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
|
{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7 : A General Strategy for Solving Material Balance Problems"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 7.1 Page no.169\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Understanding the Problem\n",
"# Solution\n",
"\n",
"# Variables\n",
"v_ts = 105.0 ; # velocity of train wrt station-[cm/s]\n",
"v_mt = 30.0 ; # velocity of man wrt train-[cm/s]\n",
"v_hm = 2.0 ; # velocity of hot dough wrt man-[cm/s]\n",
"v_am = 1.0 ; # velocity of ant wrt man- [cm/s]\n",
"# By careful reading of problem you can see that ant is moving away from man's mouth at 1 cm/s , so ant's velocity wrt station \n",
"#is say v_as\n",
"\n",
"# Calculation\n",
"v_as = v_ts + v_mt + v_am;\n",
"\n",
"# Results\n",
"print ' The ant is moving towards station at the rate of %.1f cm/s.'%v_as"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The ant is moving towards station at the rate of 136.0 cm/s.\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 7.4 Page no. 180\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Drawing a Sketch of a Mixing Process\n",
"# Solution\n",
"\n",
"# Variables\n",
"n_un= 7 ; # Number of unknowns in the given problem- 3 values of xi and 4 values Fi\n",
"n_ie = 5 ; # Number of independent equations\n",
"\n",
"# Calculations\n",
"d_o_f = n_un-n_ie ; # No. of degree of freedom\n",
"\n",
"# Results\n",
"print 'Number of degree of freedom for the given system is %i .'%d_o_f"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Number of degree of freedom for the given system is 2 .\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 7.5 Page no. 182\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Placing the unknown Information on the Diagram\n",
"# Solution\n",
"\n",
"# variables\n",
"n_un=8 ; # Number of unknowns in the given problem- 8 values of mole fractions\n",
"n_ie =6 ; # Number of independent equations- six elemental balances \n",
"\n",
"# Calculation\n",
"d_o_f= n_un-n_ie ;# Number of degree of freedom\n",
"\n",
"# Results\n",
"print 'Number of degree of freedom for the given system is %i .'%d_o_f\n",
"#Note: Experiments show that the change in CH1.8O.5N.16S.0045P.0055 and the change in C(alpha)H(beta)O(gamma) prove to be \n",
"#related by amount of biomass present and the maintenance coefficient(the moles of substrate per mole of biomass per second) \n",
"#so the respective quantities cannot be chosen independently.Consequently with this extra constraint,only one degree of freedom \n",
"#remains to be specified, the basis"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Number of degree of freedom for the given system is 2 .\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "code",
"collapsed": true,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|