ME 608 Numerical Methods in Heat, Mass, and Momentum Transfer

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1 ME 608 Numerical Methods in Heat, Mass, and Momentum Transfer Solution to Final Examination Date: May 4, :00 3:00 PM Instructor: J. Murthy Open Book, Open Notes Total: 100 points Use the finite volume method in all problems. 1. Consider steady convection and diffusion of a scalar φ in a square domain, as shown in Fig.??. The governing equation is given by: (ρvφ) = (Γ φ) + (1 φ) The left and bottom boundaries are at φ = 0 and φ = 1 respectively, while the other two boundaries are outflow boundaries. The flow field is given by V = i + j. The density ρ = 1 and Γ = 1.0. The side of the square is given by L = 2. (a) Derive the discrete algebraic equations for φ for the four cell centroids shown at the finest mesh level in Fig.??. Use a first-order upwind difference scheme for the convective operator. (b) Now derive the discrete algebraic equations for the corrections for the coarse levels 1 and 2 in Fig.?? using the algebraic multigrid procedure developed in class. (c) Assuming an initial guess at the finest level of φ = 0.5, execute one V cycle with ν 1 = 0 and ν 2 = 1. Show your answers at each level clearly. All units may be assumed to be consistent, so that no unit conversion is required. 1

2 =0 Outflow 4 3 y 1 2 V = i + j x =1 Outflow L=2 Level Level 1 Level 2 Figure 1: Computational Domain for Problem 1 2

3 (a) Discrete Equations at Level 0 Cell 1 Cell 2 Cell 3 Γ y x (φ 2 φ 1 ) + (ρu y) e φ 1 + 2Γ y ( ) φ1 φ le ft (ρu y)w φ le ft x Γ x y (φ 4 φ 1 ) + (ρv x) n φ 1 + 2Γ x y (φ 1 φ bot ) (ρv x) s φ bot = (1 φ 1 ) x y (ρu y) e φ 2 + Γ y x (φ 2 φ 1 ) (ρu y) w φ 1 Γ x y (φ 3 φ 2 )+ (ρv x) n φ 2 + 2Γ x y (φ 2 φ bot ) (ρv x) s φ bot = (1 φ 2 ) x y Cell 4 (ρu y) e φ 3 + Γ y x (φ 3 φ 4 ) (ρu y) w φ 4 + Γ x y (φ 3 φ 2 ) (ρv x) s φ 2 + (ρv x) n φ 3 = (1 φ 3 ) x y (ρu y) e φ 4 Γ y x (φ 3 φ 4 ) + (ρv x) n φ 4 + 2Γ y ( ) φ4 φ le ft (ρu y)w φ le ft + x Γ x y (φ 4 φ 1 ) (ρv x) s φ 1 = (1 φ 4 ) x y Substituting values, we obtain: (b) Discrete Equations at Level 1 9φ 1 = φ 2 + φ φ 2 = 2φ 1 + φ φ 3 = 2φ 4 + 2φ φ 4 = φ 3 + 2φ (c) Discrete Equations at Level 2 13x 1 1 = 2x r 1 1 9x 1 1 = 4x b 1 2 (d) V-cycle Computation Evaluating residuals at level 0 with φ = 0.5 we obtain 16x 2 1 = b 2 1 r 0 1 = 0.5 r 0 2 = 2.0 r 0 3 = 0.5 r 0 4 = 1.0 3

4 Restricting residuals to level 1, we obtain r 1 1 = r r0 4 = 0.5 r 1 2 = r r0 3 = 2.5 Restricting residuals to level 2, we obtain r 2 1 = r r 1 2 = 2.0 The final equation set is: Level 0 9φ 1 = φ 2 + φ φ 2 = 2φ 1 + φ φ 3 = 2φ 4 + 2φ φ 4 = φ 3 + 2φ Level 1 13x 1 1 = 2x x 1 1 = 4x Level 2 16x 2 1 = 2 Performing ν 2 = 1 relaxation sweeps at level 2 we obtain x 2 1 = Prolongating errors to level 1, and using a zero initial guess for level 1 error, we obtain x 1 1 = x 1 2 = Performing ν 2 = 1 relaxation sweeps at level 1, we obtain x 1 1 = x 1 2 = = ( ) = Prolongating errors to level 0, and assuming an intial guess of φ = 0.5, we obtain x 0 1 = x 0 2 = x 0 3 = x 0 4 =

