel walls. The temperature profile should also be parabolic, but with different values than the velocity profile.
08

Verify the reasonability of the calculated values of \(u_m\) and \(T_m\)

It is reasonable to expect the mean velocity, \(u_m\), to be greater than zero since the velocity at the channel's center should be the highest. Additionally, the mean temperature, \(T_m\), should also be a positive value greater than \(5.0 °C\) since the initial temperature in the channel is \(5.0 °C\), and the temperature increases across the channel width. In our case, \(u_m=0.5y_o\) and \(T_m=27.324y_o\), which are both reasonable and consistent with the expected behaviour of the velocity and temperature profiles in a laminar flow in a parallel plate channel.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mean Velocity Calculation
In the context of laminar flow heat transfer, the mean velocity is a fundamental concept in fluid dynamics, particularly when examining flow through channels. The mean velocity, often represented as \(u_m\), is crucial for determining how quickly a fluid travels on average across a section of a channel. It's calculated by integrating the velocity profile across the channel width and dividing by the channel width itself.

For our specific problem, where the velocity profile \(u(y)\) is given by a quadratic function, the integration process simplifies into finding the area under the curve of the velocity profile. This area corresponds to the total momentum carried by the fluid per unit width of the channel. To normalize this quantity and thus find the mean velocity, we divide by the channel width \(y_o\), which yields the average velocity experienced by a fluid particle as it moves through the channel.

After integration and applying the limits, we arrive at the equation \[u_m = 0.5 y_o\], which encapsulates the mean or average velocity of the fluid in the channel. The calculation of \(u_m\) is fundamental in analyzing flow characteristics and plays a significant role in applications such as heat exchangers and fluid transport systems.
Temperature Profile Integration
Integrating the temperature profile is a crucial step when working with heat transfer in laminar flow through channels. It enables us to determine the mean or bulk temperature, denoted as \(T_m\), which represents the average thermal energy of the fluid particles. This concept is vital in understanding how heat is distributed and carried out by the fluid flow.

The process is quite similar to finding the mean velocity; however, it considers the temperature distribution across the channel, \(T(y)\). The mean temperature is computed by integrating the given temperature profile over the channel width \(y_o\) and normalizing by the same width. The integration reveals how the temperature changes along the channel width, considering the fluctuations due to the channel's boundaries.

By conducting the steps of integration and applying the appropriate limits, we discover \[T_m = 27.324 y_o\], which supplies us with a comprehensive picture of the thermal behavior of the fluid. This mean temperature is a fundamental aspect of assessing a system's thermal performance, especially in processes like coolant system design and analysis.
Parallel Plate Channel
A parallel plate channel is a model used to describe the flow between two flat plates that are parallel to each other and separated by a distance \(y_o\). This configuration is commonly employed in engineering applications due to its simplicity and ease of mathematical modeling.

In our example, which delineates laminar flow, the fluid moves in layers through the channel, with no cross-mixing of the layers. The flow is characterized by smooth streamlines and uniform flow parameters such as velocity and temperature variations across the channel height. The velocity and temperature profiles are typically parabolic in shape due to the no-slip boundary condition at the walls and low Reynolds number conditions, implying laminar rather than turbulent flow.

Key Features of Laminar Flow in a Parallel Plate Channel:

  • Flow is unidirectional and consistent along the channel's length.
  • The flow regime avoids cross-currents and turbulence, maintaining a stable, laminar pattern.
  • When heat is applied, conduction dominates transverse to the flow, while convection affects in the direction of the flow.
  • The simplicity of the flow dynamics allows for analytical solutions to the Navier-Stokes equations, which describe fluid motion.

Understanding the behavior of laminar flow within such a channel is fundamental for designing efficient fluid transport and heating systems, as well as for predicting the thermal and fluid dynamic performance in various applications like microfluidics and cooling systems of electronic devices.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Air at \(200 \mathrm{kPa}\) enters a 2 -m-long, thin-walled tube of \(25-\mathrm{mm}\) diameter at \(150^{\circ} \mathrm{C}\) and \(6 \mathrm{~m} / \mathrm{s}\). Steam at 20 bars condenses on the outer surface. (a) Determine the outlet temperature and pressure drop of the air, as well as the rate of heat transfer to the air. (b) Calculate the parameters of part (a) if the pressure of the air is doubled.

Heated air required for a food-drying process is generated by passing ambient air at \(20^{\circ} \mathrm{C}\) through long, circular tubes \((D=50 \mathrm{~mm}, L=5 \mathrm{~m})\) housed in a steam condenser. Saturated steam at atmospheric pressure condenses on the outer surface of the tubes, maintaining a uniform surface temperature of \(100^{\circ} \mathrm{C}\). (a) If an airflow rate of \(0.01 \mathrm{~kg} / \mathrm{s}\) is maintained in each tube, determine the air outlet temperature \(T_{m, o}\) and the total heat rate \(q\) for the tube. (b) The air outlet temperature may be controlled by adjusting the tube mass flow rate. Compute and plot \(T_{m \rho}\) as a function of \(\dot{m}\) for \(0.005 \leq \dot{m} \leq\) \(0.050 \mathrm{~kg} / \mathrm{s}\). If a particular drying process requires approximately \(1 \mathrm{~kg} / \mathrm{s}\) of air at \(75^{\circ} \mathrm{C}\), what design and operating conditions should be prescribed for the air heater, subject to the constraint that the tube diameter and length be fixed at \(50 \mathrm{~mm}\) and \(5 \mathrm{~m}\), respectively?

Consider pressurized liquid water flowing at \(\dot{m}=0.1 \mathrm{~kg} / \mathrm{s}\) in a circular tube of diameter \(D=0.1 \mathrm{~m}\) and length \(L=6 \mathrm{~m}\). (a) If the water enters at \(T_{m, i}=500 \mathrm{~K}\) and the surface temperature of the tube is \(T_{s}=510 \mathrm{~K}\), determine the water outlet temperature \(T_{\text {m,o. }}\). (b) If the water enters at \(T_{m, i}=300 \mathrm{~K}\) and the surface temperature of the tube is \(T_{s}=310 \mathrm{~K}\), determine the water outlet temperature \(T_{\text {m, } \sigma}\). (c) If the water enters at \(T_{m, i}=300 \mathrm{~K}\) and the surface temperature of the tube is \(T_{s}=647 \mathrm{~K}\), discuss whether the flow is laminar or turbulent.

A thick-walled, stainless steel (AISI 316) pipe of inside and outside diameters \(D_{i}=20 \mathrm{~mm}\) and \(D_{o}=40 \mathrm{~mm}\) is heated electrically to provide a uniform heat generation rate of \(\dot{q}=10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The outer surface of the pipe is insulated, while water flows through the pipe at a rate of \(\dot{m}=0.1 \mathrm{~kg} / \mathrm{s}\).

In Chapter 1, it was stated that for incompressible liquids, flow work could usually be neglected in the steady-flow energy equation (Equation 1.12d). In the trans-Alaska pipeline, the high viscosity of the oil and long distances cause significant pressure drops, and it is reasonable to question whether flow work would be significant. Consider an \(L=100 \mathrm{~km}\) length of pipe of diameter \(D=1.2 \mathrm{~m}\), with oil flow rate \(\dot{m}=500 \mathrm{~kg} / \mathrm{s}\). The oil properties are \(\rho=900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=2000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=0.765\) \(\mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\). Calculate the pressure drop, the flow work, and the temperature rise caused by the flow work.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.