Page 162 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

154
On substituting the spatial approximation from Equation 6.11, Equation 6.13 finally
becomes

TRANSIENT HEAT CONDUCTION ANALYSIS
∂N i ∂N j ∂N i ∂N j ∂N i ∂N j
− k x (T ) T j (t) + k y (T ) T j (t) + k z (T ) T j (t) d

∂x ∂x ∂y ∂y ∂z ∂z

∂N j
+ N i G − N i ρc p T j (t) d
− N i qd q − N i h(T − T a )d q = 0 (6.15)

∂t q q
where i and j represent the nodes. Equation 6.15 can be written in a more convenient form
as
∂T

[C] + [K]{T}={f} (6.16)
∂t
or

∂T j
[C ij ] + [K ij ]{T j }= {f i } (6.17)
∂t
where

[C ij ] = ρc p N i N j d
(6.18)

∂N i ∂N j ∂N i ∂N j ∂N i ∂N j
[K ij ] = k x (T ) {T j }+ k y (T ) {T j }+ k z (T ) {T j } d

∂x ∂x ∂y ∂y ∂z ∂z

+ hN i N j d (6.19)

and
{f i }= N i Gd
− qN i d q + N i hT a d (6.20)

q q
In matrix form,

T
[C] = ρc p [N] [N]d
(6.21)

T T
[K] = [B] [D][B]d
+ h[N] [N]d (6.22)

and
T T T
{f}= G[N] d
− q[N] d q + hT a [N] d (6.23)

q
Since k x (T ), k y (T ) and k z (T ) are functions of temperature, Equation 6.16 is non-linear
and requires an iterative solution. If k x ,k y and k z are independent of temperature, then
Equation 6.16 is linear in form.

6.4 One-dimensional Transient State Problem

The relation derived in Equation 6.16 is employed here to illustrate the application to a
one-dimensional transient problem using a linear element as shown in Figure 6.2