{
 "metadata": {
  "name": "",
  "signature": "sha256:fb68f886b3c2cdc28623218c177a66cd2a93d17c7bf83ed92891b0a04da2f4c6"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Ch-4, Tariffs & Power Factor Improvement"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.1 page 59"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "day=30 #days \n",
      "pll=40 ;nll=5; tll=3 #light load \n",
      "pfl=100; nfl=3 ;tfl=5 #fan load\n",
      "prl=1*1000 #refrigerator\n",
      "pml=1*1000; nml=1 #misc. load \n",
      "t1=2.74 ;t11=15#tariff\n",
      "t2=2.70 ;t22=25  #tariff on 25 units\n",
      "tr=2.32  #reamaining units\n",
      "tc=7.00 #constant charge\n",
      "dis=0.05#discount for prompt payment\n",
      "te=(pll*nll*tll+pfl*nfl*tfl)*day+prl*day+pml*day\n",
      "tee=te/1000\n",
      "mb=tc+tr*(tee-t11-t22)+t1*t11+t2*t22\n",
      "nmb=mb*(1-dis)\n",
      "print \"total energy consumption in %d day %dunits \\nthe monthly bill Rs%.2f \\nnet monthly bill Rs%.2f\"%(day,tee,mb,nmb)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "total energy consumption in 30 day 123units \n",
        "the monthly bill Rs308.16 \n",
        "net monthly bill Rs292.75\n"
       ]
      }
     ],
     "prompt_number": 75
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.2 Page 59"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "l=100 #connected load\n",
      "md=80 #maximum demand\n",
      "wt=0.6  #working time\n",
      "c=6000  #constant cost\n",
      "t=700  #cost on per kW\n",
      "re=1.8 #rate\n",
      "ec=l*wt*8760#electricity consumption per year\n",
      "teb=c+md*t+re*ec #total electricity bill per year\n",
      "print \" energy consumption %dkWh \\n total electricity bill per year Rs%d\"%(ec,teb)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " energy consumption 525600kWh \n",
        " total electricity bill per year Rs1008080\n"
       ]
      }
     ],
     "prompt_number": 76
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.3 Page 60"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "md=160 ;lff=0.7 ;dfc=1.7#maximum demand #load factor#diversity factor bt consumers\n",
      "ic=200 #installed capacity\n",
      "ccp=30000#capital cost of plant per kW\n",
      "ctds=1800*10**6 #capital cost of transmission and distribution\n",
      "idi=0.11 #interest,depreciation insurance and taxes on capital investiment\n",
      "fmc=30*10**6  #fixed managerial and general maintanance cost\n",
      "ol=236*10**6 #operating labour,maintanance and suppies\n",
      "cm=90*10**6 #cost of metering,billing and collection\n",
      "eca=0.05  #energy consumed by auxillary\n",
      "el=0.15#energy loss and maintanance\n",
      "p=0.25\n",
      "lf=0.8#load factor\n",
      "ap=0.5 #addition energy for profit\n",
      "print 'a'\n",
      "print \" capital cost of plant Rs%e \\n total capital cost Rs%e\\n interest,depereiation system Rs%e \"%(ccp*ic*10**3,ccp*ic*10**3+ctds,(ccp*ic*10**3+ctds)*idi)\n",
      "print \"\\n sum of maximum demand of consumers energy prodused %dMW \\n energy produced %ekWh \\n energy consumed by auxilliries %ekWh\\n energy output %ekWH \\n energy sold to consumer %ekWh\\n\"%(md*dfc,md*8760*lff*10**3,md*8760*lff*eca*10**3,md*8760*lff*10**3*(1-eca),md*8760*lff*10**3*(1-eca)*(1-el))\n",
      "print '(b)fixed cost'\n",
      "idetc=(ccp*ic*10**3+ctds)*idi\n",
      "tot=idetc+fmc \n",
      "print \" interest, deprecition etc Rs%e per year\\n managerial and maintence Rs%.eper year \\n total \\t Rs%e \"%(idetc,fmc,tot)\n",
      "pro=p*tot\n",
      "gtot=tot+pro\n",
      "print \"\\n profit@%d \\tRs%eper year \\n grand total Rs%e per year\"%(p*100,pro,gtot)\n",
      "print 'Operating cost'\n",
      "tot2=ol+cm\n",
      "pro2=tot2*p\n",
      "gtot2=tot2+pro2\n",
      "print \" Operating labour,supplies maintenance etc Rs.%eper year \\n metering,billing etc Rs%eper year\\n total\\t\\tRs%e per year\\n profit \\t Rs%eper year \\n grand total \\t Rs%e per year\"%(ol,cm,tot2,pro2,gtot2)\n",
      "print 'tariff'\n",
      "co=gtot/(md*dfc*1000)\n",
      "es=md*8760*lff*10**3*(1-eca)*(1-el)\n",
      "cs=gtot2/es\n",
      "print \" cost per kW \\tRs%e \\n cost per kWh \\tRs%e\"%(co,cs)\n",
      "print '(b)'\n",
      "ep=md*1000*8760*lf\n",
      "print \" energy produced %ekWh \\n energy consumed by auxiliaries %ekWh/year \\n energy output of plant %ekWh \\n energy sold to consumer %ekWh\"%(ep,ep*eca,ep*(1-eca),ep*(1-eca)*(1-el))\n",
