# Analytical Solutions for Steady Heat Transfer in Longitudinal Fins with Temperature-Dependent Properties

Authors: Partner L. Ndlovu; Raseelo J. Moitsheki
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

Explicit analytical expressions for the temperature profile, fin efficiency, and heat flux in a longitudinal fin are derived. Here, thermal conductivity and heat transfer coefficient depend on the temperature. The differential transform method (DTM) is employed to construct the analytical (series) solutions. Thermal conductivity is considered to be given by the power law in one case and by the linear function of temperature in the other, whereas heat transfer coefficient is only given by the power law. The analytical solutions constructed by the DTM agree very well with the exact solutions even when both the thermal conductivity and the heat transfer coefficient are given by the power law. The analytical solutions are obtained for the problems which cannot be solved exactly. The effects of some physical parameters such as the thermogeometric fin parameter and thermal conductivity gradient on temperature distribution are illustrated and explained.


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## Body

## 1. Introduction

Fins are surfaces that extend from a hot object (body) to increase the rate of heat transfer to the surrounding fluid. In particular, fins are used extensively in various industrial applications such as the cooling of computer processors, air conditioning, and oil carrying pipe lines. A well-documented review of heat transfer in extended surfaces is presented by Kraus et al. [1]. The problems on heat transfer particularly in fins continue to be of scientific interest. These problems are modeled by highly nonlinear differential equations which are difficult to solve exactly. However, Moitsheki et al. [2–4] have attempted to construct exact solutions for the steady state problems arising in heat flow through fins. A number of techniques, for example, Lie symmetry analysis [2], He’s variational iteration method [5], Adomain decomposition methods [6], homotopy perturbation methods [7], homotopy analysis methods [8], methods of successive approximations [9], and other approximation methods [10] have been used to determine solutions of the nonlinear differential equations describing heat transfer in fins.Recently, the solutions of the nonlinear ordinary differential equations (ODEs) arising in extended surface heat transfer have been constructed using the DTM [11–19]. The DTM is an analytical method based on the Taylor series expansion and was first introduced by Zhou [20] in 1986. The DTM approximates the exact solution by a polynomial, and previous studies have shown that it is an efficient means of solving nonlinear problems or systems with varying parameters [21]. Furthermore, DTM is a computational inexpensive tool for obtaining analytical solution, and it generalizes the Taylor method to problems involving procedures such as fractional derivative (see e.g., [22–24]). Also, this method converges rapidly (see e.g., [25]).Models arising in heat transfer through fins may contain temperature-dependent properties such as thermal conductivity and heat transfer coefficient. The dependency of thermal conductivity and heat transfer coefficient on temperature renders such problems highly nonlinear and difficult to solve, particularly exactly. Thermal conductivity may be modeled for many engineering applications by the power law and by linear dependency on temperature. On the other hand, heat transfer coefficient can be expressed as a power law for which values of the exponent represent different phenomena (see e.g., [26]).In this paper, the DTM is employed to determine the analytical solutions to the nonlinear boundary value problem describing heat transfer in longitudinal fins of rectangular, exponential, and convex parabolic profiles. Both thermal conductivity and heat transfer coefficient are temperature dependent. We adopt the terminology exact solutions to refer to solutions given in terms of fundamental expressions such as logarithmic, trigonometric, and exponential. However, analytical solutions will be series solutions and in particular, those constructed using the DTM. The mathematical modelling of the problem under consideration is described in Section2. A brief discussion on the fundamentals of the DTM is be provided in Section 3. The comparison of the exact and analytical solutions constructed by DTM is given in Section 4. In Section 5, we provide analytical solutions for the heat transfer in longitudinal fins of various profiles. Here, heat transfer coefficient is given by the power law, and we consider two cases of the thermal conductivity, namely, the power law and the linear function of temperature. Furthermore, we describe the fin efficiency and the heat flux in Section 6. Some exciting results are discussed in Section 7. Lastly, the concluding remarks are provided in Section 8.


## 2. Mathematical Models

We consider a longitudinal one dimensional fin of cross-sectional areaAc. The perimeter of the fin is denoted by P and its length by L. The fin is attached to a fixed prime surface of temperature Tb and extends to an ambient fluid of temperature Ta. The fin thickness at the prime surface is given by δb and its profile is given by F(X). Based on the one dimensional heat conduction, the energy balance equation is then given by (see e.g., [1])
(1)AcddX(δb2F(X)K(T)dTdX)=PH(T)(T-Ta)0≤X≤L,
where K and H are nonuniform temperature-dependent thermal conductivity and heat transfer coefficients, respectively, T is the temperature distribution, F(X) is the fin profile, and X is the space variable. The length of the fin is measured from the tip to the prime surface as shown in Figure 1. Assuming that the fin tip is adiabatic (insulated) and the base temperature is kept constant, then the boundary conditions are given by
(2)T(L)=Tb,dTdX|X=0=0.
Introducing the following dimensionless variables (see e.g., [1]):
(3)x=XL,θ=T-TaTb-Ta,h=Hhb,k=Kka,M2=PhbL2Acka,f(x)=δb2F(X),
with ka being defined as the thermal conductivity at the ambient temperature and hb as the heat transfer at the prime surface (fin base), reduces (1) to
(4)ddx[f(x)k(θ)dθdx]-M2θh(θ)=0,0≤x≤1.
Here θ is the dimensionless temperature, x is the dimensionless space variable, f(x) is the dimensionless fin profile, k is the dimensionless thermal conductivity, h is the dimensionless heat transfer coefficient, and M is the thermogeometric fin parameter. The dimensionless boundary conditions then become
(5)θ(1)=1,attheprimesurface(6)dθdx|x=0=0,atthefintip.
We assume that the heat transfer coefficient is given by the power law
(7)H(T)=hb(T-TaTb-Ta)n,
where the exponent n is a real constant. In fact the values of n may vary as between -6.6 and 5. However, in most practical applications, it lies between -3 and 3 [27]. In dimensionless variables, we have h(θ)=θn. We consider the two distinct cases of the thermal conductivity as follows: (a)
the power law(8)K(T)=ka(T-TaTb-Ta)m,
with m being a constant and(b)
the linear function(9)K(T)=ka[1+γ(T-Ta)].The dimensionless thermal conductivity given by the power law and the linear function of temperature is k(θ)=θm and k(θ)=1+βθ, respectively. Here the thermal conductivity gradient is β=γ(Tb-Ta). Furthermore, we consider a various fin profiles including the longitudinal rectangular f(x)=1, the longitudinal convex parabolic f(x)=x, and the exponential profile f(x)=eax with a being the constant (see also [16]).Figure 1
Schematic representation of a longitudinal fin of an unspecified profile.


