College of Engineering and Computer Science Mechanical Engineering Department Mechanical Engineering 501B Seminar in Engineering Analysis Spring 2009 Class 14443 Instructor Larry Caretto April 20 Homework Solutions 1 Hoffman page 771 problem 23 Solve example 12 3 with T 0 0 0 oC T 1 0 200oC and Ta 100oC using equation 12 98 Compare results with the results obtained in Example 12 3 The solution to this problem with a uniform grid uses equation 12 98 on page 731 of Hoffman Q i 1 x 2 Q i 1 Q i x 2 Q i x 2 F i 1 F i x 2 2 yi 1 yi 1 1 yi 1 1 6 6 6 2 This equation solves the differential equation dy dx Qy F which is equation 12 65 on page 724 of Hoffman In the problem we are solving dT dx 2T 2Ta To do this we substitute T for y in the finite difference equation and we see that Q 2 and F 2Ta This gives the following equation 2 x 2 Ti 1 1 6 2 x 2 2Ti 1 3 2 x 2 Ti 1 1 6 2Ta x 2 In this problem 2 16 cm 2 and Ta 100oC giving the following numerical values for the terms in the finite element equation with the given grid spacing x 0 25 cm 2 x 2 1 16 1 0 25 cm 2 5 2 6 6 cm 6 2 2 x 1 16 2 1 2 1 0 25 cm 2 8 2 3 3 cm 3 16 2 2Ta x 2 2 100o C 0 25 cm 100o C cm 1 Substituting these values into our nodal equation gives 5 8 5 Ti 1 Ti Ti 1 100o C 6 3 6 At node 1 we have T1 0oC and at node 5 we have T5 200oC This gives the following equations for the first node in from these boundaries 5 o 8 5 0 C T2 T3 100 o C 6 3 6 8 5 T2 T3 100 o C 3 6 5 8 5 T3 T4 200o C 100o C 6 3 6 8 5 1600o C T3 T4 3 6 6 Jacaranda Engineering Room 3333 E mail lcaretto csun edu Mail Code 8295 Phone 818 677 6448 Fax 818 677 7062 The final set of nodal equations was solved by the matrix functions of Excel A copy of the spreadsheet with the input data and solutions is shown below Node 2 3 4 Matrix Coefficients 2 666667 0 833333 0 0 833333 2 666667 0 833333 0 0 833333 2 666667 RHS 100 100 266 66667 Solution 68 75 100 131 25 Example 4 306 13 29 36 709 The table above also shows the results from the original example with boundary conditions of 0 oC at x 0 and 100oC at x 1 and a value of Ta 0 If we only changed the right boundary condition from 100oC to 200oC we would expect the results to double We have obtained higher values than a simple doubling of the original results because the value of T a 100 acts as a source term for the equation increasing all values of T 2 Hoffman page 771 problem 25 Solve example 12 4 with T 0 0 0 oC T 1 0 200oC and Ta 100oC using equation 12 97 for a nonuniform grid Compare results with the results obtained in Example 12 4 Equation 12 97 gives the following finite element equation 1 1 Q i 1 xi 1 1 Q i 1 xi 1 Q i xi yi 1 2 yi x x x 6 3 3 i i 1 i 1 i i 1 i 1 Q xi F xi 1 F xi yi 1 x 6 2 i This equation solves the differential equation dy dx Qy F which is equation 12 65 on page 724 of Hoffman In the problem we are solving dT dx 2T 2Ta To do this we substitute T for y in the finite difference equation and we see that Q 2 and F 2Ta This gives the following equation 1 1 2 xi 1 1 2 xi 1 xi Ti 1 2Ti x x x 6 3 i i 1 i 1 2 2 1 xi Ta xi 1 xi Ti 1 x 6 2 i We can write this equation as follows using the definitions immediately below the rewritten equation 2Ta xi 1 xi 2 2 1 1 xi 1 xi 2 Ai 1 Ai Bi 2 x x 3 i i 1 Ai 1Ti 1 BiTi AiTi 1 Di 1 2 xi Ai xi 6 Jacaranda Engineering Room 3333 E mail lcaretto csun edu Mail Code 8295 Phone 818 677 6448 Fax 818 677 7062 The grid specified for this problem has five nodes The Dirichlet boundary conditions give us the values of T1 0oC and T5 200oC We have to solve for T2 T3 and T4 The x values and the corresponding coefficients are shown in the table below i 1 2 3 4 xi cm 0 375 0 29166667 0 20833333 0 125 Ai cm 1 1 66666667 2 65079365 4 24444444 7 66666667 Bi cm 1 9 650794 10 89524 14 57778 Di oC cm 533 33333 400 266 66667 We can use these coefficients in the finite element equations In the equation for node 2 the temperature T1 is zero by the boundary condition In the equation for node 4 the term A 4T4 7 666667 cm 1 200 oC because of the boundary condition T5 200oC This term is moved to the right side of the equation and subtracted from the value of D 4 266 667 oC cm giving a value of 1800 oC cm for the right side of the equation The coefficients in the nodal equations are shown below in a copy of the Excel spreadsheet used to obtain the solutions to this problem Node 2 3 4 Matrix Coefficients 9 650794 2 650794 0 2 650794 10 895238 4 244444 0 4 244444 14 57778 RHS 533 333 400 1800 Solution 88 18728 119 86747 158 37605 Example 6 865 24 9931 59 8684 The results from the original example with Ta 0 and T x 1 0 100oC are also shown in the table above As in the previous problem we see that the addition of the source term increased the results above those expected for the change in the boundary condition alone 3 Redo example 12 5 on page 738 of Hoffman using the following boundary condition at x 1 Keep all the other data and grid spacing the same dT bT c dx a 1 cm b 1 dimensionless a c 1o C For this example the finite element equation with the right side boundary condition is given by equation 12 133 on page 737 of Hoffman Q N x 2 y N 1 1 2 Q N x 2 y N 1 2 F N x 2 dy x 2 dx x N For the example Q 2 16 cm 2 F 0 N …