/*! 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 20 A thick-walled cylindrical tube ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A thick-walled cylindrical tube has inner and outer diameters of \(2 \mathrm{~cm}\) and \(5 \mathrm{~cm}\), respectively. The tube is evacuated and contains a high-temperature radiation source along its axis giving a net radiant heat flux into the inner surface of the tube of \(10^{5} \mathrm{~W} / \mathrm{m}^{2}\). The outer surface of the tube is convectively cooled by a coolant at \(300 \mathrm{~K}\) with a convective heat transfer coefficient of \(120 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the conductivity of the tube material is \(2.2 \mathrm{~W} / \mathrm{m} \mathrm{K}\), determine the temperature distribution \(T(r)\) in the tube wall. Also determine the inner- surface temperature.

Short Answer

Expert verified
The temperature distribution is given by substituting boundary conditions. Inner surface temperature can be derived after knowing constant C.

Step by step solution

01

Identify Known Parameters

Let's start by identifying all the known information in the problem:- Inner radius, \( r_i = 1 \text{ cm} = 0.01 \text{ m} \) (since diameter is \(2 \text{ cm}\))- Outer radius, \( r_o = 2.5 \text{ cm} = 0.025 \text{ m} \) (since diameter is \(5 \text{ cm}\))- Net radiant heat flux into the inner surface, \( q = 10^5 \text{ W/m}^2 \)- Convective heat transfer coefficient, \( h = 120 \text{ W/(m}^2 \text{ K)} \)- Coolant temperature, \( T_{ ext{coolant}} = 300 \text{ K} \)- Thermal conductivity, \( k = 2.2 \text{ W/(m} \cdot \text{ K)} \)
02

Establish Heat Balance Equation

For steady-state heat conduction through a cylindrical wall, the heat balance equation is based on Fourier's Law. The heat conduction rate per unit length is\[ q'' = -k \frac{dT}{dr}. \]This equation implies that the radial temperature gradient drives the heat flow.
03

Solve Fourier's Law Differential Equation

Integrating the differential equation derived from Fourier's law, we have:\[ \int q'' dr = -\int k \frac{dT}{dr} dr. \]Assuming that the heat flux \( q'' \) is consistent across the material, and considering that it equals the net radiant heat flux into the inner surface, we can solve for the temperature distribution.
04

Express Temperature Distribution Function

The temperature distribution in the cylindrical tube can be determined from the integrated form:\[ T(r) = -\frac{q}{2\pi k} \ln \left( \frac{r}{r_i} \right) + C, \]where \( C \) is a constant for integration.
05

Apply Boundary Conditions

Calculate the constant \( C \) using boundary conditions. At the outer surface, the convective heat transfer is related to the excess temperature:\[ q = h (T(r_o) - T_{\text{coolant}}). \]Substituting \( q = 10^5 \text{ W/m}^2 \), solve for \( T(r_o) \) and \( C \).
06

Determine Inner Surface Temperature

Using the temperature distribution function, calculate the inner surface temperature \( T(r_i) \) with known values:\[ T(r_i) = -\frac{10^5}{2\pi (2.2)} \ln \left( \frac{0.01}{0.01} \right) + C. \]Substitute the computed \( C \) from step 5 to find \( T(r_i) \).

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.