## 3. Fundamentals of the Differential Transform Method

In this section, the basic idea underlying the DTM is briefly introduced. Lety(t) be an analytic function in a domain 𝒟. The Taylor series expansion function of y(t) with the center located at t=tj is given by [20]
(10)y(t)=∑κ=0∞‍(t-tj)κκ![dκy(t)dtκ]t=tj,∀t∈𝒟.The particular case of (10) when tj=0 is referred to as the Maclaurin series expansion of y(t) and is expressed as
(11)y(t)=∑κ=0∞‍tκκ![dκy(t)dtκ]t=0,∀t∈𝒟.
The differential transform of y(t) is defined as follows:
(12)Y(t)=∑κ=0∞‍ℋκκ![dκy(t)dtκ]t=0,
where y(t) is the original analytic function and Y(t) is the transformed function. The differential spectrum of Y(t) is confined within the interval t∈[0,ℋ], where ℋ is a constant. From (11) and (12), the differential inverse transform of Y(t) is defined as follows:
(13)y(t)=∑κ=0∞‍(tℋ)κY(κ),
and if y(t) is expressed by a finite series, then
(14)y(t)=∑κ=0r‍(tℋ)κY(κ).Some of the useful mathematical operations performed by the differential transform method are listed in Table1.Table 1
Fundamental operations of the differential transform method.
Original function
Transformed function
y
(
t
)
=
x
(
t
)
±
z
(
t
)
Y
(
t
)
=
X
(
t
)
±
Z
(
t
)
y
(
t
)
=
α
x
(
t
)
Y
(
t
)
=
α
X
(
t
)
y
(
t
)
=
d
y
(
t
)
d
t
Y
(
t
)
=
(
κ
+
1
)
Y
(
κ
+
1
)
y
(
t
)
=
d
2
y
(
t
)
d
t
2
Y
(
t
)
=
(
κ
+
1
)
(
κ
+
2
)
Y
(
κ
+
2
)
y
(
t
)
=
d
s
y
(
t
)
d
t
s
Y
(
t
)
=
(
κ
+
1
)
(
κ
+
2
)
⋯
(
κ
+
s
)
Y
(
κ
+
s
)
y
(
t
)
=
x
(
t
)
z
(
t
)
Y
(
t
)
=
∑
i
=
0
κ
X
(
i
)
Z
(
κ
-
i
)
y
(
t
)
=
1
Y
(
t
)
=
δ
(
κ
)
y
(
t
)
=
t
Y
(
t
)
=
δ
(
κ
-
1
)
y
(
t
)
=
t
s
Y
(
t
)
=
δ
(
κ
-
s
)
y
(
t
)
=
exp
(
λ
t
)
Y
(
t
)
=
λ
κ
κ
!
y
(
t
)
=
(
1
+
t
)
s
Y
(
t
)
=
s
(
s
-
1
)
⋯
(
s
-
κ
+
1
)
κ
!
y
(
t
)
=
sin
(
ω
t
+
α
)
Y
(
t
)
=
ω
κ
κ
!
sin
(
π
κ
2
!
+
α
)
y
(
t
)
=
cos
(
ω
t
+
α
)
Y
(
t
)
=
ω
κ
κ
!
cos
(
π
κ
2
!
+
α
)The delta functionδ(κ-s) is given by
(15)δ(κ-s)={1ifκ=s,0ifκ≠s.


## 4. Comparison of Exact and Analytical Solutions

In this section, we consider a model describing temperature distribution in a longitudinal rectangular fin with both thermal conductivity and heat transfer coefficient being functions of temperature given by the power law (see e.g., [2]). The exact solution of (4) when both the power laws are given by the same exponent is given by [2]
(16)θ(x)=[cosh(Mn+1x)cosh(Mn+1)]1/(n+1).
We use this exact solution as a benchmark or validation of the DTM. The effectiveness of the DTM is determined by comparing the exact and the analytical solutions. We compare the results for the cases n=1 and n=2 with fixed values of M.


### 4.1. Casen=1

Applying the DTM to (4) with the power law thermal conductivity, f(x)=1 (rectangular profile) and given ℋ=1, one obtains the following recurrence relation:
(17)∑i=0κ‍[M2Θ(i)(κ-i+1)(κ-i+2)Θ(κ-i+2)∑i=0κ‍+(i+1)Θ(i+1)(κ-i+1)Θ(κ-i+1)∑i=0κ‍-M2Θ(i)Θ(κ-i)]=0.
Exerting the transformation to the boundary condition (6) at a point x=0,
(18)Θ(1)=0.
The other boundary conditions are considered as follows:
(19)Θ(0)=c,
where c is a constant. Equation (17) is an iterative formula of constructing the power series solution as follows:
(20)Θ(2)=M2c2(21)Θ(3)=0(22)Θ(4)=-M4c24(23)Θ(5)=0(24)Θ(6)=19M6c720(25)Θ(7)=0(26)Θ(8)=-559M8c40320(27)Θ(9)=0(28)Θ(10)=29161M10c3628800(29)Θ(11)=0(30)Θ(12)=-2368081M12c479001600(31)Θ(13)=0(32)Θ(14)=276580459M14c87178291200(33)Θ(15)=0⋮These terms may be taken as far as desired. Substituting (18) to (32) into (13), we obtain the following analytical solution:
(34)θ(x)=c+M2c2x2-M4c24x4+19M6c720x6θ(x)=-559M8c40320x8+29161M10c3628800x10-2368081M12c479001600x12θ(x)=+276580459M14c87178291200x14+⋯To obtain the value ofc, we substitute the boundary condition (5) into (34) at the point x=1. Thus, we have
(35)θ(1)=c+M2c2-M4c24+19M6c720-559M8c40320θ(x)=+29161M10c3628800-2368081M12c479001600θ(x)=+276580459M14c87178291200+⋯=1.
We omit presenting the tedious process of finding c. However, one may obtain the expression for θ(x) upon substituting the obtained value of c into (34).


