NUMERICAL SOLUTION OF THE 1- D DIFFUSION EQUATION (39)

Transcription

1 4/4/15 NUMERICAL SOLUTION OF THE 1- D DIFFUSION EQUATION (39) I Main Topics A MoCvaCon for using a numerical technique B Non- dimensionalizing the diffusion (heat flow) equacon C Finite- difference solucon to the 1- D heat equacon (diffusion equacon) 4/4/15 GG454 1 II MoCvaCon for using a numerical technique A Insight into the second order PDE governing transient flow B Insight into effect of inical condicons and boundary condicons on the solucon C To solve for a wide range of inical value/ boundary value combinacons and geometries D Finite- difference method a good learning tool 4/4/15 GG454 1

2 4/4/15 III Non- dimensionalizing the heat flow equacon A Start with the heat flow equacon, where T = temperature (variable) t = Cme (variable) =thermal diffusivity (constant) x = posicon (variable) B "Nondimensionalizing" (or scaling) eliminates t = T x 4/4/15 GG454 3 III Non- dimensionalizing the heat flow equacon C Select constant scaling terms ( dimensions:length) T max ( dimensions:! K) ( dimensions:time) D Define dimensionless terms x* = x T * = T T max t* = t t = T x 4/4/15 GG454 4

3 4/4/15 III Non- dimensionalizing the heat flow equacon t = T x C Select constant scaling terms ( dimensions:length) T max ( dimensions:! K) ( dimensions:time) D Define dimensionless terms x* = x T * = T T max t* = t E Recast dimensioned terms x = x * T = T *T max t = t * F Take derivacves of x*, t*, and T dx * dx = 1 dt * dt = dt dt * = T max 4/4/15 GG454 5 III Non- dimensionalizing the heat flow equacon F DerivaCves of x*, t*, and T (from previous slide) dx * dx = 1 G Recast derivacves in heat flow equacon using chain rule t = dt * dt t * dt dt * = t * ( ) T max x = dx * dt x * dx dt * = 1 T max x * T x dt * dt = = x x = x x * dx * dx = t = T x dt dt * = T max 1 T t * x max max x * 1 = T * ( ) x * T max 4/4/15 GG

4 4/4/15 III Non- dimensionalizing the heat flow equacon t = T x G Dimensioned derivacves in heat flow equacon t = t * T max T ( ) x = T * T max x * H SubsCtute for dimensioned derivacves in heat flow equacon to obtain dimensionless form t * T max ( ) = T * T max x * t * = T * x * ( ) All terms here are dimensionless ( ) 4/4/15 GG454 7 A Set up a dimensionless grid 1 Let Δx* = Δt* =1 The row number I gives the Cme step 3 The column number j gives the posicon 4/4/15 GG

5 4/4/15 B Explicit method 1 Approximate */ t* [T *(x*,t *+Δt*) T *(x*,t*)] t * Δt * [T * T * ] i+1, j i, j t * i+1/, j Δt * Approximate T*/ x* T * x * = T * x * i, j x * x * x * T * i, j+1 T * i, j i, j+1/ x * T * i, j T * i, j 1 i, j 1/ Row i+1 Row i = T * i, j+1 T * i, j T * i, j 1 ( ) t * = T * x * Problem: approximacons are at two different points, so equacng them is problemacc. 4/4/15 GG454 9 B Explicit method 1 Approximate */ t* [T * T * i+1, j i, ] j t * i+1/, j Δt * Approximate T*/ x* T * x * i, j T * i, j+1 T * i, j T * i, j 1 ( ) 3 Set the derivacves equal 4 Solve for T* i+1,j 5 SoluCon prone to numerical error: approximacons are at two different points Row i+1 Row i t * = T * x * 4/4/15 GG

6 4/4/15 C Crank- Nicolson method 1 Evaluate T/ x at by averaging the second derivacves at! 4/4/15 GG Approximate */ t* at " T * + i, j t * i+1/, j Δt * 3 Approximate T*/ x* at by averaging values at! T * x * i+1/, j T * x * i+1/, j T * x * i+1/, j 1 T * + T * x * i+1, j x * i, j T * T * T * i, j+1 i, j i, j 1 ( ) ( ) T * i, j T * i, j 1 ( ) 4/4/15 GG

7 4/4/15 4 Now equate the dimensionless parcal derivacves, selng dx* = dt* = 1 ( ) T * i, j T * i, j +T * i, j 1 t * = T * x * 5 Solve for T* i+1,j 1 ( T * T * +T * +T * ++1 i+1, j i+1, j 1 i, j+1 i, j 1) T * i, j T * i, j T * i, j 1 4 4/4/15 GG T * i, j The value of T at any node is equal to the average value of the two adjacent nodes at the same Cme step and the two nodes at the preceding Cme step 4/4/15 GG

8 4/4/15 D Recap 1 Convert dimensioned problem to dimensionless problem Solve the dimensionless problem 3 Convert dimensionless temperature solucon back to dimensioned solucons by mulcplying by the scaling factor x* = x T * = T T max t* = t T * i, j 1 4 x = x * T = T *T max t = t * 4/4/15 GG E Closing comments 1 The finite- difference method is essencally an averaging procedure that describes how "informacon" propagates within a system SoluCons for - D Laplace equacon and 1- D heat equacon have revealing similarices and revealing differences T x + T y = 0 t = T x The value of T at the red node is approximately the average of the T values at the blue nodes 4/4/15 GG