An alloy having composition W-20 wt.% Ni is prepared by mixing elemental powders. The resulting powder mixture is liquid phase sintered at \(1550^{\circ}\text{C}\) for 1 hour. The liquid phase sintered microstructure consists of interconnected, spherical tungsten grains dispersed in nickel. The tungsten grain size is 70 \(\mu\text{m}\) and the W-W interparticle neck diameter is 35 \(\mu\text{m}\). If W-Ni interfacial energy is 0.30 J/m\(^2\), the W-W interfacial energy (in J/m\(^2\)), rounded off to two decimal places is ______.
(Given: melting point of W: \(3410^{\circ}\text{C}\) and melting point of Ni: \(1455^{\circ}\text{C}\))
Show Hint
Use \(\sin(\phi/2)=X/D\) from the neck geometry to get the dihedral angle, then \(\gamma_{WW}=2\gamma_{WNi}\cos(\phi/2)\).
Step 1: Set up the neck size ratio.
The ratio of the interparticle neck diameter to the grain diameter fixes the dihedral angle at the neck through $\sin(\phi/2)=X/D$. Here
\[ \frac{X}{D}=\frac{35}{70}=\frac{1}{2} \]
so $\sin(\phi/2)=\dfrac{1}{2}$.
Step 2: Get the cosine without going through the angle itself.
Instead of finding $\phi/2$ in degrees first, use the identity $\cos^2\theta+\sin^2\theta=1$ directly on the half angle:
\[ \cos\left(\frac{\phi}{2}\right)=\sqrt{1-\sin^2\left(\frac{\phi}{2}\right)}=\sqrt{1-\left(\frac{1}{2}\right)^2}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \]
This is the exact value $\dfrac{\sqrt3}{2}\approx0.866$, the same number a 30-60-90 triangle gives, but reached purely algebraically.
Step 3: Apply the interfacial energy balance at the grain boundary groove.
At the W-W grain boundary edge, two W-Ni interfaces pull against the W-W boundary, giving
\[ \gamma_{WW}=2\gamma_{WNi}\cos\left(\frac{\phi}{2}\right) \]
Step 4: Substitute the known values.
\[ \gamma_{WW}=2(0.30)\left(\frac{\sqrt3}{2}\right)=0.30\sqrt3 \]
Step 5: Evaluate.
\[ 0.30\sqrt3\approx0.30\times1.732=0.5196 \]
Rounded to two decimal places, this is $0.52\ \text{J/m}^2$. The melting point data simply confirm nickel is the liquid matrix (its melting point $1455^{\circ}\text{C}$ is below the $1550^{\circ}\text{C}$ sintering temperature) while tungsten (melting point $3410^{\circ}\text{C}$) remains solid, matching the described microstructure.
Step 6: Sanity check the size of the answer.
A solid-solid grain boundary energy of about $0.52\ \text{J/m}^2$ being larger than the solid-liquid energy of $0.30\ \text{J/m}^2$ makes physical sense here, since the neck has grown to only half the grain diameter, giving a fairly open dihedral angle of $60^{\circ}$ rather than a fully wetted boundary. If the W-W boundary energy had come out lower than $2\gamma_{WNi}$, the liquid nickel would tend to penetrate and separate the tungsten grains instead of letting them sinter together into a rigid skeleton.
\[ \boxed{\gamma_{WW}\approx0.52\ \text{J/m}^2} \]