      "estc=ep*(1-eca)*(1-el)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a\n",
        " capital cost of plant Rs6.000000e+09 \n",
        " total capital cost Rs7.800000e+09\n",
        " interest,depereiation system Rs8.580000e+08 \n",
        "\n",
        " sum of maximum demand of consumers energy prodused 272MW \n",
        " energy produced 9.811200e+08kWh \n",
        " energy consumed by auxilliries 4.905600e+07kWh\n",
        " energy output 9.320640e+08kWH \n",
        " energy sold to consumer 7.922544e+08kWh\n",
        "\n",
        "(b)fixed cost\n",
        " interest, deprecition etc Rs8.580000e+08 per year\n",
        " managerial and maintence Rs3e+07per year \n",
        " total \t Rs8.880000e+08 \n",
        "\n",
        " profit@25 \tRs2.220000e+08per year \n",
        " grand total Rs1.110000e+09 per year\n",
        "Operating cost\n",
        " Operating labour,supplies maintenance etc Rs.2.360000e+08per year \n",
        " metering,billing etc Rs9.000000e+07per year\n",
        " total\t\tRs3.260000e+08 per year\n",
        " profit \t Rs8.150000e+07per year \n",
        " grand total \t Rs4.075000e+08 per year\n",
        "tariff\n",
        " cost per kW \tRs4.080882e+03 \n",
        " cost per kWh \tRs5.143550e-01\n",
        "(b)\n",
        " energy produced 1.121280e+09kWh \n",
        " energy consumed by auxiliaries 5.606400e+07kWh/year \n",
        " energy output of plant 1.065216e+09kWh \n",
        " energy sold to consumer 9.054336e+08kWh\n"
       ]
      }
     ],
     "prompt_number": 77
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.4 Page 61"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "v=230 ;ec=2020 #voltage #energy consumption\n",
      "i=40; pf=1 ;t=2; c=3.5; rc=1.8 ;mon=30 #current/power factor/time/cost/reamining cost/month\n",
      "ecd=v*i*pf*t*mon/1000 #energy corresponding to maximum demand\n",
      "cost=ecd*c\n",
      "ren=ec-ecd\n",
      "rcost=ren*rc\n",
      "tmb=cost+rcost\n",
      "at=tmb/ec\n",
      "print \" energy corresponding to maximum demand %dkWh \\n cost of above energy Rs%d \\n remaining energy %dkWh \\n cost of reamaining energy Rs%.1f \\n total monthly bill Rs.%.1f\\n avarage tariff Rs%.3fper kWh\"%(ecd,cost,ren,rcost,tmb,at)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " energy corresponding to maximum demand 552kWh \n",
        " cost of above energy Rs1932 \n",
        " remaining energy 1468kWh \n",
        " cost of reamaining energy Rs2642.4 \n",
        " total monthly bill Rs.4574.4\n",
        " avarage tariff Rs2.265per kWh\n"
       ]
      }
     ],
     "prompt_number": 78
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.5 Page 62"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "t1=3000 ;t11=0.9 #cost equation \n",
      "t2=3  #rate\n",
      "x=t1/(t2-t11)\n",
      "print \"if energy consumption per month is more than %.1fkWh,\\ntariff is more suitable\"%(x)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "if energy consumption per month is more than 1428.6kWh,\n",
        "tariff is more suitable\n"
       ]
      }
     ],
     "prompt_number": 79
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.6 Page 62"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "aec=201500 #annual energy consumption\n",
      "lf=0.35#load factor constnt\n",
      "t=4000#tariff\n",
      "tmd=1200#tariff for maximum demand\n",
      "t3=2.2\n",
      "lfb=0.55 #load factor improved\n",
      "ecd=0.25#energy consumption reduced\n",
      "md=aec/(8760*lf)\n",
      "yb=t+md*tmd+t3*aec\n",
      "mdb=aec/(8760*lfb)\n",
      "ybb=t+mdb*tmd+t3*aec\n",
      "ne=aec*(1-ecd)\n",
      "md3=ne/(8760*lf)\n",
      "ybc=t+md3*tmd+t3*ne\n",
      "aeca=yb/aec\n",
      "aecb=ybb/aec\n",
      "aecc=ybc/ne\n",
      "print 'a'\n",
      "print \"maximum demand %.2fkW \\n yearly bill Rs.%d per year \\n(b)\\n maximum demand %.2fkW \\n yearly bill Rs.%dper year\"%(md,yb,mdb,ybb)\n",
      "print \"c\"\n",
      "print \" new energy %dkWh \\n maximum demand %.2fkW \\n yearly bill Rs.%dper year \\n average energy cost in case a Rs%.4fper kWh \\n average energy cost in case b Rs%.3fper kWh\\n average energy cost in case c Rs%.3fper kWh \"%(ne,md3,ybc,aeca,aecb,aecc)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a\n",
        "maximum demand 65.72kW \n",
        " yearly bill Rs.526164 per year \n",
        "(b)\n",
        " maximum demand 41.82kW \n",
