Question 7 of 7: Solidification Defects; Chvorinov's Rule for a Bronze Casting
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2017. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Notes on the paper state that
any five questions constitute a complete paper and only the first five questions appearing in the
answer book are marked, with all questions of equal value. All seven questions are solved below
for completeness.
Reference texts: Callister & Rethwisch, Materials Science and
Engineering: An Introduction, 9th ed. (crystal structure and defects, diffusion, mechanical
behaviour and tensile testing, polymers, phase diagrams and the lever rule, precipitation
hardening, corrosion, casting and solidification).
Question 7: Solidification Defects; Chvorinov's Rule for a Bronze Casting (20 marks)
Given. (b) A 2-in cube of bronze solidifies in $t_1=8.2$ min; a $1\times1
\times6$-in bar of the same bronze, cast "under similar conditions" (same mold material and
pouring practice, so the same Chvorinov constant $B$ applies).
Find. (a) The two most common solidification defects and how to minimize them.
(b) The solidification time for the $1\times1\times6$-in bar.
Approach
(a) is descriptive: shrinkage (both macroscopic cavity/pipe and microporosity) and gas porosity
are the two classic solidification defects. (b) uses Chvorinov's rule, $t_s=B(V/A)^n$ (given on
page 1, $n=2$): the cube's data first back out the mold/alloy-specific constant $B$, which is then
applied to the bar's own $V/A$ ratio.
(a) Shrinkage (cavity/pipe and microporosity). Liquid metal is denser as a
solid than as a liquid (for most alloys), so the casting contracts during (i) liquid cooling, (ii)
solidification itself, and (iii) solid cooling to room temperature. If there is no reservoir of
liquid metal available to feed the last region to solidify (typically the thermal centre of the
casting, farthest from the mold walls), that region is left with a shrinkage cavity ("pipe," if
open to the surface, or an internal void) or, on a finer scale, scattered microporosity between
dendrite arms. Precaution: design the casting/gating so solidification progresses
directionally toward a riser (an extra reservoir of liquid metal, itself the last region to
solidify, sacrificially absorbing the shrinkage), and place risers at the thermally heaviest
sections.
(a) Gas porosity. Molten metal can dissolve far more gas (hydrogen in
aluminum and copper alloys; a variety of gases in steels) than solid metal can. As the metal
solidifies, dissolved gas is rejected from the solidifying dendrites and, if it cannot escape fast
enough, nucleates as trapped gas bubbles/pores within the casting. Precaution: degas the
melt before pouring (e.g. purging with an inert or reactive gas to strip dissolved hydrogen),
control pouring temperature and turbulence (excess turbulence entrains additional gas), and design
adequate venting in the mold so entrapped mold-cavity gas can escape ahead of the rising metal.
(b) Chvorinov constant from the cube. For the 2-in cube:
$$V_1=2^3=8\ \text{in}^3,\qquad A_1=6(2)^2=24\ \text{in}^2,\qquad \left(\frac{V}{A}\right)_1=\frac13\ \text{in}$$
$$t_1=B\left(\frac{V}{A}\right)_1^2\ \Rightarrow\ B=\frac{t_1}{(V/A)_1^2}=\frac{8.2}{(1/3)^2}=8.2(9)$$
$$\boxed{B\approx73.8\ \text{min/in}^2}$$
(b) Apply $B$ to the $1\times1\times6$-in bar.
$$V_2=(1)(1)(6)=6\ \text{in}^3,\qquad A_2=2(1\times1)+4(1\times6)=2+24=26\ \text{in}^2$$
$$\left(\frac{V}{A}\right)_2=\frac{6}{26}=0.2308\ \text{in}$$
$$t_2=B\left(\frac{V}{A}\right)_2^2=73.8(0.2308)^2$$
$$\boxed{t_2\approx3.93\ \text{min}}$$
The slender bar, with much more surface area per unit volume than the compact cube, loses heat
much faster and so solidifies in well under half the cube's time, even though it contains less than
the cube's volume of metal to begin with.
