Essentials Of Materials Science And Engineering
Essentials Of Materials Science And Engineering
4th Edition
ISBN: 9781337385497
Author: WRIGHT, Wendelin J.
Publisher: Cengage,
Question
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Chapter 17, Problem 17.18P
Interpretation Introduction

(a)

Interpretation:

The volume fraction of silicon carbide for given grinding wheel should be calculated.

Concept introduction:

According to the rules of mixtures for particulate composites:

  ρc=( f i ρ i )for{i=1..........n}ρc=f1ρ1+f2ρ2...................fnρn

Where

  ρc= density of composite

  fi= volume fraction of i component

  ρi= density of i component

Volume fraction is defined as:

  f=V1V1+V2f=volumefractionV1=volumeoffirstcomponentV2=volumeofsecondcomponentWhereV1andV2V1=weightoffirstcomponentdensityoffirstcomponentV2=weightofsecondcomponentdensityofsecondcomponentintermsofweightanddensity

  f=V1=weightoffirstcomponentdensityoffirstcomponentV1=weightoffirstcomponentdensityoffirstcomponent+V2=weightofsecondcomponentdensityofsecondcomponent

Expert Solution
Check Mark

Answer to Problem 17.18P

The volume fraction of silicon carbide is 0.336571.

Explanation of Solution

Given information:

Diameter of grinding wheel = 9inch

Thickness of grinding wheel = 1inch

Weight of grinding wheel = 6lb

Density of silicon carbide = 3.2g/cm3

Wheel is 5% porous

Cubes of silicon carbide = 0.04 cm

Based on given information:

Applying rule of mixing,

  ρc=f1ρ1+f2ρ2+f3ρ3..............(1)

  ρc = density of wheel

  f1 = the volume fraction based on porosity

  f2 = the volume fraction of silicon carbide

  f3 = the volume fraction of silica glass

  ρ1 = density of porous material is 0

  ρ2 = density of silicon carbide is 3.2g/cm3

  ρ3 = density of vulcanized rubber is 2.5g/cm3

Calculation of volume fraction of wheel with the use of porosity:

  f1=(5 100)f1=0.05

Let the volume of grinding wheel be V:

  VWheel=π4d2td=9inch1inch=2.54cmt=1inchVWheel=π4(9 inch× 2.54cm 1inch)2×1 inch×2.54cm1inchVWheel=1402.499cm3

Calculation of density using weight of grinding wheel and volume of grinding wheel

  ρgrindingwheel=weightofgrindingwheelvolumeofgrindingwheelρgrindingwheel=6lb1042.499 cm3

Conversion of lb/cm3 to g/cm3:

  1lb=0.45359kg1kg=1000gram

  ρgrindingwheel=6lb×0.45359kg1lb×1000gram1kg1042.499cm3

  ρgrindingwheel=2.6106g/cm3

The density of grinding wheel is 2.6106g/cm3.

Applying rule of mixing from equation (1):

  ρc=f1ρ1+f2ρ2+f3ρ3

The value of volume fraction of silicon carbide and silica glass is unknown.

From the rule of mixing,

  f1+f2+f3=1f3=1f1f2f3=10.05f2

Substituting the volume fraction of silica glass in terms of volume fraction of silicon carbide and porosity:

  2.6106=(0.05×0)+(3.2×f2)+(2.5)×(10.05f2)

  f2=0.366571

The value of volume fraction of silicon carbide = 0.366571

Interpretation Introduction

(b)

Interpretation:

The number of particles of silicon carbide lost by wheel when diameter is worn to 8 inch should be calculated

Concept introduction:

Volume fraction is defined as the ratio of volume of specified component to the sum of total volume of components.

  f=V1V1+V2f=volumefractionV1=volumeoffirstcomponentV2=volumeofsecondcomponent

Volume of cubic particle is defined as:

  Vcubicparticle=(a)3

Expert Solution
Check Mark

Answer to Problem 17.18P

The number of particles of silicon carbide = 1150646.830.

Explanation of Solution

Given information:

Diameter of grinding wheel = 9inch

Thickness of grinding wheel = 1inch

Weight of grinding wheel = 6lb

Density of silicon carbide = 3.2g/cm3

Density of silicon glass = 2.5g/cm3

Wheel is 5% porous

Cubes of silicon carbide = 0.04

Number of particles per cubic centimeter is defined as the ratio of number of silicon carbide particle to volume of silicon carbide particle:

  n=numberofsiliconcarbideparticlevolumeofsiliconcarbideparticle

Volume of silicon carbide particle = VinitialofSiCparticleVfinalofSiCparticle

Calculation of initial volume of particle:

  VinitialvolumeofSiCparticle=π4(9)2×1×2.54 cm31inchVinitialvolumeofSiCparticle=1402.499cm3

Calculation of final volume of particle:

  VfinalofSiCparticle=π4×d2×tVfinalofSiCparticle=π4×(8)2×1×2.54cm1VfinalofSiCparticle=823.7036cm3

Volume of silicon carbide = 1042.499-823.7=218.799cm3

For cubic particle of 0.04 cubes of silicon carbide:

  Vcubicparticle=(a)3Vcubicparticle=(0.04)3Vcubicparticle=6.4×105cm3

Number of silica particle lost is defined as,

  nlost= (Number of silicon carbide particle per cubic centimeter) × (Volume of wheel lost in wear cubic centimeter)

For 1 cubic centimeter of particle of silicon carbide:

  nSiC=f2×1cm3nSiC=0.336571cm3

Number of silicon carbide particle per cubic centimeter = 0.3365716.4×105=5258.9218

  nlost=5258.9218×218.799nlost=1150646.830

The number of silicon carbide lost particle = 1150646.830

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