        " yearly bill Rs.497486per year\n",
        "c\n",
        " new energy 151125kWh \n",
        " maximum demand 49.29kW \n",
        " yearly bill Rs.395623per year \n",
        " average energy cost in case a Rs2.6112per kWh \n",
        " average energy cost in case b Rs2.469per kWh\n",
        " average energy cost in case c Rs2.618per kWh \n"
       ]
      }
     ],
     "prompt_number": 80
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.7 Page 66"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "pl1=20 ;pf1=0.8; t1=2000#load in MVA #power factor #duration\n",
      "pl2=10 ;pf2=0.8 ;t2=1000#load in MVA #power factor #duration\n",
      "pl3=2 ;pf3=0.8 ;t3=500#load in MVA #power factor #duration\n",
      "pt=20 #/transformar power rating\n",
      "fte=0.985 ;ste=0.99 #/full load efficiency for first and second transformer\n",
      "ftl=120 ;stl=90 #core  loss inKW for first and  second transformer\n",
      "cst=200000 #cost of second transformer with compared with first transformer\n",
      "aid=0.15 #annual  interest and depreciation\n",
      "ce=0.8 #cost of energy\n",
      "tfl=pt*(1-fte)*1000#total full load\n",
      "fle=tfl-ftl #full load copper loss\n",
      "elc=fle*t1+(fle*t2/(pt/pl2)**2)+(fle*t3/(pt/pl3)**2) #energy loss due to copper loss\n",
      "eli=ftl*(t1+t2+t3)#energy loss due to iron loss\n",
      "celo=(elc+eli)*ce #cost of energy loss\n",
      "print \" \\nfirst transformer : \"\n",
      "print \" total full load losses %dkW \\n full load copper losses %dkW \\n energy loss due to copper losses %dkWh/year\\n energy loss due to iron losses %dkWh/year \\n cost of energy losses Rs%dper year\"%(tfl,fle,elc,eli,celo)\n",
      "stfl=pt*(1-ste)*1000#total full load\n",
      "sle=stfl-stl#full load copper loss\n",
      "selc=sle*t1+(sle*t2/(pt/pl2)**2)+(sle*t3/(pt/pl3)**2)#energy loss due to copper loss\n",
      "seli=stl*(t1+t2+t3)#energy loss due to iron loss\n",
      "scelo=(selc+seli)*ce#cost of energy loss\n",
      "print \" \\nsecond transformer  :\"\n",
      "print \" total full load losses %dkW \\n full load copper losses %dkW \\n energy loss due to copper losses %dkWh/year\\n energy loss due to iron losses %dkWh/year \\n cost of energy losses Rs%dper year\"%(stfl,sle,selc,seli,scelo)\n",
      "aidc=stfl*aid*1000\n",
      "tybc=aidc+scelo\n",
      "print \" additional interest and depreciation due to higher cost of second transformer Rs%d \\n total yearly charges for second transformer Rs%d per year\"%(aidc,tybc)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " \n",
        "first transformer : \n",
        " total full load losses 300kW \n",
        " full load copper losses 180kW \n",
        " energy loss due to copper losses 405900kWh/year\n",
        " energy loss due to iron losses 420000kWh/year \n",
        " cost of energy losses Rs660720per year\n",
        " \n",
        "second transformer  :\n",
        " total full load losses 200kW \n",
        " full load copper losses 110kW \n",
        " energy loss due to copper losses 248050kWh/year\n",
        " energy loss due to iron losses 315000kWh/year \n",
        " cost of energy losses Rs450440per year\n",
        " additional interest and depreciation due to higher cost of second transformer Rs30000 \n",
        " total yearly charges for second transformer Rs480440 per year\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.8 Page 67"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import asin, acos, tan, cos\n",
      "p=500 #load\n",
      "pf=0.8#power factor\n",
      "t=400 #tariff\n",
      "md=100 #maximum demand tariff\n",
      "ccb=600 #cost of capacitor bank\n",
      "id=0.11#interest and deprecistion\n",
      "sd=ccb*id/t#sin(ph2)\n",
      "d2=asin(sd)\n",
      "pf2=cos(d2)\n",
      "kva=p*(tan(acos(pf))-tan(d2))\n",
      "print \" the most economic power factor %.3f lagging \\n kvar requirement %.2fkVAR\"%(pf2,kva)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " the most economic power factor 0.986 lagging \n",
        " kvar requirement 291.35kVAR\n"
       ]
      }
     ],
     "prompt_number": 82
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.9 Page 68"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "l1=300 #load and power factor for three different loads\n",
      "pf1=1 \n",
      "l2=1000 \n",
      "pf2=0.9 \n",
      "l3=1500 \n",
      "pf3=0.8\n",
