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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Ch-20, Energy Audit"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:20.1 Page:461"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
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{
"name": "stdout",
"output_type": "stream",
"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"
]
}
],
"source": [
"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)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:20.2 Page:468"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"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"
]
}
],
"source": [
"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)"
]
}
],
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|
