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| author | hardythe1 | 2015-06-11 17:31:11 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-06-11 17:31:11 +0530 |
| commit | 251a07c4cbed1a5a960f5ed416ce6ac13c8152b7 (patch) | |
| tree | cb7f084fad6d7ee6ae89e586fad0e909b5408319 /Electrical_Power_System_by_C.L._Wadhwa | |
| parent | 47d7279a724246ef7aa0f5359cf417992ed04449 (diff) | |
| download | Python-Textbook-Companions-251a07c4cbed1a5a960f5ed416ce6ac13c8152b7.tar.gz Python-Textbook-Companions-251a07c4cbed1a5a960f5ed416ce6ac13c8152b7.tar.bz2 Python-Textbook-Companions-251a07c4cbed1a5a960f5ed416ce6ac13c8152b7.zip | |
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Diffstat (limited to 'Electrical_Power_System_by_C.L._Wadhwa')
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diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter01.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter01.ipynb new file mode 100755 index 00000000..bf7fc600 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter01.ipynb @@ -0,0 +1,77 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 01 : Fundamentals of power System"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1, Page No 11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Sb=100.0\t\t# base value of power(MVA)\n",
+ "Vb=33.0\t\t\t# base value of voltage (Kv)\n",
+ "\n",
+ "#Calculations\n",
+ "Vbl=Vb*110.0/32\n",
+ "Vbm=Vbl*32.0/110\n",
+ "Zp_ut=0.08*100*32*32/(110*33*33)\n",
+ "Zp_u_l=50*100/(Vbl**2)\n",
+ "Zp_um1=0.2*100*30*30.0/(30*33*33)\n",
+ "Zp_um2=0.2*100*30*30.0/(20*33*33)\n",
+ "Zp_um3=0.2*100*30*30.0/(50*33*33)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Base value of voltage in line = %.2f kV\" %Vbl)\n",
+ "print(\"Base value of voltage in motor circuit=%.0f kV\" %Vbm)\n",
+ "print(\"p.u value of reactance transformer =%.5f p.u\" %Zp_ut)\n",
+ "print(\"p.u value of impedence of line=%.4f p.u\" %Zp_u_l)\n",
+ "print(\"p.u value of reactance of motor 1 =%.4f p.u\" %Zp_um1)\n",
+ "print(\"p.u value of reactance of motor 2 =%.3f p.u\" %Zp_um2)\n",
+ "print(\"p.u value of reactance of motor 3 =%.4f p.u\" %Zp_um3)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Base value of voltage in line = 113.44 kV\n",
+ "Base value of voltage in motor circuit=33 kV\n",
+ "p.u value of reactance transformer =0.06839 p.u\n",
+ "p.u value of impedence of line=0.3886 p.u\n",
+ "p.u value of reactance of motor 1 =0.5510 p.u\n",
+ "p.u value of reactance of motor 2 =0.826 p.u\n",
+ "p.u value of reactance of motor 3 =0.3306 p.u\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter02.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter02.ipynb new file mode 100755 index 00000000..269565a5 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter02.ipynb @@ -0,0 +1,281 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 02 : Line Constant Calculations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2, Page No 29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "GMD=0.7788*0.8/(2*100)\n",
+ "Mgmd=((1.6*3.2*1.6)**(1.0/3))\n",
+ "\n",
+ "#Calculations\n",
+ "Z=2*(10**-4)*1000*math.log(2.015/.003115)\n",
+ "\n",
+ "#Results\n",
+ "print(\"The self GMD of the conductor =%.6f metres\" %GMD)\n",
+ "print(\"The mutual GMD of the conductor =%.3f metres \" %Mgmd)\n",
+ "print(\"Inductance =%.3f mH/km\\n\" %Z)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The self GMD of the conductor =0.003115 metres\n",
+ "The mutual GMD of the conductor =2.016 metres \n",
+ "Inductance =1.294 mH/km\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3, Page No 29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "r=1\n",
+ "D11=r**1\n",
+ "D12=2*r\n",
+ "D14=4*r\n",
+ "\n",
+ "#Calculations\n",
+ "D13=math.sqrt(16-4)*r\n",
+ "Ds1=((1*2*2*math.sqrt(3)*4*2*math.sqrt(3)*2*2)**(1.0/7))*r\n",
+ "Ds7=((2*1*2*2**2*2*2)**(1.0/7))*r #we get this after Taking r outside the 1/7th root\n",
+ "Ds=((((1*2*2*math.sqrt(3)*4*2*math.sqrt(3)*2*2)**(1.0/7))**6)*((2*1*2*2**2*2*2)**(1.0/7)))**(1.0/7)*r\n",
+ "Dseq=((.7788)**(1.0/7))*Ds\n",
+ "\n",
+ "#Results\n",
+ "print(\"Dseq.= %.2fr\" %Dseq)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dseq.= 2.18r\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4, Page No 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "GMDa=0.001947\t\t\t\t# GMD of conductor in group A\n",
+ "\n",
+ "#Calculations\n",
+ "DSA=((.001947*6*12*.001947*6*6*0.001947*6*12)**(1.0/9))\n",
+ "DSB=math.sqrt(5*(10**-3)*.7788*6)\n",
+ "Dae=math.sqrt((9**2)+6**2)\n",
+ "Dcd=math.sqrt((12**2)+9**2)\n",
+ "DMA=((9*10.81*10.81*9*15*10.81)**(1.0/6))\n",
+ "LA=2*(10**-7)*(10**6)*math.log(DMA/DSA)\n",
+ "LB=2*(10**-7)*(10**6)*math.log(DMA/DSB)\n",
+ "Tot=LA+LB\n",
+ "\n",
+ "#Results\n",
+ "print(\"inductance of line A,LA=%.3f mH/km\" %LA)\t\t#Answers don't match due to difference in rounding off of digits\n",
+ "print(\"inductance of line B,LB=%.1f mH/km\" %LB)\t\t#Answers don't match due to difference in rounding off of digits\n",
+ "print(\"total inductance of line =%.2f mH/km\" %Tot)\t#Answers don't match due to difference in rounding off of digits\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "inductance of line A,LA=0.621 mH/km\n",
+ "inductance of line B,LB=0.9 mH/km\n",
+ "total inductance of line =1.47 mH/km\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 Page No 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "GMDc=1.266*0.7788*(10**-2)\t\t# self GMD of each conductor\n",
+ "Dbc=math.sqrt((4**2)+(.75**2))\n",
+ "Dab=Dbc\n",
+ "\n",
+ "#Calculations\n",
+ "Dab=math.sqrt((4**2)+(8.25**2))\n",
+ "Daa=math.sqrt((8**2)+(7.5**2))\n",
+ "Dm1=(Dbc*8*7.5*9.1685)**(1.0/4)\n",
+ "Dm2=(Dbc*Dbc*9.1685*9.1685)**(1.0/4)\n",
+ "Dm3=Dm1\n",
+ "Dm=((Dm1*Dm2*Dm3)**(1.0/3))\n",
+ "Ds1=math.sqrt(GMDc*Daa)\t\t# self GMD of each phase\n",
+ "Ds3=Ds1\n",
+ "Ds2=math.sqrt(GMDc*9)\n",
+ "Ds=((Ds1*Ds2*Ds3)**(1.0/3))\n",
+ "Z=2*(10**-4)*(1000)*math.log(Dm/Ds)\n",
+ "\n",
+ "#Results\n",
+ "print(\"inductance=%.3f mH/km/phase\\n\" %Z)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "inductance=0.607 mH/km/phase\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6, Page No 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "GMDs=.0069\t\t#self GMD of the conductor\n",
+ "Dab=math.sqrt((3**2)+.5**2)\n",
+ "Dbc=Dab\n",
+ "Dac=6.0\n",
+ "\n",
+ "#Calculations\n",
+ "Dab=math.sqrt((3**2)+6**2)\n",
+ "Daa=math.sqrt((6**2)+5.5**2)\n",
+ "Dm1=((3.04*6*5.5*6.708)**.25)\n",
+ "Dm2=((3.04*3.04*6.708*6.708)**.25)\n",
+ "Dm=4.89\n",
+ "Ds1=math.sqrt(GMDs*Daa)\n",
+ "Ds2=0.2217\n",
+ "Ds=.228\n",
+ "Z=2*(10**-7)*(10**6)*math.log(Dm/Ds)\n",
+ "\n",
+ "#Results\n",
+ "print(\"inductance =%.3f mH/km\" %Z)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "inductance =0.613 mH/km\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7, Page No 34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ds=math.sqrt(0.025*.4*.7788)\n",
+ "\n",
+ "#Calculations\n",
+ "Dm=((6.5*13.0*6.5)**(1.0/3))\n",
+ "Z=2*(10**-4)*1000*math.log(Dm/Ds)\n",
+ "\n",
+ "#Results\n",
+ "print(\"inductance =%.3f mH/km/phase\" %Z)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "inductance =0.906 mH/km/phase\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter03.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter03.ipynb new file mode 100755 index 00000000..e0c0fd23 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter03.ipynb @@ -0,0 +1,142 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 03 : Capacitance of Transmission Lines"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1, Page No 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Dm=2.015 \t\t# mutual GMD of conductors(m)\n",
+ "r=0.4 \t\t\t# radius of conductor(cm)\n",
+ "\n",
+ "#Calculations\n",
+ "C=10**-9*1000.0/(18*math.log(201.5/.4)) \n",
+ "Ic=132*1000.0*8.928*314*(10**-9)/math.sqrt(3.0)\n",
+ "\n",
+ "#Results\n",
+ "print(\"capacitance =%.13f F/km\" %C) #Answers don't match due to different representation\n",
+ "print(\"charging current=%.4f amp/km\" %Ic) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "capacitance =0.0000000089288 F/km\n",
+ "charging current=0.2136 amp/km\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2, Page No 52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "GMDm=6.61 \t\t#mutual GMD(m)\n",
+ "\n",
+ "#Calculations\n",
+ "Ds1=math.sqrt(1.25*(10**-2)*10.965) \n",
+ "Ds3=Ds1 \n",
+ "Ds2=math.sqrt(1.25*(10**-2)*9) \n",
+ "Ds=((Ds1*Ds2*Ds3)**.333333) \n",
+ "C=1/(18*math.log(GMDm/Ds)) \n",
+ "Ic=220*1000*314*.01905*(10**-6)/math.sqrt(3) \n",
+ "\n",
+ "#Results\n",
+ "print(\"capacitance =%.6f micro-Farad/km\" %C) \n",
+ "print(\"charging current =%.2f amp/km\" %Ic)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "capacitance =0.019057 micro-Farad/km\n",
+ "charging current =0.76 amp/km\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3, Page No 53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "GMD=8.19 \n",
+ "\n",
+ "#Calculations\n",
+ "Ds=math.sqrt(2.25*(10**-2)*.4) \n",
+ "C=1/(18*math.log(GMD/Ds)) \n",
+ "Ic=220*1000*314*C*(10**-6)/math.sqrt(3) \n",
+ "\n",
+ "#Results\n",
+ "print(\"capacitance per km =%.5f micro-Farad\" %C) \n",
+ "print(\"charging current =%.3f amp\" %Ic) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "capacitance per km =0.01246 micro-Farad\n",
+ "charging current =0.497 amp\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter04.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter04.ipynb new file mode 100755 index 00000000..e3b7281e --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter04.ipynb @@ -0,0 +1,589 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4 : Performance of Lines"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1, Page No 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "R=0.496\t\t# resistance\n",
+ "X=1.536\n",
+ "Vr=2000.0\n",
+ "\n",
+ "#Calculations\n",
+ "Z=(10*2*2/(11*11)) + complex(30)*2*2/(11*11)\n",
+ "Zt=(.04+(1.3*2*2/(11*11))) + complex(0.125,(4.5*2*2/(11*11)))#Transformer impedence\n",
+ "Il=250*1000.0/2000\t# line current(amps.)\n",
+ "Pl=Il*Il*R\t\t\t#line loss(kW)\n",
+ "Po=250*0.8\t\t\t# output(kW)\n",
+ "cosr=0.8\t\t\t# power factor\n",
+ "sinr=0.6\n",
+ "n=200*100.0/(200+7.7)\n",
+ "Vs=(Vr*cosr+Il*R)+complex(Vr*sinr+Il*X)\n",
+ "V=math.sqrt((1662**2)+ (1392**2))\n",
+ "\n",
+ "#Results\n",
+ "print(\"efficiency= %.1f percent \" %n)\n",
+ "print(\"Sending end voltage,|Vs|=%.0f volts\" %V)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "efficiency= 96.3 percent \n",
+ "Sending end voltage,|Vs|=2168 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2, Page No 66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "print(\"when load is star connected\")\n",
+ "Vln=400/math.sqrt(3.0)\t\t# Line to neutral voltage(V)\n",
+ "Z=complex(7,11)\t\t\t\t\t#Impedence per phase\n",
+ "Il=231/Z\t\t\t\t\t\t# line current(amp.)\n",
+ "I=abs(231/Z)\n",
+ "Pi=3*I*I*7\n",
