Question

In casting experiments performed using a certain alloy and type of sand mold, it took 155 sec for a cube-shaped casting to solidify. The cube was 50 mm on a side. (a) Determine the value of the mold constant Cm 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 is 30 mm and length is 50 mm.

Answer 1

Answer:

A) Cm = 2.232 s/mm²

B) Time taken to solidify = 74.3 seconds

Explanation:

(A) Since a side is 50mm and all sides of a cube are equal, thus, Volume of the cube is;V = 50 x 50 x 50 = 125,000 mm³

There are 6 faces of the cube, thus Surface Area A = 6 x (50 x 50) = 15,000 mm²

So, Volume/Area = (V/A) = 125,000/15,000 = 8.333 mm

Cm is given by the formula; Cm =[Tts] /(V/A)² where Tts is time taken to solidify and it's 155 seconds in the question. Thus;

Cm = 155/(8.333)²= 2.232 s/mm²

(B) For;Cylindrical casting with D = 30 mm and L = 50 mm.;

Volume of cylinder is;

V = (πD²L) /4

So,V = (π x 30² x 50)/4 = 35,343mm³

Surface area of cylinder is;

A = (2πD²)/4 + (πDL)

Thus, A = ((π x 30²)/2) + (π x 30 x 50) = 6126 mm²

Volume/Area is;

V/A = 35,343/6126 = 5.77 mm

Same alloy and mold type was hsed as in a above, thus, Cm is still 2.232 s/mm²

Since Cm =[Tts] /(V/A)²

Making Tts the subject, we have;

Tts =Cm x (V/A)²

Tts = 2.232 x (5.77)² = 74.3 seconds