Heat Transfer Fins (and Handles)

Abstract

This chapter introduces a lumped capacity approach to steady-state conduction problems using the important practical application of heat transfer fins (and handles). The approach is applied to the coffee/mug problem, this time with half a mug, and the heat transfer into the rim and out to ambient is detailed. Extended surfaces are widely used as either heat transfer fins (devices designed to increase the overall rate of heat transfer between a solid and a fluid) or handles (something you can grab without burning your hand). For fin applications, since the rate of heat transfer by convection (and possibly radiation in parallel) is proportional to the exposed surface area, the basic idea is to increase that exposed surface area. However, because the temperature in the fin decreases with the distance from the wall (x), there is a limit to how far the fin can extend and still be effective. A “perfect” fin is one in which the temperature of the fin is the same everywhere as its base. In a real fin, the average temperature of the fin exposed to the fluid will be less than that, and therefore the fin transfers heat at a lower rate. If the fin is long enough, the temperature at the end of the fin approaches the ambient temperature. That’s a handle you can grab. The approach taken is to develop a finite element approach. First, the fin is considered as a single lumped element, a 1-node model. This approach yields a simple closed-form solution that is surprisingly versatile and captures all the key functional dependencies of the exact solution. Next, a multi-element approach (“few node”) is developed considering the fin broken into three equal elements. This method is easily extended to two or three dimensions. Finally, a detailed methodology of the 1-node model is applied to the case of the rim of a mug of coffee.

Appendices

Appendix 1. Constant-Area Fins: Differential Solution

Figure 7.2 shows an extended surface (fin) that has a base area A and perimeter p. The shape of the surface remains constant in the x direction, perpendicular to the base. The total length of the surface is L, and the material has a thermal conductivity k. The fin is exposed to a fluid at temperature T_∞ with an effective heat transfer coefficient, h. The fin is initially at the fluid temperature, and the base temperature is suddenly changed and maintained at a temperature T₀.

Model Development

A detailed derivation of the governing PDE for this case is developed for a general transient fin case, including lateral conduction. A cross-section of a differential slice of thickness dx at an arbitrary distance x placed at the center of the element is shown in Fig. 7.22. Heat enters the left face (at x – dx/2) by conduction, and splits into two channels, one leaving the surface into the fluid, and the other by conduction (at x + dx/2) into the adjacent slice.

Fig. 7.22

Energy flows in a differential element in an extended surface. The right sketch shows the element in a resistance network, which allows conductive resistance between the center and the surface exposed to the fluid

A verbally stated energy balance applied to the slice, consistent with the assumed directions shown by the arrows of Fig. 7.22, is:

$$ \begin{array}{l}\left(\begin{array}{c}\mathrm{Rate}\kern0.5em \mathrm{of}\kern0.5em \mathrm{Change}\kern0.5em \mathrm{of}\\ {}\mathrm{Energy}\kern0.5em \mathrm{Stored}\kern0.5em \mathrm{in}\kern0.5em \mathrm{slice}\end{array}\right)=\left(\begin{array}{c}\mathrm{Rate}\kern0.5em \mathrm{of}\kern0.5em \mathrm{Conduction}\\ {}\mathrm{in}\mathrm{to}\kern0.5em \mathrm{left}\kern0.5em \mathrm{face}\end{array}\right)-\left(\begin{array}{c}\mathrm{Rate}\kern0.5em \mathrm{of}\kern0.5em \mathrm{Conduction}\\ {}\mathrm{out}\kern0.5em \mathrm{of}\kern0.5em \mathrm{right}\kern0.5em \mathrm{face}\end{array}\right)\\ {}\kern13em -\left(\begin{array}{c}\mathrm{Rate}\kern0.5em \mathrm{of}\kern0.5em \mathrm{Convection}\\ {}\mathrm{out}\kern0.5em \mathrm{of}\kern0.5em \mathrm{sides}\end{array}\right)\end{array} $$

Expressed mathematically:

$$ \left(\rho cAdx\frac{\partial \overline{T}}{\partial t}\right)={\left[-kA\frac{\partial \overline{T}\;}{\partial x}\right]}_{x-dx/2}-{\left[-kA\frac{\partial \overline{T}\;}{\partial x}\right]}_{x+dx/2}- hpdx\left({T}_{surface}-{T}_{\infty}\right) $$

In the transient and conduction terms, the temperature, $\overline{T}$, represents the average temperature of the cross-section. In the convection term, the temperature is the surface temperature.

The conduction at x + Δx (where Δx = – dx/2 or + dx/2) can be related to that at the center (x) using a Taylor expansion:

$$ {\left[-kA\frac{\partial \overline{T}\;}{\partial x}\right]}_{x+\varDelta x}={\left[-kA\frac{\partial \overline{T}\;}{\partial x}\right]}_x+\frac{\partial }{\partial x}{\left[-kA\frac{\partial \overline{T}}{\partial x}\kern0.02em \right]}_x\Delta x+\cdots $$

The net conduction into the differential element is therefore:

$$ \begin{array}{l}{\left[-kA\frac{\partial \overline{T}}{\partial x}\right]}_{x-dx/2}-{\left[-kA\frac{\partial \overline{T}}{\partial x}\right]}_{x+dx/2}=\left({\left[-kA\frac{\partial \overline{T}}{\partial x}\right]}_x+\frac{\partial }{\partial x}\left[-kA\frac{\partial \overline{T}\;}{\partial x}\right]\left(\frac{-dx}{2}\right)\right)\\ {}\kern15em -\left({\left[-kA\frac{\partial \overline{T}}{\partial x}\right]}_x+\frac{\partial }{\partial x}\left[-kA\frac{\partial \overline{T}}{\partial x}\right]\left(\frac{dx}{2}\right)\right)\\ {}{\left[-kA\frac{\partial \overline{T}}{\partial x}\right]}_{x-dx/2}-{\left[-kA\frac{\partial \overline{T}}{\partial x}\right]}_{x+dx/2}=\frac{\partial }{\partial x}{\left[kA\frac{\partial T}{\partial x}\right]}_xdx\end{array} $$

