Question

Iron powder is compacted at room temperature to 75% of its theoretical density. The as-pressed iron compact is sintered in inert-gas atmosphere at \(1200^{\circ}\text{C}\) to 90% of its theoretical density. Assuming isotropic shrinkage, the linear shrinkage (in percent) undergone by the iron compact during sintering (rounded off to one decimal place) is ______ %.

Hint:

Use \(V\propto 1/D\) for constant mass, then \(L_1/L_0=(D_0/D_1)^{1/3}\) since the shrinkage is isotropic.

Answer 1

Correct Answer: 5.8

Step 1: Relate relative density to volume.
The mass of the compact stays the same throughout compaction and sintering, only its volume changes. If \(D\) is the fraction of theoretical density, then for a fixed mass \(m\) and theoretical density \(\rho_{th}\),
\[ D=\frac{\rho}{\rho_{th}}=\frac{m}{V\rho_{th}} \]
so
\[ V\propto\frac{1}{D} \]

Step 2: Write the volume ratio between the green and sintered states.
Green density fraction \(D_0=0.75\), sintered density fraction \(D_1=0.90\). Using \(V\propto1/D\),
\[ \frac{V_1}{V_0}=\frac{D_0}{D_1}=\frac{0.75}{0.90}=0.8333 \]
So the sintered volume is about \(83.33\%\) of the green volume, because the compact densifies and shrinks.

Step 3: Convert the volume ratio to a linear dimension ratio.
Because the shrinkage is isotropic (the same in all three directions), volume scales as the cube of any linear dimension \(L\):
\[ V\propto L^3 \quad\Rightarrow\quad \frac{L_1}{L_0}=\left(\frac{V_1}{V_0}\right)^{1/3} \]
\[ \frac{L_1}{L_0}=(0.8333)^{1/3} \]

Step 4: Evaluate the cube root.
\[ (0.8333)^{1/3}\approx0.9410 \]
So the sintered length is about \(94.10\%\) of the green length.

Step 5: Compute the linear shrinkage.
Linear shrinkage is the fractional decrease in length:
\[ \text{Linear shrinkage}=\left(1-\frac{L_1}{L_0}\right)\times100=(1-0.9410)\times100 \]
\[ =5.9\% \]

Final Answer:
The linear shrinkage during sintering, rounded to one decimal place, is
\[ \boxed{5.9\%} \]