Finite Volume Method for Diffusion Problem ENT 352

Finite Volume Method for Diffusion Problem ENT 352 CAED Dr. Rakhmad Arief Siregar 1

Introduction n In this section we develop the numerical method based on finite volume method by considering the simplest transport process (pure diffusion in the steady state. The governing equation of steady diffusion: n The control volume integration yield in form: n 2

Finite volume method for one dimensional steady state diffusion n Finite volume method becomes the method of choice for computational fluid dynamics software packages because: n n Applicable to unstructured mesh Relatively faster to calculate compared with FEM Better conservative properties Physically intuitive for fluid dynamixs 3

Finite volume method for one dimensional steady state diffusion n is the diffusion coefficient and S is the source term Boundary value of at point A and B are prescribe Control volume boundaries P A E B W Control volume Nodal points B = constant n Consider the steady state diffusion of a property in one dimensional domain A = constant n 4

Finite volume method for one dimensional steady state diffusion n Step 1. Grid generation x. WP x. PE xw. P W x. Pe w e E x= xwe 5

Finite volume method for one dimensional steady state diffusion n Step 2: discretisation Control volume to a discretised equation: n In a uniform grid linearly interpolated value for: n 6

Finite volume method for one dimensional steady state diffusion n Step 2: discretisation Diffusive flux n The source term S n 7

Finite volume method for one dimensional steady state diffusion n Step 2: discretisation Substituting eqs. (4. 6), (4. 7) and (4. 8) into eq. (4. 4) gives: n After re-arranged: n 8

Finite volume method for one dimensional steady state diffusion n Step 2: discretisation 9

Ex. 4. 1 n n Consider the problem of source-free heat conduction in an insulated rod whose ends are maintained at constant temperature 100 C and 500 C respectively Calculate the state temperature distribution in the rod. Thermal conductivity k equals 1000 W/m/K, A = 10 x 10 -3 m 2 0. 5 m A B A = 10 x 10 -3 m 2 TA = 100 C TB = 500 C 10

Solution n Step 1: Grid generation 1 2 3 4 5 TA TB x/2 x x x/2 11

Solution n n Step 1: Grid generation For node 2 (the same for 3 and 4) x. WP x. PE xw. P W=1 w x. Pe P =2 e E=3 x= xwe 12

Solution x. WP x. PE xw. P W=1 w x. Pe P =2 e E=3 x= xwe 13

Solution x. WP x. PE xw. P W=1 w x. Pe P =2 e E=3 x= xwe 14

Solution n n Step 1: Grid generation For node 1 15

Solution n n Step 1: Grid generation For node 5 16

Solution n For 17

Solution Node 2, 3, 4 Node 1, 5 Node 2 Node 1 Node 3 Node 5 Node 4 18

Solution By rearranging to matrix form Node 1, 5 Node 1 Node 2 Node 3 Node 4 Node 5 19

Solution n In Matrix form n T 1=140 C; T 2=220 C; T 3=300 C; T 4=380 C; T 5=460 C 20

Ex. 4. 2 n As. The same as 4. 1 but having uniform heat generation q=1000 k. W/m 3 21

Solution x. WP x. PE xw. P W=1 w x. Pe P =2 e E=3 x= xwe 22

Solution n n Step 1: Grid generation For node 1 23

Solution n n Step 1: Grid generation For node 5 24

Solution n For 25

Solution n In Matrix form n T 1=150 C; T 2=218 C; T 3=254 C; T 4=258 C; T 5=230 C 26

Ex. 4. 1 x n n Consider the problem of source-free heat conduction in an insulated rod whose ends are maintained at constant temperature 100 C and 500 C respectively Calculate the state temperature distribution in the rod. Thermal conductivity k equals 1500 W/m/K, A = 5 x 10 -3 m 2 0. 7 m A B A = 5 x 10 -3 m 2 TA = 150 C TB = 550 C 27

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