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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Ch-10, Economic Operations of Steam Plants"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.2 Page 193"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "mp=250     #from example 10.1\n",
      "def unit1(p1):\n",
      "    ic=0.2*p1+30\n",
      "    return ic\n",
      "def unit2(p2):\n",
      "    ic=0.15*p2+40\n",
      "    return ic\n",
      "mil=20\n",
      "ttt=225\n",
      "def un(ic):\n",
      "    p1=(ic-30)/0.2\n",
      "    p2=(ic-40)/0.15\n",
      "    return [p1, p2]\n",
      "from numpy import arange\n",
      "for x in arange(40,61,5):\n",
      "    [e,r]=un(x)\n",
      "    if ttt==e+r:\n",
      "        print \"for the same incremental costs unit1 should supply %dMW and unit 2 shold supply %dMW,for equal sharing each unit should supply %3.1fMW\"%(e,r,ttt/2)\n",
      "        break\n",
      "    \n",
      "\n",
      "opo=ttt/2\n",
      "from sympy.mpmath import quad\n",
      "u1=quad(unit1,[opo,e])\n",
      "u2=quad(unit2,[r,opo])\n",
      "uuu=(u1+u2)*8760\n",
      "print \"\\nyearly extra cost is (%3.2f-%3.2f)8760 =%dper year\"%(u1,u2,uuu)\n",
      "print \"\\nthis if the load is equally shared by the two units an extra cost of Rs.%d will be incurred.in other words economic loading would result in saving of Rs.%dper year\"%(uuu,uuu)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "for the same incremental costs unit1 should supply 125MW and unit 2 shold supply 100MW,for equal sharing each unit should supply 112.0MW\n",
        "\n",
        "yearly extra cost is (698.10-670.80)8760 =11991564per year\n",
        "\n",
        "this if the load is equally shared by the two units an extra cost of Rs.11991564 will be incurred.in other words economic loading would result in saving of Rs.11991564per year\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.3 Page 198"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "def unit1(p1):\n",
      "    ic=0.2*p1+30\n",
      "    return ic\n",
      "def unit2(p2):\n",
      "    ic=0.15*p2+40\n",
      "    return ic\n",
      "tol=400\n",
      "pd=50\n",
      "u1c=5\n",
      "u2c=1/0.15#from example10_1\n",
      "p1pd=u1c/(u1c+u2c)\n",
      "p2pd=u2c/(u1c+u2c)\n",
      "pi=p1pd*pd\n",
      "pt=p2pd*pd\n",
      "print \"p1=%1.5fMW\\np2=%1.5fMW\"%(pi,pt)\n",
      "p11=pi+tol/2\n",
      "p22=pt+tol/2\n",
      "up1=unit1(p11)\n",
      "up2=unit2(p22)\n",
      "print \"\\nthe total load on 2 units would be %3.2fMW and %3.2fMW respectevily. it is easy to check that incremental cost will be same for two units at these loading.\\n incremental cost of unit1 is %3.2fRs.MW,\\n incremantal cost of unit 2 is %3.2fRs./MW\"%(p11,p22,up1,up2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "p1=21.42857MW\n",
        "p2=28.57143MW\n",
        "\n",
        "the total load on 2 units would be 221.43MW and 228.57MW respectevily. it is easy to check that incremental cost will be same for two units at these loading.\n",
        " incremental cost of unit1 is 74.29Rs.MW,\n",
        " incremantal cost of unit 2 is 74.29Rs./MW\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example10.5 Page 200"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "i1=0.8\n",
