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Chapter 3: Problem 42

Consider a tube wall of inner and outer radii \(r_{i}\) and \(r_{o}\), whose temperatures are maintained at \(T_{i}\) and \(T_{o}\), respectively. The thermal conductivity of the cylinder is temperature dependent and may be represented by an expression of the form \(k=k_{o}(1+a T)\), where \(k_{o}\) and \(a\) are constants. Obtain an expression for the heat transfer per unit length of the tube. What is the thermal resistance of the tube wall?

Short Answer

Expert verified
The heat transfer per unit length of the tube is given by \(\frac{Q}{L} = k_o(1 + aT)2\pi r (T_i - T_o)\), and the thermal resistance of the tube wall is \(R_{th} = \frac{\ln{\frac{r_o}{r_i}}}{2\pi L k_o(1 + aT)}\).

Step by step solution

01

Set up the heat transfer equation

We know that heat transfer through the cylindrical wall can be represented as \(Q=\frac{k}{L}(A\Delta T)\), where \(Q\) is the heat transfer, \(k\) is the thermal conductivity, \(L\) is the length of the wall, \(A\) is the wall's surface area, and \(\Delta T\) is the temperature difference. In our case, \(k = k_o(1 + aT)\) and we need an expression for \(Q\) per unit length of the tube, i.e., \(Q/L\).
02

Find the surface area and temperature difference for the cylindrical wall

The surface area of the cylindrical wall can be given as \(A = 2 \pi r L\), with \(r\) being the average radius, i.e., \(r=\frac{r_i + r_o}{2}\). The temperature difference, \(\Delta T\), is given as \(T_i - T_o\).
03

Combine the terms and solve for heat transfer per unit length

Substitute values for \(A\) and \(\Delta T\) in the heat transfer equation: \(\frac{Q}{L} = \frac{k_o(1 + aT)2\pi r (T_i - T_o)}{L}\) Since we are looking for a formula of Q per unit length, we can eliminate the \(Ls\) in the equation. Hence the final expression for the heat transfer per unit length is: \(\frac{Q}{L} = k_o(1 + aT)2\pi r (T_i - T_o)\) Now let's find the thermal resistance of the tube wall.
04

Use the formula for thermal resistance in cylindrical systems

The thermal resistance formula for cylindrical systems is given by: \(R_{th} = \frac{\ln{\frac{r_o}{r_i}}}{2\pi Lk}\) We need to find the expression for \(R_{th}\) in terms of the given variables.
05

Substitute temperature-dependent thermal conductivity and solve for thermal resistance

Replace \(k\) with \(k_o(1 + aT)\) and simplify the expression for thermal resistance: \(R_{th} = \frac{\ln{\frac{r_o}{r_i}}}{2\pi L k_o(1 + aT)}\) This is the expression for the thermal resistance of the tube wall. In conclusion, the heat transfer per unit length of the tube is given by \(\frac{Q}{L} = k_o(1 + aT)2\pi r (T_i - T_o)\), and the thermal resistance of the tube wall is \(R_{th} = \frac{\ln{\frac{r_o}{r_i}}}{2\pi L k_o(1 + aT)}\).

