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
|
{
"metadata": {
"name": "",
"signature": "sha256:cd162bdb1fe196cb47bf3eb77e7ebc05430fce626f0c2cb37a8e99707eb3d120"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Ch-20, Energy Audit"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example:20.1 Page:461"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import asin, cos, acos, tan\n",
"lod=1 #industrial installation load\n",
"pf=0.78 #power factor\n",
"tf=200 #tariff \n",
"md=3.5 #extra maximum demand\n",
"ic=500 #installation of capacitor\n",
"id=0.15 #interest and depreciation\n",
"lf=0.8 #load factor\n",
"sinp=ic*id/tf\n",
"ph2=asin(sinp)\n",
"epf2=cos(ph2)\n",
"ph1=acos(pf)\n",
"ph1=round(ph1*10**2)/10**2\n",
"ph2=round(ph2*10**2)/10**2\n",
"q=lod*(tan(ph1)-tan(ph2))\n",
"q=round(q*10**4)/10**4\n",
"ikva=lod/pf\n",
"ikv=round(ikva*(10**5))/10**2\n",
"aeu=lod*lf*8760*10**6\n",
"eb=ikv*tf+aeu*md\n",
"print \"(a)\\neconomic power factor %.3flagging \\n(b) \\ncapacitor kVAr to improve the power factor %.4f \\n(c) \\ninitial kVA %.2fKVA \\nannual energy used %0.3ekWh \\nelectrical bill Rs%e per year\"%(epf2,q,ikv,aeu,eb)\n",
"kvc=round((lod*10**3/(round(epf2*1000)/10**3))*10**2)/10**2\n",
"ebc=kvc*tf+aeu*md\n",
"aidc=q*10**3*ic*id\n",
"te=ebc+aidc\n",
"asc=eb-te\n",
"print \"(d)\\nKVA after installation of capacitors %.2fKVA \\n\"%kvc\n",
"print \"energy bill after installation of capacitor Rs%e per year\"%ebc\n",
"print \"annual interest and depreciation of capacitor bank Rs%.1fper year \\ntotal expendition after installation of capacitors Rs%e per year \\nannual savings due to installation of capacitors Rs%d per year\"%(aidc,te,asc)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)\n",
"economic power factor 0.927lagging \n",
"(b) \n",
"capacitor kVAr to improve the power factor 0.4092 \n",
"(c) \n",
"initial kVA 1282.05KVA \n",
"annual energy used 7.008e+09kWh \n",
"electrical bill Rs2.452826e+10 per year\n",
"(d)\n",
"KVA after installation of capacitors 1078.75KVA \n",
"\n",
"energy bill after installation of capacitor Rs2.452822e+10 per year\n",
"annual interest and depreciation of capacitor bank Rs30690.0per year \n",
"total expendition after installation of capacitors Rs2.452825e+10 per year \n",
"annual savings due to installation of capacitors Rs9970 per year\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example:20.2 Page:468"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"ee=5*10**16 #electrical energy requirement\n",
"eer=0.1 #energy requirement\n",
"i=5*10**6 #investement\n",
"n=20 #life time\n",
"ec=4.1 #energy cost\n",
"r=0.13 #interest rate\n",
"dr=r/((1+r)**(n)-1) #depreciation rate\n",
"dr=round(dr*10**5)/10**5\n",
"tfc=r+dr #total fixed cost\n",
"ace=i*tfc #annual cost\n",
"ace=round(ace/10**2)*10**2\n",
"eb=i*ec #electrical bill with present motor\n",
"teb=eb*(1-eer) #electrical bill with efficiency motor\n",
"tac=teb+ace #total annual cost with efficiency cost\n",
"As=eb-tac #annual saving\n",
"print 'part (a)'\n",
"print \" depreciation rate %.5f \\n total fixed charge rate %f\\n annual cost of efficiency motor Rs%eper year \\n total electrical bill with present motors Rs%eper year \\n total electrical bill with efficiency motor Rs.%e \\n total annual cost if motors are replaced by high efficiency motors Rs%e per year \\n annual saving Rs%d per year\"%(dr,tfc,ace,eb,teb,tac,As)\n",
"print 'part (b)'\n",
"pwf=r/(1-((1+r)**-n)) #present worth factor\n",
"pwf=round(pwf*10**5)/10**5\n",
"pwm=teb/pwf #present worth annual cost with existing motors\n",
"pwm=round(pwm/10**4)*10**4 #present worth with existing motors\n",
"pwem=eb/pwf #present worth with efficiency motor\n",
"pwem=round(pwem/10**4)*10**4\n",
"pwam=teb/pwf\n",
"pwam=round(pwam/10**4)*10**4\n",
"tpw=pwam+i #total persent worth\n",
"print \" present worth factor %.5f \\n present worth of annual cost with existing motors Rs%e \\n present worth of annual cost with new motor Rs%e \\n total present worth %e per year\"%(pwf,pwem,pwam,tpw)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"part (a)\n",
" depreciation rate 0.01235 \n",
" total fixed charge rate 0.142350\n",
" annual cost of efficiency motor Rs7.118000e+05per year \n",
" total electrical bill with present motors Rs2.050000e+07per year \n",
" total electrical bill with efficiency motor Rs.1.845000e+07 \n",
" total annual cost if motors are replaced by high efficiency motors Rs1.916180e+07 per year \n",
" annual saving Rs1338200 per year\n",
"part (b)\n",
" present worth factor 0.14235 \n",
" present worth of annual cost with existing motors Rs1.440100e+08 \n",
" present worth of annual cost with new motor Rs1.296100e+08 \n",
" total present worth 1.346100e+08 per year\n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}
|