### 4.2. Casen=2

Following a similar approach given in Section4.1 and given n=2, one obtains the analytical solution
(36)θ(x)=c+M2c2x2-M4c8x4+23M6c240x6θ(x)=-1069M8c13440x8+9643M10c134400x10-1211729M12c17740800x12θ(x)=+217994167M14c3228825600x14+⋯.
The constant c may be obtained using the boundary condition at the fin base. The comparison of the DTM and the exact solutions are reflected in Tables 2 and 3 for different values of n. Furthermore, the comparison of the exact and the analytical solutions is depicted in Figures 2(a) and 2(b).Table 2
Results of the DTM and exact solutions forn=1, M=0.7.
x
DTM
Exact
Error
0
0.808093014
0.80809644
0.000003426
0.1
0.810072036
0.81007547
0.000003434
0.2
0.815999549
0.81600301
0.000003459
0.3
0.825847770
0.82585127
0.000003500
0.4
0.839573173
0.83957673
0.000003559
0.5
0.857120287
0.85712392
0.000003633
0.6
0.878426299
0.87843002
0.000003722
0.7
0.903426003
0.90342982
0.000003814
0.8
0.932056648
0.93206047
0.000003820
0.9
0.964262558
0.96426582
0.000003263
1.0
1.000000000
1.00000000
0.000000000Table 3
Results of the DTM and Exact Solutions forn=2, M=0.5.
x
DTM
Exact
Error
0
0.894109126
0.894109793
0.000000665
0.1
0.895226066
0.895226732
0.000000666
0.2
0.898568581
0.898569249
0.000000668
0.3
0.904112232
0.904112905
0.000000672
0.4
0.911817796
0.911818474
0.000000678
0.5
0.921633352
0.921634038
0.000000685
0.6
0.933496900
0.933497594
0.000000694
0.7
0.947339244
0.947339946
0.000000702
0.8
0.963086933
0.963087627
0.000000694
0.9
0.980665073
0.980665659
0.000000586
1.0
1.000000000
1.000000000
0.000000000Comparison of analytical and exact solutions. In (a)n=1, M=0.7 and (b) n=2, M=0.5.
(a)
(b)


## 4.1. Casen=1

Applying the DTM to (4) with the power law thermal conductivity, f(x)=1 (rectangular profile) and given ℋ=1, one obtains the following recurrence relation:
(17)∑i=0κ‍[M2Θ(i)(κ-i+1)(κ-i+2)Θ(κ-i+2)∑i=0κ‍+(i+1)Θ(i+1)(κ-i+1)Θ(κ-i+1)∑i=0κ‍-M2Θ(i)Θ(κ-i)]=0.
Exerting the transformation to the boundary condition (6) at a point x=0,
(18)Θ(1)=0.
The other boundary conditions are considered as follows:
(19)Θ(0)=c,
where c is a constant. Equation (17) is an iterative formula of constructing the power series solution as follows:
(20)Θ(2)=M2c2(21)Θ(3)=0(22)Θ(4)=-M4c24(23)Θ(5)=0(24)Θ(6)=19M6c720(25)Θ(7)=0(26)Θ(8)=-559M8c40320(27)Θ(9)=0(28)Θ(10)=29161M10c3628800(29)Θ(11)=0(30)Θ(12)=-2368081M12c479001600(31)Θ(13)=0(32)Θ(14)=276580459M14c87178291200(33)Θ(15)=0⋮These terms may be taken as far as desired. Substituting (18) to (32) into (13), we obtain the following analytical solution:
(34)θ(x)=c+M2c2x2-M4c24x4+19M6c720x6θ(x)=-559M8c40320x8+29161M10c3628800x10-2368081M12c479001600x12θ(x)=+276580459M14c87178291200x14+⋯To obtain the value ofc, we substitute the boundary condition (5) into (34) at the point x=1. Thus, we have
(35)θ(1)=c+M2c2-M4c24+19M6c720-559M8c40320θ(x)=+29161M10c3628800-2368081M12c479001600θ(x)=+276580459M14c87178291200+⋯=1.
We omit presenting the tedious process of finding c. However, one may obtain the expression for θ(x) upon substituting the obtained value of c into (34).


## 4.2. Casen=2

Following a similar approach given in Section4.1 and given n=2, one obtains the analytical solution
(36)θ(x)=c+M2c2x2-M4c8x4+23M6c240x6θ(x)=-1069M8c13440x8+9643M10c134400x10-1211729M12c17740800x12θ(x)=+217994167M14c3228825600x14+⋯.
The constant c may be obtained using the boundary condition at the fin base. The comparison of the DTM and the exact solutions are reflected in Tables 2 and 3 for different values of n. Furthermore, the comparison of the exact and the analytical solutions is depicted in Figures 2(a) and 2(b).Table 2
Results of the DTM and exact solutions forn=1, M=0.7.
x
DTM
Exact
Error
0
0.808093014
0.80809644
0.000003426
0.1
0.810072036
0.81007547
0.000003434
0.2
0.815999549
0.81600301
0.000003459
0.3
0.825847770
0.82585127
0.000003500
0.4
0.839573173
0.83957673
0.000003559
0.5
0.857120287
0.85712392
0.000003633
0.6
0.878426299
0.87843002
0.000003722
0.7
0.903426003
0.90342982
0.000003814
0.8
0.932056648
0.93206047
0.000003820
0.9
0.964262558
0.96426582
0.000003263
1.0
1.000000000
1.00000000
0.000000000Table 3
Results of the DTM and Exact Solutions forn=2, M=0.5.
x
DTM
Exact
Error
0
0.894109126
0.894109793
0.000000665
0.1
0.895226066
0.895226732
0.000000666
0.2
0.898568581
0.898569249
0.000000668
0.3
0.904112232
0.904112905
0.000000672
0.4
0.911817796
0.911818474
0.000000678
0.5
0.921633352
0.921634038
0.000000685
0.6
0.933496900
0.933497594
0.000000694
0.7
0.947339244
0.947339946
0.000000702
0.8
0.963086933
0.963087627
0.000000694
0.9
0.980665073
0.980665659
0.000000586
1.0
1.000000000
1.000000000
0.000000000Comparison of analytical and exact solutions. In (a)n=1, M=0.7 and (b) n=2, M=0.5.
(a)
(b)


## 5. Analytical Solutions

It is well known that exact solutions for ODEs such as (4) exist only when thermal conductivity and the term containing heat transfer coefficient are connected by differentiation (or simply if the ODE such as (4) is linearizable) [4]. In this section we determine the analytical solutions for the nonlinearizable (4), firstly when thermal conductivity is given by the power law and secondly as a linear function of temperature. In both cases and throughout this paper, the heat transfer coefficient is assumed to be a power law function of temperature. These assumptions of the thermal properties are physical realistic. We have noticed that DTM runs into difficulty when the exponent of the power law of the thermal conductivity is given by fractional values and also when the function f(x) is given in terms of fractional power law. One may follow Moradi and Ahmadikia [16] by introducing a new variable to deal with fractional powers of f(x), and on the other hand, it is possible to remove the fractional exponent of the heat transfer coefficient by fundamental laws of exponent and binomial expansion.Proposition 1.
A nonlinear ODE such as (4) may admit the DTM solution if f(x) is a constant or exponential function. However, if f(x) is a power law, then such an equation admits a DTM solution if the product,
(37)f(x)·f′(x)=α,
holds. Here α is a real constant.Proof.
Introducing the new variableξ=f(x), it follows from chain rule that (4) becomes
(38)αdξdxddξ[k(θ)dθdξ]-M2θn+1=0.
The most general solution of condition (37) is
(39)f(x)=(2αx+γ)1/2,
where γ is an integration constant.Example 2.
(a)  Iff(x)=xσ, then ξ=xσ transforms (4) into
(40)ddξ[k(θ)dθdξ]-4M2ξθn+1=0,
only if σ=1/2.
This example implies that the DTM may only be applicable to problems describing heat transfer in fins with convex parabolic profile. In the next subsections, we present analytical solutions for (4) with various functions f(x) and k(θ).


