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Chapter 4: Problem 1

A laminar flow in a \(2 \mathrm{~cm}-\mathrm{I}\).D. tube has the following velocity and temperature profiles: $$ \begin{aligned} u &=0.1\left[1-(r / 0.01)^{2}\right] \mathrm{m} / \mathrm{s} \\ T &=400-3 \times 10^{6}\left(1.875 \times 10^{-5}-0.25 r^{2}+624 r^{4}\right) \mathrm{K} \end{aligned} $$ for \(r\) in meters. Determine the bulk temperature.

Short Answer

Expert verified
Integrate to find bulk temperature, then divide by flow rate (numerical solution).

Step by step solution

01

Understand Bulk Temperature

The bulk temperature, often denoted as \( T_b \), is the average temperature across the cross-section of the flow. It's calculated by integrating the temperature profile weighted by the velocity profile over the entire cross-section and then dividing by the integral of the velocity profile (i.e., the volumetric flow rate). Mathematically, it is given by \( T_b = \frac{\int_A T u \, dA}{\int_A u \, dA} \).
02

Define Flow and Temperature Functions

Here, the velocity profile is \( u(r) = 0.1[1-(r/0.01)^2] \) m/s, and the temperature profile is \( T(r) = 400 - 3 \times 10^{6} \left( 1.875 \times 10^{-5} - 0.25r^2 + 624r^4 \right) \) K. These functions depend on the radius \( r \) and are used to compute the bulk temperature.
03

Set Up Area Element

The tube has a circular cross-section with radius \( R = 0.01 \) m. We consider differential area elements \( dA = 2\pi r \, dr \), which accounts for the circular symmetry of the tube.
04

Calculate Numerator Integral

The numerator of the expression for \( T_b \) is \( \int_0^R T u \, dA = \int_0^{0.01} T(r) u(r) \, 2\pi r \, dr \). This requires substituting the expressions for \( T(r) \) and \( u(r) \), and then integrating over the radius from 0 to 0.01 m.
05

Calculate Denominator Integral

The denominator integral is the volumetric flow rate: \( \int_0^R u \, dA = \int_0^{0.01} u(r) \, 2\pi r \, dr \). This integral uses the velocity expression \( u(r) = 0.1[1-(r/0.01)^2] \) and is from 0 to 0.01 m.
06

Evaluate Integrals

Perform the integrations from 0 to 0.01 m using integration techniques suitable for polynomials. This involves integrating function products that result from substituting for \( T(r) \) and \( u(r) \). Analytical or numerical integration (depending on complexity) yields specific values for the numerator and denominator.
07

Calculate Bulk Temperature

Divide the result of the numerator integral by the result of the denominator integral to obtain the bulk temperature \( T_b \). Mathematically, \( T_b = \frac{\int_0^R T u \, 2\pi r \, dr}{\int_0^R u \, 2\pi r \, dr} \).

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Key Concepts

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

Understanding Laminar Flow
In fluid dynamics, a laminar flow is a type of flow where the fluid moves in parallel layers, with no disruption between them. It's often found in situations where the fluid is moving slowly, or in small diameter tubes. In a laminar flow, the velocity is consistent along the direction of flow, making calculations simpler for specific parameters like velocity and temperature.

Key characteristics include:
  • Predictable and steady flow pattern
  • Low turbulence
  • Smooth layers of flow

When dealing with laminar flow in tubes, it's crucial to understand that the flow is more controlled and calculable, which simplifies the process of measuring flow-related metrics like bulk temperature.
Exploring the Velocity Profile
The velocity profile describes how the speed of the fluid changes across the cross-section of a tube. For laminar flow in a circular tube, the velocity typically has a parabolic shape. In our problem, this is expressed by the formula: \[ u(r) = 0.1\left[1-\left(\frac{r}{0.01}\right)^2\right] \] This means:

  • The velocity is highest at the center of the tube (\[r = 0\])
  • The velocity decreases to zero as it reaches the tube walls (\[r = R\])
  • The shape is parabolic, which is a signature of laminar flow

Understanding this pattern is key to calculating integral equations, as the velocity profile affects the flow of temperature within the tube. To find parameters like the bulk temperature, you need to know how velocity varies with radius.
Analyzing the Temperature Profile
The temperature profile tells us how temperature varies across the cross-section of the tube. In our scenario, the formula is given by: \[ T(r) = 400 - 3 \times 10^{6} \left( 1.875 \times 10^{-5} - 0.25r^2 + 624r^4 \right) \] This describes a more complex variation:

  • The central temperature might be different from the temperature at the edges
  • The temperature changes are polynomial in nature
  • This variation must be considered when integrating to find total flow temperatures

To determine the average (bulk) temperature of the flow, it is necessary to multiply this temperature profile by the velocity profile during integration. This gives an accurate picture of the flow's thermal characteristics.
Mastering Integration Techniques
To find the bulk temperature, we need to carry out integration. Integration allows us to calculate the cumulative effect of the velocity and temperature variations across the tube's section. The specific task is to integrate the product of the velocity and temperature across the tube's area, which is not always straightforward.

