Question

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: \(1455^{\circ}\text{C}\))

Hint:

Use \(\sin(\phi/2)=X/D\) from the neck geometry to get the dihedral angle, then \(\gamma_{WW}=2\gamma_{WNi}\cos(\phi/2)\).

Answer 1

Correct Answer: 0.5

Step 1: Understand the dihedral angle equilibrium at the neck.
In liquid phase sintering, two solid tungsten grains touch and share a solid-solid (W-W) grain boundary that sits inside the liquid nickel. At the edge of that grain boundary, where it meets the liquid, the surface tension forces must balance, just like a grain boundary groove. This balance gives the dihedral angle relation
\[ \gamma_{ss}=2\gamma_{sl}\cos\left(\frac{\phi}{2}\right) \]
where \(\gamma_{ss}\) is the solid-solid (W-W) interfacial energy, \(\gamma_{sl}\) is the solid-liquid (W-Ni) interfacial energy, and \(\phi\) is the dihedral angle at the neck.

Step 2: Relate the dihedral angle to the neck and grain geometry.
For two spherical grains of diameter \(D\) joined by a flat neck of diameter \(X\), the half dihedral angle is set by how far the neck has grown into the grain:
\[ \sin\left(\frac{\phi}{2}\right)=\frac{X}{D} \]
Here the grain size is \(D=70\ \mu\text{m}\) and the neck diameter is \(X=35\ \mu\text{m}\), so
\[ \sin\left(\frac{\phi}{2}\right)=\frac{35}{70}=0.5 \]

Step 3: Solve for the dihedral angle.
\[ \frac{\phi}{2}=\sin^{-1}(0.5)=30^{\circ} \]
so
\[ \phi=60^{\circ} \]

Step 4: Find the cosine of the half angle.
\[ \cos\left(\frac{\phi}{2}\right)=\cos(30^{\circ})=0.866 \]

Step 5: Compute the W-W interfacial energy.
Using the dihedral angle relation from Step 1 with \(\gamma_{sl}=0.30\ \text{J/m}^2\):
\[ \gamma_{ss}=2(0.30)(0.866) \]
\[ \gamma_{ss}\approx0.52\ \text{J/m}^2 \]
The melting points of W and Ni are given only to confirm that at \(1550^{\circ}\text{C}\), which lies above the melting point of Ni (\(1455^{\circ}\text{C}\)) but well below that of W (\(3410^{\circ}\text{C}\)), the nickel forms the liquid phase while the tungsten grains stay solid, which is exactly the microstructure described.

Final Answer:
The W-W interfacial energy, rounded to two decimal places, is
\[ \boxed{0.52\ \text{J/m}^2} \]