### 5.1. The Exponential Profile and Power Law Thermal Conductivity

In this section, we present solutions for equation describing heat transfer in a fin with exponential profile and power law thermal conductivity and heat transfer coefficient. That is, given (4) with f(x)=eσx, and both heat transfer coefficient and thermal conductivity being power law functions of temperature, we construct analytical solutions. In our analysis, we consider n=2 and 3 indicating the fin subject to nucleate boiling and radiation into free space at zero absolute temperature, respectively. Firstly, given n=3 and m=2 and applying the DTM, one obtains the following recurrence revelation:
(41)∑i=0κ‍∑l=0κ-i∑p=0κ-i-l‍[σpp!Θ(l)Θ(i)(κ-i-l-p+1)‍∑i=0κ‍∑i=0κ‍∑i=0κ‍×(κ-i-l-p+2)Θ(κ-i-l-p+2)∑i=0κ‍∑i=0κ‍∑i=0κ‍+2σpp!Θ(l)(i+1)Θ(i+1)∑i=0κ‍∑i=0κ‍∑i=0κ‍×(κ-i-l-p+1)Θ(κ-i-l-p+1)∑i=0κ‍∑i=0κ‍∑i=0κ‍+2σσpp!Θ(l)Θ(i)(κ-i-l-p+1)∑i=0κ‍∑i=0κ‍∑i=0κ‍×Θ(κ-i-l-p+1)∑i=0κ‍∑i=0κ‍∑i=0κ‍-M2Θ(p)Θ(l)Θ(i)Θ(κ-i-l-p)σpp!]=0.One may recall the transformed prescribed boundary conditions (18) and (19). Equation (41) is an iterative formula of constructing the power series solution as follows:
(42)Θ(2)=M2c22Θ(3)=-σM2c23Θ(4)=M2c2(3σ2-2M2c)24Θ(5)=-M2c2(σ3-4σM2c)30Θ(6)=M2c2(5σ4-78σ2M2c+58M4c2)720⋮The pervious process is continuous and one may consider as many terms as desired (but bearing in mind that DTM converges quite fast). Substituting (18) to (19) and (42) into (13), we obtain the following closed form of the solution:
(43)θ(x)=c+M2c22x2-σM2c23x3+M2c2(3σ2-2M2c)24x4-M2c2(σ3-4σM2c)30x5+M2c2(5σ4-78σ2M2c+58M4c2)720x6+⋯To obtain the value ofc, we substitute the boundary condition (5) into (43) at the point x=1. That is,
(44)θ(1)=c+M2c22-σM2c23+M2c2(3σ2-2M2c)24θ(x)=-M2c2(σ3-4σM2c)30θ(x)=+M2c2(5σ4-78σ2M2c+58M4c2)720+⋯=1.Substituting this value ofc into (43), one finds the expression for θ(x). On the other hand, given (n,m)=(2,3), one obtains the solution
(45)θ(x)=c+M22x2-σM23x3θ(x)=+M2(σ2c-2M2)8cx4-σM2(2σ2c-21M2)60cx5θ(x)=+M2(180M4-186σ2M2c+5σ4c2)720x6+⋯
Here, the constant c maybe obtained by evaluating the boundary condition θ(1)=1. The solutions (43) and (45) are depicted in Figures 3(b) and 3(a), respectively.Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity. In (a) the exponents are(n,m)=(2,3) and (b) (n,m)=(3,2).
(a)
(b)


### 5.2. The Rectangular Profile and Power Law Thermal Conductivity

In this section, we provide a detailed construction of analytical solutions for the heat transfer in a longitudinal rectangular fin with a power law thermal conductivity; that is, we consider (4) with f(x)=1 and k(θ)=θm. The analytical solutions are given in the following expressions.(a) Case(n,m)=(2,3)(46)θ(x)=c+M22x2-M44cx4+M64c2x6θ(x)=-33M8122c3x8+127M10336c4x10+⋯. (b) Case (n,m)=(3,2)(47)θ(x)=c+c2M22x2-c3M412x4+29c4M6360x6θ(x)=-307c5M85040x8+23483c6M10453600x10+⋯.
The constant c is obtained by solving the appropriate θ(x) at the fin base boundary condition. The analytical solutions in (46) and (47) are depicted in Figures 4(a) and 4(b), respectively.Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity. In (a) the exponents are(n,m)=(2,3) and (b) (n,m)=(3,2).
(a)
(b)


### 5.3. The Convex Parabolic Profile and Power Law Thermal Conductivity

In this section, we present solutions for the equation describing the heat transfer in a fin with convex parabolic profile and power law thermal conductivity. Equation (40) is considered. Here we consider the values {(n,m)=(2,3);(3,2)}. The final analytical solution is given by(a) Case(n,m)=(2,3)(48)θ(x)=c+2M23x3/2-2M45cx3+23M645c2x9/2θ(x)=-1909M82475c3x6+329222M10259875c4x15/2+⋯. (b) Case (n,m)=(3,2)(49)θ(x)=c+2c2M23x3/2-4c3M445x3+4c4M627x9/2θ(x)=-992c5M87425x6+30064c6M10212625x15/2+⋯.
The constant c is obtained by evaluating the appropriate θ(x) at the fin base boundary condition. The solutions in (48) and (49) are depicted in Figures 5(a) and 5(b), respectively.Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity. In (a) the exponents are(n,m)=(2,3) and (b) (n,m)=(3,2).
(a)
(b)


