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diff --git a/Generation_Of_Electrical_Energy_by_B._R._Gupta/Chapter10.ipynb b/Generation_Of_Electrical_Energy_by_B._R._Gupta/Chapter10.ipynb new file mode 100644 index 00000000..cdaa827a --- /dev/null +++ b/Generation_Of_Electrical_Energy_by_B._R._Gupta/Chapter10.ipynb @@ -0,0 +1,674 @@ +{ + "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": {} + } + ] +} |