      "print \" for %dkW unit power factor load \\n power factor angle %.f\\n reactive power %.fkvr\"%(l1,acos(pf1),l1*(tan(acos(pf1))))\n",
      "print \" \\nfor %dkW unit power factor load \\n power factor angle %.2f\\n reactive power %.2fkvr\"%(l2,acos(pf2),l2*(tan(acos(pf2))))\n",
      "print \" \\nfor %dkW unit power factor load \\n power factor angle %.2f\\n reactive power %.2fkvr\"%(l3,acos(pf3),l3*(tan(acos(pf3))))\n",
      "tl=l1+l2+l3\n",
      "tt=l3*(tan(acos(pf3)))+l2*(tan(acos(pf2)))+l1*(tan(acos(pf1)))\n",
      "print \"\\n total kW \\t%dkW\\n total kVAR %.1fkVAR \\n total kVA %.2fkVA \\n overall power factor %.3flagging\"%(tl,tt,(tl**2+tt**2)**0.5,tl/(tl**2+tt**2)**0.5)\n",
      "print \"\\n the maximum unity power factor load which yhe station can supply is equal to the kVA i.e.%.2fkVR\"%((tl**2+tt**2)**0.5)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " for 300kW unit power factor load \n",
        " power factor angle 0\n",
        " reactive power 0kvr\n",
        " \n",
        "for 1000kW unit power factor load \n",
        " power factor angle 0.45\n",
        " reactive power 484.32kvr\n",
        " \n",
        "for 1500kW unit power factor load \n",
        " power factor angle 0.64\n",
        " reactive power 1125.00kvr\n",
        "\n",
        " total kW \t2800kW\n",
        " total kVAR 1609.3kVAR \n",
        " total kVA 3229.54kVA \n",
        " overall power factor 0.867lagging\n",
        "\n",
        " the maximum unity power factor load which yhe station can supply is equal to the kVA i.e.3229.54kVR\n"
       ]
      }
     ],
     "prompt_number": 83
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.10 Page 68"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt, pi\n",
      "v=400#voltage\n",
      "i=25#/current\n",
      "pf=0.8#at power factor\n",
      "pf2=0.9#over all power factor\n",
      "kw=v*i*pf*sqrt(3)/1000\n",
      "print \"kw rating of induction motor %.2fkW\"%(kw)\n",
      "dm=acos(pf)\n",
      "rp=kw*tan(dm)\n",
      "print \"\\n power factor angle %.2f \\n reactive power %.2fkVR\"%(dm,rp)\n",
      "fdm=acos(pf2)\n",
      "rp2=kw*tan(fdm)\n",
      "print \"\\n final power factor %.2f \\n final reactance power %.2fkVR\"%(fdm,rp2)\n",
      "ckvb=rp-rp2\n",
      "cc=ckvb*1000/(sqrt(3)*v)\n",
      "vc=v/sqrt(3)\n",
      "xc=vc/cc\n",
      "f=50\n",
      "cec=1*10**(6)/(xc*2*pi*f)\n",
      "print \"\\n kvar rating of capacitor bank %.4f \\n current through each capacitor %.2fA\\n voltage across each capacitor %.2f \\n reactance of each capacitor %.2fohm \\n capacitance of each capacitance %.2fuf\"%(ckvb,cc,vc,xc,cec)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "kw rating of induction motor 13.86kW\n",
        "\n",
        " power factor angle 0.64 \n",
        " reactive power 10.39kVR\n",
        "\n",
        " final power factor 0.45 \n",
        " final reactance power 6.71kVR\n",
        "\n",
        " kvar rating of capacitor bank 3.6813 \n",
        " current through each capacitor 5.31A\n",
        " voltage across each capacitor 230.94 \n",
        " reactance of each capacitor 43.46ohm \n",
        " capacitance of each capacitance 73.24uf\n"
       ]
      }
     ],
     "prompt_number": 84
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.11 Page 69"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "v=400#line voltage\n",
      "i=50 #line current\n",
      "pf=0.8 #at power factor\n",
      "pf2=0.95 # overall power factor\n",
      "sm=25 #hp of synchronous motor \n",
      "e=0.9#efficiency\n",
      "kwri=v*i*pf*sqrt(3)/1000\n",
      "kvari=v*i*sqrt(3)/1000\n",
      "karri=(-kwri**2+kvari**2)**0.5\n",
      "kwsm=sm*735.5/(e*1000)\n",
      "tkw=kwri+kwsm\n",
      "print \" kw rating of installation %.1fkW \\n kVA rating of installation %.2fkva \\n kVAR rating %.2fkvar \\n kw input to synchrounous motor %.2fkw \\n total kw=%.2f\\n\"%(kwri,kvari,karri,kwsm,tkw)\n",
      "pd=acos(pf2)\n",
      "tkr=tkw*tan(pd)\n",
      "krsm=tkr-karri\n",
      "kasm=(kwsm**2+krsm**2)**0.5\n",
      "pfsm=kwsm/kasm\n",
      "if krsm<0:\n",
      "    ch='capacitor'\n",
      "    ich='leading'\n",
      "else:\n",
      "    ch='inductive'\n",
      "    ich='lagging'\n",
      "\n",
      "print \" overall power factor angle %.2fkw \\n total kvar %.2fkvar \\n kvar of synchrounous motor %.2fkvar %s \\n kva of synchrounous motor %.2fkva \\n power factor of synchrounous motor %.2f %s\"%(pd,tkr,krsm,ch,kasm,pfsm,ich)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " kw rating of installation 27.7kW \n",