The surface temperature can be related to the average temperature using a voltage divider branch between the center of the fin and its surface:

$$ \frac{\overline{T}-{T}_{\infty }}{R_k+{R}_h}=\frac{T_{surface}-{T}_{\infty }}{R_h} $$

The resistances are in the lateral direction (y and z). Solving for the surface temperature:

$$ {T}_{surface}-{T}_{\infty }=\left(\frac{R_h}{R_h+{R}_k}\right)\left(\overline{T}-{T}_{\infty}\right)=\frac{\overline{T}-{T}_{\infty }}{1+\raisebox{1ex}{${R}_k$}\!\left/ \!\raisebox{-1ex}{${R}_h$}\right.} $$

The conductive resistance is modeled as:

$$ {R}_k=\frac{\left(A/p\right)/2}{k\left(p\Delta x/2\right)} $$

The numerator, (A/p)/2, is half of the characteristic half thickness of the section, which is the average conduction distance from the surface to the center. The area in the denominator is the average between the area at the center (zero) and at the surface. The convective resistance is straightforward:

$$ {R}_h=\frac{1}{hp\varDelta x} $$

The ratio is:

$$ \frac{R_k}{R_h}=\frac{h\left(A/p\right)}{k}\equiv B $$

The governing partial differential equation (energy balance) is therefore:

$$ \rho cA\frac{\partial \overline{T}}{\partial t}=\frac{\partial }{\partial x}{\left[kA\frac{\partial \overline{T}\kern0.15em }{\partial x}\right]}_x-\frac{hp}{1+B}\left(\overline{T}-{T}_{\infty}\right) $$

The factor “B”, a Biot number, accounts for temperature variation in the lateral direction, an effect that is usually neglected (and justifiably so) for fins.

For a rectangular fin, with w ≫ t:

$$ \frac{A}{p}=\frac{tw}{2\left(t+w\right)}=\frac{t}{2\left(1+t/w\right)}\approx \frac{t}{2} $$

Introducing a dimensionless temperature that has a value of zero at the base, and approaches unity as it approaches ambient:

$$ \theta =\frac{\overline{T}-{T}_0}{T_{\infty }-{T}_0} $$

Introducing a dimensionless time and distance with normalization constants to be determined shortly:

$$ \widehat{t}=\frac{t}{\tau}\kern3em \widehat{x}=\frac{x}{L_t} $$

Assuming constant thermal conductivity and area and putting it together:

$$ \frac{\rho cA\left({T}_{\infty }-{T}_0\right)}{\tau}\frac{\partial \theta }{\partial \widehat{t}}=\frac{kA\left({T}_{\infty }-{T}_0\right)}{L_t^2}\frac{\partial^2\theta }{\partial {\widehat{x}}^2}-\frac{hp}{1+B}\left({T}_0+\theta \left({T}_{\infty }-{T}_0\right)-{T}_{\infty}\right) $$

Simplifying and rearranging:

$$ \frac{\rho c{L}_t^2}{k\tau}\frac{\partial \theta }{\partial \widehat{t}}=\frac{\partial^2\theta }{\partial {\widehat{x}}^2}+\frac{hp{L}_t^2}{\left(1+B\right)kA}\left(1-\theta \right) $$

The time constant and characteristic dimension can be defined by setting the coefficients to unity. For the convection term:

$$ {L}_t=\sqrt{\frac{\left(1+B\right)kA}{hp}} $$

For the transient term:

$$ \tau =\frac{\rho c{L}_t^2}{k} $$

The governing partial differential equation becomes:

$$ \frac{\partial \theta }{\partial \widehat{t}}=\frac{\partial^2\theta }{\partial {\widehat{x}}^2}-\theta +1 $$

Since the governing equation is first order with respect to time, a single initial condition must be specified, for problems that are in fact transient. For example, in the problem statement, the fin is initially at the fluid temperature (T_∞), and the base temperature is suddenly changed and maintained at T₀. Therefore, the initial condition (in dimensions) is:

$$ T\left(x,0\right)={T}_{\infty } $$

In dimensionless variables:

$$ \theta \left(\widehat{x},0\right)=1 $$

The governing equation is second order with respect to x, and therefore two conditions are required. One condition is that the base temperature is specified:

$$ T\left(0,t\right)={T}_0 $$

In dimensionless variables:

$$ \theta \left(0,\widehat{t}\right)=0 $$

The second condition on x is trickier. The precise statement of it is that the rate of heat conduction to the end (at x = L) must equal the rate of convection away from the end:

$$ -kA{\left.\frac{\partial T}{\partial x}\right|}_{x=L}=hA\left({T}_{\left(L,t\right)}-{T}_{\infty}\right) $$

In terms of dimensionless variables, with a dimensionless fin length $\widehat{L}\equiv L/{L}_t$:

$$ \frac{-k\left({T}_{\infty }-{T}_0\right)}{L_t}{\left.\frac{\partial \theta }{\partial \widehat{x}}\right|}_{\widehat{x}=\widehat{L}}=h\left({T}_{\infty }-{T}_0\right){\theta}_{\left({\tilde{L}}_t,\widehat{t}\right)} $$

$$ {\left.\frac{\partial \theta }{\partial \widehat{x}}\right|}_{\widehat{x}=\widehat{L}}=-\left(\frac{h{L}_t}{k}\right){\theta}_{\left(\widehat{L},\widehat{t}\right)} $$

This type of boundary condition, a convection boundary condition, involves a gradient of the variable and the variable itself. It dramatically increases the algebra required to solve the problem and results in a complicated solution.