+ "Po=3*I*I*6\n",
+ "\n",
+ "#Calculations\n",
+ "print(\"power input =%.0f watts\\n\" %Pi)\t\t#Answers don't match due to difference in rounding off of digits\n",
+ "print(\"power output=%.0f watts\\n\" %Po)\t#Answers don't match due to difference in rounding off of digits\n",
+ "print(\"when load is delta connected\\n\")\n",
+ "Ze=complex(2,3)\t\t# equivalent impedence(ohm)\n",
+ "Zp=complex(3,5)\t\t# impedence per phase\n",
+ "il=231/Zp\t\t\t#Line current(amps.)\n",
+ "IL=abs(il)\n",
+ "pi=3*IL*IL*3\n",
+ "po=3*IL*IL*2\n",
+ "\n",
+ "#Results\n",
+ "print(\"power input=%.1f watts\" %pi)\t\t\t#Answers don't match due to difference in rounding off of digits\n",
+ "print(\"power output = %.0f watts \" %po)\t\t#Answers don't match due to difference in rounding off of digits"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "when load is star connected\n",
+ "power input =6592 watts\n",
+ "\n",
+ "power output=5650 watts\n",
+ "\n",
+ "when load is delta connected\n",
+ "\n",
+ "power input=14125.0 watts\n",
+ "power output = 9417 watts \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3, Page No 66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "a=100/.5\n",
+ "Xl=2*(10**-7)*math.log(100/.5)\t\t#inductance(H/meter)\n",
+ "XL=20*(1000)*Xl\t\t\t\t\t\t# inductance of 20 km length \n",
+ "R=6.65\t\t\t\t\t\t\t\t# resistance(ohm)\n",
+ "Rc=20*1000/(58.0*90)\t\t\t\t# resistance of copper(ohm)\n",
+ "I=10*1000/(33*.8*math.sqrt(3))\t\t# the current(amps.)\n",
+ "\n",
+ "#Calculations\n",
+ "Pl=3*I*I*Rc/(10**6)\t\t\t\t\t#loss (MW)\n",
+ "n=10.0/(10+Pl)\n",
+ "print(\"Efficiency=%.4f percent \" %n)\n",
+ "Vr=19052\n",
+ "cosr=.8\t\t\t\t#power factor\n",
+ "sinr=.6\n",
+ "Vs=abs(((Vr*cosr+I*Rc) +complex(Vr*sinr+ I*R)))\n",
+ "\n",
+ "#Results\n",
+ "print(\"Vs =%.0f volts\\n\" %Vs)\t#Answer don't match due to difference in rounding off of digits\n",
+ "Reg=(Vs-Vr)*100/Vr\n",
+ "print(\"Regulation =%.2f percent\" %Reg)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency=0.9479 percent \n",
+ "Vs =28965 volts\n",
+ "\n",
+ "Regulation =52.03 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 Page No 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "IR=(400)/((math.sqrt(3)*complex(6.3,9)))\n",
+ "\n",
+ "#Calculations\n",
+ "IY=231*complex(math.cos(math.radians(-120)),math.sin(math.radians(-120)))/8.3\n",
+ "IB=231*complex(math.cos(math.radians(120)),math.sin(math.radians(120)))/complex(6.3,-8)\n",
+ "In=abs((IR +IY +IB))\t\t#Neutral current\n",
+ "print(\"Neutral current =%.2f amps\\n\" %In)\n",
+ "VR=abs(IR*complex(6,9))\n",
+ "VY=abs(IY*(8))\n",
+ "VB=abs(IB*complex(6,-8))\n",
+ "\n",
+ "#Results\n",
+ "print(\"Voltage across Phase R =%.1f volts \\n\" %VR)\n",
+ "print(\"Voltage across Phase Y =%.2f volts \\n\" %VY)\n",
+ "print(\"Voltage across Phase B =%.0f volts \\n\" %VB)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Neutral current =45.18 amps\n",
+ "\n",
+ "Voltage across Phase R =227.4 volts \n",
+ "\n",
+ "Voltage across Phase Y =222.65 volts \n",
+ "\n",
+ "Voltage across Phase B =227 volts \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5, Page No 73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "R=100*.1\t\t#Resistance of line (ohm)\n",
+ "Xl=2*(10**-7)*100*1000*math.log(200/.75)\t#inductance of line\n",
+ "X2=Xl*314\t\t\t\t\t\t\t\t\t#inductive reactance\n",
+ "C=2*(math.pi*100)*8.854*(10**-12)*100*1000*(10**6)/(math.log(200/.75))\t# capacitance per phase (micro farad)\n",
+ "\n",
+ "#Calculations\n",
+ "print(\"Using Nominal-T method\\n\")\n",
+ "Ir=20*1000.0/(math.sqrt(3)*66*.8)\n",
+ "Vr=66*1000/math.sqrt(3)\n",
+ "Vc=complex((38104*.8+ Ir*5),(38104*.6+ Ir*17.55))\t# voltage across condenser\n",
+ "Ic=complex(314*(Vc)*.9954*(10**-6))\n",
+ "ise=Ir+Ic\n",
+ "Is=abs(Ir+Ic)\n",
+ "Vs=abs(Vc + (ise*complex(5,17.53)))\n",
+ "VR=abs(Vs*complex(-3199)/complex(5,-3181))# no load recieving end voltage\n",
+ "Reg=(VR-Vr)*100.0/Vr\n",
+ "Pl=3*(Ir*Ir*5 + Is*Is*5)/1000000\n",
+ "n=20*100/(20+Pl)\n",
+ "print(\"percent regulation=%.1f \" %Reg)\n",
+ "print(\"percent efficiency=%.1f \\n\" %n)\n",
+ "print(\"Using Nominal-pi method\\n\")\n",
+ "Ir1=218.68*complex(.8,-.6)\n",
+ "Ic1=complex(314*.4977*(10**-6)*Vr)\n",
+ "Il=Ir1+Ic1\n",
+ "vs1=Vr+Il*complex(10,35.1)\n",
+ "Vs1=abs(vs1)\n",
+ "Vr1=Vs1*complex(-6398)/complex(10,-6363)\n",
+ "VR1=abs(Vr1)\t\t\t# no load recieving end voltage\n",
+ "Reg2=(VR1-Vr)*100/Vr\n",
+ "IL=abs(Ir1+Ic1)\n",
+ "Loss=3*IL*IL*10\n",
+ "n=20*100/21.388\n",
+ "\n",
+ "#Results\n",
+ "print(\"percent regulation=%.2f \" %Reg2)\n",
+ "print(\"percent efficiency=%.1f \" %n)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Using Nominal-T method\n",
+ "\n",
+ "percent regulation=18.2 \n",
+ "percent efficiency=93.0 \n",
+ "\n",
+ "Using Nominal-pi method\n",
+ "\n",
+ "percent regulation=18.23 \n",
+ "percent efficiency=93.5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6, Page No 78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "#initialisation of variables\n",
+ "R=0.2\n",
+ "L=1.3\n",
+ "C=0.01*(10**-6)\n",
+ "\n",
+ "#Calculations\n",
+ "z=complex(R,(L*314.0*(10**-3)))\t\t# serie impedence\n",
+ "y=complex(314.0*C)\t\t# shunt admittance\n",
+ "Zc=cmath.sqrt(z/y)\t\t# characterstic impedence\n",
+ "Y=cmath.sqrt(y*z)\n",
+ "Vr=132*1000/math.sqrt(3.0)\n",
+ "Ir=0\n",
+ "Vin=(Vr + Ir*Zc)/2\t\t\t# incident voltage to neutral at the recieving end\n",
+ "\n",
+ "#Results\n",
+ "print(\"Vr =%.3f volts \\n\" %Vr)\t\t#Answer don't match due to difference in rounding off of digits\n",
+ "print(\"(i)The incident voltage to neutral at the recieving end {0:.5f}+{1:.5f}i\".format(Vin.real, Vin.imag))\t#Answer don't match due to difference in rounding off of digits\n",
+ "Vin2=(Vr - Ir*Zc)/2\t\t\t\t\t\t# The reflected voltage to neutral at the recieving end\n",
+ "print(\"(ii)The reflected voltage to neutral at the recieving end{0:.5f}+{1:.5f}i\".format(Vin2.real, Vin2.imag))\t\t#Answer don't match due to difference inrounding off of digits\n",
+ "Vrp=Vr*cmath.exp(.2714*120*(10**-3))*cmath.exp(complex(1.169*120*(10**-3))/1000.0)#Taking Vrp=Vr+\n",
+ "Vrm=Vr*cmath.exp(-0.0325)*cmath.exp(complex(-.140))/1000\t\t\t#Taking Vrm=Vr-\n",
+ "v1=Vrm/2\t\t\t\t # reflected voltage to neutral at 120 km from the recieving end\n",
+ "phase_v1=math.degrees(math.atan(v1.imag/v1.real))\n",
+ "v2=Vrp/2\t\t\t #incident voltage to neutral at 120 km from the recieving end\n",
+ "phase_v2=math.degrees(math.atan(v2.imag/v2.real))#Phase angle of v2\n",
+ "print(\"(iii) reflected voltage to neutral at 120 km from the recieving end =%.2f at angle of %.2f\" %(abs(v1),phase_v1))\n",
+ "print(\"incident voltage to neutral at 120 km from the recieving end = %.2f at angle of %.2f\" %(abs(v2),phase_v2))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Vr =76210.236 volts \n",
+ "\n",
+ "(i)The incident voltage to neutral at the recieving end 38105.11777+0.00000i\n",
+ "(ii)The reflected voltage to neutral at the recieving end38105.11777+0.00000i\n",
+ "(iii) reflected voltage to neutral at 120 km from the recieving end =32.07 at angle of 0.00\n",
+ "incident voltage to neutral at 120 km from the recieving end = 39372.08 at angle of 0.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 Page No 79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ir=40.0*1000/(math.sqrt(3)*132*.8)\n",
+ "Vr=132.0*1000/math.sqrt(3)\n",
+ "\n",
+ "#Calculations\n",
+ "Zc=380*complex(math.cos(math.radians(-13.06)),math.sin(math.radians(-13.06)))\n",
+ "IR=Ir*complex(math.cos(math.radians(-38.8)),math.sin(math.radians(-38.8)))\n",
+ "Vsp=(Vr+IR*Zc)*(1.033*complex(math.cos(math.radians(8.02)),math.sin(math.radians(8.02))))/2\n",
+ "Vsm=(Vr-IR*Zc)*(0.968*complex(math.cos(math.radians(8.02)),math.sin(math.radians(8.02))))/2\n",
+ "vs=Vsp+ Vsm\n",
+ "Vs=abs(vs)\n",
+ "ise=(Vsp-Vsm)/Zc\n",
+ "Is=abs(ise)\n",
+ "P=3*Vs*Is*math.cos(math.radians(33.72))/10**6\n",
+ "n=40*100/P\n",
+ "\n",
+ "#Results\n",
+ "print(\"efficiency=%.1f\" %n)\t\t#Answer don't match due to difference in rounding off of digits"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "efficiency=92.3\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8, Page No 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "#initialisation of variables\n",
+ "yl=complex(0.2714,1.169)*120*(10**-3)\n",
+ "Ir=40*1000/(math.sqrt(3)*132*.8)\n",
+ "A=cmath.cosh(yl)\n",
+ "\n",
+ "#Calculations\n",
+ "phase_A=math.degrees(math.atan(A.imag/A.real)) #Phase angle of A\n",
+ "IR=Ir*complex(math.cos(math.radians(-38.8)),math.sin(math.radians(-38.8)))\n",
+ "Vr=132*1000/math.sqrt(3)\n",
+ "Zc=380*complex(math.cos(math.radians(-13.06)),math.sin(math.radians(-13.06)))\n",
+ "B=Zc*cmath.sinh(yl)\n",
+ "phase_B=math.degrees(math.atan(B.imag/B.real)) #Phase angle of B\n",
+ "Vs=(A*Vr+B*IR)\n",
+ "f=abs(B)\n",
+ "d=abs(Vs)\n",
+ "C=cmath.sinh(yl)/Zc\n",
+ "phase_C=math.degrees(math.atan(C.imag/C.real)) #Phase angle of C\n",
+ "D=cmath.cosh(yl)\n",
+ "phase_D=math.degrees(math.atan(D.imag/D.real))\t\t#Phase angle of D\n",
+ "\n",
+ "#Results\n",
+ "print(\"A=%.2f at an angle of %.2f \" %(abs(A),phase_A))\n",
+ "print(\"B=%.1f at an angle of %.0f \" %(abs(B),phase_B))\n",
+ "print(\"C=%.2f at an angle of %.2f \" %(abs(C),phase_C))\n",
+ "print(\"D=%.2f at an angle of %.2f \" %(abs(D),phase_D))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "A=0.99 at an angle of 0.26 \n",
+ "B=54.6 at an angle of 64 \n",
+ "C=0.00 at an angle of -89.92 \n",
+ "D=0.99 at an angle of 0.26 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 Page No 81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ir=218.7*complex(0.8,-0.6)\n",
+ "Ic1=complex(314)*0.6*(10**-6)*76200\n",
+ "Il=Ic1+Ir\n",
+ "Vs=76200 + Il*complex(24,48.38)\n",
+ "\n",
+ "#Calculations\n",
+ "phase_Vs=math.degrees(math.atan(Vs.imag/Vs.real))\n",
+ "Pl=3*24*abs(Il)*abs(Il)/1000000.0 #The Loss(MW)\n",
+ "n=40*100/(40+Pl)\n",
+ "print(\"Using Nominal- pi method\")\n",
+ "print(\"Vs=%.0f volts at an angle of %.2f \\n\" %(abs(Vs),phase_Vs))\n",
+ "print(\"efficiency=%.2f percent\" %n)\n",
+ "print(\"\\nUsing Nominal-T method\")\n",
+ "Vc=76200*complex(0.8,0.6)+218.7*complex(12,24.49)\n",
+ "Ic=complex(314)*1.2*(10**-6)*complex(63584,51076.0)\n",
+ "Is=complex(199.46,23.95)\n",
+ "Vs=(Vc + Is*complex(12,24.49))/1000.0\n",
+ "phase_Vs=math.degrees(math.atan(Vs.imag/Vs.real))\t#Phase angle of Vs\n",
+ "Pl1=3*12*((200.89**2)+ 218.7**2)/1000000 #The loss(MW)\n",
+ "n1=40*100/(40+Pl1)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Vs=%.2f at an angle of %.2f \" %(abs(Vs),phase_Vs))\n",
+ "print(\"efficiency=%.2f percent\\n\" %n1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Using Nominal- pi method\n",
+ "Vs=87299 volts at an angle of 3.95 \n",
+ "\n",
+ "efficiency=91.28 percent\n",
+ "\n",
+ "Using Nominal-T method\n",
+ "Vs=86.25 at an angle of 40.70 \n",