        " kVA rating of installation 34.64kva \n",
        " kVAR rating 20.78kvar \n",
        " kw input to synchrounous motor 20.43kw \n",
        " total kw=48.14\n",
        "\n",
        " overall power factor angle 0.32kw \n",
        " total kvar 15.82kvar \n",
        " kvar of synchrounous motor -4.96kvar capacitor \n",
        " kva of synchrounous motor 21.02kva \n",
        " power factor of synchrounous motor 0.97 leading\n"
       ]
      }
     ],
     "prompt_number": 85
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.12 page 69"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sin, atan\n",
      "psm=100 #power of synchrounous motors\n",
      "pim=200 #power of inducion motor\n",
      "v=400 #voltage \n",
      "pff=0.71; pp=-1#power factor\n",
      "rsm=0.1 #resistance of synchrounous motor\n",
      "rt=0.03 #resistance of cable\n",
      "#pf(1)=1 ;p(1)=1 #power factor in a\n",
      "#pf(2)=0.8 ;p(2)=1 #power factor in b\n",
      "#pf(3)=0.6 ;p(3)=1 #power factor in c\n",
      "pf =[1,0.8,0.6]\n",
      "p = [1,1,1]\n",
      "\n",
      "i1=pim*1000/(v*pff*sqrt(3))\n",
      "i11=i1*(complex(pff,pp*sin(acos(pff))))\n",
      "i2f=psm*1000/(v*sqrt(3))\n",
      "ch=['a', 'b', 'c']\n",
      "it = range(0,3)\n",
      "opf = range(0,3)\n",
      "for i in range(0,3):\n",
      "    print \"\\n (%s)\"%(ch[i])\n",
      "    d=acos(pf[i])\n",
      "    #it(i)=i11(1)+complex(i2f,(p(i)*i2f*tand(d)))\n",
      "    it[i]=i11+complex(i2f,(p[i]*i2f*tan(d)))\n",
      "    opf[i]=cos(atan((it[i]).imag/(it[i]).real))\n",
      "    clsm=3*((i2f)**2)*rsm\n",
      "    clt=3*(abs(it[i])**2)*rt/1000\n",
      "    print \"\\n total current =\",it[i]\n",
      "    print \"\\n overall power factor %.3f lagging \\n copper losses in synchrounous motor %.fW \\n copper losses in cable %.2fKW\"%(opf[i],clsm,clt)\n",
      "\n",
      "print \"(d)\"\n",
      "print \"copper loss of synchronous motor this is evidently minimum when tand=%d cosd=%d\"%(0,1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " (a)\n",
        "\n",
        " total current = (433.012701892-286.317624651j)\n",
        "\n",
        " overall power factor 0.834 lagging \n",
        " copper losses in synchrounous motor 6250W \n",
        " copper losses in cable 24.25KW\n",
        "\n",
        " (b)\n",
        "\n",
        " total current = (433.012701892-178.064449178j)\n",
        "\n",
        " overall power factor 0.925 lagging \n",
        " copper losses in synchrounous motor 6250W \n",
        " copper losses in cable 19.73KW\n",
        "\n",
        " (c)\n",
        "\n",
        " total current = (433.012701892-93.8675349216j)\n",
        "\n",
        " overall power factor 0.977 lagging \n",
        " copper losses in synchrounous motor 6250W \n",
        " copper losses in cable 17.67KW\n",
        "(d)\n",
        "copper loss of synchronous motor this is evidently minimum when tand=0 cosd=1\n"
       ]
      }
     ],
     "prompt_number": 86
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.13 Page 71"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import atan\n",
      "p=2#constant output in MW\n",
      "pf=0.9#power factor\n",
      "pa=10#load\n",
      "pb=5\n",
      "pfb=0.8#power factor at load of 5MW\n",
      "td=tan(acos(pf))\n",
      "go=p*(1-td*1J)\n",
      "op=0.8\n",
      "tp=tan(acos(pfb))\n",
      "print \"power factor of indection generator is leading therefor induction generator output %d%.2fiMVA /n (a) \\n\"%(go.real,go.imag)\n",
      "tl=pa*(1+tp*1J)\n",
      "sg=tl-go\n",
      "da=atan(sg.imag/sg.real)\n",
      "print \" total load %d+%.1fiMW \\n synchronous generator load %d+%.3fiMW \\n\\t\\t=%.2fMW at angle %.2f \\n power factor of synchronous generator is %.2flagging\"%(tl.real,tl.imag,sg.real,sg.imag,abs(sg),da,cos(da))\n",
      "tl1=pb*(1+tp*1J)\n",
      "sg1=tl1-go\n",
      "da1=atan(sg1.imag/sg1.real)\n",
      "print \"(b)\"\n",
      "print \" total load %d+%.1fiMW \\n synchronous generator load %d+%.3fiMW \\n\\t\\t=%.2fMW at angle %.2f \\n power factor of synchronous generator is %.2flagging\"%(tl1.real,(tl1).imag,sg1.real,(sg1).imag,abs(sg1),da1,cos(da1))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "power factor of indection generator is leading therefor induction generator output 2-0.97iMVA /n (a) \n",
        "\n",
