# Heat transfer direction in fins

1. Jul 26, 2016

### roughwinds

1. The problem statement, all variables and given/known data
I'm unsure of what exactly is changing the heat transfer direction in the triangular fin.

2. Relevant equations
$$q_{x} = -kA(x)\frac{dT(x)}{dx} (1)$$
$$q_{x+dx} = -kA(x)\frac{dT(x)}{dx} - k\frac{d}{dx}[A(x)\frac{dT(x)}{dx}] (2)$$
$$dq_{conv} = h(x)dS(x)P[T(x) - T_{∞}] (3)$$
A = cross-sectional area m²
q = heat transfer rate W
h = convection heat transfer coefficient W/m²K
k = thermal conductivity W/mK
T = temperature K
dq = convection heat transfer rate
dS = surface area of the differential element
Tb = base surface temperature K
A0 = base area
L = length

3. The attempt at a solution
Energy balance:
$$q_{x} = dq_{conv} + - q_{x+dx} (4)$$
Substitute (1), (2) and (3) in (4):
$$\frac{d}{dx}[A(x)\frac{dT(x)}{dx}] - \frac{h(x)dS(x)}{kdx}[T(x) - T_{∞}] (5)$$
$$X = \frac{x}{L}$$
$$\theta(X) = \frac{T(X) - T_{∞}}{T_{b} - T_{∞}}$$
$$K(X) = \frac{A(X)}{A_{0}}$$
$$W(X) = \frac{h(X)dS(X)}{P_{0}h_{av}dX}$$
$$M = mL$$
$$m² = \frac{h_{av}P_{0}}{kA_{0}}$$
$$\frac{dS(X)}{dX} = p(X)$$
Substitute into (5) and rearrange:
$$\frac{d}{dX}[K(X)\frac{d\theta(X)}{dX}] - W(X)M²\theta(X) = 0 (6)$$

Rectangular profile:

A(X) = A0, K(X) = 1, W(X) = 1
$$\frac{d}{dX}[\frac{d\theta(X)}{dX}] - M²\theta(X) = 0$$

Triangular profile:

A(X) =/= A0, K(X) = ~ X, W(X) = 1
$$\frac{d}{dX}[X\frac{d\theta(X)}{dX}] - M²\theta(X) = 0$$

I did the energy balance thinking of the heat going from left to right like it's shown on the rectangular fin, but on the triangular fin it goes on the opposite direction. Which is fine for what I want to do, but why exactly does this happen? Due to the equations involved the heat will only move from a bigger to an equal or smaller area? Which adjustments would I have to make to change the heat transfer direction and make a fin like this:

2. Jul 26, 2016

### jambaugh

Could you be a bit clearer as to the original problem you are trying to solve (rather than the problem you're having with it...). I think I could figure that out with some effort but I'm not inclined to exert that effort right now.

3. Jul 27, 2016

### Staff: Mentor

The direction that the heat flows has nothing to do with how the area is varying. It has everything to do with how the temperature is varying. This is determined the the boundary conditions on the fin. What are the boundary conditions being applied on the two ends of your fin?

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