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

`(T_r)/(T_c) = ((V_r)/(V_c)*(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,

`(T_r)/(T_c) = ((V_r)/(V_c)*(SA_c)/(SA_r))^2`

`(T_r)/(T_c) = ((64)/(64)*(96)/(168))^2`

`(T_r)/(T_c) = ((96)/(168))^2`

`(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`