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
|
{
"metadata": {
"name": "",
"signature": "sha256:ba05265a254158c2105561a6062dfe30873e9384b5393c6f2ca64b6f28ed143f"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 14 : The Diesel Engine"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.1 Page no : 258"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"r = 14.3\t\t\t\t\t#Compression ratio\n",
"c = 5\t\t\t\t\t#Fuel cutoff in percent of stroke\n",
"w = 0.006\t\t\t\t\t#Weight of charge in kg\n",
"T4 = 912\t\t\t\t\t#Final temperature in degree C abs\n",
"q = 8300\t\t\t\t\t#Heat in kcal\n",
"x = [0.258,0.000048]\t\t\t\t\t#Temperature expression is 0.258T+0.000048T**2, where T is in degree C abs\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"v4 = 1. #Assuming clearance volume as unity\n",
"v1 = 1.665\t\t\t\t\t#v1 from fig. 14.2 on page no. 352\n",
"T1 = (T4*v1)/v4\t\t\t\t\t#Temperature in degree C abs\n",
"qp1 = (x[0]*T1+x[1]*T1**2)\t\t\t\t\t#constant pressure heat of mixture at temperature T1 in kcal/kg\n",
"qp4 = (x[0]*T4+x[1]*T4**2)\t\t\t\t\t#constant pressure heat of mixture at temperature T4 in kcal/kg\n",
"qre = (qp1-qp4)\t\t\t\t\t#Heat required by the mixture in kcal/kg\n",
"wf = (w*qre)/q\t\t\t\t\t#Weight of oil in kg\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The weight of oil that must be injected is %3.6f kg'%(wf)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The weight of oil that must be injected is 0.000164 kg\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.2 Page no : 261"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"r = 14\t\t\t\t\t#Compression ratio\n",
"p = 1.2\t\t\t\t\t#Induction pipe pressure in kg/cm**2\n",
"bp = 0.65\t\t\t\t\t#Exhaust back pressure in kg/cm**2\n",
"Tc = 87+273\t\t\t\t\t#Charge temperature in K\n",
"Te = 850+273\t\t\t\t\t#Exhaust temperature in K\n",
"T1 = 111+273\t\t\t\t\t#Temperature at the beginning of compression in K\n",
"g = 1.2\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Cw1 = ((bp*10**4)/Te)\t\t\t\t\t#specific heat in kJ/kg.K\n",
"Cw2 = ((p*10**4*(r-1))/Tc)\t\t\t\t\t#specific heat in kJ/kg.K\n",
"T3 = ((g*Te*Cw1+Cw2*Tc)/(Cw1*g+Cw2))\t\t\t\t\t#Temperature in K\n",
"t3 = T3-273\t\t\t\t\t#Temperature in degree C\n",
"rw = (Cw1/Cw2)\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The ratio of the mass residuals to fresh charge is %3.4f'%(rw)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The ratio of the mass residuals to fresh charge is 0.0134\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