+ "efficiency=92.65 percent\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.10 Page No 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "#initialisation of variables\n",
+ "R=0.1557*160\n",
+ "GMD=(3.7*6.475*7.4)**(1.0/3)\n",
+ "Z1=2*(10**-7)*math.log(560/0.978)*160*1000\n",
+ "XL=63.8\n",
+ "\n",
+ "#Calculations\n",
+ "C=(10**-9)*2*(10**6)*math.pi*160*1000.0/(36*math.pi*math.log(560/.978))\n",
+ "Z=math.sqrt((0.1557**2)+.39875**2)*complex(math.cos(math.radians(68.67)),math.sin(math.radians(68.67)))\n",
+ "jwC=complex(314)*1.399*(10**-6)/160.0\n",
+ "Zc=cmath.sqrt(Z/jwC)\n",
+ "y=cmath.sqrt(Z*jwC)\n",
+ "yl=y*160\n",
+ "A=cmath.cosh(yl)\n",
+ "B=Zc*cmath.sinh(yl)\n",
+ "C=cmath.sinh(yl)/Zc\n",
+ "Ir=50000/(math.sqrt(3)*132)\n",
+ "Vs=(A*76.208) +(B*(10**-3)*Ir*complex(math.cos(math.radians(-36.87)),math.sin(math.radians(-36.87))))\n",
+ "VS=152.34\n",
+ "Is=C*76.208*(10**3) +(A*Ir*complex(math.cos(math.radians(-36.87)),math.sin(math.radians(-36.87))))\n",
+ "Ps=3*abs(Vs)*abs(Is)*math.cos(math.radians(33.96))\n",
+ "pf=math.cos(math.radians(33.96))\n",
+ "Vnl=abs(Vs)/abs(A)\n",
+ "reg=(Vnl-76.208)*100/76.208\n",
+ "n=50000*.8*100/abs(Ps)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Vs line to line =%.2f kV\\n\" %VS)\n",
+ "print(\"sending end current Is(A){0:.5f}+{1:.5f}i\".format(Is.real, Is.imag)) #Answer don't match due to difference in rounding off of digits\n",
+ "print(\"sending end power=%.0f kW\" %Ps)\n",
+ "print(\"sending end p.f =%.3f\" %pf)\n",
+ "print(\"percent regulation=%.1f \" %reg)\n",
+ "print(\"percent efficency=%.1f \" %n)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Vs line to line =152.34 kV\n",
+ "\n",
+ "sending end current Is(A)211.28696+-129.31797i\n",
+ "sending end power=55350 kW\n",
+ "sending end p.f =0.829\n",
+ "percent regulation=17.2 \n",
+ "percent efficency=72.3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter05.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter05.ipynb new file mode 100755 index 00000000..15a3f816 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter05.ipynb @@ -0,0 +1,175 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 05 : High Voltage DC Transmission"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1, Page No 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vo=3*math.sqrt(2)*110/math.pi\n",
+ "\n",
+ "#Calculations \n",
+ "Vd=Vo*(math.cos(math.radians(0))+math.cos(math.radians(15)))/2.0 \n",
+ "Vd1=Vo*(math.cos(math.radians(30)) + math.cos(math.radians(45)))/2.0 \n",
+ "Vd2=Vo*(math.cos(math.radians(45)) + math.cos(math.radians(60)))/2.0 \n",
+ "\n",
+ "#Results\n",
+ "print(\"(a)For a=0, Vd=%.2f kV\" %Vd) \n",
+ "print(\"(b)For a=30,Vd=%.2f kV\" %Vd1) \n",
+ "print(\"(c)For a=45,Vd=%.2f kV\" %Vd2) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)For a=0, Vd=146.02 kV\n",
+ "(b)For a=30,Vd=116.85 kV\n",
+ "(c)For a=45,Vd=89.66 kV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2, Page No 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "#Calculations\n",
+ "VL=(100.0*2*math.pi)/(3*math.sqrt(2.0)*(math.cos(math.radians(30)) + math.cos(math.radians(45)))) \n",
+ "print(\"VL=%.2f kV\" %VL) #Answers don't match due to difference in rounding off of digits\n",
+ "Tr=VL/110.0\n",
+ "\n",
+ "#Results \n",
+ "print(\"tap ratio=%.2f \" %Tr) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "VL=94.14 kV\n",
+ "tap ratio=0.86 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3, Page No 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variablesVd=100000\n",
+ "Id=800.0 # current\n",
+ "\n",
+ "#Calculations\n",
+ "X=((3*math.sqrt(2.0)*94.115*0.866*1000.0/math.pi)-Vd)*math.pi/(3.0*Id)\n",
+ "\n",
+ "#Results\n",
+ "print(\"effective reactance per phase , X=%.2f ohm\\n\" %X)\t#Answer don't match due to difference in rounding off of digits"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "effective reactance per phase , X=143.89 ohm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4 Page No 112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "a=15.0\n",
+ "d0=10.0\n",
+ "y=15.0\n",
+ "X=15.0\n",
+ "R=10.0\n",
+ "\n",
+ "#Calculations\n",
+ "Id=(3*math.sqrt(2)*120*(math.cos(math.radians(a))-math.cos(math.radians(d0+y)))*1000.0)/((R+(3.0*2*X)/math.pi)*math.pi)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Id=%.2f amp.\" %Id)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Id=249.99 amp.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter06.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter06.ipynb new file mode 100755 index 00000000..7c65aca2 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter06.ipynb @@ -0,0 +1,248 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 06 : Corona"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1, Page No 142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "t=21.0 # air temperature\n",
+ "b=73.6 # air pressure\n",
+ "do=3.92*73.6/(273+t)\n",
+ "m=0.85\n",
+ "r=0.52\n",
+ "d=250.0\n",
+ "\n",
+ "#Calculations\n",
+ "Vd=21.1*m *do*r*math.log(250/.52)\n",
+ "vd=math.sqrt(3)*Vd\n",
+ "m=0.7\n",
+ "vv=21.1*m*do*r*(1+(0.3/math.sqrt(r*do)))*math.log(250/0.52)\n",
+ "Vv=vv*math.sqrt(3)\n",
+ "Vvg=Vv*0.8/0.7\n",
+ "\n",
+ "#Results\n",
+ "print(\"critical disruptive line to line voltage=%.2f kV \" %vd)\n",
+ "print(\"visual critical voltage for local corona=%.2f kV \" %vv)\n",
+ "print(\"visual critical voltage for general corona=%.2f kV \" %Vvg)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "critical disruptive line to line voltage=97.89 kV \n",
+ "visual critical voltage for local corona=66.09 kV \n",
+ "visual critical voltage for general corona=130.83 kV \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.2, Page No 142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables \n",
+ "d=2.5 \n",
+ "di=3.0 # internal diameter\n",
+ "do=9.0 # external diameter\n",
+ "ri=di/2.0 # internal radius\n",
+ "ro=do/2.0 # external diameter\n",
+ "\n",
+ "#Calculations\n",
+ "g1max=20/(1.25*math.log(ri/(d/2))+0.208*1.5*math.log(ro/ri)) \n",
+ "\n",
+ "#Results\n",
+ "print(\"g1max=%.0f kV/cm\" %g1max) \n",
+ "print(\"Since the gradient exceeds 21.1/kV/cm , corona will be present.\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "g1max=35 kV/cm\n",
+ "Since the gradient exceeds 21.1/kV/cm , corona will be present.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.3, Page No 143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "m=1.07\n",
+ "r=0.625\n",
+ "\n",
+ "#Calculations\n",
+ "V=21*m *r*math.log(305.0/0.625)\n",
+ "Vl=V*math.sqrt(3.0)\n",
+ "\n",
+ "#Results\n",
+ "print(\"critical disruptive voltage=%.0f kV \" %V)\n",
+ "print(\"since operating voltage is 110 kV , corona loss= 0 \")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "critical disruptive voltage=87 kV \n",
+ "since operating voltage is 110 kV , corona loss= 0 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4 Page No 143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "r=0.5\n",
+ "\n",
+ "#Calculations\n",
+ "V=21*r*math.log(100.0/0.5)\n",
+ "\n",
+ "#Results\n",
+ "print(\"critical disruptive voltage=%.1f kV\" %V)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "critical disruptive voltage=55.6 kV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5, Page No 146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "D=1.036 \t # conductor diameter(cm)\n",
+ "d=2.44\t #delta spacing(m)\n",
+ "r=D/2 \t #radius(cm)\n",
+ "\n",
+ "#Calculations\n",
+ "Ratio=d*100.0/r \n",
+ "j=r/(d*100.0) \n",
+ "Rat2=math.sqrt(j) \n",
+ "t=26.67 \t #temperature\n",
+ "b=73.15 # barometric pressure\n",
+ "mv=0.72 \n",
+ "V=63.5 \n",
+ "f=50.0\t #frequency\n",
+ "do=3.92*b/(273+t) #do=dell\n",
+ "vd=21.1*.85*do*r*math.log(Ratio) \n",
+ "print(\"critical disruptive voltage=%.2f kV\" %vd) \n",
+ "Vv=21.1*mv*do*r*(1+(0.3/math.sqrt(r*do)))*math.log(Ratio) \n",
+ "Pl=241*(10**-5)*(f+25)*Rat2*((V-vd)**2)/do #power loss\n",
+ "Vd=0.8*vd \n",
+ "Pl2=241*(10**-5)*(f+25)*Rat2*((V-Vd)**2)*160/do #loss per phase /km\n",
+ "Total=3.0*Pl2\n",
+ "\n",
+ "#Results \n",
+ "print(\"visual critical voltage=%.0f kV\" %Vv) \n",
+ "print(\"Power loss=%.3f kW/phase/km\" %Pl) \n",
+ "print(\"under foul weather condition ,\") \n",
+ "print(\"critical disruptive voltage=%.2f kV\" %Vd) \n",
+ "print(\"Total loss=%.0f kW\\n\"%Total) \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "critical disruptive voltage=54.72 kV\n",
+ "visual critical voltage=66 kV\n",
+ "Power loss=0.672 kW/phase/km\n",
+ "under foul weather condition ,\n",
+ "critical disruptive voltage=43.77 kV\n",
+ "Total loss=1626 kW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter07.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter07.ipynb new file mode 100755 index 00000000..6d47fe5f --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter07.ipynb @@ -0,0 +1,214 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 07 : Mechanical Design of Transmission Lines"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1, Page No 161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "sf=5.0\t #Factor of safety\n",
+ "d=0.95\t # conductor dia(cm)\n",
+ "\n",
+ "#Calculations\n",
+ "Ws=4250.0/sf # working stress(kg/cm_2)\n",
+ "A=math.pi*(d**2)/4.0 # area (cm_2)\n",
+ "Wp=40.0*d*(10**-2) #wind pressure (kg/cm)\n",
+ "W=math.sqrt((.65**2)+(0.38**2)) # Total effective weight(kg/m)\n",
+ "T=850.0*A # working tension (kg)\n",
+ "c=T/W\n",
+ "l=160.0\n",
+ "d=l**2/(8*800)\n",
+ "\n",
+ "#Results\n",
+ "print(\"sag, d=%.0f metres \" %d)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "sag, d=4 metres \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2, Page No 161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "D=1.95 + 2.6\t\t\t# overall diameter(cm)\n",
+ "A=4.55*(10**-2)\t\t\t# area(m_2)\n",
+ "d=19.5\t\t\t\t\t#diameter of conductor(mm)\n",
+ "r=d/2.0\t\t\t\t\t#radius of conductor(mm)\n",
+ "\n",
+ "#Calculations\n",
+ "Wp=A*39 #wind pressure(kg/m_2)\n",
+ "t=13 #ice coating(mm)\n",
+ "US=8000.0 # ultimate strength(kg)\n",
+ "Aice=math.pi*(10**-6)*((r+t)**2 - r**2)#area section of ice (m_2)\n",
+ "Wice=Aice*910\n",
+ "W=(math.sqrt((.85+Wice)**2 + Wp**2))# total weight of ice (kg/m)\n",
+ "T=US/2.0 # working teansion (kg)\n",
+ "c=T/W\n",
+ "l=275 #length of span(m)\n",
+ "Smax=l*l/(8*c)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Maximum sag=%.1f metres\\n\" %Smax)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum sag=6.4 metres\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3, Page No 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "#initialisation of variables\n",
+ "A=13.2\t\t\t\t# cross section of conductor (mm_2)\n",
+ "Ar=4.1*(10**-3)\t\t# projected area\n",
+ "Wp=Ar*48.82\t\t\t# wind loadind /m(kg/m)\n",
+ "w=0.115\n",
+ "\n",
+ "#Calculations\n",
+ "W=math.sqrt((.1157**2)+(Wp**2))# effective loading per metre(kg)\n",
+ "q1=W/0.115\n",
+ "b=w/A\n",
+ "f1=21.0\t\t#working stress\n",
+ "T1=f1*A\n",
+ "c=T1/W\n",
+ "l=45.7\n",
+ "S=l*l/(8*c)\n",
