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陶瓷材料 (1021) 
Homework 6 – Solution 
 
 
1. For solid-state sintering to occur, gb needs to be smaller than 2sv. Why? If 
this were true, why this means that the dihedral angle  < 180o? 
 
Solution: 
(a) At equilibrium, Young’s equation gives gb = 2sv cos(/2). When gb is 
smaller than 2sv, formation of new grain boundary (gb) becomes 
energetically favorable so that grain surface (sv) would then disappear. 
Sintering is just the process of formation new grain boundaries instead of 
existing surfaces. 
(b) Since gb = 2sv cos(/2), we then know that (gb /2sv) = cos(/2) <1. 
Therefore,  < 180o. 
 
2. Shown below is a dihedral angle at a pore-grain junction within a polycrystal. If 
sintering is meant to occur, the pore needs to be removed from the polycrystal at 
elevated temperatures. In this regard, if there are three oxide ceramics with 
dihedral angles of 150o, 120 o, and 60 o, respectively, which one would densify 
more rapidly? Explain after calculation. Neglect the impurity effect and 
assume the oxides have comparable surface energies. 
 
Solution: 
Since gb = 2sv cos(/2), input the dihedral angles respectively. We find gb 
(150o) is the smallest; therefore,  = 150o gives rise to the fastest sintering when 
everything else being held equal. 
 
3. It takes 0.2 hour for the relative density of a 0.1 m average diameter powder to 
increase from 55 to 60 percent. Estimate the time it would take for a powder of 
10 m average diameter to achieve the same degree of densification if the 
rate-controlling mechanism is the same and is a lattice-diffusion process. 
 
Solution: 
For lattice diffusion at the initial stage of sintering, we know that 
t
kTr
D
r
x l 


 


3
5 80
 
We may assume that the ratio x/r does not vary much as the density increases from 
55 to 60%, therefore, it can be treated as a constant (i.e., C’ as shown below) and 
the above equation can be rewritten as 
3
3
80
rC
kTr
D
Ct
l




 
  
Where C is a constant including C’. For the 0.1 m average diameter powder, 
we can find the constant C by 
21
363 106.1)1005.0(
2.0  r
tC h/m3 
The 10 m average diameter powder would then need 
536213 102)105(106.1  Crt hours 
to achieve the same density. 
 
4. The activation energy for viscosity of pure silica decreases from 565 to 163 
kJ/mole upon addition of 0.5 mole fraction MgO or CaO. The addition of alkali 
oxide (e.g. Na2O) has an even more dramatic effect in lowering the activation 
energy to 96 kJ/mole for the same additive fraction. Why is so? 
 
Solution: 
In all the cases, the formation of non-bridging Si and O bonds because of the 
additive cations is responsible for the dramatic decrease in the activation energy of 
viscosity. The addition of Na oxide is more effective than that of Ca oxide (or 
Mg oxide) is because of its lower single-bond strength (see page 88 for their 
values) 
 
5. If now we have two silicate structures, Na2Si2O5 and Na2SiO3, respectively. 
Which one would show a lower viscosity? Which one should be a better glass 
former, i.e., present a stronger single-bond strength? 
 
Solution: 
The melt with the smaller fraction of non-bridging oxygen at the melting point 
should be the better glass former. Thus, Na2Si2O5 with an O/Si ratio of 2.5 
should be the better glass former than Na2SiO3 with its higher O/Si ratio. 
Remember with an O/Si ratio of 2, the viscosity is the highest. 
 
6. For the two-particle model in solid-state sintering, (a) calculate the capillary 
pressure involved in two contacting glass spheres both with a particle diameter of 
10 m sintered at 1000oC for an isothermal holding of 10 min. The x/a ratio at 
the neck is 0.28. The surface energy is 300 ergs/cm2, and the radius of curvature 
at the neck is 10-7 m. Express your calculated capillary pressure by unit of MPa. 
 
Solution: 
mx
a
x 66 104.1)105)(28.0(28.0   
MPa
x
P mJmmJ 8.2108.2)
104.1
1
10
1()1010300()11(
32 /6/1
67
/47  


 
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during the office hours.)