1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
|
{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 20 : Performance of Materials in Service ( Fracture, Fatigue, and Corrosion and its Control)"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 20.1 page no : 537"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"# Variables\n",
"l = 1.5*10**-6;\t\t\t#crack length in m\n",
"e = 70*10**9;\t\t\t#Young's modulous in N/m**2\n",
"y_e = 1.05;\t\t\t#specific surface energy in j/m**2\n",
"\n",
"# Calculations\n",
"a_f = math.sqrt((2*y_e*e)/(3.14*l));\n",
"a_f1 = a_f*10**-6;\t\t\t#in MPa\n",
"r = a_f/e;\t\t\t#ratio\n",
"\n",
"# Results\n",
"print \"Fracture strength (in MPa) = %.2f\"%a_f1\n",
"print \"Ratio of fracture strength to Youngs modulous = %.2e\"%r\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Fracture strength (in MPa) = 176.66\n",
"Ratio of fracture strength to Youngs modulous = 2.52e-03\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 20.2 pageno : 547"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"#Variables\n",
"m_m = 58.71;\t\t\t#molecular weight of ni\n",
"m_c = 74.71;\t\t\t#molecular weight of nio\n",
"p_m = 8900;\t\t\t#density of ni in kg/m**3\n",
"p_c = 7080;\t\t\t#desity of nio in kg/m**3\n",
"\n",
"#Calculations & Reults\n",
"x = m_m/p_m;\t\t\t#molar volume of ni in m**3/mol\n",
"print \"Mc/Pc (in m**3/mol) %.2e m**3/mol\"%x\n",
"y = m_c/p_c;\t\t\t#molar volume of nio in m**3/mol\n",
"print \"Mm/Pm (in m**3/mol) %.2e m**3/mol\"%y\n",
"print \" Mc/Pc > Mm/Pm Hence protective layer of NiO will form over Ni \";\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Mc/Pc (in m**3/mol) 6.60e-03 m**3/mol\n",
"Mm/Pm (in m**3/mol) 1.06e-02 m**3/mol\n",
" Mc/Pc > Mm/Pm Hence protective layer of NiO will form over Ni \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 20.3 pageno : 548"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"# Variables\n",
"x1 = 0.1;\t\t\t#in mm\n",
"t1 = 25.;\t\t\t#in hours\n",
"t2 = 300.;\t\t\t#in hours\n",
"\n",
"# Calculations\n",
"x2 = x1*math.sqrt(t2/t1);\t\t\t#in mm\n",
"\n",
"# Results\n",
"print \"Oxidation loss in 300 hours (in mm) = %.3f mm\"%x2\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Oxidation loss in 300 hours (in mm) = 0.346 mm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 20.4 pageno : 551"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"p_mg = 1.74;\t\t\t#density of magnesium in gm/cm**3\n",
"p_mgo = 3.65;\t\t\t#density of magnesium oxide in gm/cm**3\n",
"m_mg = 24.;\t\t\t#mol wt ogf mg\n",
"m_mgo = 40.3;\t\t\t#mol wt of mgo\n",
"\n",
"# Calculations\n",
"PBR = (m_mgo/p_mgo)/(m_mg/p_mg);\n",
"\n",
"# Results\n",
"print \"PBR = %.2f\"%PBR\n",
"print \"Since PBR < 1. So porous film will form which will be non protective\";\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"PBR = 0.80\n",
"Since PBR < 1. So porous film will form which will be non protective\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 20.5 pageno : 562"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"m = 0.0243;\t\t\t#one mole of magnesium in kg\n",
"\n",
"# Calculations\n",
"c = 2.*96490;\t\t\t#in C\n",
"j = 20.*10**-3;\t\t\t#in A/m**2\n",
"t = 15.*365*24*3600;\t\t\t#in sec\n",
"x = j*t; \t\t\t#in A s\n",
"w_mg = m*x/c;\t \t\t#in kg/sqm\n",
"\n",
"# Results\n",
"print \"Amount of Magnesium needed (in Kg/m**2) = %.2f kg/m**2\"%w_mg\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Amount of Magnesium needed (in Kg/m**2) = 1.19 kg/m**2\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