There is an alternative approach to the second boundary condition which will be developed here, namely the corrected length method. In it, as sketched in Fig. 7.3, the heat transfer out the end of the fin is considered by making the fin slightly longer than the actual fin, with an insulated end.

The length of the fin used is called a “corrected” length, L_c, expressed as the original length plus the added length:

$$ {L}_c=L+\Delta L $$

The area added onto the sides by the added length (the perimeter times the added length ΔL) is the same as the end of the fin. That is:

$$ p\Delta L=A $$

The corrected length is therefore:

$$ {L}_c=L+\frac{A}{p} $$

The boundary condition on the end of the extended fin is determined by an energy balance at the end, in which heat conducted to the end from the left is set to the heat convected away, which is set to zero:

$$ -kA{\left.\frac{\partial T}{\partial x}\right|}_{x={L}_c}=0 $$

Introducing dimensionless variables and cancelling factors results in a simpler boundary condition:

$$ {\left.\frac{\partial \theta }{\partial \widehat{x}}\right|}_{\widehat{x}={L}_c/{L}_t}=0 $$

To restate the result of the model development, the governing equation and initial and boundary conditions are:

$$ \begin{array}{l}\mathrm{Governing}\kern0.5em \mathrm{Equation}:\frac{\partial \theta }{\partial \widehat{t}}=\frac{\partial^2\theta }{\partial {\widehat{x}}^2}-\theta +1\\ {}\mathrm{Initial}\kern0.5em \mathrm{Condition}:\theta \left(\widehat{x},0\right)=1\\ {}\mathrm{Boundary}\kern0.5em \mathrm{Condition}\kern0.5em \mathrm{at}\kern0.5em \mathrm{base}:\theta \left(0,\widehat{t}\right)=0\\ {}\mathrm{Boundary}\kern0.5em \mathrm{Condition}\kern0.5em \mathrm{at}\kern0.5em \mathrm{end}:{\left.\frac{\partial \theta }{\partial \widehat{x}}\right|}_{\widehat{x}={L}_c/{L}_t}=0\end{array} $$

The governing equation is a linear partial differential equation with three terms (transient, conduction, convection) and no parameters (they are “buried” in the dimensionless variables). The second boundary condition does contain a parameter (its location, namely the dimensionless thermal length of the fin). The use of dimensionless variables has reduced the number of parameters in the system from seven (A, p, L, k, h, T₀, T_∞) to one.

As is often the case in heat transfer applications, it is not so much the temperature that is sought, but the heat transfer rate. For a fin, the rate at which heat transfer enters the base of the fin is:

$$ {q}_{fin}=-kA{\left.\frac{\partial T}{\partial x}\right|}_{x=0} $$

In dimensionless variables:

$$ {q}_{fin}=-\frac{kA\left({T}_{\infty }-{T}_0\right)}{L_t}{\left.\frac{\partial \theta }{\partial \widehat{x}}\right|}_{x=0} $$

Introducing the thermal length:

$$ {q}_{fin}={\left.\frac{\partial \theta }{\partial \widehat{x}}\right|}_{x=0}\sqrt{\frac{hpkA}{1+B}}\left({T}_0-{T}_{\infty}\right) $$

The derivative of dimensionless temperature with respect to dimensionless distance is all that is required to determine the heat transfer rate.

1.1 Model Implementation: Analytical Solution

The general solution can be expressed as the sum of a transient and steady-state solutions:

$$ \theta ={\theta}_{transient}\left(\widehat{x},\widehat{t}\right)+{\theta}_{ss}\left(\widehat{x}\right) $$

The transient solution involves all three terms and requires a lot of work to develop a complete solution that satisfies the initial and boundary conditions. The full solution is developed here, mainly to demonstrate the basic method of solving linear partial differential equations. The steady-state solution is much easier to obtain, and if the problem can be effectively modeled as being at steady-state, it can be used without bothering with the transient case.

In practice, however, the transient response of a heat fin may be important, for example, for electronic equipment that is cycled on and off. During the initial start-up period, the rate of heat transfer into a fin can start high, as heat is going to change the energy stored, and gradually decay to its steady-state response and dissipate heat to ambient. A good design might take advantage of that behavior.

1.2 Transient Solution

The governing equation for the transient solution is:

$$ \frac{\partial {\theta}_{transient}}{\partial \widehat{t}}=\frac{\partial^2{\theta}_{transient}}{\partial {\widehat{x}}^2}-{\theta}_{transient} $$

Note that the term “1” in the governing equation will be grouped into the steady state solution. Using the separation of variables technique, the solution is assumed to be expressible as two separable functions: one depending on position, and the other on time:

$$ {\theta}_{transient}=\mathrm{T}T\left(\widehat{t}\right)X\left(\widehat{x}\right) $$

Insertion into the governing equation:

$$ X\frac{dTT}{d\widehat{t}}=\mathrm{T}T\left(\frac{d^2X}{d{\widehat{x}}^2}-X\right) $$

Dividing through by XT:

$$ \frac{1}{\mathrm{T}}\frac{dT}{d\widehat{t}}=\left(\frac{1}{X}\frac{d^2X}{d{\widehat{x}}^2}-1\right)=-{\lambda}^2 $$

In this expression, the term on the left is a function of time only, the term in the middle is a function of position only, and it is recognized that that can only be possible if they are both equal to at most a constant, called a separation constant, and given the symbol λ, whose value will be determined by application of the initial and boundary conditions. The negative sign avoids complex numbers and squaring makes for a cleaner final solution (not obvious why at this point).

