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Purdue ME 606: View Factors

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View Factors (or Shape Factors)

$F 1\to2$ is the fraction of energy leaving surface 1 that arrives at surface 2

Assume black surfaces

flux leaving 1: $E b 1$
total leaving 1: $A 1 E b 1$
amount arriving at 2: $A 1 E b 1 F 1\to2$
absorbed at 2: $A 1 E b 1 F 1\to2$
amount leaving 2 that arrives at 1: $A 2 E b 2 F 2\to1$
net heat flow: $q_{1 \Leftrightarrow 2}=A_1 E_{b1}F_{1 \rightarrow 2}-A_2 E_{b2}F_{2 \rightarrow 1}$
for isothermal, $T 1 = T 2$ :
- $q_{1 \leftrightarrow 2} = 0$
- $A 1 F 1\to2 = A 2 F 2\to1$
$q_{1 \leftrightarrow 2}=A_1 E_{b1} F_{1 \rightarrow 2}=A_2 E_{b2} F_{2 \rightarrow 1}$

View Factor Details

Assume:

Emission and reflection are diffuse
Intensities are uniform over each surface (i.e. $T =$ constant
Arriving energies are uniform over surfaces
Non participating medium between surfaces
Black surfaces (later relaxed)

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$I(\theta,\phi)=\frac{d^2q}{dAcos(\theta)d\omega}$

$d^2q_{dA_1 \rightarrow dA_2} = I_1 cos(\theta_1)dA_1d\omega_{1-2}$

where

$d\omega_{1-2}=\frac{dA_2cos(\theta_2)}{r^2}$
$E b 1 = π I$

$d^2q_{dA_1 \rightarrow dA_2} = E_{b1} \underbrace{\frac{cos(\theta_1) cos(\theta_2) dA_2}{\pi r^2}}_{dF_{dA_1-dA_2}} dA_1$

similarly

$d^2q_{dA_2 \rightarrow dA_1} = E_{b2} \underbrace{\frac{cos(\theta_1) cos(\theta_2) dA_1}{\pi r^2}}_{dF_{dA_2-dA_1}} dA_2$

$dF_{dA_2-dA_1} dA_2 = dF_{dA_1-dA_2} dA_1$

From $d A 1$ to $A 2$

$dq_{dA_1 \rightarrow A_2} = \int_{A_2} d^2q_{dA_1 \rightarrow dA_2} = (E_{b1}-E_{b2}) \underbrace{ \int_{A_2} (\frac{cos(\theta_1)cos(\theta_2)dA_1}{\pi r^2})dA_2 }_{A_2dF_{A_2-A_1}=dA_1F_{dA_1-A_2}}$

For finite areas $A 1$ & $A 2$

$q_{1 \rightarrow 2} = \int dq_{dA_1 \rightarrow A_2} = (E_{b1}-E_{b2}) \underbrace{ \int_{A_1} \int_{A_2} \frac{cos(\theta_1)cos(\theta_2)}{\pi r^2}dA_2dA_1}_{A_2F_{2-1}=A_1F_{1-2}}$

Properties of View Factors

consider an arbitrary number of surfaces (1, 2, …, N)

$F 1\to(1 + 2 + ... + N) = F 11 + F 12 + ... + F 1 N$
$F 11 = 0$ if 1 is convex or flat

Enclosures

Summation equation: $\sum_{k=1}^NF_{1k}=1$

The view factors can be arranged in an N x N matrix

Observations
- $N 2$ shape factors
- $N$ summation equations
- $\frac{N(N-1)}{2}$ view factors must be found from first principles

Example:

Given a 3 surface enclosure where one surface 3 is a hemispherical cap, surface 1 is a part of the base and surface 2 is the remainder of the base.

Find all the view factors.

From inspection:

$F 12 = F 21 = 0$
$F 11 = F 22 = 0$
$F 13 = F 23 = 1$

Use reciprocity:

$A_1F_{13}=A_3F_{31} \Rightarrow F_{31}=\frac{A_1}{A_3}$
$A_2F_{23}=A_3F_{32} \Rightarrow F_{32}=\frac{A_2}{A_3}$

Use summation equation:

$F_{31} + F_{32} + F_{33} = 1 \Rightarrow F_{33}=1-\frac{A_1+A_2}{A_3}= 1-\frac{\pi r^2}{2 \pi r^2} = \frac{1}{2}$

Methods of Evaluating View Factors

Area Integration

Given 2 surfaces i and j, the vector from $d A i$ going to $d A j$ is given as

$\bar{s}_{ij}=-\bar{s}_{ji}=(x_j-x_i)\hat{i} +(y_j-y_i)\hat{j} +(z_j-z_i)\hat{k}$

The magnitude (length) of the vector is

$|\bar{s}_{ij}|^2= |\bar{s}_{ji}|^2 =r^2= (x_i-x_j)^2 + (y_i-y_j)^2+ (z_i-z_j)^2$

$c o s (θ i)$ and $c o s (θ j)$ can then be evaluated as

$cos(\theta_i)= \frac{\hat{n}_i \cdot \bar{s}_{ij}}{r}$

$cos(\theta_j)= \frac{\hat{n}_j \cdot \bar{s}_{ji}}{r}$

where $\hat{n}_i$ and $\hat{n}_j$ are the unit normals from surfaces i and j, respectivley.

The view factors are then calculated as

$dF_{dA_i-dA_j}=\frac{cos(\theta_i)cos{\theta_j}}{\pi r^2} dA_j$

$F_{dA_i-A_j}=\int_{A_j} \frac{cos(\theta_i)cos{\theta_j}}{\pi r^2} dA_j$

$dF_{A_i-dA_j}=\frac{1}{A_j} \int_{A_j} \frac{cos(\theta_i)cos{\theta_j}}{\pi r^2} dA_jdA_i$

$F_{A_i-A_j}=\frac{1}{A_i} \int_{A_i} \int_{A_j} \frac{cos(\theta_i)cos{\theta_j}}{\pi r^2} dA_jdA_i$

Contour Integration

Stokes’ theorem may be applied to convert the previous area integrals to an equivalent contour integral.

The view factor from a differential area $d A 1$ to an area $A 2$ can be computed

$F_{d1-2}=\frac{1}{2\pi} \oint \frac{(\bar{s}_{12} \times \hat{n}_1) \cdot d\bar{s}_2}{r^2}$

The view factor from two finite areas is

$A_1 F_{1-2}= \frac{1}{2\pi}\oint_1\oint_2 ln(r)~d\bar{s}_2 \cdot d\bar{s}_1$

where $d\bar{s}=dx\hat{i}+dy\hat{j}+dz\hat{k}$

Crossed-String Method

Valid for infinitely long parallel surfaces. From Incropera and Dewitt, Ed. 5, Problem 13.6.

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$F_{12}=\frac{1}{2 w_1}[(ac+bd)-(ad+bc)]$

Last modified on 15 Oct, 2008