Question

Cooling Fin A cooling fin is an outward projection from a mechanical or electronic device from which heat can be radiated away from the device into the surrounding medium (such as air). See Figure 6.R.1. An annular, or ring-shaped, cooling fin is normally used on cylindrical surfaces such as a circular heating pipe. See Figure 6.R.2. In the latter case, let r denote the radial distance measured from the center line of the pipe and T(r) the temperature within the fin defined for $r_0\le r\le r_1$ It can be shown that T(r) satisfies the differential equation

role="math" localid="1663927167728" $\frac{d}{dr}\left(r\frac{dT}{dr}\right)=a^{2}r(T-T_m)$

where a²is a constant and T_mis the constant air temperature.

Suppose $r_{-}\left\{0\right\}=1,r_{-}\left\{1\right\}=3,$ ,and T_m=70. Use the substitution w(r) =T(r)^_70to show that the solution of the given differential equation subject to the boundary conditions

T(1)=160, T(3)=0 is

role="math" localid="1663926265607" $T\left(r\right)=70+90\frac{K_1\left(3a\right)I_v\left(ar\right)+I_1\left(3a\right)K_v\left(ar\right)}{K_1\left(3a\right)I_0\left(a\right)+I_{!}\left(3a\right)K_0\left(a\right)}$where and I₀(x) and K₀(x)are the modified Bessel functions of the first and second kind. You will also have to use the derivatives given in (25) of Section 6.4.

Short answer

We introduce the substituting w(r)=T(r)-70 and get a problem whose general solution is given by modified Bessel functions of order 0 . Solving this problem we get

$T\left(r\right)=70+90\frac{K_1\left(3a\right)I_v\left(ar\right)+I_1\left(3a\right)K_v\left(ar\right)}{K_1\left(3a\right)I_0\left(a\right)+I_{!}\left(3a\right)K_0\left(a\right)}$

Step-by-step solution

Definition of Power series solutions

If X=X₀is an ordinary point of the differential equation , we can always find two linearly independent solutions in the form of a power series centered at X₀, that is,

$y=\sum _{n=0}^x{C}_n{(X-X_0)}^n$

A power series solution converges at least on some interval defined by $\left|X-X_0\right|<R$, where R is the distance from X₀to the closest singular point.

Using boundary conditions to solve the problem

For the given parameters r₁=1,r₂=3 and T_m =70 our equation becomes

$\frac{d}{dr}\left(r\frac{dT}{dr}\right)=a^{2}r(T-70)$

where the boundary conditions are

T(1)=160,T(3)=0

As suggested we will introduce the substitution w(r) = T(r)^_70 . For this choice of

w(r) we have that

$\frac{dw}{dr}=\frac{d}{dr}\left(T\right(r)-70)=\frac{dT}{dr}$

Furthermore, the new boundary conditions are

w(1)=T(1)^_70=90,w(3)=T(3)=0.

The new problem we have to solve is

$\frac{d}{dr}\left(r\frac{dw}{dr}\right)=a^{2}rw,w\left(1\right)=90,w\left(3\right)=0$

The equation itself can be written as

role="math" localid="1663928493854" $r\left(\frac{dw}{dr}+r\frac{d^{2}w}{d{r}^2}\right)-a^2{r}^{2}w=0$

which is equivalent

$r^2\frac{d^{2}w}{d{r}^2}+r\frac{dw}{dr}-a^2{r}^{2}w=0$

Using Bessel functions

From Section 6.4 we know that the general solution of this equation can be expressed via modified Bessel functions of order zero, and is given by

w(r)=c₁I₀(ar)+c²K₀(ar)

From the initial conditions we get:

90 = w(1)=c₁I₀(a)+c₂K₀(a)

0 = w (3) = ac₁I₀(3a)+ac₂K₀(3a)

Using the formulas

$\frac{d}{dx}{I}_0\left(x\right)=I_1\left(x\right)and\frac{d}{dx}{K}_0\left(x\right)=-K_1\left(x\right)$

we see that the system of equations from above can be written as

c₁I₀(a)+c₂K₀(a)=90

c₁I₁(3a)^_c₂K₁(3a)=0

From the second equation we see that

c₁ = c₂role="math" localid="1663931455568" $\frac{K_1\left(3a\right)}{I_1\left(3a\right)}$

Substitution

'

After substituting this in the first equation we get

$C_2\left(\frac{k_1\left(3A\right)}{i_1\left(3A\right)}+K_0\left(A\right)\right)=90$

Solving for c₂ we get

role="math" localid="1663931503706" $c_2=\frac{90i_1\left(3a\right)}{K_1\left(3a\right)I_0\left(a\right)+I_1\left(3a\right)K_0\left(a\right)}$

Returning to C₁ we get

C₁=$\frac{K_1\left(3a\right)}{I_1\left(3a\right)}$role="math" localid="1663931704583" $\frac{90i_1\left(3a\right)}{K_1\left(3a\right)I_0\left(a\right)+I_1\left(3a\right)K_0\left(a\right)}$=role="math" localid="1663931642204" $\frac{90K_1\left(3a\right)}{K_1\left(3a\right)I_0\left(a\right)+I_1\left(3a\right)K_0\left(a\right)}$

We conclude that the solution of the boundary value problem for w(r) is

role="math" localid="1663931966573" $w\left(r\right)=90\frac{K_1\left(3a\right)I_0\left(ar\right)+I_1\left(3a\right)K_0\left(ar\right)}{K_1\left(3a\right)I_0\left(a\right)+I_1\left(3a\right)K_0\left(a\right)}$

Reversing the substitution we get that the solution of the original problem is

$T\left(r\right)=70+90\frac{K_1\left(3a\right)I_0\left(ar\right)+I_1\left(3a\right)K_0\left(ar\right)}{K_1\left(3a\right)I_0\left(a\right)+I_1\left(3a\right)K_0\left(a\right)}$

which is what we wanted to show.