The time-dependent equation is:

$$ \frac{dT}{d\widehat{t}}+{\lambda}^2T\mathrm{T}=0 $$

The general solution is:

$$ \mathrm{T}T={c}_0{e}^{-{\lambda}^2\widehat{t}} $$

The space-dependent equation is:

$$ \frac{d^2X}{d{\widehat{x}}^2}+{\lambda}^2X=0 $$

The general solution is:

$$ X={c}_1 \sin \lambda \widehat{x}+{c}_2 \cos \lambda \widehat{x} $$

The full general solution, in terms of undetermined coefficients A = c₀c₁ and B = c₀c₂ is:

$$ {\theta}_{transient}={e}^{-{\lambda}^2\widehat{t}}\left(A \sin \lambda \widehat{x}+B \cos \lambda \widehat{x}\right) $$

The values of A, B, and λ are determined by application of the initial and boundary conditions. First, applying the boundary condition at the base of the fin, $\theta \left(0,\widehat{t}\right)=0$

$$ 0={e}^{-{\lambda}^2\widehat{t}}\left(A \sin \lambda 0+B \cos \lambda 0\right)=B{e}^{-{\lambda}^2\widehat{t}} $$

The only way this condition can be satisfied is if the coefficient B is set to zero. The general solution, after application of boundary condition at the wall, has been reduced to:

$$ {\theta}_{transient}=A{e}^{-{\lambda}^2\widehat{t}} \sin \lambda \widehat{x} $$

Application of the second boundary condition:

$$ {\left.\frac{\partial \theta }{\partial \widehat{x}}\right|}_{\widehat{x}=\widehat{L}={L}_c/{L}_t}=0=A\lambda {e}^{-{\lambda}^2\widehat{t}} \cos \lambda \widehat{L} $$

Where a dimensionless length (the thermal length of the fin) has been defined as:

$$ \widehat{L}=\frac{L_c}{L_t} $$

The only way this boundary condition can be met (without setting A or λ to zero, which would make satisfying the subsequent initial condition impossible) is to set the separation constant to specific values, namely:

$$ \cos \lambda \widehat{L}=0 $$

This constraint is true for the following infinite number of discrete values of the separation constant, namely:

$$ \lambda \widehat{L}=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\dots, \frac{\left(2n+1\right)\pi }{2},\dots $$

These “quantized” values of l are called eigenvalues.

The solution can now be expressed as the sum of all possible solutions (prior to satisfying the initial condition) for these discrete values of λ, with a different coefficient (A) possible for each term:

$$ {\theta}_{transient}={A}_0{e}^{-{\left(\frac{\pi }{2\widehat{L}}\right)}^2\widehat{t}} \sin \frac{\pi \widehat{x}}{2\widehat{L}}+{A}_1{e}^{-{\left(\frac{3\pi }{2\widehat{L}}\right)}^2\widehat{t}} \sin \frac{3\pi \widehat{x}}{2\widehat{L}}+\cdots +{A}_n{e}^{-{\left(\frac{\left(2\pi +1\right)}{2\widehat{L}}\right)}^2\widehat{t}} \sin \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}+\cdots $$

Expressed in terms of an infinite series:

$$ {\theta}_{transient}={\displaystyle \sum_{n=0}^{\infty }{A}_n{e}^{-\left(\frac{2\pi +1}{2{\widehat{L}}^2}\right)\widehat{t}} \sin \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}} $$

Notice that the transient solution decays to zero as time goes to infinity. Last, but not least, the initial condition must be satisfied:

$$ \begin{array}{l}\theta \left(\widehat{x},0\right)=1\\ {}1={\displaystyle \sum_{n=1}^{\infty }{A}_n \sin \frac{\left(2n+1\right)\pi \widehat{x}}{\widehat{L}}}\end{array} $$

To determine the coefficients A_n, the final step in this tortured affair, multiply both sides by a sin term with an integer “m” as a frequency, and integrate over the range of x:

$$ {\displaystyle \underset{\widehat{x}=0}{\overset{\widehat{L}}{\int }} \sin \frac{\left(2m+1\right)\pi \widehat{x}}{2\widehat{L}}d\widehat{x}}={\displaystyle \sum_{n=1}^{\infty}\left[{\displaystyle \underset{\widehat{x}=0}{\overset{\widehat{L}}{\int }}{A}_n \sin \frac{\left(2m+1\right)\pi \widehat{x}}{2\widehat{L}} \sin \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}d\widehat{x}}\right]} $$

The integral on the right is zero (sketch two sinusoids of different frequencies, and look at their areas) when m is not equal to n. That is, for every value of n, all but one of the terms in the infinite series are identically zero. Therefore:

$$ {\displaystyle \underset{\widehat{x}=0}{\overset{\widehat{L}}{\int }} \sin \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}d\widehat{x}}={\displaystyle \underset{\widehat{x}=0}{\overset{\widehat{L}}{\int }}{A}_n{\left( \sin \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}\right)}^2d\widehat{x}} $$

Performing the integration on the left hand side:

$$ {\displaystyle \underset{\widehat{x}=0}{\overset{\widehat{L}}{\int }} \sin \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}d\widehat{x}}=\frac{2\widehat{L}}{\left(2n+1\right)\pi }{\left|- \cos \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}\right|}_0^{\widehat{L}}=\frac{-2\widehat{L}}{\left(2n+1\right)\pi}\left( \cos \frac{\left(2n+1\right)\pi }{2}-1\right)=\frac{2\widehat{L}}{\left(2n+1\right)\pi } $$

The right hand side, with the help of a trigonometric identity for sin²x = (1 − cos x)/2, evaluates to a single number (noting that the second term involves a cosine integrated over a full cycle, and its integral is thus zero):

$$ {\displaystyle \underset{\widehat{x}=0}{\overset{\widehat{L}}{\int }}{A}_n{\left(\kern-0.25em , \sin \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}\right)}^2d\widehat{x}}={A}_n{\displaystyle \underset{\widehat{x}=0}{\overset{\widehat{L}}{\int }}\frac{1}{2}}\left(1- \cos \frac{2\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}\right)dx=\frac{A_n\widehat{L}}{2} $$