        " total load 10+7.5iMW \n",
        " synchronous generator load 8+8.469iMW \n",
        "\t\t=11.65MW at angle 0.81 \n",
        " power factor of synchronous generator is 0.69lagging\n",
        "(b)\n",
        " total load 5+3.7iMW \n",
        " synchronous generator load 3+4.719iMW \n",
        "\t\t=5.59MW at angle 1.00 \n",
        " power factor of synchronous generator is 0.54lagging\n"
       ]
      }
     ],
     "prompt_number": 87
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.14 page 71"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "c=40*10**(-6) #bank of capacitors in farads\n",
      "v=400 #line voltage\n",
      "i=40#/line current\n",
      "pf=0.8#power factor\n",
      "f=50#line frequency\n",
      "xc=1/(2*pi*f*c)\n",
      "ic=v/(sqrt(3)*xc)\n",
      "il=i*(pf-sin(acos(pf))*1J)\n",
      "til=il+1J*ic\n",
      "od=atan(til.imag/til.real)\n",
      "opf=cos(od)\n",
      "nlol=(abs(od)/i)**2\n",
      "print \"(a)\"\n",
      "print \" line current of capacitor bank %.1fA \\n load current %d%diA \\n total line current %d%.1fjA \\n overall p.f %.3f \\n new line loss to old line loss  %.3f\"%(ic,il.real,il.imag,(til).real,(til).imag,opf,nlol)\n",
      "pcb=(v/xc)\n",
      "print \"\\n phase current of capacitor bank %.3fA\"%(pcb)\n",
      "lcb=pcb*sqrt(3)\n",
      "print \"\\n line current of capacitor bank %.1fA\"%(lcb)\n",
      "tcu=il+lcb*1J\n",
      "print \"\\n total current =\",tcu\n",
      "print \"%.1fjA =%.2fA at an angle %.2f\"%(tcu.imag,abs(tcu),atan(tcu.imag/tcu.real))\n",
      "pf2=cos(atan(tcu.imag/(tcu).real))\n",
      "print \"\\n power factor %.1f \\n ratio of new line loss to original loss %.3f\"%(pf2,(abs(tcu)/i)**2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)\n",
        " line current of capacitor bank 2.9A \n",
        " load current 32-23iA \n",
        " total line current 32-21.1jA \n",
        " overall p.f 0.835 \n",
        " new line loss to old line loss  0.000\n",
        "\n",
        " phase current of capacitor bank 5.027A\n",
        "\n",
        " line current of capacitor bank 8.7A\n",
        "\n",
        " total current = (32-15.2937630517j)\n",
        "-15.3jA =35.47A at an angle -0.45\n",
        "\n",
        " power factor 0.9 \n",
        " ratio of new line loss to original loss 0.786\n"
       ]
      }
     ],
     "prompt_number": 88
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.15 Page 72"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "p=30 #b.h.p of induction motor\n",
      "f=50#line frequency\n",
      "v=400#line voltage\n",
      "e=0.85#effiency\n",
      "pf=0.8 #power factor\n",
      "i=p*746/(v*e*pf*sqrt(3))\n",
      "i=i*complex(pf,-sin(acos(pf)))\n",
      "ccb=(i).imag/sqrt(3)\n",
      "xc=v/ccb\n",
      "c=10**6/(2*f*pi*xc)\n",
      "prl=((abs(i)**2-i.real**2)/abs(i)**2)*100\n",
      "print \" current drawn by motor is %.1fA \\n \"%abs(i)\n",
      "print \"the line loss will be minimum when i is munimum.the minimum value of i is\",i,\"A\"\n",
      "print \" and occurs when the capacitor bank draws a line current of %djA \\n capacitor C %.2fuf \\n percentage loss reduction %d\"%(i.imag,abs(c),prl)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " current drawn by motor is 47.5A \n",
        " \n",
        "the line loss will be minimum when i is munimum.the minimum value of i is (38.0032324249-28.5024243187j) A\n",
        " and occurs when the capacitor bank draws a line current of -28jA \n",
        " capacitor C 130.95uf \n",
        " percentage loss reduction 35\n"
       ]
      }
     ],
     "prompt_number": 89
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.16 page 72"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "po=666.66 #power\n",
      "f=50 #frequency\n",
      "v=400 #voltage\n",
      "pf=0.8  ;p=-1#power factor\n",
      "pf2=0.95 ;p2=-1#improved power factor\n",
      "vc=2200 #capacitor voltage \n",
      "rc=vc\n",
      "il=po*1000/(v*pf*sqrt(3))\n",
      "il1=il*(complex(pf,p*sin(acos(pf))))\n",
      "i2c=il*pf\n",
      "tad=tan(acos(pf2))\n",
      "i2=complex(i2c,i2c*tad*p2)\n",
      "print \" load current i1 =\",il1,\"A \"\n",
      "print \"\\n load current current on improved power factor =\",i2,\"A\"\n",
      "print \"(a)\"\n",
      "ic=abs(il1-i2)\n",
      "ilc=ic*v/vc\n",
      "pic=ilc/sqrt(3)\n",
      "xc=vc/pic\n",
      "ca=10**6/(2*pi*f*xc)\n",
