MEEN 5323 Advanced Heat Transfer and Applications Project 1, Spring 2023 1. Project statement A solid copper column is attached to two surfaces on the bottom and top sides, whose temperatures are fixed at 50C and 500C, respectively. Force convection is used to cool down the column using the room temperature air at 25C at the heat transfer coefficient of 25 W/mK. The width and height of the column as well as the thermal conductivity are shown in the following figure. What is the temperature distribution of the copper column along the vertical direction in the steady state? What is the total heat transfer rate from the column to air? 2. Required operations Students need to Simulate the problem described in Sect. 1 using commercial or open-source software Use screen shots or texts of configuration files to present how is the simulation case being constructed and run T = 500C T = 25C L = 2m h = 25 W/m. K Copper k=400 W/m K a = 0.2 m T = 50C a Choose an appropriate mesh size, e.g., 100,000 computational cells Visualize the evolution of the temperature field from the beginning to the steady state, and generate a video of that visualization to be attached with the report Run on-the-fly processing or post-processing to obtain the temperature distribution along the vertical direction and the overall heat transfer rate from the column to air Compare the simulation results with the analytical solutions as shown in the attachment and plot a figure of T(z) with two curves, (1) simulation results, and (2) analytical solution 3. Deliverables Students need to submit a technical report in WORD/PDF before 11:59pm, Sunday, March 5. The report shall include the required contents stated in Sect. 2. TABLE 3.4 Temperature distribution and heat loss for fins of uniform cross section Case Tip Condition (x = L) Temperature Distribution 0/0 Fin Heat Transfer Rate q A Convection heat transfer: cosh m(L x) + (h/mk) sinh m(L - x) sinh mL + (h/mk) cosh mL M ho(L)=-kdo/dxx-L cosh mL + (h/mk) sinh mL (3.75) cosh mL + (h/mk) sinh mL (3.77) B Adiabatic: cosh m(L-x) M tanh mL do/dxx=L=0 cosh mL (3.80) (3.81) C Prescribed temperature: 0(L) = OL (0/0,) sinh mx + sinh m(L x) (cosh mL-010) M sinh mL sinh mL (3.82) (3.83) D Infinite fin (L ): 0(L) = 0 -mx (3.84) (3.85) e M 0=T-T 0 = 0(0) = T - T m = hP/kAc M = hPKA.0b 1