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We found that the thermal resistance of a cylinder was \(R_{t_{\mathrm{cyl}}}=\) \((1 / 2 \pi k l) \ln \left(r_{o} / r_{i}\right)\). If \(r_{o}=r_{i}+\delta\), show that the thermal resistance of a thin-walled cylinder \(\left(\delta \ll r_{i}\right)\) can be approximated by that for a slab of thickness \(\delta\). Thus, \(R_{t_{\text {thin }}}=\delta /\left(k A_{i}\right)\), where \(A_{i}=2 \pi r_{i} l\) is the inside surface area of the cylinder. How much error is introduced by this approximation if \(\delta / r_{i}=0.2 ?\) [Hint: Use a Taylor series.]

Short Answer

Expert verified
The error is approximately 4%.

Step by step solution

01

Write the Original Formula

The formula for the thermal resistance of a cylinder is given by:\[ R_{t_{\mathrm{cyl}}} = \frac{1}{2 \pi k l} \ln \left(\frac{r_{o}}{r_{i}}\right) \]Where \( r_{o} = r_{i} + \delta \).
02

Substituting for Outer Radius

Substitute \( r_{o} = r_{i} + \delta \) into the equation:\[ R_{t_{\mathrm{cyl}}} = \frac{1}{2 \pi k l} \ln \left(\frac{r_{i} + \delta}{r_{i}}\right) \]
03

Simplify the Logarithm

Rewrite the logarithm using the properties of logarithms:\[ \ln \left(\frac{r_{i} + \delta}{r_{i}}\right) = \ln \left(1 + \frac{\delta}{r_{i}}\right) \]
04

Apply Taylor Series Approximation

For small \( \frac{\delta}{r_{i}} \), use the Taylor series expansion for \( \ln(1 + x) \approx x \) when \( x \) is small:\[ \ln \left(1 + \frac{\delta}{r_{i}}\right) \approx \frac{\delta}{r_{i}} \]
05

Substitute Taylor Expansion into Resistance Formula

Replace the logarithm in the thermal resistance formula with the Taylor expansion:\[ R_{t_{\mathrm{cyl}}} \approx \frac{1}{2 \pi k l} \cdot \frac{\delta}{r_{i}} \]
06

Calculate the Approximate Resistance of the Thin-Walled Cylinder

The thermal resistance of a thin-walled cylinder is:\[ R_{t_{\text{thin}}} = \frac{\delta}{k A_{i}} \]where \( A_{i} = 2 \pi r_{i} l \). Substitute \( A_{i} \) into the formula:\[ R_{t_{\text{thin}}} = \frac{\delta}{k \cdot 2 \pi r_{i} l} \]
07

Calculate the Error of Approximation

Calculate the difference between the exact and approximate thermal resistances for \( \frac{\delta}{r_{i}} = 0.2 \):\[ \text{Error} = \left| \left(1 - \left(1 + \frac{\delta}{r_{i}}\right)^{-1} \right) \right| \approx \frac{\delta}{r_{i}}^2 \]With \( \left( \frac{\delta}{r_{i}} \right)^2 = 0.04 \).
08

Conclude the Error Analysis

The error introduced by this approximation when \( \frac{\delta}{r_{i}} = 0.2 \) is approximately 4%.

