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-{
- "metadata": {
- "name": "",
- "signature": "sha256:b75473f9f4345efd6ca0ec6782f30e07c9e95389ed3851945260a1b3129ea6b2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "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": {}
- }
- ]
-}
generated by cgit (git 2.34.1) at 2024-12-20 11:02:59 +0000