Question

A liquid metal is poured in a mold cavity of size 200 mm \(\times\) 200 mm \(\times\) 200 mm. The cooling is uniform in all directions with NO additional compensation for shrinkage. Considering the volumetric shrinkage during solidification and solid contraction as 7% and 8%, respectively, the length of the cube edge after cooling down to the room temperature is _________ mm (rounded off to 1 decimal place).

Hint:

For volumetric shrinkage, use the cube of the edge length and apply the combined shrinkage percentage to the edge length. The total shrinkage is the combined effect of both solidification and solid contraction.

Answer 1

We are given the following:
Initial size of the cube: 200 mm × 200 mm × 200 mm
Volumetric shrinkage during solidification: 7%
Solid contraction after cooling: 8%
Step 1: Understanding Shrinkage.
Volumetric shrinkage during solidification means the total volume of the liquid metal decreases due to phase change from liquid to solid, and solid contraction occurs when the material cools down. These shrinkage effects combine to reduce the dimensions of the cube.
Volumetric shrinkage is 7%, meaning the volume decreases by 7% during solidification.
Solid contraction is 8%, meaning the volume decreases by 8% after cooling to room temperature.
Step 2: Total Shrinkage.
The total shrinkage is the combined effect of both volumetric shrinkage and solid contraction. We can calculate the combined shrinkage using the formula for the total volume shrinkage: \[ \text{Total Shrinkage} = 1 - (1 - \text{Solidification Shrinkage})(1 - \text{Solid Contraction}) \] Substituting the given values: \[ \text{Total Shrinkage} = 1 - (1 - 0.07)(1 - 0.08) = 1 - (0.93 \times 0.92) = 1 - 0.8556 = 0.1444 \] Thus, the total shrinkage is 14.44%.
Step 3: Calculate the Final Length.
Since the shrinkage is volumetric, the edge length of the cube also shrinks by the cube root of the total volume shrinkage. The final edge length after cooling can be calculated as: \[ L_{\text{final}} = L_{\text{initial}} \times (1 - \text{Total Shrinkage}) \] Substituting the values: \[ L_{\text{final}} = 200 \times (1 - 0.1444) = 200 \times 0.8556 = 171.12 \, \text{mm} \] Thus, the final length of the cube edge after cooling down to room temperature is 171.1 mm (rounded to 1 decimal place).