+ "dT=32.2-4.5# difference in temperature\n",
+ "E=1.26*(10000)\n",
+ "a=16.6*(10**-6)\n",
+ "d=8.765*(10**-3)\n",
+ "K=f1-((l*d*q1)**2)*E/(24*f1*f1)\n",
+ "p=numpy.polynomial.polynomial.polyval3d(-84.23,0,-14.44,1)\n",
+ "r=numpy.roots(p)\n",
+ "f2= 14.823332# accepted value of f2\n",
+ "T=f2*A\n",
+ "c=T/w\n",
+ "d1=l*l/(8*c)\n",
+ "\n",
+ "#Results\n",
+ "print(\"sag at 32.2 Celsius , d=%.4f metres\" %d1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "sag at 32.2 Celsius , d=0.1534 metres\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4 Page No 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "T=2000.0 # working tension (kg)\n",
+ "w=1.0 \n",
+ "c=T/w \n",
+ "h=90-30 \n",
+ "l=300.0 #span(m)\n",
+ "\n",
+ "#Calculations\n",
+ "a=(l/2)-(c*h/l) \n",
+ "b=550.0 \n",
+ "d1=a*a/(2*c) \n",
+ "d2=(400**2)/(2*c) # sag at 400 metres(m)\n",
+ "Hm=d2-d1 #height of mid point with respect to A\n",
+ "Cl=30+Hm\n",
+ "\n",
+ "#Results \n",
+ "print(\"The clearance between the conductor and water level midway between the towers= %.3f metres \" %Cl) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The clearance between the conductor and water level midway between the towers= 54.375 metres \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter08.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter08.ipynb new file mode 100755 index 00000000..c168b80f --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter08.ipynb @@ -0,0 +1,60 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 08 : Overhead Line Insulators"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1, Page No 183"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "E3=17.5\n",
+ "\n",
+ "#Calculations\n",
+ "E1=64*E3/89.0\n",
+ "E2=9*E1/8.0\n",
+ "E=E1+E2+E3\n",
+ "\n",
+ "#Results\n",
+ "print(\"the maximum voltage that the string of the suspension insulators can withstand=%.2f kV\\n\" %E)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum voltage that the string of the suspension insulators can withstand=44.24 kV\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter09.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter09.ipynb new file mode 100755 index 00000000..c026f288 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter09.ipynb @@ -0,0 +1,358 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 09 : Insulated Cables"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.1, Page No 196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "V=85.0 # working voltage (kV)\n",
+ "gmax=65.0 # dielectric strength of insulating material (kV/cm)\n",
+ "\n",
+ "#Calculations\n",
+ "r=V/gmax\n",
+ "d=2*r\n",
+ "D=2.6*math.e\n",
+ "\n",
+ "#Results\n",
+ "print(\"Diameter of the sheath =%.2f cm\\n\" %D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of the sheath =7.07 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2, Page No 200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "e1=4.0\n",
+ "e2=4.0\n",
+ "e3=2.5\n",
+ "g1max=50.0\n",
+ "g2max=40.0\n",
+ "g3max=30.0\n",
+ "r=0.5\t\t# radius (cm)\n",
+ "\n",
+ "#Calculations\n",
+ "r1=r*e1*g1max/(e2*g2max)\n",
+ "r2=r1*e2*g2max/(e3*g3max)\n",
+ "V=66.0\n",
+ "lnc=(V-((r*g1max*math.log(r1/r))+(r1*g2max*math.log(r2/r1))))\n",
+ "m=lnc/(r2*g3max)\n",
+ "R=r2*(math.e**m)\n",
+ "D=2*R\n",
+ "\n",
+ "#Results\n",
+ "print(\"minimum internal diameter of the lead sheath,D=%.2f cms \" %D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "minimum internal diameter of the lead sheath,D=7.52 cms \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3, Page No 202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "r=0.5 \t#radius of conductor(cm)\n",
+ "g1max=34.0\n",
+ "er=5.0\n",
+ "r1=1.0\n",
+ "R=7.0/2 #external dia(cm)\n",
+ "\n",
+ "#Calculations\n",
+ "g2max=(r*g1max)/(er*r1)\n",
+ "V=((r*g1max*math.log(r1/r))+(r1*g2max*math.log(R/r1)))\n",
+ "V=V/(math.sqrt(2.0))\n",
+ "\n",
+ "#Results\n",
+ "print(\"Maximum safe working volltage ,V =%.2f kV r.m.s\\n\" %V)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum safe working volltage ,V =11.34 kV r.m.s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4 Page No 202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "r=0.9\n",
+ "r1=1.25\n",
+ "\n",
+ "#Calculations\n",
+ "r2=r1+.35\n",
+ "r3=r2+.35 # radius of outermost layer\n",
+ "Vd=20.0 # voltage difference (kV)\n",
+ "g1max=Vd/(r*math.log(r1/r))\n",
+ "g2max=Vd/(r1*math.log(r2/r1))\n",
+ "g3max=(66-40)/(r2*math.log(r3/r2))\n",
+ "\n",
+ "#Results\n",
+ "print(\"g1max =%.1f kV/cm\" %g1max)\n",
+ "print(\"g2max =%.2f kV/cm\" %g2max)\n",
+ "print(\"g3max =%.0f kV/cm\" %g3max)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "g1max =67.6 kV/cm\n",
+ "g2max =64.81 kV/cm\n",
+ "g3max =82 kV/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5, Page No 206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "V=20.0 #voltage (kV)\n",
+ "w=314.0\n",
+ "\n",
+ "#Calculations\n",
+ "C=2*3.04*10**-6\t\t\t#capacitance per phase(micro-farad)\n",
+ "KVA=V*V*w*C*1000.0\n",
+ "\n",
+ "#Results\n",
+ "print(\"3-phase kVA required =%.0f kVA\" %KVA) \t#Answer don't match due to difference in rounding off of digits"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "3-phase kVA required =764 kVA\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6, Page No 206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "C1=0.208\n",
+ "C2=0.096\n",
+ "Cx=3.0*C1\n",
+ "w=314.0\n",
+ "V=10.0\n",
+ "\n",
+ "#Calculations\n",
+ "Cy=(C1+ 2*C2)\n",
+ "Co=((1.5*Cy)-(Cx/6))\n",
+ "C=Co/2.0\n",
+ "\n",
+ "#Results\n",
+ "print(\"(i)Capacitance between any two conductors=%.3f micro-Farad/km\" %C)\n",
+ "c=((2*C2 + ((2/3)*C1)))\n",
+ "print(\"(ii)Capacitance between any two bunched conductors and the third conductor=%.2f micro-Farad/km\" %c)\n",
+ "I=V*w*Co*1000*(10**-6)/math.sqrt(3)\n",
+ "print(\"(iii)the charging current per phase per km =%.3f A\" %I)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)Capacitance between any two conductors=0.248 micro-Farad/km\n",
+ "(ii)Capacitance between any two bunched conductors and the third conductor=0.19 micro-Farad/km\n",
+ "(iii)the charging current per phase per km =0.899 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7 Page No 213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "rm=(2.28/2)-(.152/2)# mean radius of sheath (cm)\n",
+ "d=5.08\n",
+ "a=d/rm\n",
+ "w=314.0\n",
+ "\n",
+ "#Calculations\n",
+ "Xm=2*(10**-7)*math.log(a) # mutual inductance (H/m)\n",
+ "Xm2=2000*Xm\n",
+ "V=w*Xm2*400\n",
+ "\n",
+ "#Results\n",
+ "print(\"Voltage induced =%.2f volts \\n\" %V)#Answer don't match exactly due to difference in rounding off of digits i between calculations"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage induced =78.54 volts \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8, Page No 214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "R=2*0.1625\n",
+ "Rs=2*2.14\n",
+ "M=314.0\n",
+ "\n",
+ "#Calculations\n",
+ "w=6.268*10**-4\n",
+ "r=Rs*M*M*w*w/(R*((Rs**2)+(M*M*w*w)))\n",
+ "\n",
+ "#Results\n",
+ "print(\"ratio=%.4f \" %r)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ratio=0.0278 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter10.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter10.ipynb new file mode 100755 index 00000000..4e68db1a --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter10.ipynb @@ -0,0 +1,230 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10 : Voltage Control "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1, Page No 235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "V=1.0 \t#voltage (p.u)\n",
+ "Pa=0.5\t\t#active power at A (p.u)\n",
+ "Pr=0.375\t# reactive power at A(p.u)\n",
+ "\n",
+ "#Calculations\n",
+ "Xca=0.075+0.04\t# reactance between C and A \n",
+ "Pl=((Pa**2)+(Pr**2))*Xca/(V**2)\n",
+ "pac=1.5\n",
+ "prc=2.0\n",
+ "Pta=0.5+1.5\t\t# total active power between E and C \n",
+ "Ptr=Pr+Pl+2.0\t# reactive power between E and C \n",
+ "Xt=.05+.025\t\t#total reactance beteween E an C \n",
+ "Pl2=((2*2)+(2.4199**2))# loss (p.u)\n",
+ "Pat=200.0\n",
+ "Prt=315.9\n",
+ "pf=0.5349\n",
+ "\n",
+ "#Results\n",
+ "print(\"Total active power supplied by generator =%.0f MW \" %Pat)\n",
+ "print(\"Total reactive power supplied by generator =%.1f MW \" %Prt)\n",
+ "print(\"p.f of the generator =%.4f \\n\" %pf)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total active power supplied by generator =200 MW \n",
+ "Total reactive power supplied by generator =315.9 MW \n",
+ "p.f of the generator =0.5349 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2, Page No 236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "l1=150\n",
+ "tstr=1\n",
+ "load2=72.65\n",
+ "R=30.0\n",
+ "\n",
+ "#Calculations\n",
+ "P=(l1*(10**6))/3\n",
+ "X=80\n",
+ "Q=(load2*(10**6))/3\n",
+ "Vs=(230*(10**3))/math.sqrt(3)\n",
+ "Vr=Vs\n",
+ "ts2=1/(1-(((R*P)+(X*Q))/(Vs*Vr)))\n",
+ "ts=math.sqrt(ts2)\n",
+ "\n",
+ "#Results\n",
+ "print(\"ts=%.2f p.u\\n\" %ts)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ts=1.11 p.u\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.3, Page No 242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "A=0.895\n",
+ "Vr=215.0\n",
+ "B=182.5\n",
+ "\n",
+ "#Calculations\n",
+ "x=A*(Vr**2)/B\n",
+ "y=78.6-1.4//B-A\n",
+ "p=math.degrees(math.acos(0.9))\n",
+ "X1=x/50.0\n",
+ "Vs=265*182.5/215\n",
+ "Vr1=Vs/A\n",
+ "Reg=100*(Vr1-Vr)/Vr\n",
+ "print(\"(i) sending end voltage (kV)=%.1f kV\" %Vs)\n",
+ "print(\"recieving end voltage =%.0f kV\" %Vr1)\n",
+ "print(\"Regulation = %.2f percent\" %Reg)\n",
+ "Vs1=236.0\n",
+ "Q=Vs1*Vr/B\n",
+ "QP=0.25*50\n",
+ "PR=0.50*50\n",
+ "cosQ=0.958\n",
+ "\n",
+ "#Results\n",
+ "print(\"(ii)QP(MVAr)=%.1f MV Ar\" %QP)\n",
+ "print(\" PR(MVAr)=%.0f MV Ar\" %PR)\n",
+ "print(\"CosQ=%.3f \\n\" %cosQ)\n",
+ "MN=4.55\n",
+ "Sbmax=MN*50\n",
+ "print(\"maximum power transmitted =%.1f MW\" %Sbmax)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) sending end voltage (kV)=224.9 kV\n",
+ "recieving end voltage =251 kV\n",
+ "Regulation = 16.90 percent\n",
+ "(ii)QP(MVAr)=12.5 MV Ar\n",
+ " PR(MVAr)=25 MV Ar\n",
+ "CosQ=0.958 \n",
+ "\n",
+ "maximum power transmitted =227.5 MW\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.4, Page No 244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "#initialisation of variables\n",
+ "a=0\n",
+ "b=73.3\n",
+ "A=1.0\n",
+ "B=20.88\n",
+ "Vs=66.0\n",
+ "Vr=66.0\n",
+ "Load=75.0\n",
+ "\n",
+ "#Calculations\n",
+ "p=numpy.polynomial.polynomial.polyval2d(14624,400,1)\n",
+ "r=numpy.roots(p)\n",
+ "Qr=- 40.701538\n",
+ "C=-Qr + (75*.6/.8)\n",
+ "Smax=(Vr**2)*(1-math.cos(math.radians(b)))/B\n",
+ "\n",
+ "#Results\n",
+ "print(\"The phase modifier capacity =%.2f MV Ar\" %C)\n",
+ "print(\"Maximum power transmitted ,Pmax =%.2f MW\" %Smax)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The phase modifier capacity =96.95 MV Ar\n",