Here's why integration is important:
  • It helps find the total flow characteristics through an entire cross-section
  • Allows calculation of bulk properties like average temperature
  • In this context, involves both polynomial and non-linear elements

By integrating the product of the velocity and temperature profiles over the radius, we can compute the numerator of the bulk temperature formula. Similarly, integrating the velocity profile alone gives us the denominator. Both integrals are essential for obtaining the bulk temperature.

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Most popular questions from this chapter

A \(95 \mathrm{~mm}\)-high Styrofoam cup has \(1.5 \mathrm{~mm}\)-thick walls, and its outside diameter varies from \(77 \mathrm{~mm}\) at the top to \(43 \mathrm{~mm}\) at the base. It is filled with \(200 \mathrm{ml}\) of coffee at \(80^{\circ} \mathrm{C}\), sealed with a \(0.5 \mathrm{~mm}\)-thick plastic lid, and placed on a wooden table. If the ambient air is \(24^{\circ} \mathrm{C}\) and 1000 mbar, estimate the time for coffee to cool to \(60^{\circ} \mathrm{C}\). Take \(k=0.033 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and \(0.33 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the Styrofoam and plastic lid, respectively, and \(\varepsilon=0.85\) for both. Make reasonable assumptions but discuss their validity.

A vertical circuit board is \(3 \mathrm{~mm}\) thick, \(12 \mathrm{~cm}\) high, and \(18 \mathrm{~cm}\) long. On one side are 100 closely spaced logic chips, each dissipating \(0.04 \mathrm{~W}\). The board is a composite containing copper and has an effective thermal conductivity of \(15 \mathrm{~W} / \mathrm{m}\) \(\mathrm{K}\). The heat generated by the chips is conducted across the board and dissipated from the backside to ambient air at \(305 \mathrm{~K}, 1 \mathrm{~atm}\), which flows along the board at \(5 \mathrm{~m} / \mathrm{s}\). Determine the maximum temperature of the front side of the board. Use a transition Reynolds number of \(1 \times 10^{5}\).

In a material processing experiment on a space station, a \(1 \mathrm{~cm}\)-diameter alloy sphere spins at \(3000 \mathrm{rpm}\) in a nitrogen-filled enclosure at \(800 \mathrm{~K}\) and \(1 \mathrm{~atm}\). The sphere is maintained at \(1000 \mathrm{~K}\) by focusing a beam of infrared radiation on the sphere. At what rate must radiant energy be absorbed by the sphere at steady state? The emittance of the sphere is \(0.15\).

Liquid metals are attractive for high-temperature heat transfer applications owing to their characteristic high convective heat transfer coefficients. Liquid potassium flows at \(4 \mathrm{~m} / \mathrm{s}\) in a long, \(2 \mathrm{~cm}\)-I.D. tube. Calculate the Nusselt number and heat transfer coefficient (i) for a uniform wall temperature. (ii) for a uniform wall heat flux. Evaluate properties at \(900 \mathrm{~K}\). Repeat for liquid sodium.

The following table gives velocity and temperature profiles for a turbulent flow of air between parallel plates \(4 \mathrm{~cm}\) apart. The coordinate \(y\) is measured from the wall, and the profiles are symmetrical about the centerplane. Determine the bulk velocity and the bulk temperature. $$ \begin{aligned} &\begin{array}{ccc} y, \mathrm{~cm} & u, \mathrm{~m} / \mathrm{s} & T, \mathrm{~K} \\ \hline 0.0 & 0.0 & 373.0 \\ 0.2 & 16.3 & 322.4 \\ 0.4 & 18.2 & 315.2 \\ 0.6 & 19.4 & 310.7 \\ 0.8 & 20.2 & 307.3 \\ 1.0 & 20.9 & 304.7 \end{array}\\\ &\begin{array}{ccc} y, \mathrm{~cm} & u, \mathrm{~m} / \mathrm{s} & T, \mathrm{~K} \\ \hline 1.2 & 21.4 & 302.6 \\ 1.4 & 21.8 & 301.0 \\ 1.6 & 22.1 & 299.8 \\ 1.8 & 22.2 & 299.1 \\ 2.0 & 22.3 & 298.8 \end{array} \end{aligned} $$
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