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Key Concepts

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

Thermal Resistance
Thermal resistance measures how well a system resists the flow of heat. Just like electrical resistance impedes the flow of electricity, thermal resistance prevents heat from passing through a material. For a cylindrical tube, this concept is vital as it helps us understand the efficiency of the tube wall in preventing heat loss or transfer.
Using the formula for cylindrical thermal resistance, \( R_{th} = \frac{\ln{\frac{r_o}{r_i}}}{2\pi L k} \), we can ascertain the resistance of the tube wall. Here, \( r_o \) and \( r_i \) are the outer and inner radii, respectively, \( L \) is the length of the tube, and \( k \) is the thermal conductivity. This formula highlights the relationship between resistance and the tube's dimensions and properties.
In practice, thermal resistance will depend on factors such as:
  • Material's thermal conductivity
  • Difference in radii \((r_o - r_i)\)
  • Temperature differences across the tube
This concept helps engineers design efficient systems by selecting appropriate materials and dimensions to control heat flow. This ensures energy conservation and optimizes thermal management.
Cylindrical System
Understanding a cylindrical system's unique geometry is crucial for analyzing heat transfer. Unlike flat surfaces, cylindrical systems have radial dimensions that influence heat distribution. They are commonly found in tubes, pipes, and other circular structures.
The surface area in a cylindrical system is different from a flat plate. It is calculated using the formula \( A = 2 \pi r L \). Here, \( r \) is often taken as the average of the inner and outer radii (\( r = \frac{r_i + r_o}{2} \)), and \( L \) is the length of the cylinder. This formula accounts for the curvature and allows us to calculate the heat transfer accurately across the curved surface.
A cylindrical system's distinctive features are:
  • Radial symmetry, meaning the same properties extend all around the tube
  • Varying surface areas as a function of radius
  • Differential heat flow due to radial thermal gradients
These attributes make solving heat transfer problems more complex in cylindrical systems, requiring specific formulas and approaches to ensure accuracy.
Temperature Dependent Thermal Conductivity
In many cases, the conductivity of materials changes with temperature. This is especially relevant in heat transfer studies where precise calculations are critical. For a cylindrical system, using a temperature-dependent thermal conductivity can enhance the accuracy of heat transfer analysis.
The expression \( k = k_o(1 + a T) \) represents a linear dependency where \( k_o \) is the base thermal conductivity at a reference temperature, and \( a \) is a constant that indicates the degree of dependency on temperature. This relation helps in considering real-time variations in how the material conducts heat as the tube's temperature changes.
Key points about temperature-dependent conductivity include:
  • More accurate predictions of heat transfer rates
  • Better representation of material properties at different temperatures
  • Consideration of non-linear thermal effects in system design
Understanding this concept allows engineers to design systems that maintain effective heat management under varying thermal conditions, increasing the reliability and efficiency of thermal systems.

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Most popular questions from this chapter

Measurements show that steady-state conduction through a plane wall without heat generation produced a convex temperature distribution such that the midpoint temperature was \(\Delta T_{o}\) higher than expected for a linear temperature distribution. Assuming that the thermal conductivity has a linear dependence on temperature, \(k=k_{o}(1+\alpha T)\), where \(\alpha\) is a constant, develop a relationship to evaluate \(\alpha\) in terms of \(\Delta T_{o}, T_{1}\), and \(T_{2}\).

A storage tank consists of a cylindrical section that has a length and inner diameter of \(L=2 \mathrm{~m}\) and \(D_{i}=1 \mathrm{~m}\), respectively, and two hemispherical end sections. The tank is constructed from 20-mm-thick glass (Pyrex) and is exposed to ambient air for which the temperature is \(300 \mathrm{~K}\) and the convection coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The tank is used to store heated oil, which maintains the inner surface at a temperature of \(400 \mathrm{~K}\). Determine the electrical power that must be supplied to a heater submerged in the oil if the prescribed conditions are to be maintained. Radiation effects may be neglected, and the Pyrex may be assumed to have a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate \(\dot{q}\). As shown in the sketch for Case 1 , the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{o}\). For Case 2 , a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(\mathrm{A}\) and \(\mathrm{B}\). The thermal resistance of the strip is \(R_{t}^{\prime \prime}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=\) \(20 \mathrm{~mm}, \dot{q}=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{o}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for Case 1 on \(T-x\) coordinates. Describe the key features of this distribution. Identify the location of the maximum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Case 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2 ? (d) What is the location of the maximum temperature in the composite wall of Case 2 ? Calculate this temperature.

An air heater consists of a steel tube \((k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), with inner and outer radii of \(r_{1}=13 \mathrm{~mm}\) and \(r_{2}=16\) \(\mathrm{mm}\), respectively, and eight integrally machined longitudinal fins, each of thickness \(t=3 \mathrm{~mm}\). The fins extend to a concentric tube, which is of radius \(r_{3}=\) \(40 \mathrm{~mm}\) and insulated on its outer surface. Water at a temperature \(T_{\infty, i}=90^{\circ} \mathrm{C}\) flows through the inner tube, while air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) flows through the annular region formed by the larger concentric tube. (a) Sketch the equivalent thermal circuit of the heater and relate each thermal resistance to appropriate system parameters. (b) If \(h_{i}=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the heat rate per unit length? (c) Assess the effect of increasing the number of fins \(N\) and/or the fin thickness \(t\) on the heat rate, subject to the constraint that \(N t<50 \mathrm{~mm}\).

A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of \(5^{\circ} \mathrm{C}\). The tube wall has inner and outer radii of 25 and \(75 \mathrm{~mm}\), respectively, and a thermal conductivity of \(10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is \(R_{t, c}^{\prime}=\) \(0.01 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\). The outer surface of the heater is exposed to a fluid with \(T_{\infty}=-10^{\circ} \mathrm{C}\) and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heater power per unit length of tube required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C} .\)
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