Essentials Of Materials Science And Engineering
4th Edition
ISBN: 9781337670845
Author: ASKELAND
Publisher: Cengage
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Question
Chapter 15, Problem 15.3P
Interpretation Introduction
(a)
Interpretation:
The application of alumina should be determined.
Concept Introduction:
Alumina is short and convenient name given to aluminum oxide. It is designated by Al2O3.
It is also known as aloxite or alundum.
Properties of aluminum are:
- It is white in color.
- It is odorless in nature.
- It is insoluble in water.
- It is having a density of 3.987 g/cm3.
Interpretation Introduction
(b)
Interpretation:
The application of silica should be determined
Concept Introduction:
Silica is a short name provided for silicon dioxide. Its chemical designation is SiO2. Silica is a major constituent of sand.
Properties of silica are:
- It is transparent.
- Its melting point is 1,713 degree Celsius.
- The boiling point is 2950 degree Celsius.
- It appears in white color.
- It affects immune system.
Interpretation Introduction
(c)
Interpretation:
The applications barium titanate should be determined.
Concept Introduction:
Barium titanate whose chemical formula is BaTiO3. It is an organic compound.
The properties are:
- It appears in powder from.
- It appears white in color.
- It is odorless in nature.
- It is insoluble in water.
- It dissolves in concentrated HF acid.
Interpretation Introduction
(d)
Interpretation:
The applications of zirconia should be determined.
Concept Introduction:
Zirconia is short form termed to zirconium dioxide.It is naturally occurring form with a monoclinic crystalline structure.
Properties of zirconium are:
- It appears white in color.
- It appears in powdered form.
- Its density is 5.68 g/cm3.
- It's melting point is 2,715 degree Celsius.
- Its boiling point is 4,300 degree Celsius.
Interpretation Introduction
(e)
Interpretation:
The applications of diamonds should be determined.
Concept Introduction:
Diamond is a crystal which is typically used as gemstone among which diamonds age 1 billion to 3.5 billion years old.
Properties of the diamond are as follows:
- They have high hardness.
- They are chemically stable.
- They appear in various colors like yellow, colorless, brown and grey as well.
- They are isotropic in nature.
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Figure 2
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a) Determine IDQ and VSDQ.
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Figure 1
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4- In the system shown in the figure, the water velocity in the 12 in. diameter
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Chapter 15 Solutions
Essentials Of Materials Science And Engineering
Ch. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Prob. 15.3PCh. 15 - Prob. 15.4PCh. 15 - Prob. 15.5PCh. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Prob. 15.8PCh. 15 - Prob. 15.9PCh. 15 - Prob. 15.10P
Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42PCh. 15 - Prob. 15.43PCh. 15 - Prob. 15.44PCh. 15 - Prob. 15.45PCh. 15 - Prob. 15.46PCh. 15 - Prob. 15.47PCh. 15 - Prob. 15.48PCh. 15 - Prob. 15.49PCh. 15 - Prob. 15.50DPCh. 15 - Prob. 15.51CPCh. 15 - Prob. K15.1KPCh. 15 - Prob. K15.2KPCh. 15 - Prob. K15.3KP
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