      "i2=1.0\n",
      "l1=complex(0.04,0.12)\n",
      "l2=complex(0.03,0.1)\n",
      "l3=complex(0.03,0.12)\n",
      "vl=1\n",
      "\n",
      "i3=i1+i2\n",
      "v1=vl+i3*(l1)+i1*(l2)\n",
      "v2=vl+i3*(l1)+i2*(l3)\n",
      "p1=(i1*v1).real\n",
      "p2=(i2*v2).real\n",
      "cos1=(v1).real/abs(v1)\n",
      "cos2=(v2).real/abs(v2)\n",
      "b11=abs(((l1).real+(l2).real)/(v1**2*cos1**2))\n",
      "b22=abs(((l1).real+(l3).real)/(v2**2*cos2**2))\n",
      "b12=abs(((l1).real/(v1*v2*cos1*cos2)))\n",
      "pl=(p1**2)*b11+(p2**2)*b22+2*p1*p2*b12\n",
      "print \"i1+i3=%dpu\\nv1=%1.3f+%1.3fp.u\\nv2=%1.3f+%1.3fp.u\\np1=%1.3fp.u\\np2=%1.3fp.u\\ncos(ph1)=%1.3f\\ncos(ph2)=%1.3f\\nb11=%1.5fp.u\\nb22=%1.5fp.u\\nb12=%1.5fp.u\\npl=%1.6fp.u\"%(i3,v1.real,(v1).imag,v2.real,(v2).imag,p1,p2,cos1,cos2,b11,b22,b12,pl)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i1+i3=1pu\n",
        "v1=1.096+0.296p.u\n",
        "v2=1.102+0.336p.u\n",
        "p1=0.877p.u\n",
        "p2=1.102p.u\n",
        "cos(ph1)=0.965\n",
        "cos(ph2)=0.957\n",
        "b11=0.05827p.u\n",
        "b22=0.05764p.u\n",
        "b12=0.03312p.u\n",
        "pl=0.178800p.u\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example10.7 Page 206"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import atan, cos\n",
      "za=complex(0.03,0.09)\n",
      "zb=complex(0.1,0.3)\n",
      "zc=complex(0.03,0.09)\n",
      "zd=complex(0.04,0.12)\n",
      "ze=complex(0.04,0.12)\n",
      "ia=complex(1.5,-0.4)\n",
      "ib=complex(0.5,-0.2)\n",
      "ic=complex(1,-0.1)\n",
      "id=complex(1,-0.2)\n",
      "ie=complex(1.5,-0.3)\n",
      "il1=.4\n",
      "il2=.6\n",
      "na1=1 ;nb1=0.6; nc1=0; nd1=.4; ne1=.6\n",
      "na2=0 ;nb2=-0.4; nc2=1 ;nd2=.4; ne2=.6\n",
      "vl=1\n",
      "#some thing is messed\n",
      "v1=vl+za*ia\n",
      "v2=vl-zb*ib+zc*ic\n",
      "a1=atan((ia).imag/(ia).real)\n",
      "a2=atan((ic).imag/(ic).real)\n",
      "cosa=cos(a1-a2)\n",
      "cosph1=cos(atan((v1).imag/(v1).real)-a1)\n",
      "cosph2=cos(atan((v2).imag/(v2).real)-a2)\n",
      "b11=(na1**2*(za).real+nb1**2*(zb).real+nc1**2*(zc).real+nd1**2*(zd).real+ne1**2*(ze).real)/(abs(v1)**2*cosph1)\n",
      "b22=(na2**2*(za).real+nb2**2*(zb).real+nc2**2*(zc).real+nd2**2*(zd).real+ne2**2*(ze).real)/((abs(v2)**2)*cosph2)\n",
      "bb12=(abs(v1)*abs(v2)*cosph1*cosph2)\n",
      "ab12=(na2*na1*(za).real+nb2*nb1*(zb).real+nc1*nc2*(zc).real+nd2*nd1*(zd).real+ne2*ne1*0.03)\n",
      "b12=cosa*ab12/bb12\n",
      "print \"bus voltages at 2 buses are \\nv1=%1.3f+i%1.3f,\\nv2=%1.3f+i%1.3f\"%((v1).real,(v1).imag,(v2).real,(v2).imag)\n",
      "print \"\\nloss coffecients are \\nb11=%1.5fp.u\\nb22=%1.5fp.u\\nb12=%1.5fp.u \\n\"%(b11,b22,b12)\n",
      "print \"loss coffecients in actual values is \\nb11=%eM(W)-1\\nb22=%eM(W)-1\\nb12=%eM(W)-1\\n\"%(b11/100,b22/100,b12/100)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "bus voltages at 2 buses are \n",
        "v1=1.081+i0.123,\n",
        "v2=0.929+i-0.043\n",
        "\n",
        "loss coffecients are \n",
        "b11=0.07877p.u\n",
        "b22=0.07735p.u\n",
        "b12=-0.00714p.u \n",
