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
226
227
|
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 10: Combustion Processes"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 10.1"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Mass of O2 required (lbm) = 96.00\n",
"press enter key to exit\n"
]
},
{
"data": {
"text/plain": [
"''"
]
},
"execution_count": 1,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#If 3 mol carbon are completely burned in a process, determine \n",
"#(a) the number of moles of O2 required and\n",
"#(b) the mass of O2 required\n",
"#initialisation of variables\n",
"n= 3 \t\t\t\t#lbm mol\n",
"Mo2= 32 \t\t\t#lbm/lbm mol\n",
"#CALCULATIONS\n",
"m= n*Mo2 \t\t\t#Mass of O2 required\n",
"#RESULTS\n",
"print '%s %.2f' % ('Mass of O2 required (lbm) = ',m)\n",
"raw_input('press enter key to exit')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 10.3"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Air-fuel ratio (gm air/gm fuel) = 15.12\n",
"press enter key to exit\n"
]
},
{
"data": {
"text/plain": [
"''"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#Determine the air-fuel ratio on a gram-molal and a mass basis if octane is \n",
"#completely burned in the presence of theoretical air\n",
"#initialisation of variables\n",
"n= 12.5 \t\t\t\t\t#mol\n",
"n1= 3.76 \t\t\t\t\t#mol\n",
"M= 114 \t\t\t\t\t\t#gm/gm mol\n",
"M1= 28.96 \t\t\t\t\t#gm/gm mol\n",
"#CALCULATIONS\n",
"n2= n*(1+n1) \t\t\t\t#Moles\n",
"m= n2*M1/M \t\t\t\t\t#Air-fuel ratio \n",
"#RESULTS\n",
"print '%s %.2f' % ('Air-fuel ratio (gm air/gm fuel) = ',m)\n",
"raw_input('press enter key to exit')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 10.4"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Air-fuel ratio (kg mol air/kg mol fuel) = 89.25\n",
"press enter key to exit\n"
]
},
{
"data": {
"text/plain": [
"''"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#Develop the stoichiometric equation that expresses the process of octane \n",
"#being burned with 150% theoretical air. What is the air-fuel ratio on a \n",
"#kilogram molal basis?\n",
"#initialisation of variables\n",
"p= 150\n",
"nO2= 12.5 \t\t\t\t\t\t\t\t#mol\n",
"n1= 3.76\n",
"#CALCULATIONS\n",
"n2= (nO2*(p/100.))+(nO2*n1*(p/100.))\t#ir-fuel ratio \n",
"#RESULTS\n",
"print '%s %.2f' % ('Air-fuel ratio (kg mol air/kg mol fuel) = ',n2)\n",
"raw_input('press enter key to exit')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 10.5"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Benzene required (gm) = 568.40\n",
"press enter key to exit\n"
]
},
{
"data": {
"text/plain": [
"''"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#Benzene is used in a heating process that is 65% efficient. How much benzene\n",
"#is required to heat 185 liter tank of water from 10 to 30C?\n",
"#initialisation of variables\n",
"P= 65 \n",
"T= 30 \t\t\t\t#C\n",
"T1= 10 \t\t\t\t#C\n",
"c= 4.19 \t\t\t#J/gm C\n",
"h= 41961\n",
"m= 185 \t\t\t\t#lt\n",
"#CALCULATIONS\n",
"Q= m*1000*c*(T-T1)\t#Heat required\n",
"M= (Q*100.)/(h*P) \t#Mass of benzene required\n",
"#RESULTS\n",
"print '%s %.2f' % ('Benzene required (gm) = ',M)\n",
"raw_input('press enter key to exit')"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.6"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
|