      "print \" line current of %dV capacitor bank %.2fA\\n line current of %d capacitor bank %.2fA \\n phase current of capacitor bank %.2fA \\n reactance %.2f \\n capacitance %.2fF*10**(-6)\"%(v,ic,vc,ilc,pic,xc,ca)\n",
      "print \"(b)\"\n",
      "kr=3*vc*pic/1000\n",
      "print \" kVA rating %.1fkVA \\n kVA rating of transformer to convert %dV to %dV will be the same as the kVA rating of capacitor bank\"%(kr,v,vc)\n",
      "pl=100*(abs(il1)**2-abs(i2)**2)/abs(il1)**2\n",
      "print \"percentage reduction in losses %d percent\"%(pl)\n",
      "print \"(d)\"\n",
      "pi=ic/sqrt(3)\n",
      "xcc=v/pi\n",
      "cc=1*10**6/(2*pi*f*xcc)\n",
      "roc=ca/cc\n",
      "print \" phase current %.1fA \\n reactance %.2fohm \\n capasitance %.2f*10**-6F \\n ratio of capacitance %.3f\"%(pi,xcc,cc,roc)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " load current i1 = (962.240826145-721.680619609j) A \n",
        "\n",
        " load current current on improved power factor = (962.240826145-316.273264908j) A\n",
        "(a)\n",
        " line current of 400V capacitor bank 405.41A\n",
        " line current of 2200 capacitor bank 73.71A \n",
        " phase current of capacitor bank 42.56A \n",
        " reactance 51.70 \n",
        " capacitance 61.57F*10**(-6)\n",
        "(b)\n",
        " kVA rating 280.9kVA \n",
        " kVA rating of transformer to convert 400V to 2200V will be the same as the kVA rating of capacitor bank\n",
        "percentage reduction in losses 29 percent\n",
        "(d)\n",
        " phase current 234.1A \n",
        " reactance 1.71ohm \n",
        " capasitance 25.00*10**-6F \n",
        " ratio of capacitance 2.463\n"
       ]
      }
     ],
     "prompt_number": 90
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.17 Page 74"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "v1=132#line voltage at primary\n",
      "v2=11#line voltage at secondary\n",
      "p=10 #power\n",
      "pf=0.8 #power factor\n",
      "mva=p*(complex(pf,sin(acos(pf))))\n",
      "print \" MVA rating of secondary = %dMVA =%d+%djMVA \\n \"%(p,abs(mva),mva.imag)\n",
      "print \"\\n since the power factor at primary terminals is unity,rating of primary need be %dMVA only \\n the tertiary will supply capacitor curren.since p.f is to be raised to 1 ,the mav compensation needed is 6MVA so rating of teritiary is %dMVA\"%(abs(mva),mva.imag)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " MVA rating of secondary = 10MVA =9+5jMVA \n",
        " \n",
        "\n",
        " since the power factor at primary terminals is unity,rating of primary need be 9MVA only \n",
        " the tertiary will supply capacitor curren.since p.f is to be raised to 1 ,the mav compensation needed is 6MVA so rating of teritiary is 5MVA\n"
       ]
      }
     ],
     "prompt_number": 91
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.18 Page 74"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "v=11 #line voltage\n",
      "f=50#line frequency\n",
      "l=400 #load of alternator\n",
      "pf=0.8 #power factor\n",
      "e=0.85#efficiency\n",
      "p=l/pf\n",
      "lo=l+p*sin(acos(pf))*1J\n",
      "print \"a\"\n",
      "print \"when pf is rased to 1 the alternator can supply %dkW for the same value of armture current hence it can supply %dKW to synchronous motor\"%(p,p-l)\n",
      "print \"b\"\n",
      "print \"b.h.p =%.2fHP\"%(100*e/0.746)\n",
      "kvam=p-lo\n",
      "td=atan((kvam).imag/(kvam).real)\n",
      "pff=cos(td)\n",
      "print \"\\ncosd=%.3fleading\"%(pff)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a\n",
        "when pf is rased to 1 the alternator can supply 500kW for the same value of armture current hence it can supply 100KW to synchronous motor\n",
        "b\n",
        "b.h.p =113.94HP\n",
        "\n",
        "cosd=0.316leading\n"
       ]
      }
     ],
     "prompt_number": 92
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.19 Page 74"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "kw=100  #let kw=100kw\n",
      "pf=0.6 #power foctor\n",
      "pf2=0.8 #power factor\n",
      "kvar=kw*tan(acos(pf))\n",
      "kvar2=kw*tan(acos(pf2))\n",
      "ckar=((kvar-kvar2))/10\n",
      "ck=round(ckar)*10\n",
      "print \" capacitor kVAR required for %dkW\\n load for same power factor improvement %dKVAR\"%(round(ckar),ck)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " capacitor kVAR required for 6kW\n",
        " load for same power factor improvement 60KVAR\n"
