Question 7 of 7: Casting Solidification Defects; Chvorinov's Rule

Question 7: Casting Solidification Defects; Chvorinov's Rule (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (b) 2 in cube of bronze, $t_s=8.2$ min; a $1\times1\times6$ in bar cast under similar conditions (same mold/metal, so the same Chvorinov constant $B$ and exponent $n=2$ apply, per page 1).

Find. (a) The two most common solidification defects and precautions. (b) Solidification time for the bar.

Approach

Part (a) is conceptual: the two classic solidification defects are shrinkage porosity and gas porosity. Part (b) applies Chvorinov's rule, $t_s=B(V/A)^n$: since the mold/metal/pouring conditions are unchanged, the constant $B$ (and $n=2$) found from the cube apply directly to the bar — only the geometry-dependent volume-to-surface-area ratio changes.

1. (a) Shrinkage porosity/cavity. Liquid metal occupies more volume than the same mass of solid; as solidification proceeds inward from the mold walls, the last region to solidify (often the thermal centre or a hot spot) has no further liquid available to feed it, leaving a void. Precaution: place risers (feeder heads) at the last-to-solidify location to supply make-up liquid, promote directional solidification toward the riser (progressively larger sections, or chills to locally speed solidification elsewhere), and design gating to avoid isolated hot spots.
2. Gas porosity. Dissolved gases (e.g. hydrogen in aluminum and copper alloys) are more soluble in the liquid than in the solid; as the metal solidifies, rejected gas can be trapped as bubbles within the casting. Precaution: degas the melt before pouring, use non-turbulent gating/pouring practice to avoid entraining air, and provide adequate mold venting.
3. (b) Chvorinov constant from the cube. For the 2 in cube, $V=2^3=8\ \text{in}^3$ and $A=6(2\times2)=24\ \text{in}^2$, so $V/A=1/3=0.3333$ in. With $n=2$ (given, page 1) and $t_s=8.2$ min, $$B=\frac{t_s}{(V/A)^2}=\frac{8.2}{(0.3333)^2}=\boxed{73.8\ \text{min/in}^2}.$$
4. Solidification time for the bar. For the $1\times1\times6$ in bar, $V=1\times1\times6=6\ \text{in}^3$ and $A=2(1\times1)+2(1\times6)+2(1\times6)=2+12+12=26\ \text{in}^2$, so $V/A=6/26=0.2308$ in. Applying the same $B$, $$t_{s,bar}=B\left(\frac{V}{A}\right)^2=73.8\times(0.2308)^2=\boxed{3.93\ \text{min}}.$$ The bar solidifies more than twice as fast as the cube despite having less than 3/4 of its volume, because its thinner cross-section gives it much more surface area per unit volume to reject heat through.