Consider convection and diffusion in one dimension as in the figure with = temperature, position...

  

Transcribed Image Text:

Consider convection and diffusion in one dimension as in the figure → with = temperature, position x, convection u and the boundary conditions as indicated. A source distribution S applies as in the figure → at t>0 with a = -200; b= 100, x₁=0.6 m, x2 = 0.2 m. The transient convection diffusion is described by (ρφ) θρυφ) _ ὃ ap = Ət əx əx ????х a. with Node aw 1 2 0 Length L = 1.5 m is divided into 45 grid points (8x = 0.0333 m), while F = pu=2.0, D=1/8x = 0.9. Show that the discretisation for a general node (except x=0 and x=L) gives Ax (p=62AX+F. Op + (30E-20p-w)] - Fu [ow + (30p-20w-ww)] = Dee - #r) - D₂(p = 0w) Fc[Op where op is the previous value for op; initially op=0 for all points. For point A: mirroring for node W (where =0) gives 4-op and For point B the zero gradient boundary gives B = OP. b. Show that this gives the discretisation as apop = awow+aEE + app + Su pax dp Ar ap = aw+ag + ap+ (Fe-Fw)-Sp and discretisation parameters are as given in the table below (next page). Source S is linearised to S+Spor. ag Dw+Fw De A 3-44 Dw+Fw De x-0 6-0 45 Dw+Fw 0 0 0 0 Sp De+4-D₁+F₁) (D₁+F₁) 0₁ + / Felp - 3 2-0 u-2.0 m/s Su a+b 1 ; ap ax X-L and show that the B F. (30p-w)+F(0w +20p-30g) Fw(30p-20w-ww) +Fe(Ow+20p-30E) F (30p-20w-ww) X-L do/dx-0 PA (90p-80A-DE) (9 Consider convection and diffusion in one dimension as in the figure → with = temperature, position x, convection u and the boundary conditions as indicated. A source distribution S applies as in the figure → at t>0 with a = -200; b= 100, x₁=0.6 m, x2 = 0.2 m. The transient convection diffusion is described by (ρφ) θρυφ) _ ὃ ap = Ət əx əx ????х a. with Node aw 1 2 0 Length L = 1.5 m is divided into 45 grid points (8x = 0.0333 m), while F = pu=2.0, D=1/8x = 0.9. Show that the discretisation for a general node (except x=0 and x=L) gives Ax (p=62AX+F. Op + (30E-20p-w)] - Fu [ow + (30p-20w-ww)] = Dee - #r) - D₂(p = 0w) Fc[Op where op is the previous value for op; initially op=0 for all points. For point A: mirroring for node W (where =0) gives 4-op and For point B the zero gradient boundary gives B = OP. b. Show that this gives the discretisation as apop = awow+aEE + app + Su pax dp Ar ap = aw+ag + ap+ (Fe-Fw)-Sp and discretisation parameters are as given in the table below (next page). Source S is linearised to S+Spor. ag Dw+Fw De A 3-44 Dw+Fw De x-0 6-0 45 Dw+Fw 0 0 0 0 Sp De+4-D₁+F₁) (D₁+F₁) 0₁ + / Felp - 3 2-0 u-2.0 m/s Su a+b 1 ; ap ax X-L and show that the B F. (30p-w)+F(0w +20p-30g) Fw(30p-20w-ww) +Fe(Ow+20p-30E) F (30p-20w-ww) X-L do/dx-0 PA (90p-80A-DE) (9

Answer rating: 100% (QA)

To show the discretization for a general node except x0 and xL we can start with the given equation ddx S Applying the finite difference method we can discretize the equation using the central differe... View the full answer