**Objectives**

- Adapt the direct solve template for Lab 3 to the axisymmetric convective heat transfer problem statement for the fin tube cylinder.
- Assume the fin geometry is appropriate for a low horsepower (~3 hp) lawnmower air-cooled engine. Keep
*h*and T_{hot}from Lab 3, select the thermal conductivity for alumninum, and assume the fins see ambient air (300 K). For these data, execute the resultant FE*k*=1 solution algorithm for a sequence of uniformly refined meshes starting with M=2 in the wall and M=6 in the fin. - Recalling the Lab 5 Newton template, revise the Part A template to a Newton template. Verify one-step iterative convergence occurs for one or more meshes used in Part B.
- Redefine the Part A statement with replacement of the cylinder surface linear convection BC,
*h*(T-T_{hot}), with the radiation BC, ac(T^{4}-T_{hot}^{4}). Here, a~1 is an approximation for viewfactor times emissivity and c is the Stefan-Boltzmann constant. The fins remain under convection heat transfer with ambient air. - Revise the Part C template for the problem statement of Part D with speciic accounting for the radiation BC non-linearity.
- Repeat one or more Part C executions using the Part E template.

**Problem Statement **

(Direct Solve)

Figure 1: Axisymmetric fin tube schematic. All dimensions are in millimeters.

As shown in Figure 1, a fin geometry has been chosen appropriate for a low horsepower lawnmower air-cooled engine. The convection heat transfer coefficient and the pipe fluid temperature from Lab #3 are used. The ambient air over the fins has the same convection coefficient at a temperature of 300 K. Furthermore, the fin material is assumed to consist of pure Aluminum with a constant thermal conductivity (k=237 W/m-K).

With an element averaged thickness and with convective boundary conditions as shown, the MATLAB FEMLIB pseudo-code for the element assembled Galerkin Weak Statement becomes

per the development shown in the course test (CM1.63-68). It should be noted that the second and third terms of this statement describe the convective loading at the pipe wall. They thus only modify the first node. The fourth and fifth terms of the statement describe the convective loading along both sides of the fin. They thus only modify the fin nodes.

