Question

Consider steady-state one-dimensional flow along an inclined flat plane, with a position-dependent viscosity. The dimensionless coordinate x is measured from the surface of the plane, and flow is in the direction z along the plane. In dimensionless variables, the viscosity is given by Î¼/Î¼0 = 1 - Î±(1 - x)^2, which leads to the differential equation (1 - Î±(1 - x)^2) d^2v/dx^2 + 2Î±(1 - x) dv/dx + Ïg cosÎ² = 0, and the boundary conditions dv/dx = 0 at x = 1, v = 0 at x = 0. We will set up the FEM equations using piecewise linear trial functions, dividing the domain into 3 elements of equal length. a) Re-write the differential equation in the form - d/dx[p(x) dv/dx] = f and derive the weak form similarly to the example done in class. The main difference will be that the second derivative yields the integral âˆ« p(x)Ï†i' Ï†j' dx from 0 to 1, which will lead to the stiffness matrix. b) Form the element stiffness matrices KeJI for this problem, with Î± = 0.5. You may use the Galerkin element matrices given in the class handout and apply them to the weak form of the differential equation. For a variable coefficient differential equation, the local element stiffness integral is approximated by âˆ«P(Î·)Ï†I' Ï†J' dx from 0 to 1 = P(0.5) âˆ«Ï†I' Ï†J' dÎ· from 0 to 1. c) Form the element RHS vector FeJ if Ïg cos(Î²) = 1; assemble the element stiffness matrices and RHS vectors into a system of algebraic equations for the unknown node values. d) Solve the system of equations by any method available to you.

          Consider steady-state one-dimensional flow along an inclined flat plane, with a position-dependent viscosity. The dimensionless coordinate x is measured from the surface of the plane, and flow is in the direction z along the plane. In dimensionless variables, the viscosity is given by Î¼/Î¼0 = 1 - Î±(1 - x)^2, which leads to the differential equation (1 - Î±(1 - x)^2) d^2v/dx^2 + 2Î±(1 - x) dv/dx + Ïg cosÎ² = 0, and the boundary conditions dv/dx = 0 at x = 1, v = 0 at x = 0. We will set up the FEM equations using piecewise linear trial functions, dividing the domain into 3 elements of equal length.

a) Re-write the differential equation in the form - d/dx[p(x) dv/dx] = f and derive the weak form similarly to the example done in class. The main difference will be that the second derivative yields the integral âˆ« p(x)Ï†i' Ï†j' dx from 0 to 1, which will lead to the stiffness matrix.

b) Form the element stiffness matrices KeJI for this problem, with Î± = 0.5. You may use the Galerkin element matrices given in the class handout and apply them to the weak form of the differential equation. For a variable coefficient differential equation, the local element stiffness integral is approximated by âˆ«P(Î·)Ï†I' Ï†J' dx from 0 to 1 = P(0.5) âˆ«Ï†I' Ï†J' dÎ· from 0 to 1.

c) Form the element RHS vector FeJ if Ïg cos(Î²) = 1; assemble the element stiffness matrices and RHS vectors into a system of algebraic equations for the unknown node values.

d) Solve the system of equations by any method available to you.

Added by Becky R.

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