Fin/Extended surface differential equation for temperature

[roughwinds]

I'm trying to deduce the differential equation for temperature for a triangular fin:

I know that for a rectangular fin, such as:

I can do:
Energy entering the left:
q_x= -kA\frac{dT(x)}{dx}

Energy leaving the right:
q_{x+dx} = -kA\frac{dT(x)}{dx} - kA\frac{d² T(x)}{dx²}dx

Energy lost by convection:
dq_{conv} = h_eA(T-T_e)
dq_{conv} = h_eP(T-T_e)dx

q_x - q_{x+dx} - dq_{conv} = 0

q_x = q_{x+dx} + dq_{conv}

-kA\frac{dT(x)}{dx} = -kA\frac{dT(x)}{dx} - kA\frac{d²T(x)}{dx²}dx + h_eP(T-T_e)dx

kA\frac{d²T(x)}{dx²}dx = h_eP(T-T_e)dx

\frac{d²T(x)}{dx²} = \frac{h_eP(T-T_e)}{kA}

But I don't understand why A = Pdx on the energy lost by convection, so I don't know how to adapt that for a triangular fin.

Seems to me that T(x,y,z) still changes predominantly on the x axis, so I assume the other two equations remain unchanged.

 

[Chestermiller]

There are two areas involved here, one perpendicular to the flow of heat in the x direction (which you call A), and the other perpendicular to the flow of heat from the fin surface into the surrounding air (which is Pdx). For a fin with triangular shape, A is decreasing with x, and P is about constant. Also, since A is a function of x, when you do your heat balance on a differential section of fin, the area A must stay inside the heat flux derivative.

Chet

 

[roughwinds]

There are two areas involved here, one perpendicular to the flow of heat in the x direction (which you call A), and the other perpendicular to the flow of heat from the fin surface into the surrounding air (which is Pdx). For a fin with triangular shape, A is decreasing with x, and P is about constant. Also, since A is a function of x, when you do your heat balance on a differential section of fin, the area A must stay inside the heat flux derivative.

Chet

So this is correct

?

q_x = -kA(x)\frac{dT(x) }{dx}
q_{x+dx} = -kA(x)\frac{dT(x) }{dx} - k\frac{dA(x)}{dx}\frac{d²T(x)}{dx²}
dq_{conv} = h_eP(T(x)-T_e)dx
-kA(x)\frac{dT(x) }{dx} = -kA(x)\frac{dT(x) }{dx} - k\frac{dA(x)}{dx}\frac{d²T(x)}{dx²} + h_eP(T(x)-T_e)dx
k\frac{dA(x)}{dx}\frac{d²T(x)}{dx²} = h_eP(T(x)-T_e)dx
\frac{dA(x)}{dx}\frac{d²T(x) }{dx²} = \frac{h_eP(T(x)-T_e)}{k}
T(x)-T_e = θ(x)
m² = \frac{h_eP}{k}
\frac{d²T(x)}{dx²} = \frac{d²θ(x)}{dx²}
\frac{dA(x)}{dx}\frac{d²θ(x) }{dx²} = m²θ(x) (1)

What I've posted here is just a portion of an exercise I was doing, I would then have to solve the ODE to find an equation for T(x). So now I need to get rid of this derivative of A(x). To be honest the only thing I can think of to proceed from here is to find an equation for A(x) and derivate it.

A(x) = 2h(x)P
c² = h² + x²
h(x) = sqrt(c²-x²)
A(x) = 2sqrt(c²-x²)P
\frac{dA(x)}{dx} = \frac{-2Px}{sqrt(c²-x²)}
\frac{d²θ(x) }{dx²} = -\frac{m²θ(x)sqrt(c²-x²)}{2Px}

And this is feeling too awkward to be correct, lol. Can you integrate back at (1)?
------
Just thought of
n²(x) = -\frac{m²sqrt(c²-x²)}{2Px}
\frac{d²θ(x)}{dx²} = n²(x)θ(x)
\frac{d²θ(x)}{dx²} - n²(x)θ(x) = 0
θ(x) = c_1e^{nx} + c_2e^{-nx}
I know how to continue from here, but I'm having the sensation that this is wrong.