The MATLAB program listed below performs a uniform mesh refinement study along the fin with an initial case of 2 elements along the pipe wall and 6 elements along the fin. The fin thickness is treated as element averaged data. As already mentioned, care is taken to apply the boundary convection loads to the appropriate nodes in the diffusion and residual matrices.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % Joseph Tipton % % ES551, Midterm - Part A % % 1-D, steady-state, conduction in an axisymmetric fin w/ convective BCs % % (using direct solve with a linear Lagrange FE basis) % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; clc; global X1; % array for node coordinates % constant data h = 20/0.17612; % convection coefficient (W/m^2-K) T_hot = (5/9)*(1500 + 459.67); % pipe fluid temperature (K) T_air = 300; % air temperature (K) k = 237; % Al conduction coefficient (W/m-K) r_pipe = 25.4 / 1000; % radius of pipe (m) r_wall = 12.7 / 1000; % length of wall (m) r_fin = 38.1 / 1000; % length of fin (m) THK_wall = 12.7 / 1000; % wall thickness (m) THK_fin_ini = 3.175 / 1000; % initial fin thickness (m) THK_fin_end = 1 / 1000; % final fin thickness (m) m_wall = 2; % desired initial mesh across wall m_fin = 6; % desired initial mesh across fin for i = 1:10 % compute mesh refinement M_wall = m_wall * 2^(i-1); M_fin = m_fin * 2^(i-1); elements(i,1) = M_wall + M_fin; % non-uniform element discretization R_wall = linspace(r_pipe,r_wall+r_pipe,M_wall + 1); R_fin = linspace(r_wall+r_pipe,r_fin+r_pipe+r_wall,M_fin + 1); X1 = cat(2,R_wall,R_fin(2:end)); nnodes = size(X1,2); % element averaged thickness for j = 1:M_wall THK(1,j) = THK_wall; end for j = 1:M_fin THK(1,M_wall+j) = THK_fin_ini - ((THK_fin_ini - THK_fin_end)/r_fin/2)*(X1(j+M_wall+1) + X1(j+M_wall) - 2*r_pipe - 2*r_wall); end % load the FE matrix library load femlib; % assemble diffusion matrix DIFF = asjac1D(k,THK',X1',-1,A3011L,[]); % complete Matrix for [HBC] % HBC on fin elements only X1 = R_fin; DIFFHBC = zeros(nnodes,nnodes); DIFFHBC(M_wall+1:end,M_wall+1:end) = asjac1D(2*h,[],X1',1,A3000L,[]); DIFFHBC = DIFFHBC + DIFF; X1 = cat(2,R_wall,R_fin(2:end)); % complete Matrix for [HBC] on element 1 DIFFHBC(1,1) = DIFFHBC(1,1) + (h * r_pipe * THK(1,1)); % set up data matrix (b) % modify b for the convective BCs along the fin surface T_AIR(1,1:nnodes) = T_air; b = asres1D(2*h,[],X1',1,A3000L,T_AIR); b(1:M_wall+1,1) = 0; % modify b for convective BC at node 1 b(1,1) = h * r_pipe * T_hot * THK(1,1); % solve the linear system Q = DIFFHBC \ b; % compute Energy Norm enorm(i,1) = 0.5*Q'*DIFFHBC*Q; % save max temperatures Qmax(i,1) = max(Q); % plot results plot(X1*1000,Q,':') hold on end delta_Qmax = Qmax(2:end) - Qmax(1:end-1); slope_Qmax = log10(delta_Qmax(1:end-1) ./ delta_Qmax(2:end)) / log10(2); delta_enorm = enorm(2:end) - enorm(1:end-1); slope_enorm = log10(delta_enorm(1:end-1) ./ delta_enorm(2:end)) / log10(2); ANSWER(:,1) = elements; ANSWER(:,2) = Qmax; ANSWER(:,3) = [0;delta_Qmax]; ANSWER(:,4) = [0;0;slope_Qmax]; ANSWER(:,5) = enorm; ANSWER(:,6) = [0;delta_enorm]; ANSWER(:,7) = [0;0;slope_enorm]; fprintf('Elements Qmax (Qmax)change (Qmax)slope ||Q|| change||Q|| Slope(Q)\n\n') fprintf('%6.0f %12.6g %12.6g %12.6g %12.6g %12.6g %12.6g\n',ANSWER'); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(Newton Template Implimentation)

While Part A directly solved for the temperature, Part C desires to impliment the Newton Method to obtain the same problem solution. The Newton Method solution format is

where {FQ} signifies the Galerkin Weak Statement (GWS^{h}). The Jacobian is the partial derivative of the weak statement (WS^{h}) with respect to the temperature, Q.

The relationship above allows for an iterative solution of the temperature until the change reaches some predefined threshold.

For the previously described linear problem statement, the Jacobian matrix simply becomes