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Key Concepts

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

Taylor Series Approximation
The Taylor series is a mathematical tool used for approximating functions. When we want to estimate a complex function at a certain point, the Taylor series can offer a simpler, polynomial form of the original function.
In the context of thermal resistance of a cylinder, we use the Taylor series to approximate the natural logarithm function. Specifically, for small values of a variable \( x \), the natural logarithm \( \ln(1 + x) \) can be simplified as \( x \).
This simplification is key in making the formula for the thermal resistance of a thin-walled cylinder more manageable.
  • It allows us to express \( \ln\left(1 + \frac{\delta}{r_i}\right) \) in a simpler form for situations where \( \frac{\delta}{r_i} \) is a small fraction.
  • This results in an approximate expression that is much easier to work with, offering a quick estimation while maintaining a reasonable level of accuracy.
  • However, it is important to remember that approximations introduce some error, so the Taylor series is best applied when \( \frac{\delta}{r_i} \) is sufficiently small.
The approximation becomes less accurate when the fraction is larger, which is why we also calculate the possible error involved.
Thin-Walled Cylinder Approximation
Approximating a thick cylinder as a thin-walled one simplifies the calculation of thermal resistance.
When we assume that the wall thickness \( \delta \) is much smaller than the inner radius \( r_i \), the cylinder's resistance approximately equals that of a slab.
The formula \( R_{t_{\text{thin}}} = \frac{\delta}{k A_i} \) emerges from this approximation, where
\( A_i = 2 \pi r_i l \) is the inside surface area of the cylinder.
  • Here, the tiny increment \( \delta \) in the radius becomes the effective thickness of an equivalent slab.
  • This method is particularly useful in engineering and design as it offers an easier way to handle complex shapes.
  • Such approximations reduce the mathematical complexity and focus on achieving a practical or workable solution.
Though it is an approximation, it remains accurate enough for many engineering applications, especially when precise heat transfer calculations are not critical.
Thermal Conductivity
Thermal conductivity is a crucial property in understanding how materials conduct heat. It is defined as the rate at which heat passes through a material when there's a difference in temperature. In the context of calculating thermal resistance, the thermal conductivity \( k \) plays a central role.
The relationship is inversely proportional, meaning the higher the thermal conductivity, the lower the thermal resistance, and vice versa.
  • In materials with high thermal conductivity, heat flows more easily, resulting in smaller thermal resistance.
  • Conversely, materials with low thermal conductivity create higher resistance to heat flow.
  • This property is essential in selecting materials for thermal insulation or conduction purposes.
In the case of a cylinder or a thin-walled cylinder, thermal conductivity assists in determining how well the material can transfer heat along its length and through its wall.
When examining applications and substance efficiency, knowing thermal conductivity helps predict and control heat exchange effectively.

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Most popular questions from this chapter

A very small diameter, electrically insulated heating wire runs down the center of a \(7.5 \mathrm{~mm}\) diameter rod of type 304 stainless steel. The outside is cooled by natural convection \((\bar{h}=6.7\) \(\left.\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\right)\) in room air at \(22^{\circ} \mathrm{C}\). If the wire releases \(12 \mathrm{~W} / \mathrm{m}\), plot \(T_{\text {rod }}\) vs. radial position in the rod and give the outside temperature of the rod. (Stop and consider carefully the boundary conditions for this problem.)

An outside pipe is insulated and we measure its temperature with a thermocouple. The pipe serves as an electrical resistance heater, and \(\dot{q}\) is known from resistance and current measurements. The inside of the pipe is cooled by the flow of liquid with a known bulk temperature. Evaluate the heat transfer coefficient, \(\bar{h}\), in terms of known information. The pipe dimensions and properties are known. [Hint: Remember that \(\bar{h}\) is not known and we cannot use a boundary condition of the third kind at the inner wall to get \(T(r) .]\)

\(800 \mathrm{~W} / \mathrm{m}^{3}\) of heat is generated within a \(10 \mathrm{~cm}\) diameter nickelsteel sphere for which \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The environment is at \(20^{\circ} \mathrm{C}\) and there is a natural convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) around the outside of the sphere. What is its center temperature at the steady state? \(\left[21.37^{\circ} \mathrm{C}\right.\). \(]\)

Heat is generated at \(54,000 \mathrm{~W} / \mathrm{m}^{3}\) in a \(0.16 \mathrm{~m}\) diameter sphere. The sphere is cooled by natural convection with fluid at \(0{ }^{\circ} \mathrm{C}\), and \(\bar{h}=\left[2+6\left(T_{\text {surface }}-T_{\infty}\right)^{1 / 4}\right] \mathrm{W} / \mathrm{m}^{2} \mathrm{~K}, k_{\text {sphere }}=9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Find the surface temperature and center temperature of the sphere.

Prove that if \(k\) varies linearly with \(T\) in a slab, and if heat transfer is one-dimensional and steady, then \(q\) may be evaluated precisely using \(k\) evaluated at the mean temperature in the slab.

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