Fin optimization

Fins can come in a variety of shapes, e.g., rectangular and circular with constant cross section, and annular and triangular with variable cross section. For a given fin shape, fin material, and convection conditions, there exists an optimized design which transfers the maximum amount of heat for a given mass of the fin. The methodology to finding this optimum design is presented here.

The simplest case to examine is the rectangular fin, as illustrated in Fig.1.8. The fin is taken to be long in and out of the paper (i.e., W ≫ L, where W is the width). Since fin mass is proportional to the profile area A_P (= bL) times W , the optimization problem can be stated as finding the thickness b and length L which maximize q/W for a given A_P. Possible shapes of the fin, for a fixed profile area A_P, are illustrated (in a very exaggerated manner) in Fig.1.9.

To further simplify the problem the fin assumed to have an adiabatic tip. The heat transfer is q =p

hP kA_c(T_b− T∞) tanh N with

N² = hP L² kAc

For a long fin (W ≫ b), P ≈ 2W and A^C = bW . Thus q^′ = q

W =√

2bhk (T_b− T∞) tanh N (1.78)

where now

N² = 2hL² kb

The length L can be eliminated using A_P = bL. The formula for N becomes N² = 2hA²_P

kb³ and by inverting this equation,

b = 2hA²_P kN²

1/3

(1.79) and replacing this into the formula for the heat transfer gives

q^′ = 4h²k A_P
1/3

(T_b− T∞) N^−1/3 tanh N (1.80) Observe that we are taking the profile area Ap to be fixed. Likewise, the properties h, k, and T_b− T∞ are assumed to be constants. The above formula therefore indicates that, for these given constraints, there is an optimum value of N which will maximize the fin heat transfer rate.

Obtaining the optimum N is relatively easy at this point. From the previous equation, the heat transfer is functionally related to N via

f (N ) = N^−1/3 tanh N and a maximum q^′ implies that

df dN = 0 or, for this particular case,

cosh N sinh N − 3N = 0 (1.81)

This is a nonlinear equation for N and can be solved using standard numerical techniques. The solution is

N_opt = 1.419 = 2hA²_P kb³_opt

!1/2

(1.82) Once A_P is fixed, the above formula can be used to obtain b_opt, and L follows from L = A_P/b_opt. And by replacing N_opt into Eq. (1.80) we get

q_opt^′ = 4h²kA_P N_opt

1/3

(T_b− T∞) tanh N_opt

= 1.256 h²kA_P
1/3

(T_b− T∞) (1.83)

The same process can be applied to triangular fins (which will require information on Bessel functions that will be presented in the following chapter). Under the assumption of b²/L² ≪ 1 – which is consistent with the 1–D heat transfer approximation – the following result can be obtained:

q^′_opt= 1.422 h²kA_P
1/3

(T_b− T∞) (1.84)

Recognize that for a fixed A_P (and consequently, fin mass), an optimized triangular fin can transfer more heat than an optimized rectangular fin.

In designing a fin, one would always want to choose the optimum value of b for a fixed q^′ and AP. Fins are often placed in arrays (such as on the head of a motorcycle engine). To maximize the heat transfer coefficient, the spacing between the fins should be somewhat greater than twice the boundary layer thickness. In selecting a material for the fin, it follows from Eqs. (1.83) and (1.84) that

A_P ∝ 1 h²k

 q_opt^′ T_b− T∞

3

(1.85) For a given required heat transfer rate and fixed ‘environmental’ parameters (h, T_b, T_∞), the opti-mum profile area will be inversely proportional to the fin thermal conductivity. Consequently, the fin mass (which is ρAPW ) will be proportional to ρ/k, where ρ is the density of the fin material.

Aluminum is often a good choice for fins – since it has relatively high k and low ρ. It is also relatively cheap.

Exercises

1. Consider again the derivation of the heat conduction equation, Eq. (1.10). Say that mass transfer occurs at the boundaries and convects an energy flux ρue into/out of the system, where u is the velocity vector. Using the fact that ∂e = c∂T , and taking the mass of the system to be constant, derive the more general form of the energy equation:

1 α

 ∂T

∂t + u · (∇T )



= ∇²T +q^′′′

k

Note: this will involve use of the divergence theorem as outlined in Sec. 1. Also, use continuity (system mass = constant) along with the vector identity

∇ · (ρue) = e∇ · ρu + ρu · (∇e)

2. A 1 cm thick copper wire (kc = 400 W/m/K) conducts a large electrical current, which results in a heat dissipation within the wire of 60 W per meter of length. The wire is covered with insulation having k_ins = 1.2 W/m/K, and the outer surface of the insulation is cooled by convection, with h = 30 W/m²· K and T∞ = 25^◦C.

(a) Formally ‘pose’ the heat conduction problem for the wire and the insulation regions.

What are the boundary conditions at the interface between the wire and the insulation?

(b) Using a resistance analogy for the insulation, determine, via energy balance principles (not via solution of the DEs) the surface temperature of the wire (i.e., T at r = 1 cm) as a function of the insulation outer radius r_ins. Identify an appropriate Biot number for

the wire. Based on the magnitude of this Biot number, would you think it is necessary to solve the conduction equation for the wire to obtain the wire centerline temperature?

(c) Plot the wire surface temperature vs. r_ins. Physically interpret the minimum in T_s that occurs for a critical value of r_ins. Analytically, what is this critical thickness equal to?

