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{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Ch-8, Phase & Phase diagrams"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-8.1 page no-227"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"#given\n",
"#according to reduced phase rule, we have\n",
"#D=C-P+2\n",
"C=2 # for two component system\n",
"P=1\n",
"for P in range (1,6):\n",
" print \"P =\",P,' ',\n",
" #no of variables\n",
" V=P*(C-1)+2\n",
" #degrees of freedom\n",
" D=C-P+2\n",
" print \"D =\",D\n",
"\n",
"print \"we can see that for P=5 we have D=-1 i.e non existent so,two components cannot have more than 4 phases in equilibrium\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"P = 1 D = 3\n",
"P = 2 D = 2\n",
"P = 3 D = 1\n",
"P = 4 D = 0\n",
"P = 5 D = -1\n",
"we can see that for P=5 we have D=-1 i.e non existent so,two components cannot have more than 4 phases in equilibrium\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-8.2 page no-235"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"#given\n",
"#density of alpha and beta phases\n",
"rhoalpha=10300 #kg/m**3\n",
"rhobeta=7300 #kg/m**3\n",
"#refer to fig-8.5 in book\n",
"#at point B, the composition of lead in alpha-phase is 82% and that of tin in alpha-phase is 18%\n",
"leadalpha=82\n",
"tinalpha=18\n",
"#so we get\n",
"#82/rholead+18/rhotin=100/rhoalpha -----------(1)\n",
"#similarly at point E\n",
"#the composition of tin and lead resp are 97% and 3%\n",
"leadbeta=3\n",
"tinbeta=97\n",
"#so we get\n",
"#3/rholead+97/rhotin=100/rhobeta ------(2)\n",
"#solving 1 and 2\n",
"#we get\n",
"rholead=11364.1 #kg/m**3\n",
"rhotin=7220.14 #kg/m**3\n",
"#let density of eutectic composition is rhoe. knowing the compositions at point D, we can write\n",
"#38/rholead+62/rhotin=100/rhoe\n",
"#so \n",
"rhoe=100/(38/rholead+62/rhotin) #kg/m**3\n",
"#it is given that there is 88% eutectic composition by volume. its conversion in weight proportions yeild\n",
"W=88/100*rhoe #kgf\n",
"Wlead=38/100*W #Kgf\n",
"Wtin=62/100*W #kgf\n",
"#there is 12% beta phase by volume which on converion to weight proportion gives\n",
"Wdash=12/100*rhobeta #Kgf\n",
"Wdashlead=3/100*Wdash #kgf\n",
"Wdashtin=97/100*Wdash #kgf\n",
"#total weight of lead and tin can be estimated now as\n",
"Wddashlead=Wlead+Wdashlead #kgf\n",
"Wddashtin=Wtin+Wdashtin #kgf\n",
"#percentafe of tin\n",
"percenttin=Wddashtin/(Wddashtin+Wddashlead)*100 \n",
"print \"percentace of tin is %0.2f\"%(percenttin)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"percentace of tin is 65.72\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-8.4 page no-242"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"#melting point of A and B\n",
"MptA=1250 #degrees celcius\n",
"MptB=1900 #degrees celcius\n",
"#part(a)\n",
"#according to the conditions given in question phase diagram can be drawn as shown in fig 8.10 given in book\n",
"#part(b)\n",
"#at 1250 degrees celcius , it is a peritectic solution\n",
"#equation representing the equilibrium is given in the book which denotes forward reaction as cooling and backward as heating\n",
"#part (c)\n",
"#we are just considering a tie line just above the peritectic line at temp 1251 degrees celcius\n",
"#at this point\n",
"Cs1=80\n",
"Cl1=30 \n",
"C01=75\n",
"#point below peritectic line has temp as 1249 degrees celcius\n",
"Cs2=80\n",
"Cl2=50 \n",
"C02=75\n",
"#weight fraction of the phase present in the material of overall composition 75% B at 1251 degrees celcius and 75% B at 1249 degrees \n",
"f_alpha1=(Cs1-C01)/(Cs1-Cl1)*100 #%\n",
"f_beta1=(C01-Cl1)/(Cs1-Cl1)*100 #%\n",
"f_alpha2=(Cs2-C02)/(Cs2-Cl2)*100 #%\n",
"f_beta2=(C02-Cl2)/(Cs2-Cl2)*100 #%\n",
"#part (d)\n",
"# 75% B at room temp\n",
"Cs3=90\n",
"Cl3=30 \n",
"C03=75\n",
"f_alpha3=(Cs3-C03)/(Cs3-Cl3)*100 #%\n",
"f_beta3=(C03-Cl3)/(Cs3-Cl3)*100 #%\n",
"#the microstructure is also shown in book at page no 243\n",
"print \"weight fraction of the phase present in the material of overall composition 75 percent B at 1251 degrees celcius and 75percent B at 1249 degrees are %d, %d, %0.2f, %0.2f and at 75 percent concentration of B at room temp is %d, %d\"%(f_alpha1,f_beta1,f_alpha2,f_beta2,f_alpha3,f_beta3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"weight fraction of the phase present in the material of overall composition 75 percent B at 1251 degrees celcius and 75percent B at 1249 degrees are 10, 90, 16.67, 83.33 and at 75 percent concentration of B at room temp is 25, 75\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-8.5 page no-246"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"#according to the given conditions in the ques the graph can be drawn as shown in fig 8.12\n",
"#the given temp 576.9 degees celcius is just below the eutectic temp 577 degres celcius\n",
"#from the graph, we can have the following values\n",
"C_beta_e=100\n",
"C_e=1.65\n",
"C_0=10\n",
"#hence weight fraction is given by\n",
"W_alpha=(C_beta_e-C_0)/(C_beta_e-C_e)\n",
"print \"the weight fraction of alpha in an alloy containing 10percent Si at 576.9 degrees celcius is %0.3f \"%(W_alpha)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the weight fraction of alpha in an alloy containing 10percent Si at 576.9 degrees celcius is 0.915 \n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-8.6 page no-247"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"#weight percent of tin and lead are\n",
"W1=90\n",
"W2=10\n",
"#let the amount of tin that can be added to the crucible without changing the system's solidificarion is \"m\" \n",
"#eutectic arm extends to 97% tin\n",
"#therefore, with the addition of \"m\" gram of tin, the composition of alloy should not exceed 97% tin\n",
"#therefore\n",
"#(900+m)/(1000+m)=97/100\n",
"#so\n",
"m=(97/100*1000-900)/(1-97/100)/1000 #kg because 1kg=1000g\n",
"print \"maximum %0.2f kg of tin can be added without changing the systems temperature\"%(m)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximum 2.33 kg of tin can be added without changing the systems temperature\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-8.7 page no-250"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"#the eutectoidal mixture of pearlite consists of two phases viz. alpha (ferrite) and Fe3C (cementite). the eutectoid composition contains 0.83% carbon. the lever rule is applied in which the lever arm has ferrite (=0% carbon) at one end and cementite (6.67% carbon) at the other end. the fulcrum is taken at 0.83% carbon. hence by applying lever rule, we get\n",
"Walpha=(6.67-0.83)/(6.67-0.00)\n",
"WFe3C=(0.83-0.0)/(6.67-0.0)\n",
"print \"the weight fractions of ferrite and cementite are %0.3f and %0.3f resp\"%(Walpha,WFe3C)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the weight fractions of ferrite and cementite are 0.876 and 0.124 resp\n"
]
}
],
"prompt_number": 20
}
],
"metadata": {}
}
]
}
|