The coefficient is therefore determined:

$$ {A}_n=\frac{4}{\left(2n+1\right)\pi } $$

The complete transient solution is thereby attained and expressed as an infinite series:

$$ {\theta}_{transient}={\displaystyle \sum_{n=0}^{\infty}\frac{4}{\left(2n+1\right)\pi }{e}^{-{\left(\frac{\left(2n+1\right)\pi }{2\widehat{L}}\right)}^2\widehat{t}} \sin \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}} $$

Expressing this result for the first few terms:

$$ {\theta}_{transient}=\frac{4}{\pi}\left[\frac{1}{1}{e}^{-{\left(\frac{\pi }{2\widehat{L}}\right)}^2\widehat{t}} \sin \frac{\pi x}{2\widehat{L}}+\frac{1}{3}{e}^{-{\left(\frac{3\pi }{2\widehat{L}}\right)}^2\widehat{t}} \sin \frac{3\pi x}{2\widehat{L}}+\frac{1}{5}{e}^{-{\left(\frac{5\pi }{2\widehat{L}}\right)}^2\widehat{t}} \sin \frac{5\pi x}{2\widehat{L}}+\cdots \right] $$

Notice how each term is multiplied by an exponential factor with a negative exponent that grows in absolute value with the number of the term. Except for very early time, where t ≪ 1, the first term dominates the solution.

To determine the heat transfer rate associated with the transient solution, the derivative with respect to x is taken and evaluated at x = 0:

$$ {\left.\frac{\partial {\theta}_{transient}}{\partial \widehat{x}}\right|}_{\widehat{x}=0}={\displaystyle \sum_{n=0}^{\infty }2{e}^{-{\left(\frac{\left(2n+1\right)\pi }{2\widehat{L}}\right)}^2\widehat{t}}}=2{e}^{-{\left(\frac{\pi }{2\widehat{L}}\right)}^2\widehat{t}}+2{e}^{-{\left(\frac{3\pi }{2\widehat{L}}\right)}^2\widehat{t}}+2{e}^{-{\left(\frac{5\pi }{2\widehat{L}}\right)}^2\widehat{t}}+\cdots $$

The heat transfer rate associated with the transient solution is therefore:

$$ {q}_{fin, transient}=\sqrt{\frac{hpkA}{1+B}}\left({T}_0-{T}_{\infty}\right){\displaystyle \sum_{n=0}^{\infty }2{e}^{-{\left(\frac{\left(2n+1\right)\pi }{2\widehat{L}}\right)}^2\widehat{t}}} $$

1.3 Steady-State Solution

The steady-state solution is much easier, especially with the corrected length boundary condition. The governing equation (a linear ordinary differential equation with constant coefficients) and boundary conditions are:

$$ \begin{array}{l}\frac{d^2{\theta}_{ss}}{d{\widehat{x}}^2}-{\theta}_{ss}=-1\\ {}{\left.\frac{\partial \theta }{\partial \widehat{x}}\right|}_{\widehat{x}=\widehat{L}}=0\end{array} $$

The general solution comprises a homogeneous and particular solution:

$$ {\theta}_{ss}={\theta}_{ss,H}+{\theta}_{ss,P} $$

The homogeneous solution is that with the right hand side set to zero, and the solution is:

$$ {\theta}_{ss,H}={c}_1{e}^{-\widehat{x}}+{c}_2{e}^{\widehat{x}} $$

The particular solution is:

$$ {\theta}_{ss,P}=K $$

Substitution of these expressions into the governing equation yields K = 1, and the general solution is therefore:

$$ {\theta}_{ss}={c}_1{e}^{-\widehat{x}}+{c}_2{e}^{\widehat{x}}+1 $$

Application of the boundary condition at the base (at x = 0):

$$ 0={c}_1+{c}_2+1 $$

Application of the second boundary condition (at x = L_c):

$$ 0=-{c}_1{e}^{-\widehat{L}}+{c}_2{e}^{\widehat{L}} $$

Combining them:

$$ 0=\left({c}_2+1\right){e}^{-\widehat{L}}+{c}_2{e}^{\widehat{L}} $$

Solving for c₂:

$$ {c}_2=\frac{-{e}^{-\widehat{L}}}{e^{-\widehat{L}}+{e}^{\widehat{L}}} $$

And

$$ {c}_1=\frac{e^{-\widehat{L}}}{e^{-\widehat{L}}+{e}^{\widehat{L}}}-1=\frac{e^{-\widehat{L}}-\left({e}^{-\widehat{L}}+{e}^{\widehat{L}}\right)}{e^{-\widehat{L}}+{e}^{\widehat{L}}}=\frac{-{e}^{\widehat{L}}}{e^{-\widehat{L}}+{e}^{\widehat{L}}} $$

The steady-state solution is therefore:

$$ {\theta}_{ss}=1-\left(\frac{e^{\widehat{L}}{e}^{-\widehat{x}}+{e}^{-\widehat{L}}{e}^{\widehat{x}}}{e^{-\widehat{L}}+{e}^{\widehat{L}}}\right)=1-\left(\frac{e^{\left(\widehat{L}-\widehat{x}\right)}+{e}^{-\left(\widehat{L}-\widehat{x}\right)}}{e^{-\widehat{L}}+{e}^{\widehat{L}}}\right) $$