### 5.4. The Rectangular Profile and Linear Thermal Conductivity

In this section, we present solutions for the equation representing the heat transfer in a fin with rectangular profile and the thermal conductivity depending linearly on temperature. That is, we consider (4) with f(x)=1 and k(θ)=1+βθ. The analytical solutions for this problem for different values of n are given by(a) Casen=0(50)θ(x)=c+M2c2(1+βc)x2-M4c(-1+2βc)24(1+βc)3x4+M6c(1-16βc+28β2c2)720(1+βc)5x6-M8c(-1+78βc-600β2c2+896β3c3)40320(1+βc)7x8+M10c(1-332βc+7812β2c2-39896β3c3+51184β4c4)3628800(1+βc)9x10+⋯.(b) Casen=1(51)θ(x)=c+M2c22(1+βc)x2-M4c3(-1+2βc)24(1+βc)3x4+M6c4(10-16βc+19β2c2)720(1+βc)5x6-M8c5(-80+342βc-594β2c2+559β3c3)40320(1+βc)7x8+M10c6(1000-7820βc+24336β2c2-36908β3c3+29161β4c4)3628800(1+βc)9x10+⋯.(c) Casen=2(52)θ(x)=c+M2c32(1+βc)x2+M4c58(1+βc)3x4+M6c7(3+2β2c2)80(1+βc)5x6+M8c9(49-20βc+66β2c2-40β3c3)4480(1+βc)7x8+M10c11(427-440βc+1116β2c2-1020β3c3+672β4c4)134400(1+βc)9x10+⋯.(d) Casen=3(53)θ(x)=c+M2c42(1+βc)x2+M4c7(4+βc)24(1+βc)3x4+M6c10(52+32βc+25β2c2)720(1+βc)5x6+M8c13(1288+1020βc+1212β2c2-95β3c3)40320(1+βc)7x8+M10c16(52024+45688βc+77184β2c2-680β3c3+15025β4c4)3628800(1+βc)9x10+….The constant c may be obtained from the boundary condition on the appropriate solution. The solutions in (50), (51), (52), and (53) are depicted in Figure 6.Figure 6
Temperature profile in a longitudinal rectangular fin with linear thermal conductivity and varying values ofn. Here β=0.5 and M=1.5 are fixed.


### 5.5. The Convex Parabolic Profile and Linear Thermal Conductivity

In this section, we present solutions for the equation describing the heat transfer in a fin with convex parabolic profile and the thermal conductivity depending linearly on temperature. That is, we consider (40) with k(θ)=1+βθ. The analytical solution for this problem for different values of n is given by(a) Casen=0(54)θ(x)=c+2M2c3(1+βc)x3/2-2M4c(-2+3βc)45(1+βc)3x3+M6c(2-25βc+33β2c2)405(1+βc)5x9/2-M8c(-10+599βc-3582β2c2+4059β3c3)66825(1+βc)7x6+⋯.(b) Casen=1(55)θ(x)=c+2M2c23(1+βc)x3/2-2M4c3(-4+βc)45(1+βc)3x3+2M6c4(9-11βc+10β2c2)405(1+βc)5x9/2-2M8c5(-330+1118βc-1584β2c2+1093β3c3)66825(1+βc)7x6+⋯.(c) Casen=2(56)θ(x)=c+2M2c33(1+βc)x3/2+2M4c5(6+βc)45(1+βc)3x3+M6c7(16+3βc+7β2c2)135(1+βc)5x9/2+M8c9(1160-107βc+1116β2c2-367β3c3)22275(1+βc)7x6+⋯.(d) Casen=3(57)θ(x)=c+2M2c43(1+βc)x3/2+2M4c7(8+3βc)45(1+βc)3x3+4M6c10(23+17βc+9β2c2)405(1+βc)5x9/2+2M8c13(5000+4868βc+4356β2c2+363β3c3)66825(1+βc)7x6+⋯.The constantc may be obtained from the boundary condition on the appropriate θ(x). The solutions in (54), (55), (56) and (57) are depicted in Figure 7.Figure 7
Temperature profile in a longitudinal convex parabolic fin with linear thermal conductivity and varying values ofn. Here β=0.5 and M=1.5 are fixed.


### 5.6. The Exponential Profile and Linear Thermal Conductivity

In this section, we present solutions for equation heat transfer in a fin with convex parabolic profile and the thermal conductivity depending linearly on temperature. That is, we consider (4) with k(θ)=1+βθ and f(x)=eσx. Here σ is a constant. The analytical solutions for this problem for different values of n are given by(a) Casen=0(58)θ(x)=c+M2c2(1+βc)x2-σM2c(2+βc+β2c)6(1+βc)2x3+M2c(M2(1-2β2c)+σ2(3+3βc+β3c2+β4c2+β2c(3+c)))24(1+βc)4x4+⋯.(b) Casen=1(59)θ(x)=c+M2c22(1+βc)x2-σM2c2(2+βc+β2c)6(1+βc)2x3+M2c2(M2c(2+βc-2β2c)+σ2(3+3βc+β3c2+β4c2+β2c(3+c)))24(1+βc)3x4+⋯.(c) Casen=2(60)θ(x)=c+M2c32(1+βc)x2-σM2c3(2+βc+β2c)6(1+βc)2x3+M2c3(M2c2(3+2βc-2β2c)+σ2(3+3βc+β3c2+β4c2+β2c(3+c)))24(1+βc)3×x4+⋯.(d) Casen=3(61)θ(x)=c+M2c42(1+βc)x2-σM2c4(2+βc+β2c)6(1+βc)2x3+M2c4(M2c3(4+3βc-2β2c)+σ2(3+3βc+β3c2+β4c2+β2c(3+c)))24(1+βc)3×x4+⋯.The constantc may be obtained from the boundary condition on the appropriate θ(x). The solutions in (58), (59), (60), and (61) are depicted in Figure 8.Figure 8
Temperature profile in a longitudinal fin of exponential profile with linear thermal conductivity and varying values ofn. Here β=0.5 and M=1.5 are fixed.