+ "Maximum power transmitted ,Pmax =148.67 MW\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter11.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter11.ipynb new file mode 100755 index 00000000..6e179df1 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter11.ipynb @@ -0,0 +1,102 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 : Neutral Grounding"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.1, Page No 251"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "C1=2*math.pi*(10**-9)/(36*math.pi*math.log((4*4*8)**(1.0/3)/(10*(10**-3))))\n",
+ "\n",
+ "#Calculations\n",
+ "C=C1*192*(10**9) #capacitance per phase (micro farad)\n",
+ "L=(10)**6.0/(3*314*314*C)\n",
+ "V=132.0 #voltage (kV)\n",
+ "MVA=V*V/(3.0*314*L)\n",
+ "\n",
+ "#Results\n",
+ "print(\"inductance ,L=%.2f H\" %L)\n",
+ "print(\"MVA rating of suppressor coil =%.3f MVA per coil\" %MVA)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "inductance ,L=1.97 H\n",
+ "MVA rating of suppressor coil =9.379 MVA per coil\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.2, Page No 252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "wL=1.0/(3*314*(10)**-6)\n",
+ "\n",
+ "#Calculations\n",
+ "print(\"(i)inductive reactance for 100 percent of the length of line=%.1f ohms\" %wL)\n",
+ "wL=10.0**6/(3*314*.9)\n",
+ "\n",
+ "#Results\n",
+ "print(\"(ii)inductive reactance for 90percent of the length of line=%.1f ohms\" %wL)\n",
+ "wL=1.0/(3*314*(10)**-6)/.8\n",
+ "print(\"(iii)inductive reactance for 80 percent of the length of line=%.1f ohms\" %wL)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)inductive reactance for 100 percent of the length of line=1061.6 ohms\n",
+ "(ii)inductive reactance for 90percent of the length of line=1179.5 ohms\n",
+ "(iii)inductive reactance for 80 percent of the length of line=1327.0 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter12.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter12.ipynb new file mode 100755 index 00000000..d1d032ac --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter12.ipynb @@ -0,0 +1,302 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12 : Thyristors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.1, Page No 278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "L=2*(10**-7)*math.log(100/.75)\t\t#inductance per unit length\n",
+ "C=2*math.pi*(10**-9)/(36*math.pi*math.log(100.0/0.75))\t#Capacitance per phase per unit length (F/m)\n",
+ "Z1=math.sqrt(L/C)\n",
+ "E=11000.0\n",
+ "\n",
+ "#Calculations\n",
+ "print(\"(i)the natural impedence of line=%.0f ohms\" %Z1)\n",
+ "Il=E/(math.sqrt(3)*Z1) #line current(amps)\n",
+ "print(\"(ii)line current =%.1f amps\" %Il)\n",
+ "R=1000.0\n",
+ "Z2=R\n",
+ "E1=2*Z2*E/((Z1+Z2)*math.sqrt(3))\n",
+ "Pr=3*E1*E1/(R*1000) #Rate of power consumption\n",
+ "Vr=(Z2-Z1)*E/(math.sqrt(3)*(Z2+Z1)*1000)\t\t#Reflected voltage\n",
+ "Er=3*Vr*Vr*1000/Z1\t\t\t\t\t\t\t\t#rate of reflected voltage\n",
+ "print(\"(iii)rate of energy absorption =%.1f kW\" %Pr)\n",
+ "print(\"rate of reflected energy =%.1f kW\" %Er)\n",
+ "print(\"(iv)Terminating resistance should be equal to surge impedence of line =%.0f ohms\" %Z1)\n",
+ "L=.5*(10**-8)\n",
+ "C=10**-12\n",
+ "Z=math.sqrt(L/C)\t\t# surge impedence\n",
+ "VR=2*Z*11/((Z1+Z)*math.sqrt(3))\n",
+ "Vrl=(Z-Z1)*11/((Z1+Z)*math.sqrt(3))\n",
+ "PR1=3*VR*VR*1000/(Z)\n",
+ "d=Vrl\n",
+ "Prl=3*d*d*1000.0/Z1\n",
+ "\n",
+ "#Results\n",
+ "print(\"(v)Refracted power =%.1f kW\" %PR1)\n",
+ "print(\"Reflected power =%.1f kW\" %Prl)\n",
+ "##Answer don't match exactly due to difference in rounding off of digits i between calculations\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)the natural impedence of line=294 ohms\n",
+ "(ii)line current =21.6 amps\n",
+ "(iii)rate of energy absorption =289.2 kW\n",
+ "rate of reflected energy =122.9 kW\n",
+ "(iv)Terminating resistance should be equal to surge impedence of line =294 ohms\n",
+ "(v)Refracted power =257.9 kW\n",
+ "Reflected power =154.3 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.2, Page No 280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Xlc=0.3*(10**-3)\t\t# inductance of cable(H)\n",
+ "Xcc=0.4*(10**-6)\t\t# capacitance of cable (F)\n",
+ "Xlo=1.5*(10**-3)\t\t#inductance of overhead line(H)\n",
+ "Xco=.012*(10**-6)\t\t# capacitance of overhead line (F)\n",
+ "\n",
+ "#Calculations\n",
+ "Znc=math.sqrt((Xlc/Xcc))\n",
+ "Znl=math.sqrt((Xlo/Xco))\n",
+ "\n",
+ "#Results\n",
+ "print(\"Natural impedence of cable=%.2f ohms \" %Znc)\n",
+ "print(\"Natural impedence of overhead line=%.1f ohms \" %Znl)\n",
+ "E=2*Znl*15/(353+27)\n",
+ "print(\"voltage rise at the junction due to surge =%.2f kV \" %E)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Natural impedence of cable=27.39 ohms \n",
+ "Natural impedence of overhead line=353.6 ohms \n",
+ "voltage rise at the junction due to surge =27.91 kV \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.3, Page No 280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Z1=600.0\n",
+ "Z2=800.0\n",
+ "Z3=200.0\n",
+ "E=100.0\n",
+ "\n",
+ "#Calculations\n",
+ "E1=2*E/(Z1*((1/Z1)+(1/Z2)+(1/Z3)))\n",
+ "Iz2=E1*1000.0/Z2\n",
+ "Iz3=E1*1000.0/Z3\n",
+ "\n",
+ "#Results\n",
+ "print(\"Transmitted voltage =%.2f kV \" %E1)\n",
+ "print(\"The transmitted current in line Z2=%.2f amps \" %Iz2)\n",
+ "print(\"The transmitted current in line Z3=%.1f amps \" %Iz3)\n",
+ "\n",
+ "#Answer don't match exactly due to difference in rounding off of digits i between calculations"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Transmitted voltage =42.11 kV \n",
+ "The transmitted current in line Z2=52.63 amps \n",
+ "The transmitted current in line Z3=210.5 amps \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.4 Page No 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Z=350.0 \t#surge impedencr (ohms)\n",
+ "\n",
+ "#Calculations\n",
+ "C=3000.0*(10**-12)\t# earth capacitance(F) \n",
+ "t=2.0*(10**-6)\n",
+ "E=500.0\n",
+ "E1=2*E*(1-math.exp((-1*t/(Z*C))))\n",
+ "\n",
+ "#Results\n",
+ "print(\"the maximum value of transmitted voltage=%.2f kV \" %E1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum value of transmitted voltage=851.14 kV \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.5, Page No 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Z=350.0 \t\t#surge impedencr (ohms)\n",
+ "L=800*(10**-6) \n",
+ "t=2*(10**-6)\n",
+ "E=500.0\n",
+ "\n",
+ "#Calculations\n",
+ "E1=E*(1-math.exp((-1*t*2*Z/L)))\n",
+ "\n",
+ "#Results\n",
+ "print(\"The maximum value of transmitted voltage=%.1f kV \" %E1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum value of transmitted voltage=413.1 kV \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.6, Page No 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "eo=50.0\n",
+ "x=50.0\n",
+ "R=6.0\n",
+ "Z=400.0\n",
+ "G=0\n",
+ "v=3*(10**5)\n",
+ "e=2.68\n",
+ "\n",
+ "#Calculations\n",
+ "e1=(eo*(e**((-1/2)*R*x/Z)))\n",
+ "# answess does not match due to the difference in rounding off of digits. \n",
+ "print(\"(i)the value of the Voltage wave when it has travelled through a distance 50 Km=%.1f kV \" %e1)\n",
+ "Pl=e1*e1*1000.0/400\n",
+ "io=eo*1000.0/Z\n",
+ "t=x/v\n",
+ "H=-(50*125*400*((e**-0.75)-1))/(6.0*3*10**5)\n",
+ "\n",
+ "#Results\n",
+ "print(\"(ii)Power loss=%.3fkW \\n heat loss=%.3f kJ\" %(Pl,H))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)the value of the Voltage wave when it has travelled through a distance 50 Km=23.9 kV \n",
+ "(ii)Power loss=1424.550kW \n",
+ " heat loss=0.726 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter13.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter13.ipynb new file mode 100755 index 00000000..b99fc5a9 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter13.ipynb @@ -0,0 +1,868 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13 : Symmetrical Components and fault calculations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.1, Page No 302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Va=100.0*complex(math.cos(math.radians(0)),math.sin(math.radians(0)))\n",
+ "Vb=33.0*complex(math.cos(math.radians(-100)),math.sin(math.radians(-100)))\n",
+ "Vc=38.0*complex(math.cos(math.radians(176.5)),math.sin(math.radians(176.5)))\n",
+ "L=1.0*(math.cos(math.radians(120)) + math.sin(math.radians(120)))\n",
+ "\n",
+ "#Calculations\n",
+ "Va1=((Va + L*Vb + (L**2)*Vc))/3\n",
+ "Va2=((Va + L*Vc + (L**2)*Vb))/3\n",
+ "Vco=((Va + Vb + Vc))/3\n",
+ "\n",
+ "#Results\n",
+ "print( \"Va1= {0:.5f}+{1:.5f}i\".format(Va1.real, Va1.imag))\n",
+ "print( \"Va2= {0:.5f}+{1:.5f}i\".format(Va2.real, Va2.imag))\n",
+ "print( \"Vco= {0:.5f}+{1:.5f}i\".format(Vco.real, Vco.imag))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Va1= 30.94033+-3.86151i\n",
+ "Va2= 28.44975+-1.16829i\n",
+ "Vco= 18.78016+-10.05960i\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.2, Page No 303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ia=complex(500,150)\t\t# Line current in phase a\n",
+ "Ib=complex(100,-600)\t# Line current in phase b\n",
+ "Ic=complex(-300,600)\t# Line current in phase c\n",
+ "\n",
+ "#Calculations\n",
+ "L=complex(math.cos(math.radians(120)),math.sin(math.radians(120)))\n",
+ "Iao=(Ia+Ib+Ic)/3\n",
+ "Ia1=(Ia+Ib*L+(L**2)*Ic)/3\n",
+ "Ia2=(Ia+(L**2)*Ib +(L*Ic))/3\n",
+ "\n",
+ "#Results\n",
+ "print( \"Iao(amps)= {0:.5f}+{1:.5f}i\".format(Iao.real, Iao.imag))\n",
+ "print( \"Ia1(amps)= {0:.5f}+{1:.5f}i\".format(Ia1.real, Ia1.imag))\n",
+ "print( \"Ia2(amps)= {0:.5f}+{1:.5f}i\".format(Ia2.real, Ia2.imag))\n",
+ "# Answer in the book is not correct.wrong calculation in the book"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Iao(amps)= 100.00000+50.00000i\n",
+ "Ia1(amps)= 546.41016+165.47005i\n",
+ "Ia2(amps)= -146.41016+-65.47005i\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.3, Page No 314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ea=1.0\n",
+ "Z1=complex(0.25)\n",
+ "Z2=complex(.35)\n",
+ "Zo=complex(0.1)\n",
+ "Ia1=Ea/(Z1+Z2+Zo)\n",
+ "L=-complex(0.5,0.866)\n",
+ "Ia2=Ia1\n",
+ "Iao=Ia2\n",
+ "\n",
+ "#Calculations\n",
+ "Ia=Ia1+Ia2+Iao\n",
+ "Ib=25*1000/((math.sqrt(3)*13.2))\n",
+ "If=Ib*abs(Ia)\n",
+ "Va1=Ea-(Ia1*Z1)\n",
+ "Va2=-Ia2*Z2\n",
+ "Va0=-Iao*Zo\n",
+ "Va=Va1+Va2+Va0\n",
+ "Vb1=(L**2)*Va1\n",
+ "Vb2=L*Va2\n",
+ "Vbo=Va0\n",
+ "Vco=Va0\n",
+ "Vc1=L*Va1\n",
+ "Vc2=(L**2)*Va2\n",
+ "Vb=Vb1+Vb2+Vbo\n",
+ "Vc=Vco+Vc1+Vc2\n",
+ "Vab=Va-Vb\n",
+ "Vac=Va-Vc\n",
+ "Vbc=Vb-Vc\n",
+ "vab=(13.2*abs(Vab))/math.sqrt(3)\n",
+ "vac=(13.2*abs(Vac))/math.sqrt(3)\n",
+ "vbc=(13.2*abs(Vbc))/math.sqrt(3)\n",
+ "\n",
+ "#Results\n",
+ "print(\"fault current (amps)= %.2f\" %If)#Answer don't match due to difference in rounding off of digits\n",
+ "print(\"Vab(kV)= {0:.5f}+{1:.5f}i\" .format(Vab.real, Vab.imag))\t\t#Answer don't match due to difference in rounding off of digits\n",
+ "print(\"Vac(kV)= {0:.5f}+{1:.5f}i\" .format(Vac.real, Vac.imag))\t\t#Answer don't match due to difference in rounding off of digits\n",