The MATLAB program below follows the same structure as in Part A but impliments Newton's Method to solve for the temperature profile. The initial temperature guess is simply a linear nodal variation from the pipe fluid temperature to the ambient air temperature.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % Joseph Tipton % % ES551, Midterm - Part C % % 1-D, steady-state, conduction in an axisymmetric fin w/ convective BCs % % (using Newton's Method with a linear Lagrange FE basis) % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; clc; global X1; % array for node coordinates % constant data h = 20/0.17612; % convection coefficient (W/m^2-K) T_hot = (5/9)*(1500 + 459.67); % pipe fluid temperature (K) T_air = 300; % air temperature (K) k = 237; % Al conduction coefficient (W/m-K) r_pipe = 25.4 / 1000; % radius of pipe (m) r_wall = 12.7 / 1000; % length of wall (m) r_fin = 38.1 / 1000; % length of fin (m) THK_wall = 12.7 / 1000; % wall thickness (m) THK_fin_ini = 3.175 / 1000; % initial fin thickness (m) THK_fin_end = 1 / 1000; % final fin thickness (m) m_wall = 2; % desired initial mesh across wall m_fin = 6; % desired initial mesh across fin % set max iteration count and convergence criteria maxit = 20; % max iterations eps = .001; % convergence tolerance for i = 1:10 % compute mesh refinement M_wall = m_wall * 2^(i-1); M_fin = m_fin * 2^(i-1); elements(i,1) = M_wall + M_fin; % non-uniform element discretization R_wall = linspace(r_pipe,r_wall+r_pipe,M_wall + 1); R_fin = linspace(r_wall+r_pipe,r_fin+r_pipe+r_wall,M_fin + 1); X1 = cat(2,R_wall,R_fin(2:end)); nnodes = size(X1,2); % element averaged thickness for j = 1:M_wall THK(1,j) = THK_wall; end for j = 1:M_fin THK(1,M_wall+j) = THK_fin_ini - ((THK_fin_ini - THK_fin_end)/r_fin/2)*(X1(j+M_wall+1) + X1(j+M_wall) - 2*r_pipe - 2*r_wall); end % load the FE matrix library load femlib; % initial temperature guess Q = (linspace(T_hot,T_air,nnodes))'; % iteration loop for j = 1:maxit; % complete residual [RES] matrix % HBC on fin elements only X1 = R_fin; RES = zeros(nnodes,1); RES(M_wall+1:end,1) = asres1D(2*h,[],X1',1,A3000L,Q(M_wall+1:end)); X1 = cat(2,R_wall,R_fin(2:end)); % add DIFF matrix to the HBC matrix RES = RES + asres1D(k,THK',X1',-1,A3011L,Q); % complete Matrix for [HBC] on element 1 RES(1,1) = RES(1,1) + (h * r_pipe * THK(1,1)*Q(1,1)); % set up data matrix (b) % modify b for the convective BCs along the fin surface T_AIR(1,1:nnodes) = T_air; b = asres1D(2*h,[],X1',1,A3000L,T_AIR); b(1:M_wall+1,1) = 0; % modify b for convective BC at node 1 b(1,1) = h * r_pipe * T_hot * THK(1,1); RES = RES - b; % complete jacobian [JAC] matrix % assemble diffusion matrix DIFF = asjac1D(k,THK',X1',-1,A3011L,[]); % complete Matrix for [HBC] % HBC on fin elements only X1 = R_fin; DIFFHBC = zeros(nnodes,nnodes); DIFFHBC(M_wall+1:end,M_wall+1:end) = asjac1D(2*h,[],X1',1,A3000L,[]); DIFFHBC = DIFFHBC + DIFF; X1 = cat(2,R_wall,R_fin(2:end)); % complete Matrix for [HBC] on element 1 DIFFHBC(1,1) = DIFFHBC(1,1) + (h * r_pipe * THK(1,1)); JAC = DIFFHBC; dQ = JAC\(-1*(RES)); % linear solve for dQ Q = Q + dQ; % update Q dQmax = max(abs(dQ)); % convergence criteria if dQmax < eps % test for convergence against tolerance eps break % quit if converged end end % save iteration convergence count iteration(i,1) = j; % save max temperatures Qmax(i,1) = max(Q); % calculate and display energy norm. % res must be recalculated using the converged solution X1 = R_fin; RES = zeros(nnodes,1); RES(M_wall+1:end,1) = asres1D(2*h,[],X1',1,A3000L,Q(M_wall+1:end)); X1 = cat(2,R_wall,R_fin(2:end)); RES = RES + asres1D(k,THK',X1',-1,A3011L,Q); RES(1,1) = RES(1,1) + (h * r_pipe * THK(1,1)*Q(1,1)); T_AIR(1,1:nnodes) = T_air; b = asres1D(2*h,[],X1',1,A3000L,T_AIR); b(1:M_wall+1,1) = 0; b(1,1) = h * r_pipe * T_hot * THK(1,1); RES = RES - b; enorm(i,1) = 0.5 * Q' * RES; % plot results plot(X1*1000,Q,':') hold on end delta_Qmax = Qmax(2:end) - Qmax(1:end-1); slope_Qmax = log10(delta_Qmax(1:end-1) ./ delta_Qmax(2:end)) / log10(2); delta_enorm = enorm(2:end) - enorm(1:end-1); slope_enorm = log10(delta_enorm(1:end-1) ./ delta_enorm(2:end)) / log10(2); ANSWER(:,1) = elements; ANSWER(:,2) = iteration; ANSWER(:,3) = Qmax; ANSWER(:,4) = [0;delta_Qmax]; ANSWER(:,5) = [0;0;slope_Qmax]; ANSWER(:,6) = enorm; ANSWER(:,7) = [0;delta_enorm]; ANSWER(:,8) = [0;0;slope_enorm]; fprintf('Elements Iterations Qmax (Qmax)change (Qmax)slope ||Q|| change||Q|| Slope(Q)\n\n') fprintf('%6.0f %6.0f %12.6g %12.6g %12.6g %12.6g %12.6g %12.6g\n',ANSWER'); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Pipe Wall Radiation BC

(Newton Method Solver)

Finally, the pipe wall boundary condition is changed from convective to radiative heat transfer. Since radiation varies to the fourth power of the temperature, the boundary conditions introduce a non-linearity to the problem that prohibits a simple direct temperature solution.