This equation can be expressed in a hyperbolic cosine form:

$$ \begin{array}{l} \cosh (y)=\frac{e^y+{e}^{-y}}{2}\\ {} \sinh (y)=\frac{e^y-{e}^{-y}}{2}\\ {} \tanh (y)=\frac{ \sinh (y)}{ \cosh (y)}=\frac{e^y-{e}^{-y}}{e^y+{e}^{-y}}\\ {}{\theta}_{ss}=1-\frac{ \cosh \left(\widehat{L}-\widehat{x}\right)}{ \cosh \left(\widehat{L}\right)}\end{array} $$

Returning to dimensions:

$$ \frac{T_{ss}-{T}_0}{T_{\infty }-{T}_0}=1-\frac{ \cosh \left(\frac{L_c-x}{L_t}\right)}{ \cosh \left(\frac{L_c}{L_t}\right)} $$

The derivative evaluated at x = 0 is:

$$ {\left.\frac{\partial {\theta}_{ss}}{\partial \widehat{x}}\right|}_{\widehat{x}=0}=-{c}_1+{c}_2=\frac{e^{\widehat{L}}-{e}^{-\widehat{L}}}{e^{\widehat{L}}+{e}^{-\widehat{L}}}=\frac{ \sinh \left(\widehat{L}\right)}{ \cosh \left(\widehat{L}\right)}= \tanh \left(\widehat{L}\right) $$

The steady-state heat transfer rate, entering the definition of $\widehat{L}$, is therefore:

$$ {q}_{fin,ss}= \tanh \left(\frac{L_c}{L_t}\right)\sqrt{\frac{hpkA}{1+B}}\left({T}_0-{T}_{\infty}\right) $$

The full solution is therefore mercifully upon us:

$$ \theta =\left[{\displaystyle \sum_{n=0}^{\infty}\frac{4}{\left(2n+1\right)\pi }{e}^{-{\left(\frac{\left(2n+1\right)\pi }{2\widehat{L}}\right)}^2\widehat{t}} \sin \frac{\left(2n+1\right)\pi \widehat{x}}{2\widehat{L}}}\right]+\left[1-\frac{ \cosh \left(\widehat{L}-\widehat{x}\right)}{ \cosh \left(\widehat{L}\right)}\right] $$

where the first bracketed term is the transient solution, and the second is the steady-state solution. The heat transfer rate is:

$$ {q}_{fin}=\sqrt{\frac{hpkA}{1+B}}\left({T}_0-{T}_{\infty}\right)\left[ \tanh \left(\frac{L_c}{L_t}\right)+{\displaystyle \sum_{n=0}^{\infty }2{e}^{-{\left(\frac{\left(2n+1\right)\pi }{2\widehat{L}}\right)}^2\widehat{t}}}\right] $$

where a transverse Biot number is defined as $B=\frac{h\left(A/p\right)}{k}$ _, the corrected length is ${L}_c=L+\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$p$}\right.$ _, and the thermal length is ${L}_t=\sqrt{\frac{\left(1+B\right)kA}{hp}}$ _.

Personal comments: I consider this really elegant mathematical solution to the transient rim problem to be “Good Mathematics” and “Bad Engineering.” Engineering is about addressing a human need. To make this value judgment about whether the method is good or bad engineering requires me to revisit, and perhaps modify, the problem statement, an iteration of the modeling process. If the objective is to merely find the heat transfer rate, then the 1-node solution is really simple mathematically, and yields the same physics as the PDE solution.

The big problem I have with the PDE solution is not that it requires a sophisticated mathematical excursion. If that’s what it takes, then so be it. My problem is that the solution ties me to some tight restrictions. The base wall temperature must be fixed in time. The initial condition must be that the fin is entirely at ambient, etc. The 1-node model gives all that. An extension to an n-node solution requires some more work, but that numerical method can be applied to much more general situations quite easily, once the technique is mastered. For example, suppose a fin is designed to dissipate heat from a motor that turns on and off intermittently, but has a clear maximum time it is on. A fin might be designed that takes advantage of high initial heat rates (where heat goes in part to changing the energy stored, and the rest transferred to the fluid).

In short, the 1-node model teaches the essential physics and yields key design parameters. The multi-node numerical method yields more detail and is arguably more quantitatively accurate than the 1-node model. On the other hand, there is so much uncertainty in convection coefficient values that accuracy can be misleading.

Heat transfer modeling is indeed an art.

Appendix 2. Variable Area Fins: Finite Difference Numerical Solution

Figure 7.23 shows a circumferential fin attached to a pipe, an example of an extended surface (fin) that has a base area A_base and base perimeter p_base and a cross-sectional area A(x) and perimeter p(x) that vary in the distance perpendicular to the base (x, radial position). The total length of the fin is L, and the material has a thermal conductivity k. The fin is exposed to a fluid at temperature T_∞ with an effective heat transfer coefficient, h. The fin is initially at the fluid temperature, and the base temperature is suddenly changed and maintained at a temperature T₀.

Fig. 7.23

Circumferential fin of length L, thickness t place on a pipe of outer diameter D. The left sketch is an end view of the pipe; the right view is a cross-section from a side view of the pipe

Model Development

The governing differential equation was derived in the previous appendix by application of the energy balance to a differential element:

$$ \rho cA\frac{\partial \overline{T}}{\partial t}=\frac{\partial }{\partial x}{\left[kA\frac{\partial \overline{T}}{\partial x}\right]}_x-\frac{hp}{1+B}\left(\overline{T}-{T}_{\infty}\right) $$

With an initial condition and two boundary conditions (at the base and end of the fin). The difference between this case and the constant-area is that the area cannot be taken out of the conduction derivative term. Assuming constant thermal conductivity, using the chain rule for the conduction term and rearranging:

$$ \frac{\partial T}{\partial t}=\alpha \frac{\partial^2T}{\partial {x}^2}+\alpha \frac{1}{A}\frac{\partial A}{\partial x}\frac{\partial T}{\partial x}-\frac{hp}{\rho cA\left(1+B\right)}\left(\overline{T}-{T}_{\infty}\right) $$

In a finite difference explicit numerical solution, the time derivative is expressed as a forward finite difference approximation, and the spatial derivatives as central differences. The boundary conditions are expressed as forward for x = 0, and backward for x = L second-order finite differences. The details are not developed here.