+ "print(\"Vbc(kV)= {0:.5f}+{1:.5f}i\" .format(Vbc.real, Vbc.imag))\t\t#Answer don't match due to difference in rounding off of digits\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fault current (amps)= 4686.28\n",
+ "Vab(kV)= 0.21426+-0.98971i\n",
+ "Vac(kV)= 0.21431+0.98971i\n",
+ "Vbc(kV)= 0.00005+1.97943i\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.4 Page No 318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ea=1.0\n",
+ "L=complex(math.cos(math.radians(120)),math.sin(math.radians(120)))\n",
+ "Z1=complex(0.25)\n",
+ "Z2=complex(0.35)\n",
+ "Ia1=Ea/(Z1+Z2)\n",
+ "Ia2=-Ia1\n",
+ "Iao=0\n",
+ "\n",
+ "#Calculations\n",
+ "Ib1=(L**2)*Ia1\n",
+ "Ib2=L*Ia2\n",
+ "Ibo=0\n",
+ "Ib=Ib1+Ib2 +Ibo\n",
+ "Iba=1093\n",
+ "If=Iba*abs(Ib)\n",
+ "Va1=Ea-(Ia1*Z1)\n",
+ "Va2=-Ia2*Z2\n",
+ "Vao=0\n",
+ "Va=Va1+Va2+Vao\n",
+ "Vb=(L**2)*Va1+L*Va2\n",
+ "Vc=Vb\n",
+ "Vab=Va-Vb\n",
+ "Vac=Va-Vc\n",
+ "Vbc=Vb-Vc\n",
+ "\n",
+ "#Results\n",
+ "vab=(abs(Vab)*13.2)/math.sqrt(3)\n",
+ "vbc=(abs(Vbc)*13.2)/math.sqrt(3)\n",
+ "vac=(abs(Vac)*13.2)/math.sqrt(3)\n",
+ "print(\"fault current (amps)= %.2f\" %If)#Answer don't match due to difference in rounding off of digits\n",
+ "print(\"Vab(kV)= {0:.5f}+{1:.5f}i\" .format(vab.real, vab.imag))\t\t#Answer don't match due to difference in rounding off of digits\n",
+ "print(\"Vac(kV)= {0:.5f}+{1:.5f}i\" .format(vac.real, vac.imag))\t\t#Answer don't match due to difference in rounding off of digits\n",
+ "print(\"Vbc(kV)= {0:.5f}+{1:.5f}i\" .format(vbc.real, vbc.imag))\t\t#Answer don't match due to difference in rounding off of digits"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fault current (amps)= 3155.22\n",
+ "Vab(kV)= 13.33679+0.00000i\n",
+ "Vac(kV)= 13.33679+0.00000i\n",
+ "Vbc(kV)= 0.00000+0.00000i\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.5, Page No 322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ea=complex(1,0)\n",
+ "Zo=complex(0.1)\n",
+ "Z1=complex(0.25)\n",
+ "Z2=complex(0.35)\n",
+ "\n",
+ "#Calculations\n",
+ "Ia1=Ea/(Z1+(Zo*Z2/(Zo+Z2)))\n",
+ "Va1=Ea-Ia1*Z1\n",
+ "Va2=Va1\n",
+ "Vao=Va2\n",
+ "Ia2=-Va2/Z2\n",
+ "Iao=-Vao/Zo\n",
+ "I=Ia2+Iao\n",
+ "If=3*Iao # fault current\n",
+ "Ib=1093 # base current\n",
+ "Ifl=abs(If*Ib)\n",
+ "print(\"fault current (amps)= %.2f\" %Ifl) #Answer don't match due to difference in rounding off of digits\n",
+ "Va=3*Va1\n",
+ "Vb=0\n",
+ "Vc=0\n",
+ "Vab=abs(Va)*13.2/math.sqrt(3)\n",
+ "Vac=abs(Va)*13.2/math.sqrt(3)\n",
+ "Vbc=abs(Vb)*13.2/math.sqrt(3)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Vab(kV)= {0:.5f}+{1:.5f}i\" .format(Vab.real, Vab.imag))\t\t\n",
+ "print(\"Vac(kV)= {0:.5f}+{1:.5f}i\" .format(Vac.real, Vac.imag))\t\t\n",
+ "print(\"Vbc(kV)= {0:.5f}+{1:.5f}i\" .format(Vbc.real, Vbc.imag))\t\t"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fault current (amps)= 7780.68\n",
+ "Vab(kV)= 5.42514+0.00000i\n",
+ "Vac(kV)= 5.42514+0.00000i\n",
+ "Vbc(kV)= 0.00000+0.00000i\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.6, Page No 335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vbl=13.8*115/13.2\t\t\t\t# base voltage on the line side of transformer(kV)\n",
+ "Vbm=120*13.2/115\t\t\t\t# base voltage on the motor side of transformer(kV)\n",
+ "Xt=10*((13.2/13.8)**2)*30/35.0\t# percent reactance of transformer\n",
+ "Xm=20*((12.5/13.8)**2)*30.0/20\t# percent reactance of motor\n",
+ "Xl=80*30*100.0/(120.0*120)\t\t#percent reactance of line \n",
+ "Xn=2*3*30*100/(13.8*13.8)\t\t# neutral reactance\n",
+ "Xz=200*30*100.0/(120.0*120)\n",
+ "\n",
+ "#Calculations\n",
+ "Zn=complex(0.146)\t\t\t\t# negative sequence impedence\n",
+ "Zo=.06767\t\t\t\t\t\t# zero sequence impedence\n",
+ "Z=complex(0.3596)\t\t\t\t#total impedence\n",
+ "Ia1=1.0/Z\n",
+ "Ia2=Ia1\n",
+ "Iao=Ia2\n",
+ "If1=3*Ia1\n",
+ "Ib=30*1000/(math.sqrt(3)*13.8)\n",
+ "Ibl=30*1000/(math.sqrt(3)*120)\n",
+ "Ifc=Ibl*abs(If1)\n",
+ "Z1=complex(0.146)\n",
+ "Z2=Z1\n",
+ "IA1=1.0/(Z1+Z2)\n",
+ "IA2=-IA1\n",
+ "L=complex(math.cos(math.radians(120)),math.sin(math.radians(120)))\n",
+ "IAo=0\n",
+ "IB=(L**2)*IA1 + L*IA2\n",
+ "IC=-IB\n",
+ "IF=abs(IB)*Ibl\n",
+ "Zo=complex(0.06767)\n",
+ "ia1=1/(Z1+(Zo*Z2/(Zo+Z2)))\n",
+ "ia2=ia1*Zo/(Z2+Zo)\n",
+ "iao=complex(3.553)\n",
+ "If2=3*iao\n",
+ "IF2=abs(If2*Ibl)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Fault Current (i)L-G fault, If=%.0f amps \" %Ifc)\n",
+ "print(\"(ii)L-L fault ,If=%.1f amps\" %IF)\n",
+ "print(\"(iii)L-L-G, If =%.0f amps\" %IF2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fault Current (i)L-G fault, If=1204 amps \n",
+ "(ii)L-L fault ,If=856.2 amps\n",
+ "(iii)L-L-G, If =1538 amps\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.8 Page No 342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "vx=3.0\t\t# percent reactance of the series element\n",
+ "sinr=0.6\n",
+ "\n",
+ "#Calculations\n",
+ "V=vx*sinr\n",
+ "\n",
+ "#Results\n",
+ "print(\"Percent drop of volts=%.1f percent\" %V)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percent drop of volts=1.8 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.9, Page No 342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Sb=8.0\t\t# Base MVA\n",
+ "\n",
+ "#Calculations\n",
+ "Zeq=(complex(0.15))*(complex(0.315))/(complex(0.465))\n",
+ "Scc=abs(Sb/Zeq)\n",
+ "\n",
+ "#Results\n",
+ "print(\"short circuit capacity=%.2f MVA\" %Scc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "short circuit capacity=78.73 MVA\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.10 Page No 343"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "X=1200*100.0/800\t\t# percent reactance of other generating station\n",
+ "Xc=0.5*1200/(11.0*11)\n",
+ "\n",
+ "#Calculations\n",
+ "Sc=1200*100.0/86.59\t\t# short circuit MVA of the bus\n",
+ "Xf=119.84\t\t\t\t# equivalent fault impedence between F and neutral bus \n",
+ "MVA=1200*100.0/Xf\n",
+ "\n",
+ "#Results\n",
+ "print(\"short circuit capacity of each station=%.0f MVA\" %MVA)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "short circuit capacity of each station=1001 MVA\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.11 Page No 343"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Sb=100.0 # base power (MVA)\n",
+ "\n",
+ "#Calculations\n",
+ "SC=Sb/0.14\n",
+ "\n",
+ "#Results\n",
+ "print(\"S.C. MVA =%.2f MVA \" %SC)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "S.C. MVA =714.29 MVA \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.12 Page No 344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ib=50*1000.0/(math.sqrt(3.0)*13.2)# base current (amps.)\n",
+ "Vf=12.5/13.5\t\t# the Prefault Voltage (p.u)\n",
+ "Xf=(complex(0.3))*(complex(0.2))/(complex(0.5))\t# Fault impedence(p.u)\n",
+ "\n",
+ "#Calculations\n",
+ "If=0.9469/(Xf)\t\t#Fault current (p.u)\n",
+ "Ifl=30*1000.0/((math.sqrt(3)*12.5*.8))\t#full load current (amps)\n",
+ "Il=1732*(complex(math.cos(math.radians(36.8)),math.sin(math.radians(36.8))))/2186.0\t\t#load current(p.u)\n",
+ "Ifm=3*(If)/5.0\t\t# fault current supplied by motor (p.u)\n",
+ "Ifg=2*(If)/5.0\t\t# fault current supplied by generator (p.u)\n",
+ "Ig=abs(Ifg +Il)\t\t#Net current supplied by generator during fault(p.u)\n",
+ "Im=abs(Ifm-Il)\t\t#Net current supplied by motor during fault(p.u)\n",
+ "Igf=Ig*2186\n",
+ "Imf=Im*2186\n",
+ "Ifc=2186*If\n",
+ "\n",
+ "#Results\n",
+ "print(\"Fault current from the generator =%.3f amps\" %Igf)\n",
+ "print(\"Fault current from the motor =%.3f amps\" %Imf)\n",
+ "print(\"Fault current (amps) = {0:.5f}+{1:.5f}i\" .format(Ifc.real, Ifc.imag))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fault current from the generator =8351.308 amps\n",
+ "Fault current from the motor =9022.600 amps\n",
+ "Fault current (amps) = 17249.36167+0.00000i\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.13, Page No 345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Sb=75.0 \t\t\t# Base MVA\n",
+ "\n",
+ "#Calculations\n",
+ "Xpu=0.15*Sb/15.0\t# p.u reactance of the generator\n",
+ "Xt=complex(-0.08)\t#p.u reactanceof the transformer\n",
+ "X=9.75/112.0\n",
+ "Xa=X*33*33/75.0\n",
+ "\n",
+ "#Results\n",
+ "print(\"The reactance of the reactor =%.3f ohms\" %Xa)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reactance of the reactor =1.264 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.14, Page No 346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "Z1eq= complex((((8+5)*(8+5+12.0))/(100.0*(13+25))))\n",
+ "Z2eq=Z1eq\n",
+ "Zoeq=complex((5*45)*(10^-2)/(5+45))\n",
+ "Ea=1\n",
+ "\n",
+ "#Calculations\n",
+ "Ia1=Ea/(Z1eq+ ((Zoeq*Z2eq)/(Zoeq+Z2eq)))\n",
+ "Ia2=(-Ia1*Zoeq)/(Zoeq+Z2eq)\n",
+ "Iao=(-Ia1*Z2eq)/(Zoeq+Z2eq)\n",
+ "Va1=Ea-(Ia1*Z1eq)\n",
+ "Va2=-Ia2*Z2eq\n",
+ "Vao=Va2\n",
+ "Ia=0\n",
+ "Ib=complex(-0.5 ,0.866)*Ia1 + (complex(-0.5,0.866)*Ia2) + Iao\n",
+ "Ic=complex(-0.5 ,0.866)*Ia1 + complex(-0.5 ,0.866)*Ia2 + Iao\n",
+ "ia1=Ia1*25/38\n",
+ "IA1=complex(ia1)\n",
+ "ia2=Ia2*25/38\n",
+ "IA2=complex(-ia2)\n",
+ "IA=IA1 + IA2\n",
+ "IB=IA1*complex(-0.5 ,0.866) + IA2*complex(-0.5 ,0.866)\n",
+ "IC=IA1*complex(-0.5 ,0.866) + IA2*complex(-0.5 ,0.866)\n",
+ "Va=Va1+Va2+Vao\n",
+ "Vb=0\n",
+ "Vc=0\n",
+ "Vab=.2564-Vb\n",
+ "Vbc=Vb-Vc\n",
+ "Vca=Vc-.2564\n",
+ "VA1=Ea-IA1*complex(.05)\n",
+ "VA2=-IA2*complex(0.05)\n",
+ "VA=VA1+VA2\n",
+ "VB=((complex(-0.5 ,0.866)*VA1) +(complex(-0.5 ,0.866)*VA2))\n",
+ "VC=VA1*complex(-0.5 ,0.866) + VA2*complex(-0.5 ,0.866)\n",
+ "VAB=VA-VB\n",
+ "VBC=VB-VC\n",
+ "VCA=VC-VA\n",
+ "\n",
+ "#Results\n",
+ "#Answers don't match due to difference in rounding off of digits\n",
+ "print(\"fault currents ,Ia= %.2f\" %Ia)\n",
+ "print(\"Ib= {0:.5f}+{1:.5f}i\" .format(Ib.real, Ib.imag))\t\n",
+ "print(\"Ic= {0:.5f}+{1:.5f}i\" .format(Ic.real, Ic.imag))\t\t\t#Calculation in book is wrong.\n",
+ "print(\"IA= {0:.5f}+{1:.5f}i\" .format(IA.real, IA.imag))\t\n",
+ "print(\"IB= {0:.5f}+{1:.5f}i\" .format(IB.real, IB.imag))\t\n",
+ "print(\"IC= {0:.5f}+{1:.5f}i\" .format(IC.real, IC.imag))\t\n",
+ "print(\"Voltages at fault point\")\n",
+ "print(\"Vab(p.u)= %.2f\" %Vab)\n",
+ "print(\"Vbc(p.u)= %.2f\" %Vbc)\n",
+ "print(\"Vca(p.u)= %.2f\" %Vca)\n",
+ "print(\"VAB= {0:.5f}+{1:.5f}i\" .format(VAB.real, VAB.imag))\t\n",
+ "print(\"VBC= {0:.5f}+{1:.5f}i\" .format(VBC.real, VBC.imag))\t\n",
+ "print(\"VCA= {0:.5f}+{1:.5f}i\" .format(VCA.real, VCA.imag))\t\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fault currents ,Ia= 0.00\n",
+ "Ib= 0.01390+-0.00802i\n",
+ "Ic= 0.01390+-0.00802i\n",
+ "IA= 7.69231+0.00000i\n",
+ "IB= -3.84615+6.66154i\n",
+ "IC= -3.84615+6.66154i\n",
+ "Voltages at fault point\n",
+ "Vab(p.u)= 0.26\n",
+ "Vbc(p.u)= 0.00\n",
+ "Vca(p.u)= -0.26\n",
+ "VAB= 0.92308+-0.53292i\n",
+ "VBC= 0.00000+0.00000i\n",
+ "VCA= -0.92308+0.53292i\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.15, Page No 349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ia1=complex(-0.8,-2.6) + complex(0.8,-0.4)\n",