Cylindrical Tubes
Cylindrical tubes are a vital component in numerous engineering applications. They are characterized by a hollow, circular cross-section with a certain material thickness. A thick-walled cylindrical tube, like the one described in our problem, has both an inner and outer diameter that are distinctly different from one another. This design allows for various functions, such as containing different substances or providing structural stability.
Examples of cylindrical tubes include pipelines, industrial reactors, and even household items like water pipes. In heat transfer contexts, cylindrical tubes can facilitate the transfer of heat between substances on their inner and outer surfaces.
  • Inner and Outer Diameter: These values establish the overall geometry of the tube. It's important to convert these measurements to the same units when performing calculations.
  • Applications: These elements are frequently seen in heat exchangers where efficient heat transfer is crucial.
Cylindrical tubes also have specific formulas for calculations that differ from flat surfaces due to the shape's complexity. Understanding these is essential for anyone tackling heat conduction problems.
Temperature Distribution
Temperature distribution reveals how temperature varies across the material of an object. For a cylindrical tube, it’s important to understand where heat enters and exits during processing.
In this scenario, we must consider both the heat entering via radiation at the inner surface and the heat leaving via convection at the outer surface. This balance affects the temperature at every point in the tube's walls. To determine the temperature distribution, integration based on physical laws is often used, such as seen with Fourier’s Law. Calculating this distribution involves several factors:
  • Boundary Conditions: Settings that define how temperature behaves at the tube's surfaces.
  • Material Properties: Like thermal conductivity, which affects how easily heat traverses the tube material.
Essentially, calculating temperature distribution helps determine any temperature hotspots and can guide design decisions to prevent failures.
Fourier's Law
Fourier's Law is fundamental in describing heat conduction. The mathematical expression relates heat conduction to the temperature gradient and thermal properties of a material. In a cylindrical coordinate system, this relationship changes slightly but follows the same principle.
The law states that the heat conduction rate is proportional to the negative gradient of the temperature and the material's conductivity. Mathematically, this is expressed as \( q'' = -k \frac{dT}{dr} \),where:
  • \( q'' \) is the heat flux,
  • \( k \) is the thermal conductivity,
  • \( \frac{dT}{dr} \) is the temperature gradient in the radial direction.
The negative sign indicates heat flows from warmer to cooler areas. When solving for temperature distribution in a cylindrical tube, Fourier’s Law is applied to establish the heat balance and derive expressions that allow us to determine temperatures across the material. This understanding of heat conduction is critical for accurate thermal management in engineering designs.
Convective Cooling
Convective cooling is a process where heat is removed from a surface by the motion of fluid over it. In the context of our problem, the outer surface of the cylindrical tube is convectively cooled by a coolant.
The convective heat transfer coefficient \( h \) quantifies how effectively heat is transferred in this way and is crucial for assessing the rate of heat removal. Factors influencing convection include the fluid's velocity, viscosity, and temperature difference between the tube and fluid.
In our example, applying boundary conditions involves setting up the convective heat balance equation \( q = h (T(r_o) - T_{\text{coolant}}) \), which defines the relationship between the heat flux leaving the tube and the temperature at the outer surface. By solving this equation, we can identify how efficient the convective cooling process is, and it helps determine necessary changes to maintain desired operational temperatures. Convective cooling is paramount in various industries where heat management is essential, such as electronics cooling and automotive engine design.

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

A steel heat exchanger tube of \(2 \mathrm{~cm}\) outer diameter is fitted with a steel spiral annular fin of \(5 \mathrm{~cm}\) outer diameter and \(0.2 \mathrm{~mm}\) thickness, wound at a pitch of 3 \(\mathrm{mm}\). If the outside heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the bare tube and 15 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\) on the finned tube, determine the reduction in the outside thermal resistance achieved by adding the fins. Take \(k_{\text {steel }}=42 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

The absorber of a simple flat-plate solar collector with no coverplate consists of a \(2 \mathrm{~mm}\)-thick aluminum plate with \(6 \mathrm{~mm}\)-diameter aluminum water tubes spaced at a pitch of \(10 \mathrm{~cm}\), as shown. On a clear summer day near the ocean, the air temperature is \(20^{\circ} \mathrm{C}\), and a steady wind is blowing. The solar radiation absorbed by the plate is calculated to be 680 \(\mathrm{W} / \mathrm{m}^{2}\), and the convective heat transfer coefficient is estimated to be \(14 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If water at \(20^{\circ} \mathrm{C}\) enters the collector at \(7 \times 10^{-3} \mathrm{~kg} / \mathrm{s}\) per meter width of collector, and the collector is \(3 \mathrm{~m}\) long, estimate the outlet water temperature. For the aluminum take \(k=200 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and \(\varepsilon=0.20\). (Hint: Evaluate the heat lost by reradiation using Eq. (1.19) with a constant value of \(h_{r}\) corresponding to a guessed average plate temperature. Then apply the steady-flow equation to an elemental length of the collector, and so derive a differential equation governing the water temperature increase along the collector.)

A \(5 \mathrm{~mm} \times 2 \mathrm{~mm} \times 1 \mathrm{~mm}\)-thick semiconductor laser is mounted on a \(1 \mathrm{~cm}\) cube copper heat sink and enclosed in a Dewar flask. The laser dissipates \(2 \mathrm{~W}\), and a cryogenic refrigeration system maintains the copper block at a nearly uniform temperature of \(90 \mathrm{~K}\). Estimate the top surface temperature of the laser chip for the following models of the dissipation process: (i) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer underneath the top surface of the laser. (ii) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer at the midplane of the chip. (iii) The energy is dissipated uniformly through the chip. Take \(k=170 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the chip, and neglect parasitic heat gains from the Dewar flask.

The thermal conductivity of a solid may often be assumed to vary linearly with temperature, \(k=k_{0}\left[1+a\left(T-T_{0}\right)\right]\), where \(k=k_{0}\) at a reference temperature \(T_{0}\) and \(a\) is a constant coefficient. Consider a solid slab, \(0

Heat is generated uniformly in a \(8 \mathrm{~cm}\)-thick slab at a rate of \(450 \mathrm{~kW} / \mathrm{m}^{3} .\) One face of the slab is insulated and the other is cooled by water at \(20^{\circ} \mathrm{C}\), giving a heat transfer coefficient of \(800 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the conductivity of the slab is \(12.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\), determine the maximum temperature in the slab.

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.