Instead, Newton's Method provides the capability to solve the problem statement with the nonlinearities. The Jacobian now becomes

With the Newton's Method code already created in Part C, implimentation of the new boundary conditions is simple. The MATLAB code below details this implimentation.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % Joseph Tipton % % ES551, Midterm - Part E % % 1-D, steady-state, conduction in an axisymmetric fin w/ a radiation BC % % (using Newton's Method with a linear Lagrange FE basis) % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; clc; global X1; % array for node coordinates % constant data alpha = 1.0; % radiosity sigma = 5.670E-8; % Stefan-Boltzmann Constant (W/m^2-K^4) h = 20/0.17612; % convection coefficient (W/m^2-K) T_hot = (5/9)*(1500 + 459.67); % pipe fluid temperature (K) T_air = 300; % air temperature (K) k = 237; % Al conduction coefficient (W/m-K) r_pipe = 25.4 / 1000; % radius of pipe (m) r_wall = 12.7 / 1000; % length of wall (m) r_fin = 38.1 / 1000; % length of fin (m) THK_wall = 12.7 / 1000; % wall thickness (m) THK_fin_ini = 3.175 / 1000; % initial fin thickness (m) THK_fin_end = 1 / 1000; % final fin thickness (m) m_wall = 2; % desired initial mesh across wall m_fin = 6; % desired initial mesh across fin % set max iteration count and convergence criteria maxit = 20; % max iterations eps = .001; % convergence tolerance for i = 1:10 % compute mesh refinement M_wall = m_wall * 2^(i-1); M_fin = m_fin * 2^(i-1); elements(i,1) = M_wall + M_fin; % non-uniform element discretization R_wall = linspace(r_pipe,r_wall+r_pipe,M_wall + 1); R_fin = linspace(r_wall+r_pipe,r_fin+r_pipe+r_wall,M_fin + 1); X1 = cat(2,R_wall,R_fin(2:end)); nnodes = size(X1,2); % element averaged thickness for j = 1:M_wall THK(1,j) = THK_wall; end for j = 1:M_fin THK(1,M_wall+j) = THK_fin_ini - ((THK_fin_ini - THK_fin_end)/r_fin/2)*(X1(j+M_wall+1) + X1(j+M_wall) - 2*r_pipe - 2*r_wall); end % load the FE matrix library load femlib; % initial temperature guess Q = (linspace(T_hot,T_air,nnodes))'; % iteration loop for j = 1:maxit; % complete residual [RES] matrix % HBC on fin elements only X1 = R_fin; RES = zeros(nnodes,1); RES(M_wall+1:end,1) = asres1D(2*h,[],X1',1,A3000L,Q(M_wall+1:end)); X1 = cat(2,R_wall,R_fin(2:end)); % add DIFF matrix to the HBC matrix RES = RES + asres1D(k,THK',X1',-1,A3011L,Q); % complete Matrix for [HBC] on element 1 RES(1,1) = RES(1,1) + (alpha * sigma * r_pipe * THK(1,1) * Q(1,1)^4); % set up data matrix (b) % modify b for the convective BCs along the fin surface T_AIR(1,1:nnodes) = T_air; b = asres1D(2*h,[],X1',1,A3000L,T_AIR); b(1:M_wall+1,1) = 0; % modify b for convective BC at node 1 b(1,1) = alpha * sigma * r_pipe * T_hot^4 * THK(1,1); RES = RES - b; % complete jacobian [JAC] matrix % assemble diffusion matrix DIFF = asjac1D(k,THK',X1',-1,A3011L,[]); % complete Matrix for [HBC] % HBC on fin elements only X1 = R_fin; DIFFHBC = zeros(nnodes,nnodes); DIFFHBC(M_wall+1:end,M_wall+1:end) = asjac1D(2*h,[],X1',1,A3000L,[]); DIFFHBC = DIFFHBC + DIFF; X1 = cat(2,R_wall,R_fin(2:end)); % complete Matrix for [HBC] on element 1 DIFFHBC(1,1) = DIFFHBC(1,1) + (4 * alpha * sigma * r_pipe * THK(1,1) * Q(1,1)^3); JAC = DIFFHBC; dQ = JAC\(-1*(RES)); % linear solve for dQ Q = Q + dQ; % update Q dQmax = max(abs(dQ)); % convergence criteria if dQmax < eps % test for convergence against tolerance eps break % quit if converged end end % save iteration convergence count iteration(i,1) = j; % save max temperatures Qmax(i,1) = max(Q); % calculate and display energy norm. % res must be recalculated using the converged solution X1 = R_fin; RES = zeros(nnodes,1); RES(M_wall+1:end,1) = asres1D(2*h,[],X1',1,A3000L,Q(M_wall+1:end)); X1 = cat(2,R_wall,R_fin(2:end)); RES = RES + asres1D(k,THK',X1',-1,A3011L,Q); RES(1,1) = RES(1,1) + (alpha * sigma * r_pipe * THK(1,1) * Q(1,1)^4) T_AIR(1,1:nnodes) = T_air; b = asres1D(2*h,[],X1',1,A3000L,T_AIR); b(1:M_wall+1,1) = 0; b(1,1) = alpha * sigma * r_pipe * T_hot^4 * THK(1,1); RES = RES - b; enorm(i,1) = 0.5 * Q' * RES; % plot results plot(X1*1000,Q,':') hold on end delta_Qmax = Qmax(2:end) - Qmax(1:end-1); slope_Qmax = log10(delta_Qmax(1:end-1) ./ delta_Qmax(2:end)) / log10(2); delta_enorm = enorm(2:end) - enorm(1:end-1); slope_enorm = log10(delta_enorm(1:end-1) ./ delta_enorm(2:end)) / log10(2); ANSWER(:,1) = elements; ANSWER(:,2) = iteration; ANSWER(:,3) = Qmax; ANSWER(:,4) = [0;delta_Qmax]; ANSWER(:,5) = [0;0;slope_Qmax]; ANSWER(:,6) = enorm; ANSWER(:,7) = [0;delta_enorm]; ANSWER(:,8) = [0;0;slope_enorm]; fprintf('Elements Iterations Qmax (Qmax)change (Qmax)slope ||Q|| change||Q|| Slope(Q)\n\n') fprintf('%6.0f %6.0f %12.6g %12.6g %12.6g %12.6g %12.6g %12.6g\n',ANSWER'); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

