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Isometric view of the furnace wall. The red side represents the firebrick. The green represents the insulating brick.
Temperature distribution and thermal resistance through a composite wall. In this assignment there are only two walls so ignore the third wall "C". Source: Bergman, Theodore L., Lavine, Adrienne S., Incropera, Frank P., Dewitt, David P. (2011) Fundamentals of Heat and Mass Transfer. Seventh edition. Jefferson City (MO): John Wiley & Sons, Inc . Chapter 3, One-Dimensional, Steady-State Conduction; p. 116.
Background
Background
This is an assignment for my ME-441 Computer Simulation and Analysis class. In the Computer Simulation and Analysis class ANSYS Workbench and Discovery AIM were used extensively to simulate. Instructions were provided, and sometimes the models, either from Creo Parametric or SolidWorks, were provided as well. Some assignments however need the objects to be modelled from scratch or modified before hand.
The assignment
The assignment
In this assignment, a furnace wall undergoes a heat transfer analysis using ANSYS AIM Thermal. The temperature distribution needs to be found between a firebrick and an insulating brick of the furnace wall during the heat transfer simulation between the two.
The firebrick is nine inches or 0.75 ft in length and has a thermal conductivity of 0.8 BTU/hr.ft.F or 2.222e-4 BTU/s.ft.F. The insulating brick is five inches or 0.4167 ft in length and has a thermal conductivity of 0.1 BTU/hr.ft.F or 2.778e-5 BTU/sft.F. The temperature inside the furnace (left side) is 3,000 F, with a convection heat transfer coefficient of 12 BTU/hr.ft^2.F or 3.333e-3 BTU/s.ft^2.F. The temperature outside the furnace (right side) is 80 F, with a convection heat transfer coefficient of 2 BTU/hr.ft^2.F or 5.556e-4 BTU/s.ft^2.F.
The theoretical calculations goes:
Total thermal resistance
R_total = R_conv,1 + R_wall,1 + R_wall,2 + R_conv,2 = (1/(h_1 x A)) + (L_1/(k_1 x A)) + (L_2/(k_2 x A)) + (1/(h_2 x A)). Area (A) is assumed to be 1 ft^2.
R_total = (1/3.333 x 10^-3) + (0.75/2.222 x 10^-4) + (0.4167/2.778 x 10^-5) + (1/5.556 x 10^-4) = 20.474 x 10^3 s.F/BTU
Heat transfer
Q. = (T_inf,1 - T_inf,2)/R_total
Q. = (3000 - 80) F / 20.474 x 10^3 s.F/BTU = 142.62 x 10^-3 BTU/s
Temperatures
T_1 = T_inf,1 - (Q. x R_conv,1) = 3000 - (142.62 x 10^-3 x 0.3 x 10^3) = 2957.21 F
T_2 = T_inf,1 - (Q. x (R_conv,1 + R_wall,1)) = 3000 - (142.62 x 10^-3 x (0.3 x 10^3 + 3.375 x 10^3)) = 2475.87 F
T_3 = T_inf,2 + (Q. x R_conv,2) = 80 + (142.62 x 10^-3 x 1.799 x 10^3) = 336.57 F
Results
Results
The temperature from the left end of the firebrick is 2957.2 F, while the temperature from the right end of the insulating brick is 336.69 F. The temperature where the firebrick is in contact with the insulating brick is 2475.8 F. When calculating the %error, the temperature for T_1 is a 0.00034% error, T_2 is a 0.0028% error, and T_3 is a 0.036% error. The value for heat flux is 0.14262 BTU/s.ft^2, which matches if the theoretical heat transfer is divided by the area (1 ft^2).
Convection condition on the firebrick.
Convection condition on the insulating brick.
Temperature distribution of the furnace wall.
Temperature between the two bricks.
Heat flux vector through the two bricks.
Heat flux magnitude between the two bricks.
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