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.

Respuesta :

Answer:

A) The value of the mold constant Cm in Chvorinov’s Rule is; Cm = 2.232 s/mm²

B) Time taken to solidify; Tts = 74.3 seconds

(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

V = 125,000 mm³

There are 6 faces of the cube, thus; Surface Area is;

A = 6 x (50 x 50)

A = 15,000 mm²

Thus;

Volume/Area is;

(V/A) = 125,000/15,000

(V/A) = 8.333 mm

Formula for mold constant 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

Cm = 2.232 s/mm²

(B) We are told the Cylindrical casting has a diameter; D = 30 mm and Length; L = 50 mm.;

Volume of cylinder is;

V = (πD²L)/4

Thus;

V = (π × 30² × 50)/4 = 35,343mm³

Surface area of cylinder is;

A = (2πD²)/4 + (πDL)

Thus:

A = ((π x 30²)/2) + (π x 30 x 50)

A = 6126 mm²

Volume/Area is;

V/A = 35,343/6126 = 5.77

mm

Same alloy and mold type was used as in answer 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

Tts = 74.3 seconds

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