Appendix 3. Iterative Solutions of Nonlinear Algebraic Equations

Nonlinear algebraic equations arise in many problems of engineering interest. They cannot be solved analytically, in general. Rather, some form of iterative procedure is required. In an iterative procedure, a guess to the solution is first made. Then a recursion formula based on the governing equations is used to make another guess. Then another guess is made, then another and another… Once the initial guess is made, the algorithm takes over for all subsequent guesses (called an iteration). If the method is successful, successive iterations eventually “converge” to the exact solution. It never reaches it, but a CONVERGENCE CRITERION can be defined that means “Close Enough!” The method is said to DIVERGE if the absolute value grows to infinity. Some systems exhibit CHAOS.

Some general points…

• Many different algorithms have been “invented.” Heat transfer problems are generally solvable using the Method of Successive Substitution. It is easy to implement and usually works. It can easily be extended to systems of equations. Newton’s method is another, and will be introduced for a single nonlinear algebraic equation and outlined for systems of nonlinear equations. It is much harder to implement with systems (with more than one unknown) and is occasionally necessary in heat transfer problems.

• There is no guarantee that a given method will work, that is, find the desired solution to the problem.

• If an iterative procedure converges, it does so to a correct solution of the system of equations. However, that solution may make no physical sense (i.e., a negative absolute temperature).

• If an implementation diverges, it might converge for another initial guess.

3.1 Single Nonlinear Algebraic Equations

Solution of a single nonlinear algebraic equation is the natural place to start. The methods can be extended to systems of equations (with more than one unknown). The quadratic equation will be used to demonstrate how to implement the methods and to compare the response on a system that has an analytical solution. In addition, if the goal is not to solve the equation, but to study CHAOS, iterative techniques on the quadratic equation are a great place to start.

Quadratic equation:

$$ a{x}^2+bx+c=0 $$

The quadratic equation is about the simplest nonlinear algebraic equation there is. As a nonlinear equation, it has the rare distinction of having an exact solution. It has two solutions, its two roots x₁ and x₂:

$$ {x}_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\kern1em {x}_2=\frac{-b-\sqrt{b^2-4ac}}{2a} $$

Keep in mind that in what follows, the objective is to learn new techniques that can be implemented to solve more difficult nonlinear equations, not to solve the quadratic equation…

3.1.1 Method of Successive Substitution: x = g(x)

The method of successive substitution is easy to implement, and it works often. It can be used to solve for many systems of linear or nonlinear equations as well.

STEP 1: Rearrange equation:

The governing nonlinear equation is rearranged into the recursion form:

$$ {x}_{k+1}=g\left({x}_k\right) $$

where the subscript “k” refers to the iteration number. There are an infinite number of ways this rearrangement can be done. For example, the quadratic equation written in its conventional form is:

$$ a{x}^2+bx+c=0 $$

Three equivalent forms of the quadratic equation (with subscripts showing the iteration number) are:

$$ \begin{array}{l}\kern-6.2em \mathrm{Form}\ 1:\kern0.5em {x}_{k+1}=\frac{-1}{b}\left[c+a{x}_k^2\right]\kern3.7em \left(\mathrm{isolating}\ \mathrm{t}\mathrm{he}\ {2}^{\mathrm{nd}}\kern0.5em \mathrm{t}\mathrm{erm}\right)\\ {}\kern-6.2em \mathrm{Form}\ 2:\kern0.5em {x}_{k+1}=\pm \sqrt{\frac{-1}{a}\left[c+b{x}_k\right]}\kern2.3em \left(\mathrm{isolating}\ \mathrm{t}\mathrm{he}\ {1}^{\mathrm{st}}\kern0.5em \mathrm{t}\mathrm{erm}\right)\\ {}\kern-6.2em \mathrm{Form}\ 3:\kern0.5em {x}_{k+1}=a{x}_k^2+\left(b+1\right){x}_k+c\kern1em \left(\mathrm{adding}\kern0.5em `\mathrm{x}'\kern0.5em \mathrm{t}\mathrm{o}\kern0.5em \mathrm{both}\kern0.5em \mathrm{sides}\right)\end{array} $$

STEP 2: Make an initial guess, x_0, and ITERATE:

Examples:

A spreadsheet was used to investigate the how the method works. Three cases are shown in Figs. 7.24, 7.25, 7.26, 7.27, 7.28:

Fig. 7.24

Iteration trajectory for Case 7.1

Fig. 7.25

Iteration trajectory for Case 7.2

Fig. 7.26

First 20 iterations for Case 7.3

Fig. 7.27

First 50 iterations for Case 7.3

Fig. 7.28

First 200 iterations for Case 7.3

Case 7.1 shows a situation in which both roots are found, one of them by two forms. A plot of the “trajectory” is shown.

Case 7.1

Three forms converge to different solutions

Case 7.2 shows one form converging, one form “crashing” immediately (because the square root of a negative number was asked for), and one form diverging.

Case 7.2

Two forms diverge

Case 7.3 shows CHAOS. The iteration trajectory neither converges nor diverges. Sensitivity to initial guess is shown by plotting two trajectories with nearly identical initial guesses.

Case 7.3

CHAOS

3.1.2 Newton’s Method: f(x) = 0

Newton’s method is an iterative method that uses the rate of change of the function to help make an “intelligent” choice for the next iteration. The recursion relation can be derived with the help of a graphical analysis. The following sketch shows a general function f(x) in the vicinity of its solution, where it equals 0.