## 5.1. The Exponential Profile and Power Law Thermal Conductivity

In this section, we present solutions for equation describing heat transfer in a fin with exponential profile and power law thermal conductivity and heat transfer coefficient. That is, given (4) with f(x)=eσx, and both heat transfer coefficient and thermal conductivity being power law functions of temperature, we construct analytical solutions. In our analysis, we consider n=2 and 3 indicating the fin subject to nucleate boiling and radiation into free space at zero absolute temperature, respectively. Firstly, given n=3 and m=2 and applying the DTM, one obtains the following recurrence revelation:
(41)∑i=0κ‍∑l=0κ-i∑p=0κ-i-l‍[σpp!Θ(l)Θ(i)(κ-i-l-p+1)‍∑i=0κ‍∑i=0κ‍∑i=0κ‍×(κ-i-l-p+2)Θ(κ-i-l-p+2)∑i=0κ‍∑i=0κ‍∑i=0κ‍+2σpp!Θ(l)(i+1)Θ(i+1)∑i=0κ‍∑i=0κ‍∑i=0κ‍×(κ-i-l-p+1)Θ(κ-i-l-p+1)∑i=0κ‍∑i=0κ‍∑i=0κ‍+2σσpp!Θ(l)Θ(i)(κ-i-l-p+1)∑i=0κ‍∑i=0κ‍∑i=0κ‍×Θ(κ-i-l-p+1)∑i=0κ‍∑i=0κ‍∑i=0κ‍-M2Θ(p)Θ(l)Θ(i)Θ(κ-i-l-p)σpp!]=0.One may recall the transformed prescribed boundary conditions (18) and (19). Equation (41) is an iterative formula of constructing the power series solution as follows:
(42)Θ(2)=M2c22Θ(3)=-σM2c23Θ(4)=M2c2(3σ2-2M2c)24Θ(5)=-M2c2(σ3-4σM2c)30Θ(6)=M2c2(5σ4-78σ2M2c+58M4c2)720⋮The pervious process is continuous and one may consider as many terms as desired (but bearing in mind that DTM converges quite fast). Substituting (18) to (19) and (42) into (13), we obtain the following closed form of the solution:
(43)θ(x)=c+M2c22x2-σM2c23x3+M2c2(3σ2-2M2c)24x4-M2c2(σ3-4σM2c)30x5+M2c2(5σ4-78σ2M2c+58M4c2)720x6+⋯To obtain the value ofc, we substitute the boundary condition (5) into (43) at the point x=1. That is,
(44)θ(1)=c+M2c22-σM2c23+M2c2(3σ2-2M2c)24θ(x)=-M2c2(σ3-4σM2c)30θ(x)=+M2c2(5σ4-78σ2M2c+58M4c2)720+⋯=1.Substituting this value ofc into (43), one finds the expression for θ(x). On the other hand, given (n,m)=(2,3), one obtains the solution
(45)θ(x)=c+M22x2-σM23x3θ(x)=+M2(σ2c-2M2)8cx4-σM2(2σ2c-21M2)60cx5θ(x)=+M2(180M4-186σ2M2c+5σ4c2)720x6+⋯
Here, the constant c maybe obtained by evaluating the boundary condition θ(1)=1. The solutions (43) and (45) are depicted in Figures 3(b) and 3(a), respectively.Temperature profile in a longitudinal fin with exponential profile and power law thermal conductivity. In (a) the exponents are(n,m)=(2,3) and (b) (n,m)=(3,2).
(a)
(b)


## 5.2. The Rectangular Profile and Power Law Thermal Conductivity

In this section, we provide a detailed construction of analytical solutions for the heat transfer in a longitudinal rectangular fin with a power law thermal conductivity; that is, we consider (4) with f(x)=1 and k(θ)=θm. The analytical solutions are given in the following expressions.(a) Case(n,m)=(2,3)(46)θ(x)=c+M22x2-M44cx4+M64c2x6θ(x)=-33M8122c3x8+127M10336c4x10+⋯. (b) Case (n,m)=(3,2)(47)θ(x)=c+c2M22x2-c3M412x4+29c4M6360x6θ(x)=-307c5M85040x8+23483c6M10453600x10+⋯.
The constant c is obtained by solving the appropriate θ(x) at the fin base boundary condition. The analytical solutions in (46) and (47) are depicted in Figures 4(a) and 4(b), respectively.Temperature profile in a longitudinal fin with rectangular profile and power law thermal conductivity. In (a) the exponents are(n,m)=(2,3) and (b) (n,m)=(3,2).
(a)
(b)


## 5.3. The Convex Parabolic Profile and Power Law Thermal Conductivity

In this section, we present solutions for the equation describing the heat transfer in a fin with convex parabolic profile and power law thermal conductivity. Equation (40) is considered. Here we consider the values {(n,m)=(2,3);(3,2)}. The final analytical solution is given by(a) Case(n,m)=(2,3)(48)θ(x)=c+2M23x3/2-2M45cx3+23M645c2x9/2θ(x)=-1909M82475c3x6+329222M10259875c4x15/2+⋯. (b) Case (n,m)=(3,2)(49)θ(x)=c+2c2M23x3/2-4c3M445x3+4c4M627x9/2θ(x)=-992c5M87425x6+30064c6M10212625x15/2+⋯.
The constant c is obtained by evaluating the appropriate θ(x) at the fin base boundary condition. The solutions in (48) and (49) are depicted in Figures 5(a) and 5(b), respectively.Temperature profile in a longitudinal fin with convex parabolic profile and power law thermal conductivity. In (a) the exponents are(n,m)=(2,3) and (b) (n,m)=(3,2).
(a)
(b)


## 5.4. The Rectangular Profile and Linear Thermal Conductivity

In this section, we present solutions for the equation representing the heat transfer in a fin with rectangular profile and the thermal conductivity depending linearly on temperature. That is, we consider (4) with f(x)=1 and k(θ)=1+βθ. The analytical solutions for this problem for different values of n are given by(a) Casen=0(50)θ(x)=c+M2c2(1+βc)x2-M4c(-1+2βc)24(1+βc)3x4+M6c(1-16βc+28β2c2)720(1+βc)5x6-M8c(-1+78βc-600β2c2+896β3c3)40320(1+βc)7x8+M10c(1-332βc+7812β2c2-39896β3c3+51184β4c4)3628800(1+βc)9x10+⋯.(b) Casen=1(51)θ(x)=c+M2c22(1+βc)x2-M4c3(-1+2βc)24(1+βc)3x4+M6c4(10-16βc+19β2c2)720(1+βc)5x6-M8c5(-80+342βc-594β2c2+559β3c3)40320(1+βc)7x8+M10c6(1000-7820βc+24336β2c2-36908β3c3+29161β4c4)3628800(1+βc)9x10+⋯.(c) Casen=2(52)θ(x)=c+M2c32(1+βc)x2+M4c58(1+βc)3x4+M6c7(3+2β2c2)80(1+βc)5x6+M8c9(49-20βc+66β2c2-40β3c3)4480(1+βc)7x8+M10c11(427-440βc+1116β2c2-1020β3c3+672β4c4)134400(1+βc)9x10+⋯.(d) Casen=3(53)θ(x)=c+M2c42(1+βc)x2+M4c7(4+βc)24(1+βc)3x4+M6c10(52+32βc+25β2c2)720(1+βc)5x6+M8c13(1288+1020βc+1212β2c2-95β3c3)40320(1+βc)7x8+M10c16(52024+45688βc+77184β2c2-680β3c3+15025β4c4)3628800(1+βc)9x10+….The constant c may be obtained from the boundary condition on the appropriate solution. The solutions in (50), (51), (52), and (53) are depicted in Figure 6.Figure 6
Temperature profile in a longitudinal rectangular fin with linear thermal conductivity and varying values ofn. Here β=0.5 and M=1.5 are fixed.