+ "Ia2=complex(-3)\n",
+ "Iao=complex(-3)\n",
+ "A=complex(-0.8,-2.6) + complex(0.8,2)\n",
+ "a=.8\n",
+ "b=.6\n",
+ "\n",
+ "#Calculations\n",
+ "Ipf=complex(a,b)\n",
+ "Isfc=3*Ia1\n",
+ "iA1=complex(0.8,-4)\n",
+ "iA2=complex(-1)\n",
+ "iAo=0\n",
+ "IA1=complex(iA1)\n",
+ "IA2=complex(-iA2)\n",
+ "IA=IA1 + IA2\n",
+ "L=complex(math.cos(math.radians(120)),math.sin(math.radians(120)))\n",
+ "IB=(L**2)*IA1 + IA2*L\n",
+ "IC=(L**2)*IA2 + IA1*L\n",
+ "\n",
+ "#Results\n",
+ "print(\"(i) pre- fault current in line a = {0:.5f}+{1:.5f}i\" .format(Ipf.real, Ipf.imag))\t\n",
+ "print(\"(ii) the subtransient fault current in p.u= {0:.5f}+{1:.5f}i\" .format(Isfc.real, Isfc.imag))\t\n",
+ "print(\"IA= {0:.5f}+{1:.5f}i\" .format(IA.real, IA.imag))\t\n",
+ "print(\"IB= {0:.5f}+{1:.5f}i\" .format(IB.real, IB.imag))\t\n",
+ "print(\"IC= {0:.5f}+{1:.5f}i\" .format(IC.real, IC.imag))\t"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) pre- fault current in line a = 0.80000+0.60000i\n",
+ "(ii) the subtransient fault current in p.u= 0.00000+-9.00000i\n",
+ "IA= 1.80000+-4.00000i\n",
+ "IB= -4.36410+2.17321i\n",
+ "IC= 2.56410+1.82679i\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.16, Page No 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "S_C_MVA=0.5/.05\n",
+ "\n",
+ "#Results\n",
+ "print(\"S.C.MVA=%.2f MVA\" %S_C_MVA)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "S.C.MVA=10.00 MVA\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.17, Page No 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "vab=2000.0\n",
+ "vbc=2800.0\n",
+ "vca=2500.0\n",
+ "vb=2500.0\t\t# base voltage (V)\n",
+ "\n",
+ "#Calculations\n",
+ "Vab=vab/vb\t\t# per unit voltages \n",
+ "Vbc=vbc/vb\n",
+ "Vca=vca/vb\n",
+ "a=math.degrees(math.acos(((1.12**2)-((.8**2)+1))/(2*.8)))\n",
+ "b=136.11348\n",
+ "Vlab=Vab*complex(math.cos(math.radians(76.06)),math.sin(math.radians(76.06)))\t\t# line voltage \n",
+ "Vlca=Vca*complex(math.cos(math.radians(180)),math.sin(math.radians(180)))\t\t# line voltage \n",
+ "Vlbc=Vbc*complex(math.cos(math.radians(-43.9)),math.sin(math.radians(-43.9)))# line voltage \n",
+ "L=1*complex(math.cos(math.radians(120)),math.sin(math.radians(120)))\n",
+ "Vab1=(Vlab +(L*Vlbc) + ((L**2)*Vlca))/3.0 \t# symmetrical component of line voltage \n",
+ "Vab2=(Vlab +(L*Vlca) + ((L**2)*Vlbc))/3.0 # symmetrical component of line voltage \n",
+ "Vabo=0# symmetrical component of line voltage \n",
+ "Van1=Vab1*complex(math.cos(math.radians(-30)),math.sin(math.radians(-30)))\n",
+ "Van2=Vab2*complex(math.cos(math.radians(30)),math.sin(math.radians(30)))\n",
+ "Ia1=Van1/(1*complex(math.cos(math.radians(0)),math.sin(math.radians(0))))\n",
+ "Ia2=Van2/(1*complex(math.cos(math.radians(0)),math.sin(math.radians(0))))\n",
+ "VA1=complex(-Van1)\n",
+ "VA2=complex(Van2)\n",
+ "VA=VA1+ VA2\n",
+ "VB1=(L**2)*VA1\n",
+ "VB2=(L)*VA2\n",
+ "VB=VB1 + VB2\n",
+ "VC2=(L**2)*VA2\n",
+ "VC1=(L)*VA1\n",
+ "VC=VC1 + VC2\n",
+ "VAB=VA-VB\n",
+ "VBC=VB-VC\n",
+ "VCA=VC-VA\n",
+ "IA=VA\n",
+ "IB=VB\n",
+ "IC=VC\n",
+ "phase_IA=math.degrees(math.atan(IA.imag/IA.real))\n",
+ "phase_IB=math.degrees(math.atan(IB.imag/IB.real))\n",
+ "phase_IC=math.degrees(math.atan(IC.imag/IC.real))\n",
+ "\n",
+ "#Results\n",
+ "print(\"VAB= {0:.5f}+{1:.5f}i\" .format(VAB.real, VAB.imag))\t\n",
+ "print(\"VBC= {0:.5f}+{1:.5f}i\" .format(VBC.real, VBC.imag))\t\n",
+ "print(\"VCA= {0:.5f}+{1:.5f}i\" .format(VCA.real, VCA.imag))\t\n",
+ "print(\"IA(p.u)=%.2f at an agle of %.1f\" %(abs(IA),phase_IA))\n",
+ "print(\"IB(p.u)=%.2f at an agle of %.1f\" %(abs(IB),phase_IB))\n",
+ "print(\"IC(p.u)=%.2f at an agle of %.1f\" %(abs(IC),phase_IC))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "VAB= -0.77661+-1.80702i\n",
+ "VBC= -0.77644+1.19272i\n",
+ "VCA= 1.55305+0.61429i\n",
+ "IA(p.u)=1.12 at an agle of 46.1\n",
+ "IB(p.u)=1.00 at an agle of 90.0\n",
+ "IC(p.u)=0.80 at an agle of -13.9\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter14.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter14.ipynb new file mode 100755 index 00000000..f411d65f --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter14.ipynb @@ -0,0 +1,468 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14 : Protective Relayes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.1, Page No 366"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "If=4000.0\t# fault current \n",
+ "I=5*1.25\t# operating current of relay \n",
+ "\n",
+ "#Calculations\n",
+ "CT=400.0/5\t# CT ratio\n",
+ "PSM=If/(I*CT)\t# plug setting multiplier\n",
+ "\n",
+ "#Results\n",
+ "print(\"PSM =%.2f\" %PSM)\n",
+ "print(\"operating time for PSM=8 is 3.2sec.\")\n",
+ "print(\"actual operating time = 1.92 sec.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "PSM =8.00\n",
+ "operating time for PSM=8 is 3.2sec.\n",
+ "actual operating time = 1.92 sec.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.2, Page No 369"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Z=1000.0*complex(math.cos(math.radians(60)),math.sin(math.radians(60))) #impedence\n",
+ "X=math.cos(math.radians(50))*1000*math.cos(math.radians(60))\n",
+ "Xl=1000.0*math.cos(math.radians(60))\n",
+ "Xc=Xl-X\n",
+ "\n",
+ "#Calculations\n",
+ "C=1000000.0/(314.0*Xc)\n",
+ "\n",
+ "#Results\n",
+ "#Answers don't match due to difference in rounding off of digits\n",
+ "print(\"X= %.2f\" %X)\n",
+ "print(\"Xc= %.2f\" %Xc)\n",
+ "print(\"C(micro farads)= %.2f\" %C)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X= 321.39\n",
+ "Xc= 178.61\n",
+ "C(micro farads)= 17.83\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.3, Page No 384"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Isec1=4000.0/40# secondary current(amps)\n",
+ "PSM=100.0/5# PSM if 100% setting is used\n",
+ "Isec2=4000.0/40\n",
+ "PSM2=100.0/6.25#PSM if setting used is 125%\n",
+ "TMSb=0.72/2.5\n",
+ "\n",
+ "#Calculations\n",
+ "PSM1=5000.0/(6.25*40)\n",
+ "to=2.2\n",
+ "tb=to*TMSb\n",
+ "PSMa=5000/(6.25*80)\n",
+ "TMS=1.138/3\n",
+ "PSMa1=6000/(6.25*80)\n",
+ "ta=(2.6*.379)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Actual operating time of realy at b=%.3f sec\" %tb)\n",
+ "print(\"Actual operating time of realy at a=%.3f sec \" %ta)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Actual operating time of realy at b=0.634 sec\n",
+ "Actual operating time of realy at a=0.985 sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.4 Page No 399"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vph=6600/(math.sqrt(3))\n",
+ "Ifull=5000/(math.sqrt(3)*6.6)\n",
+ "\n",
+ "#Calculations\n",
+ "Ib=Ifull*.25\n",
+ "x=Ib*800.0/Vph\n",
+ "\n",
+ "#Results\n",
+ "print(\"Percent of the winding remains unprotected = %.2f \" %x)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percent of the winding remains unprotected = 22.96 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.5, Page No 399"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Iph=10000.0/math.sqrt(3)\t# phase voltage of alternator(V)\n",
+ "x=1.8*100*10*1000.0/(5*Iph)\n",
+ "\n",
+ "#Calculations\n",
+ "print(\"(i) percent winding which remains unprotected=%.2f \" %x)\n",
+ "Ip=Iph*.2\n",
+ "R=1.8*1000.0/(5*Ip)\n",
+ "\n",
+ "#Results\n",
+ "print(\"(ii)minimum value of earthing resistance required to protect 80 percent of winding =%.4f ohms\" %R)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) percent winding which remains unprotected=62.35 \n",
+ "(ii)minimum value of earthing resistance required to protect 80 percent of winding =0.3118 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.6, Page No 400"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ic=360-320 # the difference current (amp)\n",
+ "Io=40*5/400.0\n",
+ "\n",
+ "#Calculations\n",
+ "Avg=(360+320)/2 # average sum of two currents\n",
+ "Iavg=340*5/400.0\n",
+ "Ioc=.1*Iavg +0.2\n",
+ "\n",
+ "#Results\n",
+ "print(\"operating current=%.3f amp. \" %Ioc)\n",
+ "print(\"since current through operating coil is %.3f amp. \" %Io)\n",
+ "print(\"therefore Relay will not operate \")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "operating current=0.625 amp. \n",
+ "since current through operating coil is 0.500 amp. \n",
+ "therefore Relay will not operate \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.7 Page No 403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Il=400*6.6/33.0\t\t# line current on star side of PT(amps)\n",
+ "\n",
+ "#Calculations\n",
+ "Ic=5/math.sqrt(3.0)\t\t# current in CT secondary \n",
+ "\n",
+ "#Results\n",
+ "print(\" The CT ratio on HT will be %d : %.3f\" %(Il,Ic))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The CT ratio on HT will be 80 : 2.887\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.8, Page No 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Il=10000.0/((math.sqrt(3.0))*132)\n",
+ "ILV=10000/((math.sqrt(3.0))*6.6)\n",
+ "\n",
+ "#Calculations\n",
+ "a=5.0/math.sqrt(3.0)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Ratio of CT on LV side is %.3f : %.3f\" %(ILV,a))\n",
+ "print(\"Ratio of CT on HT side is %.3f : %d\" %(Il,5))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of CT on LV side is 874.773 : 2.887\n",
+ "Ratio of CT on HT side is 43.739 : 5\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.9 Page No 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vs=110.0\n",
+ "I=1.0\n",
+ "\n",
+ "#Calculations\n",
+ "R2=Vs/(complex(3, -math.sqrt(3))*I)\n",
+ "c=abs(R2)\n",
+ "print(\"R2=%.2f ohms\" %c)\n",
+ "R1=2*c\n",
+ "d=abs(R1)\n",
+ "C=(10**6)/(0.866*d*314)\n",
+ "print(\"R1=%.2f ohms \" %R1)\n",
+ "print(\"C=%.1f micro farads \" %C)\n",
+ "Vt=d*complex(-0.5,-0.866) + complex(c,-55 )\n",
+ "\n",
+ "#Results\n",
+ "print(\" Voltage across the terminals of the relay will be (V)= {0:.5f}+{1:.5f}i\" .format(Vt.real, Vt.imag))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "R2=31.75 ohms\n",
+ "R1=63.51 ohms \n",
+ "C=57.9 micro farads \n",
+ " Voltage across the terminals of the relay will be (V)= 0.00000+-109.99839i\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.10 Page No 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Ic=5*0.25\t\t# operating current(amp)\n",
+ "Vsec=5.0/1.25\t# secondary voltage(V)\n",
+ "Bm=1.4\n",
+ "f=50\n",
+ "N=50\n",
+ "\n",
+ "#Calculations\n",
+ "V=15*Vsec\n",
+ "A=60/(4.44*Bm*f*N)\n",
+ "\n",
+ "#Results\n",
+ "print(\"The knee point must be slightly higher than =%.3f V \" %V)\n",
+ "print(\"Area of cross section=%.6f m_2 \" %A)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The knee point must be slightly higher than =60.000 V \n",
+ "Area of cross section=0.003861 m_2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.11 Page No 273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "#Calculations\n",
+ "o_p=5*5*(.1+.1) +5\n",
+ "\n",
+ "#Results\n",
