MATH SOLVE

3 months ago

Q:
# Total solidification times of three casting geometries are to be compared: (1) a sphere with diameter = 10 cm, (2) a cylinder with diameter and length both = 10 cm, and (3) a cube with each side = 10 cm. The same casting alloy is used in all three cases. (a) Determine the relative solidification times for each geometry. (b) Based on the results of part (a), which geometric element would make the best riser? (c) If the mold constant = 3.5 min/cm2 in Chvorinov’s rule, compute the total solidification time for each casting.

Accepted Solution

A:

Answer:a) See Belowb) See Belowc) See BelowStep-by-step explanation:The relative solidification time is given by formula [tex]t_s=C_m(\frac{V}{A})^2[/tex]Where C_m is the Chvorinov's constant given as 3.5V is VolumeA is Surface Areaa)Sphere:Volume = [tex]\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5)^3 = 523.6[/tex]Surface Area = [tex]4\pi r^2=4\pi (5)^2 = 314.16[/tex]Now putting in the formula:Relative Solidification Time of Sphere = [tex]t_s=C_m(\frac{V}{A})^2\\t_s=3.5(\frac{523.6}{314.16})^2\\t_s=9.72[/tex] minutesRelatively, We can say t_s = 2.778 * C_mCube:Volume = 10^3 = 1000Surface Area = 6x^2 = 6(10)^2 = 600Relative Solidification Time of Cube = [tex]t_s=C_m(\frac{V}{A})^2\\t_s=3.5(\frac{1000}{600})^2\\t_s=9.72[/tex] minutesRelatively, We can say t_s = 2.778 * C_mCylinder:Volume = [tex]\pi r^2 h = \pi (5)^2 (10)=785.4[/tex]Surface Area = [tex]2\pi r^2 + 2\pi r h=471.24[/tex]Relative Solidification Time of Cylinder = [tex]t_s=C_m(\frac{V}{A})^2\\t_s=3.5(\frac{785.4}{471.24})^2\\t_s=9.72[/tex]Relatively, We can say t_s = 2.778 * C_mb)All of them have same relative solidification time. So we can say that all 3 of them are good for making risers.c)Since, we hadRelatively, We can say t_s = 2.778 * C_mwe get:t_s = 2.778 * C_m = 2.778 * 3.5 = 9.72 minutes