**Discussion of Results **

(Direct Solve)

Figure 2: Steady-state temperature solutions for the fin tube cylinder using direct solve method. The red curve indicates the final converged solution.

Figure 2 above shows the temperature solutions at each mesh refinement. With no Dirichlet boundary conditions to serve as an anchor, the entire solution shifts in temperature until a satisfactory convergence is reached. As expected, the temperature solution begins with a linear progression as heat conducts through the pipe wall. Along the fin, the solution becomes non-linear as the convection boundary conditions augment the conductive heat transfer.

As seen in Table 1, both the Tmax and Energy norm error convergence calculations reveal a linear slope. It would thus seem that the data non-smoothness is degrading convergence from the optimal slope of 2.

Table 1: Tmax and Energy Norm error convergence results using a Lagrange (k=1) Finite Element basis.

================================================================================ Elements Qmax (Qmax)change (Qmax)slope ||Q|| change||Q|| Slope(Q) ________________________________________________________________________________ 8 349.853 0 0 30478.9 0 0 16 360.158 10.3044 0 32048.4 1569.47 0 32 365.103 4.94554 1.05905 32819.2 770.879 1.02570 64 367.523 2.41994 1.03115 33200.8 381.576 1.01453 128 368.720 1.19663 1.01600 33390.6 189.773 1.00770 256 369.315 0.594962 1.00811 33485.2 94.6265 1.00396 512 369.611 0.296641 1.00408 33532.5 47.2475 1.00201 1024 369.760 0.14811 1.00205 33556.1 23.6072 1.00101 2048 369.834 0.0740025 1.00102 33567.9 11.7995 1.00051 4096 369.871 0.036988 1.00051 33573.8 5.8987 1.00025 ================================================================================

(Newton Template Implimentation)

Figure 3: Steady-state temperature solutions for the fin tube cylinder using Newton's Method. The red curve indicates the final converged solution.

