The rod shown below is made of medium carbon steel (p = 7800 kg/m³, k = 43 W/m.°C, c = 473 J/kg. °C). Rod dimensions are

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answerhappygod
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The rod shown below is made of medium carbon steel (p = 7800 kg/m³, k = 43 W/m.°C, c = 473 J/kg. °C). Rod dimensions are

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The Rod Shown Below Is Made Of Medium Carbon Steel P 7800 Kg M K 43 W M C C 473 J Kg C Rod Dimensions Are 1
The Rod Shown Below Is Made Of Medium Carbon Steel P 7800 Kg M K 43 W M C C 473 J Kg C Rod Dimensions Are 1 (66.53 KiB) Viewed 20 times
The Rod Shown Below Is Made Of Medium Carbon Steel P 7800 Kg M K 43 W M C C 473 J Kg C Rod Dimensions Are 2
The Rod Shown Below Is Made Of Medium Carbon Steel P 7800 Kg M K 43 W M C C 473 J Kg C Rod Dimensions Are 2 (64.56 KiB) Viewed 20 times
The Rod Shown Below Is Made Of Medium Carbon Steel P 7800 Kg M K 43 W M C C 473 J Kg C Rod Dimensions Are 3
The Rod Shown Below Is Made Of Medium Carbon Steel P 7800 Kg M K 43 W M C C 473 J Kg C Rod Dimensions Are 3 (14.64 KiB) Viewed 20 times
You have to use ansys program i dont want you to use any other
programs.
The rod shown below is made of medium carbon steel (p = 7800 kg/m³, k = 43 W/m.°C, c = 473 J/kg. °C). Rod dimensions are: R = 10 mm, L = 500 mm. Initial uniform temperature of the rod is 500 °C. The rod is to be cooled in a large medium kept at 25 °C Consider the following three cases of cooling conditions: Case#1: If the rod surface temperature is assumed constant and equal to the cooling medium temperature (i.e., Ts=25°C). Case #2: If the rod surface is exposed to the cooling medium with a heat transfer coefficient equals to h₂= 8000 W/m²-K. Case #3: If the cooling medium is changed such that the heat transfer coefficient is only h3 = 100 W/m²K. R
Conduct transient CFX simulations to determine the cooling time of the rod for the three cooling cases indicated above. As a criterion, it is assumed that the rod has cooled down when the temperature difference between its center and the cooling medium drops below 5% of the initial temperature difference. • Monitors and results curves (for only center and surface) • Cooling time for each case 4 contours figures of temperature for longitudinal section at (time=0, time =0.25, time=0.5 and at final time) for each case • Figure show change of temperature with time of the next points point X Y like radius (m) | Z (m) Center 0 0 0.25 R=2mm 0 2e-3 0.25 R=4mm 0 4e-3 0.25 R=6mm 0 6e-3 0.25 R=8mm 0 8e-3 0.25 surface 0 0.01 0.25 Compare the cooling time of case #3 simulation with an analytical analysis result which gives a cooling time of 553 seconds.
Simulate the same problem discussed in tutorial 4 for case #2 and case #3 and don't forget to change the total time in these cases to be like 600 seconds as it will take much longer time.
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