        "\n",
        "loss coffecients in actual values is \n",
        "b11=7.877236e-04M(W)-1\n",
        "b22=7.734557e-04M(W)-1\n",
        "b12=-7.136298e-05M(W)-1\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.8 Page 207"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "r1=22 ;r2=30 ;q1=0.2 ;q2=0.15\n",
      "b22=0; b12=0 ;p1=100 ;pl=15#transmission losses are 0\n",
      "b11=pl/(p1)**2\n",
      "def power(x): #mathematical computation\n",
      "    p1=(x-r1)/(q1+2*b11*x)\n",
      "    p2=(x-r2)/q2\n",
      "    return [p1, p2]\n",
      "[a,b]=power(60)\n",
      "print \"l1=1/(1-%.3f*p1)\\nl2=[1/(1-0)]=1\\ngiven lamda=60\\nsince ic1*l1=lamda  ic2*l2=lamda\\ntotal load=%dMW\"%(b11*2,a+b-(b11*a**2))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "l1=1/(1-0.000*p1)\n",
        "l2=[1/(1-0)]=1\n",
        "given lamda=60\n",
        "since ic1*l1=lamda  ic2*l2=lamda\n",
        "total load=390MW\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.9 Page 208"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "r1=22 ;r2=30 ;q1=0.2 ;q2=0.15\n",
      "b22=0; b12=0 ;p1=100 ;pl=15#transmission losses are 0\n",
      "b11=pl/(p1)**2\n",
      "def power(x): #mathematical computation\n",
      "    p1=(x-r1)/(q1+2*b11*x)\n",
      "    p2=(x-r2)/q2\n",
      "    return [p1,p2]\n",
      "[a,b]=power(60)\n",
      "pt=a+b-(b11*a**2)\n",
      "from sympy.mpmath import quad\n",
      "#z=quad('q1*u+r1','u',a,161.80)\n",
      "z=quad(lambda u:q1*u+r1,[a,161.80])\n",
      "#y=quad('q2*v+r2','v',b,162.5)\n",
      "y=quad(lambda v:q2*v+r2,[b,162.5])\n",
      "m=z+y\n",
      "print \"net change in cost =Rs.%dper hour\"%(m)\n",
      "print \"\\nthus scheduling the generation by taking transmission losses into account would mean a saving of Rs.%dper hour in fuel cost\"%(m)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "net change in cost =Rs.-3757per hour\n",
        "\n",
        "thus scheduling the generation by taking transmission losses into account would mean a saving of Rs.-3757per hour in fuel cost\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.10 Page 208"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "b11=0.001\n",
      "b12=-0.0005\n",
      "b22=0.0024\n",
      "q1=0.08\n",
      "r1=16\n",
      "q2=0.08\n",
      "r2=12\n",
      "lamda=20\n",
      "\n",
      "p2=0\n",
      "for x in range(1,5):\n",
      "    p1=(1-(r1/lamda)-(2*p2*b12))/((q1/lamda)+2*b11)\n",
      "    p2=(1-(r2/lamda)-(2*p1*b12))/((q2/lamda)+2*b22)\n",
      "\n",
      "pl=b11*p1**2+2*b12*p1*p2+b22*p2**2\n",
      "pr=p1+p2-pl\n",
      "print \"p1=%2.1fMW,p2=%2.1fMW\\npl=%1.1fMW\\npower recevied %2.1fMW\"%(p1,p2,pl,pr)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "p1=189.2MW,p2=135.1MW\n",
        "pl=54.1MW\n",
        "power recevied 270.3MW\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.11 Page 209"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "a1=561 ;b1=7.92 ;c1=0.001562\n",
      "a2=310 ;b2=7.85 ;c2=0.00194\n",
      "ce=c1*c2/(c1+c2)\n",
      "print \"ce=%e\"%(ce)\n",
      "be=((b1/c1)+(b2/c2))*ce\n",
      "print \"be=%1.4f\"%(be)\n",
      "ae=a1-((b1**2)/4*c1)+a2-((b2**2)/4*c2)+((be**2)/4*ce)\n",
