Question

In casting experiments performed using a certain alloy and type of sand mold, it took 170 sec for a cube-shaped casting to solidify. The cube was 50 mm on a side.

a. Determine the value of the mold constant in Chvorinov's rule.
b. If the same alloy and mold type were used, find the total solidification time for a cylindrical casting in which the diameter = 50 mm and length = 50 mm.

Answer 1

Answer:

Answer for the question is : Solidification time will be same i.e. 170. See attached file for explanation.

Explanation:

Answer 2

Answer:

a) the value of the mold constant in Chvorinov`s rule is 2.45 s/mm²

b) the total solidification time is 170 s

Explanation:

a) The volume is 50³ = 125000 mm³

The area is

A = 6 * 50² = 15000 mm²

The ratio V/A is

125000/15000 = 8.33 mm

The mold constant in Chvorinov`s rule is

$C_{ch } =\frac{T}{(V/A)^{2} } =\frac{170}{8.33^{2} } =2.45s/mm^{2}$

b) The volume is

$V=\frac{\pi D^{2}L }{4} =\frac{\pi 50^{2}*50 }{4} =98174.8mm^{3}$

The area is

$A=\frac{2\pi D}{4} +\pi DL=\frac{2\pi 50^{2} }{4} +\pi *50*50=11781mm^{3}$

V/A = 98174.8/11781 = 8.33

The total solidification is

T =2.45 * (8.33²) = 170 s