The profile of the fin and the dimensionless temperature are as followsDisplay Formula

$z=\lambda +(1\u2212\lambda )\phi \varphi =x\u2215H$

(4)

The dimensionless temperature, defined as

$\theta =T\u2215T\xaffl$, where

$T\xaffl$ is the average fluid temperature both measured above thee ambient temperature, is equal to

Display Formula$\theta =(1z)1\u22152[C1I1(2pz1\u22152)+C2K1(2pz1\u22152)]p=u\u2215(1\u2212\lambda )$

(5)

The foregoing equations show that the dimensionless temperature

$\theta =\theta (\phi ,u,\lambda )$, while for a given

$u$ is a function of

$\phi $ and

$\lambda $, with

$z(0)=\lambda $, and

$z(1)=1$. Note that integrating Eq.

4 from 0 to 1, we obtain the average fin diameter,

$Dave=Dend$$(1+\lambda )\u22152$, thus

$0.9163\u2a7dzave\u2a7d1$. In order to obtain the dimensionless temperature,

$\u03f4x$, the authors use Eqs. (11), (12a), (12b), (Eq. (3.73) in K-B). However, the K-B dimensionless temperature has been obtained using constant

$h$, and initial conditions (I think they mean boundary conditions) of

$Q(x=b)=Qpf$, and

$\u03f4(x=b)=\u03f4ew$. However, the temperature is not specified at

$x=b$. I believe that a more reasonable boundary condition would have been to consider an adiabatic tip,

$Q(x=b)=0$. For example, when the authors’ parameter

$C=0$, the top plate is adiabatic, while the authors also never considered the heat transfer from the tip, their Eq. (10). This new boundary condition would have simplified the equations considerably, especially the expression for the heat dissipation (see Ref.

5). Now, the boundary conditions that accompany

5 are

$Q(u)=Qpf$, and

$(d\theta \u2215dx)x=0=0$. We may now observe that, because the values of

$\lambda $, are nearly one, which result in very large values of the parameter

$p=u\u2215(1\u2212\lambda )$, it would have been be justified to use for their calculations the following much simpler expression for cylindrical spines

Display Formula$\theta =Bcosh(\xi )cosh(u),\xi =u\xd7(x\u2215H)$

(6)

In the foregoing equation the symbol

$B$ is equal to

$B=(k2Qpf\u2215\pi hBi3\u22152\theta en)$.