      "print \"ae=%3.3f \\ncost characteristics of composite unit for demand pt is \\nct=%3.3f+%1.4f*p1+%ep1**2\"%(ae,ae,be,ce)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "ce=8.652998e-04\n",
        "be=7.8888\n",
        "ae=870.959 \n",
        "cost characteristics of composite unit for demand pt is \n",
        "ct=870.959+7.8888*p1+8.652998e-04p1**2\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.12 Page 210"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "a1=7700 ;b1=52.8 ;c1=5.5*10**-3\n",
      "a2=2500; b2=15 ;c2=0.05#given eqution\n",
      "plo=200 ;pup=800\n",
      "ct=1000\n",
      "l=[500,900,1200,500] ;t=[6, 16 ,20 ,24]#from given graph\n",
      "def cost(y):\n",
      "    p1=(2*c2*y-(b1-b2))/(2*(c1+c2))\n",
      "    p2=y-p1\n",
      "    return [p1,p2]\n",
      "ma=max(l)\n",
      "mi=min(l)\n",
      "for x in range(0,3):\n",
      "    [e ,g]=cost(l[x])\n",
      "    if e<plo or g<plo or e>pup or g>pup:\n",
      "        if e<plo or g<plo:\n",
      "            v=min(e,g)\n",
      "            [u for u, j in enumerate((e,g)) if j == v]\n",
      "    if u==0:\n",
      "        e=plo\n",
      "        g=l(x)-e\n",
      "    else:\n",
      "        g=plo\n",
      "        e=l[x]-g\n",
      "    \n",
      "    \n",
      "print \"\\np1=%3.2fMW\\tp2=%3.2fMW\"%(e,g)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "p1=1000.00MW\tp2=200.00MW\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.13 Page 211"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "a1=2000 ;b1=20 ;c1=0.05; p1=350; p2=550\n",
      "a2=2750 ;b2=26 ;c2=0.03091\n",
      "def cost(a,b,c,p):\n",
      "    co=a+b*p+c*p**2\n",
      "    return co\n",
      "print \"(a)\"\n",
      "toco=cost(a1,b1,c1,p1)+cost(a2,b2,c2,p2)\n",
      "print \"total cost when each system supplies its own load Rs%.3f per hour\"%(toco)\n",
      "l=p1+p2\n",
      "p11=(b2-b1+2*c2*l)/(2*(c1+c2))\n",
      "p22=l-p11\n",
      "totco=cost(a1,b1,c1,p11)+cost(a2,b2,c2,p22)\n",
      "sav=toco-totco\n",
      "tilo=p11-p1\n",
      "print \"(b)\"\n",
      "print \"\\n total cost when load is supplied in economic load dispatch method Rs%d per hour \\n saving %.3f \\n tie line load %.3f MW\"%(totco,sav,tilo)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)\n",
        "total cost when each system supplies its own load Rs41525.275 per hour\n",
        "(b)\n",
        "\n",
        " total cost when load is supplied in economic load dispatch method Rs41447 per hour \n",
        " saving 77.277 \n",
        " tie line load 30.905 MW\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example10.14 Page 212"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "a1=5000 ;b1=450 ;c1=0.5 #for system 1 \n",
      "e1=0.02 ;e2=-0.02#error\n",
      "a1c=a1*(1-e1); b1c=b1*(1-e1) ;c1c=c1*(1-e1)\n",
      "a2c=a1*(1-e2) ;b2c=b1*(1-e2) ;c2c=c1*(1-e2)\n",
      "tl=200\n",
      "def cost(a,b,c,p):\n",
      "    co=a+b*p+c*p**2\n",
      "    return co\n",
      "p11=(b2c-b1c+2*c2c*tl)/(2*(c1c+c2c))\n",
      "p22=tl-p11\n",
      "totco=cost(a1c,b1c,c1c,p11)+cost(a2c,b2c,c2c,p22)\n",
      "print \"\\npower at station 1 is %dMW \\t power at station 2 is %dMW \\n total cost on economic critieria method Rs%d per hour\"%(p11,p22,totco)\n",