 

[Chestermiller]

If P is the dimension shown on your diagram, then the differential heat balance should read:
$$k\frac{\partial}{\partial x}\left(A(x)\frac{\partial T}{\partial x}\right)-2Ph(T-T_e)=0$$
Also, from the figure, A(x) = P t(x), where t is the thickness of the fin at location x. So:
$$k\frac{\partial}{\partial x}\left(t(x)\frac{\partial T}{\partial x}\right)-2h(T-T_e)=0$$

 

[roughwinds]

If P is the dimension shown on your diagram, then the differential heat balance should read:
$$k\frac{\partial}{\partial x}\left(A(x)\frac{\partial T}{\partial x}\right)-2Ph(T-T_e)=0$$
Also, from the figure, A(x) = P t(x), where t is the thickness of the fin at location x. So:
$$k\frac{\partial}{\partial x}\left(t(x)\frac{\partial T}{\partial x}\right)-2h(T-T_e)=0$$

I don't think I expressed myself properly before but anyway, finally managed to get it in the way I wanted:
http://i.share.pho.to/0a247f78_o.jpeg
http://i.share.pho.to/57739467_o.jpeg
T(x) - T_e = θ(x)
m = \frac{h_e(L/e)(1+(e/L)²)^{1/2}}{k}
xθ'' + θ' - mθ = 0
Just need to solve that now.

 

[Chestermiller]

I don't think I expressed myself properly before but anyway, finally managed to get it in the way I wanted:
http://i.share.pho.to/0a247f78_o.jpeg
http://i.share.pho.to/57739467_o.jpeg
T(x) - T_e = θ(x)
m = \frac{h_e(L/e)(1+(e/L)²)^{1/2}}{k}
xθ'' + θ' - mθ = 0
Just need to solve that now.

This is an Euler differential equation.

 

[roughwinds]

This is an Euler differential equation.

Wouldn't it have to be
x²θ′′+xθ′−mθ=0
for it to be an Euler differential equation?

I tried to do power series but I got stuck in the end:
y=\sum_{n=0}^\infty a_k x^k
θ'=\sum_{n=1}^\infty ka_k x^{k-1}
θ''=\sum_{n=2}^\infty k(k-1)a_k x^{k-2}
\sum_{n=2}^\infty k(k-1)a_k x^{k-1} + \sum_{n=1}^\infty ka_k x^{k-1} - \sum_{n=0}^\infty m²a_k x^k = 0
\sum_{n=1}^\infty k(k+1)a_{k+1} x^{k} + \sum_{n=0}^\infty (k+1)a_{k+1} x^{k} - \sum_{n=0}^\infty m²a_k x^k = 0
a_1 - m²a_o + \sum_{n=1}^\infty k(k+1)a_{k+1} x^{k} + \sum_{n=1}^\infty (k+1)a_{k+1} x^{k} - \sum_{n=1}^\infty m²a_k x^k = 0
a_1 - m²a_o = 0
a_1 = m²a_o
\sum_{n=1}^\infty k(k+1)a_{k+1} x^{k} + \sum_{n=1}^\infty (k+1)a_{k+1} x^{k} - \sum_{n=1}^\infty m²a_k x^k = 0
\sum_{n=1}^\infty[k(k+1)a_{k+1} x^{k} + (k+1)a_{k+1} x^{k} - m²a_k x^k]= 0
\sum_{n=1}^\infty[k(k+1)a_{k+1} + (k+1)a_{k+1} - m²a_k]x^{k}= 0
\sum_{n=1}^\infty[(k+1)(ka_{k+1} + a_{k+1}) - m²a_k]x^{k}= 0
\sum_{n=1}^\infty[(k+1)(k+1)a_{k+1} - m²a_k]x^{k}= 0
\sum_{n=1}^\infty[(k+1)²a_{k+1} - m²a_k]x^{k}= 0
(k+1)²a_{k+1} - m²a_k= 0
a_{k+1}= \frac{m²a_k}{(k+1)²}
k = 1, a_2 = \frac{m²a_1}{4} = \frac{m^{4}a_0}{2!}
k = 2, a_3 = \frac{m²a_2}{9} = \frac{m^{6}a_0}{(3!)²}
k = 3, a_4 = \frac{m²a_3}{16} = \frac{m^{8}a_0}{(4!)²}
k = 4, a_5 = \frac{m²a_4}{25} = \frac{m^{10}a_0}{(5!)²}
a_{2k}= \frac{m^{4k}a_0}{(2k!)²}
a_{2k+1}= \frac{m^{2(2k+1)}a_0}{((2k+1)!)²}
θ(x) = a_0\sum_{n=0}^\infty \frac{m^{4k}x^{2k}}{(2k!)²} + a_0\sum_{n=0}^\infty \frac{m^{2(2k+1)}x^{2k+1}}{((2k+1)!)²}

 

[Chestermiller]

Wouldn't it have to be
x²θ′′+xθ′−mθ=0
for it to be an Euler differential equation?

Ooops. Sorry. My mistake.