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• Ingrid Zetina Tejero

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14 Ceramic Materials

14–42 The specific gravity of Al2O3 is 3.96 g/cm3. A ceramic part is produced by sintering alumina powder. It weighs 80 g when dry, 92 g after it has soaked in water, and 58 g when suspended in water. Calculate the apparent porosity, the true porosity, and the closed pores. Solution:

From the problem statement, r  3.96, Wd  80 g, Ww  92, and Ws  58. From the equations, apparent porosity 

Ww  Wd 92  80  100   100  35.29% Ww  Ws 92  58

The bulk density is B  Wd(Ww  Ws)  80(92  58)  2.3529 g/cm3. Therefore: true porosity 

rB 3.96  2.3529  100  40.58%  100  r 3.96

closed porosity  40.58  35.29  5.29% 14–43 Silicon carbide (SiC) has a specific gravity of 3.1 g/cm3. A sintered SiC part is produced, occupying a volume of 500 cm3 and weighing 1200 g. After soaking in water, the part weighs 1250 g. Calculate the bulk density, the true porosity, and the volume fraction of the total porosity that consists of closed pores. Solution:

The appropriate constants required for the equations are: r  3.1 g/cm3 Ww  1250 g

B  1200 g 500 cm3  2.4 g/cm3 Wd  1200 g

Therefore: B  2.4  Wd  1Ww  Ws 2  1200 11250  Ws 2

or Ws  750 g

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apparent porosity  true porosity 

Ww  Wd 1250  1200  100   100  10% Ww  Ws 1250  750 1r  B2 13.1  2.42   100  22.58% r 3.1

closed porosity  22.58  10  12.58% fclosed  12.58 22.58  0.44 14–54 Calculate the O:Si ratio when 20 wt% Na2O is added to SiO2. Explain whether this material will provide good glass forming tendencies. Above what temperature must the ceramic be heated to be all-liquid? Solution:

MWsoda  2122.992  16  61.98 g/mol MWsilica  28.08  21162  60.08 g/mol mole fraction Na2O  OSi 

20 g 61.98 g/mol  0.1951 20 61.98  80 60.08

11 ONa2O210.19512  12 OSiO2 210.80492  2.24 11 SiSiO2 210.80492

Since the OSi ratio is less than 2.5, it should be possible to produce a glass. From the Na2O–SiO2 phase diagram (Figure 14–11), we find that, for 20 wt% Na2O, the liquidus temperature is about 1000C. We must heat the material above 1000C to begin the glass-making operation. 14–55 How many grams of BaO can be added to 1 kg of SiO2 before the O:Si ratio exceeds 2.5 and glass-forming tendencies are poor? Compare this to the case when Li2O is added to SiO2. Solution:

We can first calculate the required mole fraction of BaO required to produce an O:Si ratio of 2.5: OSi  2.5  fBaO  0.33

11 OBaO2 fBaO  12 OSiO2 211  fBaO 2 11 Si/SiO2 211  fBaO 2 and

fsilica  0.67

The molecular weight of BaO is 137.3  16  153.3 g/mol, and that of silica is 60.08 g/mol. The weight percent BaO is therefore: wt% BaO 

10.33 mol21153.3 g/mol2  100  55.69% 10.3321153.32  10.672160.082

For 1 kg of SiO2, the amount of BaO is: 0.5569 

x g BaO x g BaO  1000 g SiO2

or x  1257 g BaO

The mole fraction of Li2O required is: OSi  2.5  fLi2O  0.33

11 OLi2O2 fLi2O  12 OSiO2 211  fLi2O 2 11 Si SiO2 211  fLi2O 2

and

fsilica  0.67

CHAPTER 14

Ceramic Materials

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The molecular weight of Li2O is 2(6.94)  16  29.88 g/mol, and that of silica is 60.08 g/mol. The weight percent Li2O is therefore: wt% Li2O 

10.33 mol2129.88 g mol2  100  19.7% 10.332129.882  10.672160.082

For 1 kg of SiO2, the amount of Li2O is: 0.197 

x g Li2O x g Li2O  1000 g SiO2

or x  245 g Li2O

Much larger amounts of BaO can be added compared to Li2O and still retain the ability to form a glass. 14–56 Calculate the O:Si ratio when 30 wt% Y2O3 is added to SiO2. Will this material provide good glass-forming tendencies? Solution:

MWyttria  2188.912  31162  225.82 g/mol MWsilica  60.08 g/mol The mole fraction of yttria is (assuming a base of 100 g of ceramic): fyttria 

30 g 225.82 g/mol  0.102 30 225.82  7060.08

The OSi ratio is then: OSi 

13 OY2O3 210.1022  12 OSiO2 210.8982  2.34 11 SiSiO2 210.8982

The material will produce a glass. 14–57 Lead can be introduced into a glass either as PbO (where the Pb has a valence of 2) or as PbO2 (where the Pb has a valence of 4). Such leaded glasses are used to make what is marketed as “crystal glass” for dinnerware. Draw a sketch (similar to Figure 14–10) showing the effect of each of these oxides on the silicate network. Which oxide is a modifier and which is an intermediate? Solution:

PbO2 provides the same number of metal and oxygen atoms to the network as does silica; the PbO2 does not disrupt the silicate network; therefore the PbO2 is a intermediate. PbO does not provide enough oxygen to keep the network intact; consequently PbO is a modifier.

s

Pb Pb

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14–58 A glass composed of 65 mol% SiO2, 20 mol% CaO, and 15 mol% Na2O is prepared. Calculate the O:Si ratio and determine whether the material has good glass-forming tendencies. Estimate the liquidus temperature of the material using Figure 14–16. Solution:

Based on the mole fractions, we can determine the O:Si ratio: OSi 

12 OSiO2 210.652  11 OCaO210.202  11 ONa2O210.152 11 SiSiO2 210.652

OSi  2.54

The glass-forming tendencies are relatively poor and special attention to the cooling rate may be required. To determine the liquidus, we must find the weight percentages of each constituent. The molecular weights are: MWsilica  60.08 g/mol MWCaO  40.08  16  56.08 g/mol MWsoda  2122.992  16  61.98 g/mol 10.652160.082  100 10.652160.082  10.202156.082  10.152161.982  65.56%

wt% SiO2 

10.202156.082  100 10.652160.082  10.202156.082  10.152161.982  18.83%

wt% CaO 

10.152161.982  100 10.652160.082  10.202156.082  10.152161.982  15.61%

wt% Na2O 

From the ternary phase diagram, this overall composition gives a liquidus temperature of about 1140C.