## 5.5. The Convex Parabolic Profile and Linear Thermal Conductivity

In this section, we present solutions for the equation describing the heat transfer in a fin with convex parabolic profile and the thermal conductivity depending linearly on temperature. That is, we consider (40) with k(θ)=1+βθ. The analytical solution for this problem for different values of n is given by(a) Casen=0(54)θ(x)=c+2M2c3(1+βc)x3/2-2M4c(-2+3βc)45(1+βc)3x3+M6c(2-25βc+33β2c2)405(1+βc)5x9/2-M8c(-10+599βc-3582β2c2+4059β3c3)66825(1+βc)7x6+⋯.(b) Casen=1(55)θ(x)=c+2M2c23(1+βc)x3/2-2M4c3(-4+βc)45(1+βc)3x3+2M6c4(9-11βc+10β2c2)405(1+βc)5x9/2-2M8c5(-330+1118βc-1584β2c2+1093β3c3)66825(1+βc)7x6+⋯.(c) Casen=2(56)θ(x)=c+2M2c33(1+βc)x3/2+2M4c5(6+βc)45(1+βc)3x3+M6c7(16+3βc+7β2c2)135(1+βc)5x9/2+M8c9(1160-107βc+1116β2c2-367β3c3)22275(1+βc)7x6+⋯.(d) Casen=3(57)θ(x)=c+2M2c43(1+βc)x3/2+2M4c7(8+3βc)45(1+βc)3x3+4M6c10(23+17βc+9β2c2)405(1+βc)5x9/2+2M8c13(5000+4868βc+4356β2c2+363β3c3)66825(1+βc)7x6+⋯.The constantc may be obtained from the boundary condition on the appropriate θ(x). The solutions in (54), (55), (56) and (57) are depicted in Figure 7.Figure 7
Temperature profile in a longitudinal convex parabolic fin with linear thermal conductivity and varying values ofn. Here β=0.5 and M=1.5 are fixed.


## 5.6. The Exponential Profile and Linear Thermal Conductivity

In this section, we present solutions for equation heat transfer in a fin with convex parabolic profile and the thermal conductivity depending linearly on temperature. That is, we consider (4) with k(θ)=1+βθ and f(x)=eσx. Here σ is a constant. The analytical solutions for this problem for different values of n are given by(a) Casen=0(58)θ(x)=c+M2c2(1+βc)x2-σM2c(2+βc+β2c)6(1+βc)2x3+M2c(M2(1-2β2c)+σ2(3+3βc+β3c2+β4c2+β2c(3+c)))24(1+βc)4x4+⋯.(b) Casen=1(59)θ(x)=c+M2c22(1+βc)x2-σM2c2(2+βc+β2c)6(1+βc)2x3+M2c2(M2c(2+βc-2β2c)+σ2(3+3βc+β3c2+β4c2+β2c(3+c)))24(1+βc)3x4+⋯.(c) Casen=2(60)θ(x)=c+M2c32(1+βc)x2-σM2c3(2+βc+β2c)6(1+βc)2x3+M2c3(M2c2(3+2βc-2β2c)+σ2(3+3βc+β3c2+β4c2+β2c(3+c)))24(1+βc)3×x4+⋯.(d) Casen=3(61)θ(x)=c+M2c42(1+βc)x2-σM2c4(2+βc+β2c)6(1+βc)2x3+M2c4(M2c3(4+3βc-2β2c)+σ2(3+3βc+β3c2+β4c2+β2c(3+c)))24(1+βc)3×x4+⋯.The constantc may be obtained from the boundary condition on the appropriate θ(x). The solutions in (58), (59), (60), and (61) are depicted in Figure 8.Figure 8
Temperature profile in a longitudinal fin of exponential profile with linear thermal conductivity and varying values ofn. Here β=0.5 and M=1.5 are fixed.


## 6. Fin Efficiency and Heat Flux

### 6.1.  Fin Efficiency

The heat transfer rate from a fin is given by Newton’s second law of cooling:(62)Q=∫0L‍PH(T)(T-Ta)dx.
Fin efficiency is defined as the ratio of the fin heat transfer rate to the rate that would be if the entire fin were at the base temperature and is given by (see e.g., [1])
(63)η=QQideal=∫0L‍PH(T)(T-Ta)dXPhbL(Tb-Ta).
In dimensionless variables, we have
(64)η=∫01‍θn+1dx.
We consider the solutions (50), (51), (52), and (53) and depict the fin efficiency (63) in Figure 9.Figure 9
Fin efficiency of a longitudinal rectangular fin. Hereβ=0.75.


### 6.2. Heat Flux

The fin base heat flux is given by the Fourier’s law(65)qb=AcK(T)dTdx.
The total heat flux of the fin is given by [1]
(66)q=qbAcH(T)(Tb-Ta).
Introducing the dimensionless variable as described in Section 2 implies
(67)q=1Bik(θ)h(θ)dθdx,
where the dimensionless parameter Bi=hbL/ka is the Biot number. We consider a number of cases for the thermal conductivity and the heat transfer coefficient.


#### 6.2.1. Linear Thermal Conductivity and Power Law Heat Transfer Coefficient

In this case (67) becomes
(68)q=1Bi(1+βθ)θ-ndθdx.
The heat flux in (68) at the base of the fin is plotted in Figure 10.Figure 10
Base heat flux in a longitudinal rectangular fin with linear thermal conductivity. Hereβ=0.1.


#### 6.2.2. Power Law Thermal Conductivity and Heat Transfer Coefficient

In this case (67) becomes
(69)q=1Biθm-ndθdx.
Not surprisingly, heat flux in one-dimensional fins is higher given values Bi≪1. The heat flux in (69) is plotted in Figures 11(a), 11(b), and 11(c).Heat flux across a longitudinal rectangular fin with linear thermal conductivity. (a)n-m<0, (b) n-m>0, and (c) n-m=0. Here Bi=0.01.
(a)
(b)
(c)


## 6.1.  Fin Efficiency

The heat transfer rate from a fin is given by Newton’s second law of cooling:(62)Q=∫0L‍PH(T)(T-Ta)dx.
Fin efficiency is defined as the ratio of the fin heat transfer rate to the rate that would be if the entire fin were at the base temperature and is given by (see e.g., [1])
(63)η=QQideal=∫0L‍PH(T)(T-Ta)dXPhbL(Tb-Ta).
In dimensionless variables, we have
(64)η=∫01‍θn+1dx.
We consider the solutions (50), (51), (52), and (53) and depict the fin efficiency (63) in Figure 9.Figure 9
Fin efficiency of a longitudinal rectangular fin. Hereβ=0.75.


## 6.2. Heat Flux

The fin base heat flux is given by the Fourier’s law(65)qb=AcK(T)dTdx.
The total heat flux of the fin is given by [1]
(66)q=qbAcH(T)(Tb-Ta).
Introducing the dimensionless variable as described in Section 2 implies
(67)q=1Bik(θ)h(θ)dθdx,
where the dimensionless parameter Bi=hbL/ka is the Biot number. We consider a number of cases for the thermal conductivity and the heat transfer coefficient.


