Question

Two castings are proposed for a study. One casting is a cube 10.16 cm (4 in.) on a side; the other is a rectangle 10.64 cm (16 in.) tall × 10.16 cm (4 in.) wide × 2.45 cm (1 in.) thick. Both are cast from the same metal at the same temperature, and the mold material is the same in both cases. If the cube-shaped casting solidifies in 14.5 min, how rapidly do we expect the rectangular casting to solidify?

Answer 1

To solve the problem it is necessary to apply the concepts related to Chvorinov's Law, which states that

$\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2$

Where,

$V_c$ = Volume cube

$SA_c$ = Superficial Area from Cube

$V_r$ = Volume Rectangle

$SA_r$= Superficial Area from Rectangle

Our values are given as (I will try to develop the problem in English units for ease of calculations),

$V_c = 4^3 = 64in^3$

$SA_c = 6*4^2=96in^2$

$SA_r = 2*(1*16+4*1+16*4)=168in^2$

$V_r = 4*1*16 = 64in^3$

Applying the Chvorinov equation we have to,

$\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2$

$\frac{T_r}{T_c} = (\frac{64}{64}*\frac{96}{168})^2$

$\frac{T_r}{T_c} = (\frac{96}{168})^2$

$\frac{T_r}{T_c} = 0.3265$

The stipulated time for the cube is 14.5 then,

$T_r = 0.3265*14.5$

$T_r = 4.735min$