The first “guess” is x_o and the function value is f(x_o) at that point. The slope of the function: ${\left.\frac{df}{dx}\right|}_{x_0}$ is drawn at that point. The next iteration value chosen, x₁, is taken to be where that slope intersects the x-axis.

$$ {\left.\frac{df}{dx}\right|}_{x_0}=\frac{f\left({x}_0\right)-0}{x_0-{x}_1} $$

Solving this equation for x₁:

$$ {x}_1={x}_0-\frac{f\left({x}_0\right)}{{\left.\frac{df}{dx}\right|}_{x_0}} $$

In general, the recursion relation is:

$$ {x}_{k+1}={x}_k-\frac{f\left({x}_k\right)}{{\left.\frac{df}{dx}\right|}_{x_k}} $$

To implement the method, an initial “guess” is made, and the recursion relation is used in an iterative fashion.

3.1.3 Newton’s Method for Systems of Nonlinear Algebraic Equations

In a resistance network, if the thermal resistances are expressed as functions of temperature (which are unknown), for example, with radiation coefficients, then the system of algebraic equations that results from discretizing space will be nonlinear. In most heat transfer applications, the relatively easy to implement Method of Successive Substitutions (i.e., Gauss–Seidel method applied to 3-node fin) works well. When it fails, Newton’s method (which is harder to implement, but more likely to succeed) can be implemented. The system of n equations in n unknown temperatures can be written in the form (one for each node of a space-discretized network):

$$ \begin{array}{l}{f}_1\left({T}_{1,}{T}_2,\dots, {T}_n\right)=0\\ {}{f}_2\left({T}_{1,}{T}_2,\dots, {T}_n\right)=0\\ {}\cdots \\ {}{f}_n\left({T}_{1,}{T}_2,\dots, {T}_n\right)=0\end{array} $$

The method involves making an initial guess for all unknown temperatures, and then making an improved guess based on how the function is changing. The initial guess is given a superscript “k” and the improved guess the superscript “k + 1”. The total derivative of each nodal equation consists of the contribution from each unknown variable:

$$ \begin{array}{l}d{f}_1={\left.\frac{\partial {f}_1}{\partial {T}_1}\right|}^{(k)}\left({T}_1^{\left(k+1\right)}-{T}_1^{(k)}\right)+{\left.\frac{\partial {f}_1}{\partial {T}_2}\right|}^{(k)}\left({T}_2^{\left(k+1\right)}-{T}_2^{(k)}\right)+\cdots +{\left.\frac{\partial {f}_1}{\partial {T}_n}\right|}^{(k)}\left({T}_n^{\left(k+1\right)}-{T}_n^{(k)}\right)\\ {}\kern1.4em ={f}_1^{\left(k+1\right)}-{f}_1^{(k)}\approx 0-{f}_1^{(k)}\\ {}d{f}_2={\left.\frac{\partial {f}_2}{\partial {T}_1}\right|}^{(k)}\left({T}_1^{\left(k+1\right)}-{T}_1^{(k)}\right)+{\left.\frac{\partial {f}_2}{\partial {T}_2}\right|}^{(k)}\left({T}_2^{\left(k+1\right)}-{T}_2^{(k)}\right)+\cdots +{\left.\frac{\partial {f}_2}{\partial {T}_n}\right|}^{(k)}\left({T}_n^{\left(k+1\right)}-{T}_n^{(k)}\right)\\ {}\kern1.4em ={f}_2^{\left(k+1\right)}-{f}_2^{(k)}\approx 0-{f}_2^{(k)}\\ {}\cdots \\ {}d{f}_n={\left.\frac{\partial {f}_n}{\partial {T}_1}\right|}^{(k)}\left({T}_1^{\left(k+1\right)}-{T}_1^{(k)}\right)+{\left.\frac{\partial {f}_n}{\partial {T}_2}\right|}^{(k)}\left({T}_2^{\left(k+1\right)}-{T}_2^{(k)}\right)+\cdots +{\left.\frac{\partial {f}_n}{\partial {T}_n}\right|}^{(k)}\left({T}_n^{\left(k+1\right)}-{T}_n^{(k)}\right)\\ {}\kern1.4em ={f}_n^{\left(k+1\right)}-{f}_n^{(k)}\approx 0-{f}_n^{(k)}\end{array} $$

The right hand side represents an approximation to the total change. The zero reflects that the function evaluated at the next iteration is supposed to be zero, but it should be closer to zero than the previous iteration. Since all the functions and their derivatives (possibly by numerical approximation) can be evaluated at the “k” iteration, this represents a system of linear algebraic equations that can be solved for the temperatures at the next iteration.

Workshop 7.1: Triangular Fins

Compare the steady-state performance of triangular fins with rectangular fins with the same base thickness (t) and the same total mass of material. Assume that the depth of the fins (not shown) are large compared to the thickness. Think about it before analyzing it.

Level 1: Develop a 1-node analysis for the triangular fin (place the node at the centroid), and develop an expression for the ratio of the heat flow from the rectangular fin to that of the corresponding triangular fin. Is there a key dimensionless parameter to plot against?

Level 2: Develop a few-node model for both in spreadsheets, and compare the results with the 1-node model.

Level 3: Develop an “n” node model for both using a program code, and compare with the 1-node model.

Workshop 7.2: Circumferential Rectangular Fin

Analyze the steady-state performance of a fin constructed of a rectangular plate placed on the outer surface of a pipe, a common design. The fin is exposed to an ambient fluid (T_∞) with an effective heat transfer coefficient h. The spacing between fins (Δx) is shown but not used in this workshop in which a single fin is isolated.

• Conduct a 1-node analysis for a thermally short model.

• Develop a 1-node thermally long model.

• Develop a few-node model (identify and take advantage of symmetry).