       ]
      }
     ],
     "prompt_number": 93
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.20 Page 75"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "p=160 #kva for transformer \n",
      "pf=0.6 #power factor\n",
      "el=96 #effective load\n",
      "eli=120 #effective load increase\n",
      "rc=eli*(tan(acos(pf))-tan(acos(eli/p)))\n",
      "opf=eli/p\n",
      "print \" required capacitor kVAR %dKVAR \\n overall power factor %.2f \\n it is seen that point d is on %.2f line\"%(rc,opf,opf)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " required capacitor kVAR 54KVAR \n",
        " overall power factor 0.75 \n",
        " it is seen that point d is on 0.75 line\n"
       ]
      }
     ],
     "prompt_number": 94
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.21 page 76"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "md=800 #maximum demand\n",
      "pf=0.707 #power factor\n",
      "c=80 #cost\n",
      "p=200 #power\n",
      "e=0.99#efficiency\n",
      "pff=0.8 #fulload pf\n",
      "ikva=md/pf\n",
      "iafc=(round(ikva*100)*(c)/100)\n",
      "rsm=ikva*pf\n",
      "act=p*(0.7355)/e\n",
      "at=-act*sin(acos(pff))\n",
      "tkw=rsm+act\n",
      "tkvr=rsm+at\n",
      "tkva=(tkw**2+tkvr**2)**0.5\n",
      "ikvad=tkva-ikva\n",
      "infc=ikvad*c\n",
      "print \" initial kVA %.2fkVA \\n initial annual fixed charges Rs%.1f \\n after installation of synchronous motor reactive power of induction motor %dkVars\\n active power input of synchrounous motor %.2fkW\\n reactive power input to synchrounous motor %.2fKVAR \\n total kW %.2fKW \\n total kVars %.2fkVARS \\n total kVA %.2fkVA \\n increase in KVA demand %.2fkVA\\n increase in annual fixed charges Rs%.1f \"%(ikva,iafc,rsm,act,at,tkw,tkvr,tkva,ikvad,infc)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " initial kVA 1131.54kVA \n",
        " initial annual fixed charges Rs90523.2 \n",
        " after installation of synchronous motor reactive power of induction motor 800kVars\n",
        " active power input of synchrounous motor 148.59kW\n",
        " reactive power input to synchrounous motor -89.15KVAR \n",
        " total kW 948.59KW \n",
        " total kVars 710.85kVARS \n",
        " total kVA 1185.38kVA \n",
        " increase in KVA demand 53.84kVA\n",
        " increase in annual fixed charges Rs4306.9 \n"
       ]
      }
     ],
     "prompt_number": 95
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 4.22 Page 77"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "t=16#working  time\n",
      "d=300 #working days\n",
      "hv=1 ;hvmd=50 #tariff on high voltage\n",
      "lv=1.1 ;lvmd=60 #tariff on low voltage\n",
      "al=250#avarage load\n",
      "pf=0.8#power factor\n",
      "md=300 #maximum demand\n",
      "hvec=500#cost of hv equipment\n",
      "l=0.05 #loss of hv equipment\n",
      "id=0.12  #interest and deprecistion\n",
      "ter=al*md*t \n",
      "mdv=md/pf\n",
      "print \" total energy requirement %2.2ekWH \\n maximum demand %dKVA\"%(ter,mdv)\n",
      "print \"(a)HV supply\"\n",
      "chv=mdv*hvec\n",
      "idc=chv*id\n",
      "ere=ter/(1-l)\n",
      "dch=mdv*hvmd\n",
      "ech=round(ere*hv/1000)*1000\n",
      "tanc=ech+dch+idc\n",
      "print \" cost of HV equipment Rs%e\\n interest and depreciation charges Rs%d \\n energy received %ekWh\\n demand charges Rs%d \\n energy charges Rs%2e \\n total annual cost Rs%d\"%(chv,idc,ere,dch,ech,tanc)\n",
      "print \"(b) LV supply\"\n",
      "lvdc=mdv*lvmd\n",
      "lvec=ter*lv\n",
      "lvtac=lvec+lvdc\n",
      "lvdac=lvtac-tanc\n",
      "print \" demand charges Rs%d \\n energy charges Rs%2.e \\n total annual cost Rs%d \\n difference in annual cost Rs%d\"%(lvdc,lvec,lvtac,lvdac)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " total energy requirement 1.20e+06kWH \n",
        " maximum demand 375KVA\n",
        "(a)HV supply\n",
        " cost of HV equipment Rs1.875000e+05\n",
        " interest and depreciation charges Rs22500 \n",
        " energy received 1.263158e+06kWh\n",
        " demand charges Rs18750 \n",
        " energy charges Rs1.263000e+06 \n",
        " total annual cost Rs1304250\n",
        "(b) LV supply\n",
        " demand charges Rs22500 \n",
        " energy charges Rs1e+06 \n",
        " total annual cost Rs1342500 \n",
        " difference in annual cost Rs38250\n"
       ]
      }
     ],
     "prompt_number": 96
    }
   ],
   "metadata": {}
  }
 ]
}