Question 5 of 7: Estimating Injection-Molding Solidification Time for a Thin vs. a Thick Part
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, December 2019 — 17-Ind-B2 Manufacturing Processes. 3-hour closed-book exam; candidates may use a Casio or Sharp approved calculator. Any five questions constitute a complete paper (only the first five as they appear are marked officially); all seven are answered below as a full study resource.
Reference texts. Groover, Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 6th ed. (primary text for this subject — material selection, casting, polymer processing, machining, and composites).
Question 5: Estimating Injection-Molding Solidification Time for a Thin vs. a Thick Part (20 marks)
Given. Two flat, injection-molded sections of the same polymer and processed under the same mold/melt conditions: thin part wall thickness $h_1 = 3$ mm; thick part wall thickness $h_2 = 11$ mm. No material thermal properties (thermal diffusivity, melt temperature, mold temperature, or ejection/freeze temperature) are given in the source.
Find. An estimate of how the two parts' solidification (cooling) times compare.
Fig. Q5 — cross-sections of the two injection-molded parts (not to scale); the thicker part's core is much farther from either mold-wall face than the thin part's.
Approach. Model each part as an infinite flat plate cooling by conduction from both faces into the mold. The exact 1-D transient-conduction solution for injection-molding solidification time is $t=\dfrac{h^2}{\pi^2\alpha}\ln\!\left[\dfrac{4}{\pi}\dfrac{T_i-T_m}{T_f-T_m}\right]$, where $h$ is the part thickness, $\alpha$ the polymer's thermal diffusivity, and $T_i,T_m,T_f$ the melt, mold, and freeze (ejection) temperatures. For a FIXED material and fixed process temperatures, every factor in that expression except $h$ is unchanged between the two parts, so the solidification time scales exactly with $h^2$ — this lets the two times be compared as a ratio without needing the (unstated) thermal-property or temperature data.
Isolate the thickness dependence. Since $\alpha$, $T_i$, $T_m$, $T_f$ are identical for both parts (same polymer, same mold/melt settings), $t \propto h^2$, so $\dfrac{t_{\text{thick}}}{t_{\text{thin}}}=\left(\dfrac{h_2}{h_1}\right)^2$.
Compute the thickness ratio. $\dfrac{h_2}{h_1}=\dfrac{11}{3}=\boxed{3.67}$.
Square it to get the solidification-time ratio. $\dfrac{t_{\text{thick}}}{t_{\text{thin}}}=\left(\dfrac{11}{3}\right)^2=\boxed{13.4}$. The 11 mm part therefore takes roughly 13–14 times longer to fully solidify than the 3 mm part, everything else being equal.
Question 5 — solidification-time estimate
Quantity
Value
Thickness ratio $h_2/h_1$
3.67
Solidification-time ratio $t_{\text{thick}}/t_{\text{thin}} = (h_2/h_1)^2$
≈ 13.4
Interpretation
Thick (11 mm) part solidifies ≈13–14× slower than the thin (3 mm) part
Check. The source gives only the two thicknesses — no polymer type, melt temperature, mold temperature, or freeze temperature is stated, so an absolute solidification time (in seconds) cannot be computed without inventing data. The scaling-law ratio above ($t\propto h^2$) is the estimate the given data actually supports, and directly explains why thick injection-molded sections are disproportionately expensive to cool (see Question 4's related cost/quality discussion for this subject).
