Derive an expression for the geometric shape factor F1–2 for a rectangular surface A1, 1 by 20 m placed parallel to and centered 5 m above a 20-m-square surface A2.
GIVEN
• Rectangular surface A1 and square surface A2
• A1 is parallel to and centered 5 m above A2
• Dimensions of A1 = 1 m × 20 m
• A2 is 20 m square
FIND
• The shape factor F12
SKETCH
Given a differential element dA1, at x,y,z = (x1, y1, –5 m)
Given a differential element dA2, at x,y,z = (x2, y2 0)
The distance between elements (r) is: r2 = (x1 – x2)2 + (y1 – y2)2 + (5 m –0)2
Since the surfaces are parallel: cos ?1 = cos ?2 = (5 m)/r
The double integral can be expanded into the following quadruple integral
This can be simplified somewhat by trigonometric substitutions, however, it is fairly simple to solve numerically as it is. Let all the dx terms equal ?x, where 20/?x and 1/?x are both integers. The integral can then be approximated by
where
This is implemented in the Pascal program shown below
var dx,dx2,sum,r4,.real;
ix1,ix2,iy1,iy2,nx1,nx2,ny1,ny2.integer;
x1,x2,y1,y2:real;
begin
dx = 1.00;
dx2 = dx/2;
nx1 = trunc(20/dx);
nx2 = trunc(20/dx);
ny1 = trunc(1/dx);
ny2 = trunc(20/dx);
sum = 0.00;
for ix1 = 1 to nx1 do
begin
x1 = ix1*dx-dx2;
write1n(x 1:8.3);
for iy1 = 1 to ny 1 do
begin
y1 = iy1*dx–dx2;
for ix2 = 1 to nx2 do
begin
x2 = ix2*dx–dx2;
for iy2 = 1 to ny2 do
begin
y2 = iy2*dx–dx2;
r4 = (x1–x2)*(x1–x2) + (y1 – y2)* (y1 – y2) + 25;
r4 = 1/r4/r4;
sum = sum + r4;
end;
end;
end;
end;
sum = sum* 25/3.14159265/20*dx*dx*dx*dx;
writen(*F12 = Sum:8.4);
end.
Running this program yields the following result
F12 = 0.427
Comment: If the geometry is approximated as two-dimensional so that the view factor can be calculated using the crossed string method, we get F12 = 0.97, a significant error.
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