5 Performing ν 2 = 1 relaxation sweeps at level 0, we obtain φ 1 = φ 2 = φ 3 = φ 4 = = = = =

6 u=1, v=2 u=2, v=1 2. Consider Darcy flow through a porous medium in the square domain of side L = 2 shown in Fig.??. The equations governing the motion are given by: (ρv) = 0 p µv = 0 Here, the density ρ = 1.0, viscosity µ = 1.0 and the permeability = The left and bottom boundaries of the domain have a velocity vector specified as V = i + 2j. The top and right boundaries have a velocity vector specified as V = 2i + j. Use a staggered pressure-velocity scheme and the SIMPLE algorithm to solve this problem. (a) Develop the discrete u and v momentum equations for the appropriate faces. (b) Develop the discrete p equations for the appropriate cells. (c) Execute one iteration of the SIMPLE algorithm and report cell pressures and face velocities. (d) Do you need to iterate? Why or why not? (e) If we were to solve the Darcy-Brinkman equation for porous media, given by: (ρv) = 0 p µv + (µ V) = 0 would you have to iterate if you used the SIMPLE algorithm? Why or why not? u=2, v=1 f6 f5 f7 d 4 c b 3 f4 L=2 y f8 a 1 2 f1 f2 x u=1, v=2 f3 Figure 2: Computational Domain for Problem 2 All units may be assumed to be consistent, so that no unit conversion is required. 6

7 (a) U-momentum Equation Cell a: 0 = p x µu 0 = (p 2 p 1 ) y µu a x y a Pa u a = (p 1 p 2 ) a Pa = µ x d a = 1 a Pa u ( a = d a p 1 p ) 2 Cell c: 0 = p x µu 0 = (p 3 p 4 ) y µu c x y a Pc u a = (p 4 p 3 ) a Pc = µ x d c = 1 a Pc u ( c = d c p 4 p ) 3 V-momentum Equation Cell b: 0 = p y µv 0 = (p 3 p 2 ) x µv b x y a Pb v b = (p 2 p 3 ) a Pb = µ y d b = 1 a Pb v ( b = d b p 2 p ) 3 Cell d: 7

8 0 = p y µv 0 = (p 4 p 1 ) x µv d x y a Pd v d = (p 1 p 4 ) a Pd = µ y 1 d d = a Pd v ( d = d d p 1 p ) 4 (b) Continuity Equation Cell 1: We have all-velocity boundary conditions. Therefore we have one cell too many for continuity and the pressure level is arbitrary. Set p 1 = 0 to fix the level of pressure. Cell 2: Cell 3: Cell 4: F f 3 Fa ( ρ yd a p 1 p ) 2 +F b + ρ xd b ( p 2 p 3) Ff 2 (ρ yd a + ρ xd b ) p 2 = ρ yd a p 1 + ρ xd b p 3 + ( F a F b + F f 2 F f 3 ) (ρ yd c + ρ xd b ) p 3 = ρ yd c p 4 + ρ xd b p 2 + ( F c + F b F f 4 F f 5 ) (ρ yd c + ρ xd d ) p 4 = ρ yd c p 3 + ρ xd b p 1 + ( F d F c + F f 7 F f 6 ) (c) SIMPLE Iteration U-momentum Solution: With p = 0 everywhere, we obtain u a = u c = 0. V-momentum Solution: With p = 0 everywhere, we obtain v b = v d = 0. P Solution: All values of a P for x = y are a P = µ x = 100. Thus all d values are Substituting numerical values, we 8

9 obtain: F a = 0 F b = 0 F c = 0 F d = 0 F f 1 = 2 F f 2 = 2 F f 3 = 2 F f 4 = 2 F f 5 = 1 F f 6 = 1 F f 7 = 1 F f 8 = 2 and p 2 = 10 2 p 3 + (Fa Fb ) p 3 = 10 2 p p 2 + (Fc + Fb 3) p 4 = 10 2 p 3 + (Fd F c ) Exploiting symmetry, we set p 2 = p 4. Solving for the pressure correction, we obtain: Correcting velocities and pressures we obtain: and p 1 = 0 p 2 = 150 p 3 = 300 p 4 = 150 u a = 1.5 u c = 1.5 v b = 1.5 v d = 1.5 p 1 = 0 p 2 = 150 p 3 = 300 p 4 = 150 It is easy to see that the solution is correct simply from the symmetries in the problem. The velocities are clearly continuity-satisfying. The momentum equations are also seen to be satisfied identically. (d) We do not need to iterate, as seen above. The problem is linear, so no iteration is required because of nonlinearities. The approximations in the SIMPLE algorithm (i.e., assuming nb a nb u nb = 0 and nb a nb v nb = 0 ) are 9

10 not necessary here because we have neither convection nor diffusion terms. Furthermore, even though this is 2D problem, and we are solving u and v sequentially, the two equations are not coupled, as they usually would be through the convective terms. Therefore there is no price to pay for a sequential solution. Thus we expect to obtain the answer in one iteration. (e) The Darcy-Brinkman equation has a diffusion term. This would necessitate making the SIMPLE approximations nb a nb u nb = 0 and nb a nb v nb = 0. Therefore iteration would be required. 10

ME 608 Numerical Methods in Heat, Mass, and Momentum Transfer. q 0 = εσ ( T 4 T 4) Figure 1: Computational Domain for Problem 1.

ME 608 Numerical Methods in Heat, Mass, and Momentum Transfer. q 0 = εσ ( T 4 T 4) Figure 1: Computational Domain for Problem 1. ME 68 Numerical Methods in Heat, Mass, and Momentum Transfer Mid-Term Examination Solution Date: March 1, 21 6: 8: PM Instructor: J. Murthy Open Book, Open Notes Total: 1 points 1. Consider steady 1D conduction