      "tocoe=cost(a1c,b1c,c1c,tl/2)+cost(a2c,b2c,c2c,tl/2)\n",
      "eop=tocoe-totco\n",
      "print \"\\nextra operating cost due to erroneous scheduling Rs.%d per hour\"%(eop)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "power at station 1 is 111MW \t power at station 2 is 89MW \n",
        " total cost on economic critieria method Rs109879 per hour\n",
        "\n",
        "extra operating cost due to erroneous scheduling Rs.121 per hour\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.16 Page 215"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "ia=32 ;ib=32 ;ic=1.68; f=10**5\n",
      "wt=18; rt=24-wt\n",
      "p=30\n",
      "def inpu(a,b,c,f,t,p):\n",
      "    In=(a+b*p+c*p**2)*f*t\n",
      "    return In\n",
      "hi1=inpu(ia,ib,ic,f,wt,p); hi2=inpu(ia,ib,ic,f,rt,p/2)\n",
      "print \"(a)\"\n",
      "print \"for full load condition for %d hours is %ekj \\n for half load condition for%d s %ekj \\n total load %ekj\"%(wt,hi1,rt,hi2,hi1+hi2)\n",
      "print \"(b)\"\n",
      "te=p*wt+(p/2)*rt\n",
      "ul=te/24\n",
      "hin=inpu(ia,ib,ic,f,24,ul)\n",
      "sav=hi1+hi2-hin\n",
      "savp=sav/(te*1000)\n",
      "print \"\\n total energy produced\\t%dMW \\n unifor load\\t%dMW \\n heat input under uniform load condition %ekj \\n saving in heat energy %ekj \\n saving in heat energy per kWh %dkj/kWh\"%(te,ul,hin,sav,savp)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)\n",
        "for full load condition for 18 hours is 4.507200e+09kj \n",
        " for half load condition for6 s 5.340000e+08kj \n",
        " total load 5.041200e+09kj\n",
        "(b)\n",
        "\n",
        " total energy produced\t630MW \n",
        " unifor load\t26MW \n",
        " heat input under uniform load condition 4.799232e+09kj \n",
        " saving in heat energy 2.419680e+08kj \n",
        " saving in heat energy per kWh 384kj/kWh\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example 10.17 Page 216"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "a1=450 ;b1=6.5 ;c1=0.0013\n",
      "a2=300 ;b2=7.8 ;c2=0.0019\n",
      "a3=80 ;b3=8.1 ;c3=0.005\n",
      "tl=800#total load\n",
      "ma = [600, 400, 200]\n",
      "mi = [100, 50, 50]\n",
      "d=np.mat([[1 ,1, 1] ,[2*c1, -2*c2, 0], [0, -2*c2, 2*c3]])\n",
      "p=np.mat([[tl], [(b2-b1)], [(b2-b3)]]) \n",
      "pp=(d**-1)*p #matrix inversion method\n",
      "print \"\\nloads on generaating station by economic creatirian method isp1=%fMW,p2=%fMW,p3=%fMW\"%(pp[0],pp[1],pp[2])\n",
      "for i in range(0,3):\n",
      "    if pp[i]<mi[i]:\n",
      "        pp[i]=mi[i]\n",
      "    \n",
      "    if pp[i]>mi[i]:\n",
      "        pp[i]=ma[i]\n",
      "    \n",
      "\n",
      "pp[1] = tl-pp[0] - pp[2]\n",
      "print \"\\nloads on generating station under critical conditions p1=%dMW p2=%dMW p3=%dMW\"%(pp[0],pp[1],pp[2])"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "loads on generaating station by economic creatirian method isp1=669.734705MW,p2=116.134272MW,p3=14.131023MW\n",
        "\n",
        "loads on generating station under critical conditions p1=600MW p2=150MW p3=50MW\n"
       ]
      }
     ],
     "prompt_number": 25
    }
   ],
   "metadata": {}
  }
 ]
}