As expected, Figure 3 indicates identical temperature solutions when using Newton's method. Table 2 lists the numerical results. Since the problem statement remains linear, only one iteration step is reqired to converge upon a solution in the Newton method process. The Tmax error convergence data are identical to that of the direct solve results. Now, however, the energy norm slope data appears useless. It would seem that computer round-off errors are the main culprit.

Table 2: Tmax and Energy Norm error convergence results using a Lagrange (k=1) Finite Element basis.

================================================================================================= Elements Iteration Qmax (Qmax)change (Qmax)slope ||Q|| change||Q|| Slope(Q) _________________________________________________________________________________________________ 8 2 349.853 0 0 1.60494e-012 0 0 16 2 360.158 10.3044 0 3.82036e-013 -1.22290e-012 0 32 2 365.103 4.94554 1.05905 1.40789e-012 1.02585e-012 0.253487 64 2 367.523 2.41994 1.03115 7.19733e-013 -6.88152e-013 0.576020 128 2 368.720 1.19663 1.01600 -3.44524e-012 -4.16498e-012 -2.597510 256 2 369.315 0.594962 1.00811 -2.38385e-012 1.06140e-012 1.972350 512 2 369.611 0.296641 1.00408 7.76942e-012 1.01533e-011 -3.257910 1024 2 369.760 0.14811 1.00205 1.76922e-012 -6.00020e-012 0.758862 2048 2 369.834 0.0740025 1.00103 1.42276e-011 1.24584e-011 -1.054040 4096 2 369.871 0.0369881 1.00051 4.04675e-011 2.62399e-011 -1.074640 =================================================================================================

(Pipe Wall Radiation BC)

(Newton Method Solver)

Figure 4: Steady-state temperature solutions for the fin tube cylinder with radiative pipe BCs using direct solve method. The red curve indicates the final converged solution.

Finally, Figure 4 shows the temperature solutions when radiative HT conditions are imposed at the pipe wall. The profile is identical to the previous solutions, but the overall temperatures are lower. This makes sense since a switch from liquid convection to gas radiation would degredate heat transfer to the pipe surface.

Table 3 indicates that, at every mesh, an extra 2 iteration steps are required to converge upon a solution using the Newton's method solver. Both the Tmax and energy norm slope calculations indicate a slope of 1 which again indicates a performance degredation attributed to data non-smoothness.

Table 3: Tmax and Energy Norm error convergence results using a Lagrange (k=1) Finite Element basis.

========================================================================================= Elements Iteration Qmax (Qmax)change (Qmax)slope ||Q|| change||Q|| Slope(Q) _________________________________________________________________________________________ 8 4 345.531 0 0 -309.035 0 0 16 4 356.738 11.207 0 -252.165 56.8699 0 32 4 362.115 5.377 1.05952 -223.611 28.5546 0.993944 64 4 364.746 2.63055 1.03144 -209.346 14.2646 1.00129 128 4 366.046 1.30063 1.01615 -202.222 7.12392 1.00169 256 4 366.693 0.646637 1.00819 -198.663 3.55923 1.00111 512 4 367.015 0.322396 1.00412 -196.884 1.77885 1.00062 1024 4 367.176 0.160967 1.00207 -195.995 0.889227 1.00032 2048 4 367.257 0.0804259 1.00104 -195.551 0.444562 1.00017 4096 4 367.297 0.0401985 1.00052 -195.328 0.222268 1.00008 =========================================================================================

**Conclusion **

In conclusion, a convergence study has been completed for 1-D heat conduction in an axisymmetric fin tube cylinder. For the case of convective boundary conditions, solutions have been achieved utilizing the direct matrix solution and Newton's iterative methods. For the case of a radiative boundary condition, non-linearities in the problem statement require Newton's method.

**Objectives**

**10.31.04** *Joe - your mid-term report is excellently organized and presented!! Your discussion of results is clear, concise and to the point - very well done! Only criticism is that Conclusions content should respond one to one with objectives, which are clearly stated. Your problem statement section is the best I have seen! Score is 17/17. AJB*