3. A rectangular fin of length L, thickness t and width W is mounted on a surface that is maintained at a constant temperature T_b. The fin is exposed to a uniform source of radiant flux q^′′_R, of which a fraction α is absorbed by the fin. Heat transfer from the fin occurs by convection to an ambient at T_∞ (which is less than T_b), which is characterized by a heat transfer coefficient h. Assuming that 1) the tip of the fin is adiabatic, and 2) thermal emission from the fin is negligible in comparison to convection, determine formulas for the temperature distribution in the fin and the total heat transfer from the base. If thermal emission is not negligible, how is the problem complicated? Devise a way of approximating the effect of thermal emission on the fin by ‘linearizing’ the thermal emission rate law.

4. Shown that, for an arbitrary fin, Eqs. (1.64) and (1.65) are completely equivalent. Hint: use the DE.

5. A plane wall of thickness L has an insulated boundary at x = 0. Radiation is absorbed within the wall, which results in a heat generation characterized by κq^′′₀e^−κ(L−x), where κ is the radiative absorption coefficient of the wall material and q^′′₀ is the incident radiative flux on the wall surface. The outer surface of the wall is cooled by convection to an ambient temperature T_∞. Formulate and solve the heat conduction equation for the temperature distribution within the wall. What happens to the distribution when κL ≫ 1 (i.e., the wall becomes highly absorbing)? Show that this case becomes equivalent to a problem in which radiation absorption ‘disappears’ from the DE (as a heat generation function), yet ‘reappears’

in the problem in the boundary condition at x = L.

Mathematica solution

Given here is the Mathematica solution to the spherical coordinate heat transfer problem, presented beginning with Eq. (1.50). A more detailed explanation of the application of Mathematica to solution of ODEs will be given in Ch. 2.

The formulas given below are in nondimensional form. The inner boundary condition – which is typically posed for spherical coordinates as finite T at r → 0 – needs to be given in a more precise mathematical form. This is done by imposing

r² dT dr  

r→0

= 0

which is essentially the same as saying that the total heat transfer rate at the center is zero. The extra r² is needed to keep the solution finite at the origin. A functional form of the solution (i.e.,

T (r, Bi, a)) which can be used to generate plots for numerical values of the parameters is defined in line [6]. Pay attention to the replacement rules when such definitions are made: the quantities x, bix, and ax are simply dummy variables – problems would arise if the arguments were written as r, bi, and a since these symbols are already used in the symbolic form of the solution.

In[1]:=de=D[r^2D[t[r],r],r]/r^2+E^(-a r)==0;

bc1=Limit[x^2 D[t[x],x],x->0]==0;

bc2=-t’[1]==bi t[1];

soln=DSolve[{de,bc1,bc2},t[r],r]

Out[4]={{t[r] ->

((2 + 2*a + a^2 2*bi a*bi

-2*E^a + 2*bi*E^a)/(E^a*a^3*bi)) \ + (-(1/a^2) - 2/(a^3*r))/E^(-(-a*r)) + 2/(a^3*r)}}

In[5]:=Simplify[%]

Out[5]={{t[r] ->

-(1/a^3*((a^2 - a*(-2 + bi) + 2*(-1 + bi)*(-1 + E^a))/

(E^a*bi) - 2/r +

(2 + a*r)/(r/E^(-(-(-a*r))))))}}

In[6]:=tfunc[x_,bix_,ax_]:=t[r]/.soln[[1]]

/.r->x/.bi->bix/.a->ax

In[7]:=Plot[{tfunc[r,10,.1],tfunc[r,10,1], tfunc[r,10,10]},{r,0,1}]

0 0.2 0.4 0.6 0.8 1

0 0.025 0.050 0.075 0.100 0.125 0.150 0.175

a = 10

a = 0.1

a = 1

T

r

Advanced 1–D Analytical Methods

2.1 Introduction

In a wide variety of heat conduction problems the flow of heat occurs primarily in one direction.

Chapter 1 outlined the application of a steady, 1–D conduction analysis to relatively ‘simple’

configurations, such as the plane wall, the infinite–length cylinder and the sphere (with heat flow in the r direction), and the fin with a uniform cross sectional area. On the other hand, fins of nonuniform cross section present a more difficult analytical problem. The difficulty arises simply from the fact that the ordinary differential equations which describe the heat flow in these situations do not have ‘common’ analytical solutions.

When I took this course as a graduate student, and when I have taught it in the past, several lectures were devoted to deriving analytical solution to ODEs that are typical of general, 1–D extended surface heat transfer. Such derivations typically begin with a power series representation of the solution, which, when manipulated into the ODE, can be used to define the functional form of the solution. Perhaps you recall such methods in your differential equations courses – one (of many) analytical methods which were painful to comprehend and easy to forget.

It is, in my opinion, no longer necessary to present such derivations in an advanced conduction class. These derivations are inherently mathematical in nature, and do not reveal any of the underlying physics to the problem (such as a grasp of the temperature profile in a fin). In addition, the advanced symbolic mathematics packages (i.e., Mathematica) which are available today allow us to completely bypass the painful details to deriving the solution to an ODE – and cut to the chase.

This chapter will examine the solutions to ordinary differential equations that characterize 1–D heat flow in triangular and annular fins (which are forms of Bessel’s equation), and introduce the use of Mathematica to derive and manipulate the solutions.

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