### 6.2.1. Linear Thermal Conductivity and Power Law Heat Transfer Coefficient

In this case (67) becomes
(68)q=1Bi(1+βθ)θ-ndθdx.
The heat flux in (68) at the base of the fin is plotted in Figure 10.Figure 10
Base heat flux in a longitudinal rectangular fin with linear thermal conductivity. Hereβ=0.1.


### 6.2.2. Power Law Thermal Conductivity and Heat Transfer Coefficient

In this case (67) becomes
(69)q=1Biθm-ndθdx.
Not surprisingly, heat flux in one-dimensional fins is higher given values Bi≪1. The heat flux in (69) is plotted in Figures 11(a), 11(b), and 11(c).Heat flux across a longitudinal rectangular fin with linear thermal conductivity. (a)n-m<0, (b) n-m>0, and (c) n-m=0. Here Bi=0.01.
(a)
(b)
(c)


## 6.2.1. Linear Thermal Conductivity and Power Law Heat Transfer Coefficient

In this case (67) becomes
(68)q=1Bi(1+βθ)θ-ndθdx.
The heat flux in (68) at the base of the fin is plotted in Figure 10.Figure 10
Base heat flux in a longitudinal rectangular fin with linear thermal conductivity. Hereβ=0.1.


## 6.2.2. Power Law Thermal Conductivity and Heat Transfer Coefficient

In this case (67) becomes
(69)q=1Biθm-ndθdx.
Not surprisingly, heat flux in one-dimensional fins is higher given values Bi≪1. The heat flux in (69) is plotted in Figures 11(a), 11(b), and 11(c).Heat flux across a longitudinal rectangular fin with linear thermal conductivity. (a)n-m<0, (b) n-m>0, and (c) n-m=0. Here Bi=0.01.
(a)
(b)
(c)


## 7. Some Discussions

The DTM has resulted in some interesting observations and study. We have observed in Figures2(a) and 2(b) an excellent agreement between the analytical solutions generated by DTM and the exact solution obtained in [2]. In particular, we considered a fin problem in which both thermal conductivity and heat transfer coefficient are given by the same power law. Furthermore, we notice from Table 2 that an absolute error of approximately 3.5e-005 is produced by DTM of order O(15). In Table 3, an absolute error of approximately 6.5e-006 is produced for the same order. This confirms that the DTM converges faster and can provide accurate results with a minimum computation. As such, a tremendous confidence in the DTM in terms of the accuracy and effectiveness was built, and thus we used this method to solve other problems for which exact solutions are harder to construct.In Figures3(a), 3(b), 4(a), 4(b), 5(a), and 5(b), we observe that the fin temperature increases with the decreasing values of the thermogeometric fin parameter. Here, the values of the exponents are fixed. Also, we observe that fin temperature is higher when n-m>0, that is, when heat transfer coefficient is higher than the thermal conductivity. We observe in Figures 6, 7, and 8 that the fin temperature increases with the increasing values of n. Furthermore, it appears that the fin with exponential profile performs the least in transferring the heat from the base, since the temperature in such a fin is much higher than that of the rectangular and the convex parabolic profiles. In other words, heat dissipation to the fluid surrounding the extended surface is much faster in longitudinal fins of rectangular and convex parabolic profiles. In Figure 9, fin efficiency decreases with increasing thermogeometric fin parameter. Also, fin efficiency increases with increasing values of n. It is easy to show that the thermogeometric fin parameter is directly proportional to the aspect ratio (extension factor) with square root of the Biot number being the proportionality constant. As such, shorter fins are more efficient than longer ones. Else, the increased Biot number results in less efficient fin whenever the space is confined, that is, where the length of the fin cannot be increased. Figure 10 depicts the heat flux at the fin base. The amount of heat energy dissipated from the fin base is of immense interest in engineering [28]. We observe in Figure 10 that the base heat flux increases with the thermogeometric fin parameter for considered values of the exponent n (see also [28]). Figures 11(a), 11(b), and 11(c) display the heat flux across the fin length. We note that the heat flux across the fin length increases with increasing values of the thermogeometric fin parameter.


## 8. Concluding Remarks

In this study, we have successfully applied the DTM to highly nonlinear problems arising in heat transfer through longitudinal fins of various profiles. Both thermal conductivity and heat transfer coefficient are given as functions of temperature. The DTM agreed well with exact solutions when the thermal conductivity and heat transfer coefficient are given by the same power law. A rapid convergence to the exact solution was observed. Following the confidence in DTM built by the results mentioned, we then solved various exciting problems. The exotic results have been shown in tables and figures listed in this paper.The results obtained in this paper are significant improvements on the known results. In particular, both the heat transfer coefficient and the thermal conductivity are allowed to be given by the power law functions of temperature, and also we considered a number of fin profiles. We note that exact solutions are difficult if not impossible to construct when the exponents of these properties are distinct.Perhaps the notable advantage of the DTM is the generalization of the Taylor method to problems involving unusual derivative procedures such as fractional, fuzzy, orq-derivative [22]. Some generalizations have been made by Odibat et al. [24], and they referred to their new method as the Generalized Differential Transform Method (GDTM). This showed great improvement compared to the Fractional Differential Transform Method (FDTM) introduced by Arikoglu and Ozkol [23].We have shown with the help of an example that DTM may only be applied to fin problems involving heat transfer through fins with convex parabolic profile. Note that given an ODE such as (4) with a power law heat transfer coefficient of a fractional exponent, then one can easily remove the fraction by basic exponential rules and employment of the Binomial expansion. However, using the DTM, one runs into difficulty if the power law thermal conductivity in the same equation is given by the fractional exponent. We do not know whether these observations call for the “modified” DTM to solve problems arising in heat flow through fins with other profiles, such as longitudinal triangular and concave parabolic, and also with fractional power law thermal conductivity.The main results obtained in this paper give insight into heat transfer in boiling liquids where the heat transfer coefficient is temperature dependent and may be given by a power law. The thermal conductivity of some materials such as gallium nitride (GaN) and Aluminium Nitride (AlN) may be modeled by power law temperature dependency [29, 30]. Thus, the solutions constructed here give a better comparison of heat transfer in terms of material used since in many engineering applications thermal conductivity is given as a linear function of temperature. Furthermore, a good study in terms of performance and efficiency of fin with different profiles is undertaken. These finding could help in the design of fins. It is claimed in [14] that DTM results are more accurate than those constructed by Variational Iteration Methods (VIM) and Homotopy Perturbation Methods (HPM). However, it would be risky to use the DTM approximate solutions as benchmarks for the numerical schemes. Nevertheless, we have also shown that DTM converges rapidly in just fifteen terms to the exact solution.



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