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Code Manual for CONTAIN 2.0 - Federation of American Scientists

Code Manual for CONTAIN 2.0 - Federation of American Scientists

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The interface temperature Oihas still not been defined at this point. All interface temperatures can<br />

be eliminated from the equation by imposing the assumption that the heat flux on both sides <strong>of</strong> the ~<br />

interface between two nodes must be equal. If this is done, the temperature gradient at the interface<br />

in Equation (10-124) can be expressed in terms <strong>of</strong> differences between the node-center temperatures<br />

alone. The discretized conduction equation <strong>for</strong> a given node is then given by<br />

Ki+lAi+lAt<br />

T:+l-Tf = c(T;;’ -T;+l) - ‘iAiAtc (T:+’-T;;’) + ~ (10-125)<br />

micP,i m.c<br />

1P,]<br />

.<br />

micp,i<br />

The quantity ~ is the effective conductance between node i and i+l, derived by applying the flux<br />

continuity condition at the interface. This quantity depends upon the thermal conductivity <strong>of</strong> the two<br />

regions on either side <strong>of</strong> the interface, and the distances to the center <strong>of</strong> each node. This parameter<br />

is geometry-dependent because the distances to the node centers and the interface areas are both<br />

geometry-dependent. It can readily be shown that the conductance 1$ and the half-node thicknesses<br />

~ri <strong>for</strong> the three geometries are given by<br />

Ki .<br />

ki+l ‘1<br />

i5ri,~<br />

[$l-l;[ 1<br />

The 8ri and ~ri, ~depend on the structure geometry.<br />

For the slab geometry<br />

ihi = ~ri+l = ‘i+l - ‘i<br />

2<br />

For the cylindrical geometry<br />

where<br />

[1 i<br />

5ri = ri in –<br />

r.1<br />

[1<br />

?ki+l = ri+l in v<br />

r<br />

;=<br />

[1<br />

112<br />

ri~, + ri2<br />

2<br />

(10-126)<br />

(10-127)<br />

(10-128)<br />

(10-129)<br />

(10-130)<br />

Rev O 10-62 6/30/97

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