Question

a cubical box of length 3 cm is solidified in 5 min in a sand casting .if the box is melted and casted into sperical object ,then calculate time required tosolidify the spherical object in the same atomspherei condition.

Answer 1

Answer:

Chvorinov's Rule

Timing for solidification is C(V/A)²

V= Volume

A= Total Surface Area

Timing for solidification = C(V/A)

C is mold constant in Chvorinov's Rule.

A cubical box of length 3cm.

Volume= 3×3×3 = 27cm³

Surface area= 6(3)² = 54cm²

= 5 = C (27/54)²

= 5 = C (1/2)²

= C = 20 min/cm²

For Sphere Volume = 27cm³

= (4/3)πR³ = 27

= R³ = 27×3/4π

= R = 1.861

(V/A)² = { (4/3)πR³/4πR² }² = R²/9

= 0.3848

Time = 20 × 0.3848 = 7. 7 min

7min 42sec

So, the time required to solidify the spherical object in the same atmospheric condition

= 7 min 42 sec.

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