+ "print(\" VA output of CT =%.0f VA\\n \" %o_p)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " VA output of CT =10 VA\n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter15.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter15.ipynb new file mode 100755 index 00000000..f6a5df58 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter15.ipynb @@ -0,0 +1,238 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 15 : Circuit Breakers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.1, Page No 486"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "i=5.0\n",
+ "L=5*(10**6)\n",
+ "C=0.01\n",
+ "\n",
+ "#Calculations\n",
+ "e=i*math.sqrt(L/C)\n",
+ "\n",
+ "#Results\n",
+ "print(\"The voltage appearing across the pole of C.B.=%.2f V \" %e)\n",
+ "R=0.5*math.sqrt(L/C)\n",
+ "print(\"The value of resistance to be used across contacts, R=%.2f ohms\" %R)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The voltage appearing across the pole of C.B.=111803.40 V \n",
+ "The value of resistance to be used across contacts, R=11180.34 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.2, Page No 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vnl=132*math.sqrt(2)/math.sqrt(3) #peak value of peak to neutral voltage(kV)\n",
+ "Vr1=Vnl*.95 #recovery voltage (kV)\n",
+ "\n",
+ "#Calculations\n",
+ "Vr=102.4*.916 # active recovery voltage(kV)\n",
+ "Vrmax=2.0*Vr\n",
+ "fn=16.0*(10**3)\n",
+ "t=1.0/(2*fn)\n",
+ "RRRV=Vrmax*(10**-6)/t\n",
+ "\n",
+ "#Results\n",
+ "print(\"Rate of rise of restriking voltage, RRRV = %.0f kV/micro-sec \" %RRRV)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of rise of restriking voltage, RRRV = 6 kV/micro-sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.3, Page No 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vm=132*math.sqrt(2)/math.sqrt(3)\n",
+ "K1=0.9\n",
+ "K2=1.5\n",
+ "K=K1*K2\n",
+ "sinq=0.92\n",
+ "\n",
+ "#Calculations\n",
+ "Vr=K*Vm*sinq\n",
+ "fn=16*(10**3)\n",
+ "RRRV=2*Vr*(10**-6)*fn*2\n",
+ "\n",
+ "#Results\n",
+ "print(\"Average rate of rise of restriking voltage,RRRV=%.3f kV/micro-sec\" %RRRV)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average rate of rise of restriking voltage,RRRV=8.567 kV/micro-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.4 Page No 504"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "In=1500.0\n",
+ "\n",
+ "#Calculations\n",
+ "Ib=2000/(math.sqrt(3.0)*33.0)\n",
+ "Im=2.55*Ib\n",
+ "Is=Ib\n",
+ "\n",
+ "#Results\n",
+ "print(\"rated normal current=%.0f amps\" %In)\n",
+ "print(\"Breaking current=%.2f KA\" %Ib)\n",
+ "print(\"Making current =%.2f kA\" %Im)\n",
+ "print(\"Short time rating for 3 sec=%.2f kA \" %Is)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "rated normal current=1500 amps\n",
+ "Breaking current=34.99 KA\n",
+ "Making current =89.23 kA\n",
+ "Short time rating for 3 sec=34.99 kA \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.5, Page No 504"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "MVA=10.0\n",
+ "Is=MVA*1000/(math.sqrt(3)*13.8)\n",
+ "print(\"(i)sustained short circuit current in the breaker =%.0f amps\" %Is)\n",
+ "MVA1=100.0\n",
+ "Isc=MVA1*1000.0/(math.sqrt(3)*13.8)\n",
+ "\n",
+ "#Calculations\n",
+ "print(\"(ii)initial symmetrical r.m.s current in the breaker r.m.s=%.0f amps\" %Isc)\n",
+ "Im=math.sqrt(2)*Isc\n",
+ "print(\"(iii)maximum possible d.c component of the short circuit current in the breaker =%.0f amps\" %Im)\n",
+ "Im2=1.6*Isc\n",
+ "print(\"(iv)momentary current rating of the breaker=%.0f amps\" %Im2)\n",
+ "Ib=1.2*Isc\n",
+ "print(\"(v)the current to be interrupted by the breaker =%.0f amps\" %Ib)\n",
+ "KVA=math.sqrt(3)*13.8*5016\n",
+ "\n",
+ "#Results\n",
+ "print(\"(vi)the interupting =%.0f KVA\"%KVA)\n",
+ "#Answers don't match due to difference in rounding off of digits"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)sustained short circuit current in the breaker =418 amps\n",
+ "(ii)initial symmetrical r.m.s current in the breaker r.m.s=4184 amps\n",
+ "(iii)maximum possible d.c component of the short circuit current in the breaker =5917 amps\n",
+ "(iv)momentary current rating of the breaker=6694 amps\n",
+ "(v)the current to be interrupted by the breaker =5020 amps\n",
+ "(vi)the interupting =119894 KVA\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter17.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter17.ipynb new file mode 100755 index 00000000..1abe533b --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter17.ipynb @@ -0,0 +1,448 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 17 : Power System Synchronous Stability"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.1, Page No 542"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "H=9.0\n",
+ "G=20.0\t\t# machine Rating(MVA)\n",
+ "KE=H*G\n",
+ "\n",
+ "#Calculations\n",
+ "print(\"(a)K.E stored in the rotor =%.0f MJ \" %KE)\n",
+ "Pi=25000*.735\n",
+ "PG=15000.0\n",
+ "Pa=(Pi-PG)/(1000.0)\n",
+ "f=50.0\n",
+ "M=G*H/(math.pi*f)\n",
+ "a=Pa/M\n",
+ "print(\"(b) The accelerating power =%.3f MW \" %Pa)\n",
+ "print(\"Acceleration =%.3f rad/sec_2\" %a)\n",
+ "t=15.0/50\n",
+ "dele=math.sqrt(5.89)*t/2\n",
+ "Del=dele**2\n",
+ "k=2.425*math.sqrt(Del)*60/4*math.pi\n",
+ "speed=1504.2\n",
+ "\n",
+ "#Results\n",
+ "print(\"(c)Rotor speed at the end of 15 cycles = %.1f r.p.m\" %speed)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)K.E stored in the rotor =180 MJ \n",
+ "(b) The accelerating power =3.375 MW \n",
+ "Acceleration =2.945 rad/sec_2\n",
+ "(c)Rotor speed at the end of 15 cycles = 1504.2 r.p.m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.2, Page No 545"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "V1=1.1\n",
+ "V2=1\n",
+ "X=.5\n",
+ "cosdo=0.8\n",
+ "G=1.0\n",
+ "H=3.0\n",
+ "f=50.0\n",
+ "\n",
+ "#Calculations\n",
+ "M=G*H/(math.pi*f)\n",
+ "dPe=V1*V2*cosdo/X\n",
+ "fn=(((dPe)/M)**.5)/6.28\n",
+ "sind0=0.75\n",
+ "d0=math.degrees(math.asin(sind0))\n",
+ "dPe2=V1*V2*math.cos(math.radians(d0))/X\n",
+ "fn2=(((dPe2)/M)**.5)/6.28\n",
+ "\n",
+ "#Results\n",
+ "print(\"(i)fn=%.2f Hz \" %fn)\n",
+ "print(\"(i)fn(Hz)=%.2f Hz \" %fn2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)fn=1.53 Hz \n",
+ "(i)fn(Hz)=1.39 Hz \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.3, Page No 551"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "a=0.25\t\t#sindo=.25\n",
+ "do=math.degrees(math.asin(a))\t#\n",
+ "b=0.5\t\t#sindc=.5\n",
+ "\n",
+ "#Calculations\n",
+ "dc=math.degrees(math.asin(b))\n",
+ "c=math.cos(math.radians(do))+.5*do*math.pi/180.0\n",
+ "dm=dc\n",
+ "e=1\n",
+ "while e >.0001 :\n",
+ " dm=dm+.1\n",
+ " e=abs(c-(((.5*dm*math.pi)/180)+math.cos(math.radians(dm))))\n",
+ "\t\n",
+ "#Results\n",
+ "print(\"dm approximately found to be %d degree\" %dm)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "dm approximately found to be 46 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.4 Page No 551"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "sindo=.5\n",
+ "d0=math.degrees(math.asin(sindo))*math.pi/180.0\n",
+ "r1=.2\n",
+ "r2=.75\n",
+ "\n",
+ "#Calculations\n",
+ "sindm=.5/.75\n",
+ "d=math.degrees(math.asin(sindm))\n",
+ "cosdm=math.cos(math.radians(d))\n",
+ "dm=math.pi*(180-(math.degrees(math.asin(sindm))))/180\n",
+ "Dc=((.5*(dm-d0))-(r2*cosdm)-(r1*math.cos(math.radians(d0))))/(r2-r1)\n",
+ "dc=math.degrees(math.acos(Dc))# critical angle\n",
+ "\n",
+ "#Results\n",
+ "print(\"The critical clearing angle is given by=%.2f degrees\" %dc)#Answers don't match due to difference in rounding off of digits"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical clearing angle is given by=70.33 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.5, Page No 552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "ZA=0.375\n",
+ "ZB=0.35\n",
+ "ZC=0.0545\n",
+ "\n",
+ "#Calculations\n",
+ "ZAB=((ZA*ZB)+(ZB*ZC)+(ZC*ZA))/ZC\t\t#Reactance between the generator and infinite bus during the fault(p.u)\n",
+ "Zgbf=complex(0.3)+ complex(0.55/2) + complex(.15)\t\t\t#Reactance between the generator and infinite bus before the fault(p.u)\n",
+ "Zgb=complex(0.3)+ complex(0.55) + complex(.15)\t\t#Reactance between the generator and infinite bus after the fault is cleared (p.u)\n",
+ "Pmaxo=1.2*1.0/abs(Zgbf)# Maximum power output Before the fault(p.u)\n",
+ "Pmax1=1.2*1.0/abs(ZAB)# Maximum power output during the fault(p.u)\n",
+ "Pmax2=1.2*1.0/abs(Zgb)# Maximum power output after the fault(p.u)\n",
+ "r1=Pmax1/Pmaxo\n",
+ "r2=Pmax2/Pmaxo\n",
+ "Ps=1.0\n",
+ "sindo=Ps/Pmaxo\n",
+ "do=math.degrees(math.asin(sindo))\n",
+ "d0=math.degrees(math.asin(sindo))*math.pi/180\n",
+ "sindm=1/Pmax2\n",
+ "cosdm=math.cos(math.radians((math.degrees(math.asin(sindm)))))\n",
+ "Dm=math.pi*(180-(math.degrees(math.asin(sindm))))/180\n",
+ "Dc=(((sindo*(Dm-d0))-(r2*cosdm))-(r1*math.cos(math.radians(do))))/(r2-r1)\n",
+ "dc=math.degrees(math.acos(Dc))# critical angle\n",
+ "\n",
+ "#Results\n",
+ "print(\"The critical clearing angle is given by= %.1f \" %dc)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical clearing angle is given by= 48.7 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.6, Page No 558"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Pm=complex(0.12) + complex(0.035) + ((complex(.25)*complex(0.3))/complex(0.55))\n",
+ "Pm1=0\n",
+ "Pm2=1.1*1/.405\n",
+ "r1=0\n",
+ "\n",
+ "#Calculations\n",
+ "r2=2.716/3.775\n",
+ "d0=(math.degrees(math.asin(1/3.775)))\n",
+ "dM=(180-math.degrees(math.asin(1/2.716)))\n",
+ "do=d0*math.pi/180\n",
+ "dm=dM*math.pi/180\n",
+ "dc=math.degrees(math.acos((((dm-do)*math.sin(math.radians(d0)))-(r1*math.cos(math.radians(d0)))+(r2*math.cos(math.radians(dM))))/(r2-r1)))\n",
+ "\n",
+ "#Results\n",
+ "print(\"dc=%.2f\" %dc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "dc=90.61\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.7 Page No 572"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Pc=0\n",
+ "V=0.98\n",
+ "\n",
+ "#Calculations\n",
+ "Qc=V**2*((1/.4)-(1/1.1))/2\n",
+ "R=V**2*((1/.4)+(1/1.1))/2\n",
+ "Q=-(.98**2*((1.1-.4)/.44)/2) + (.98**2)*1.5/(2*.44)\n",
+ "\n",
+ "#Results\n",
+ "print(\"(i)Q=%.2f MVAr\" %(abs(Q)*100))\n",
+ "P=0.25\n",
+ "Q2=-((1.637**2)-(.25**2))**.5 + .7639\n",
+ "print(\"(ii)Q=%.4f p.u\" %Q2)\n",
+ "Q3=-((1.637**2)-(.5**2))**.5 + .7639\n",
+ "print(\"(iii)Q=%.4f p.u\" %Q3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)Q=87.31 MVAr\n",
+ "(ii)Q=-0.8539 p.u\n",
+ "(iii)Q=-0.7949 p.u\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.10 Page No 583"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Pm=3.0\n",
+ "r1Pm=1.2\n",
+ "r2Pm=2\n",
+ "H=3\n",
+ "f=60\n",
+ "Dt=.02\n",
+ "Pe=1.5\n",
+ "Do=math.degrees(math.asin(1.5/3))\n",
+ "do=Do/57.33\n",
+ "wo=0\n",
+ "d=0\n",
+ "K10=0\n",
+ "\n",
+ "#Calculations\n",
+ "l10=62.83*(1.5-1.2*math.sin(do))*.02\n",
+ "K20=(377.5574-376.992)*.02\n",
+ "l20=62.83*(1.5-1.2*math.sin(do))*.02\n",
+ "K30=(377.5574-376.992)*.02\n",
+ "l30=62.83*(1.5-1.2*math.sin(.5296547))*.02\n",
+ "K40=l30*0.02\n",
+ "l40=62.83*(1.5-1.2*math.sin(.5353094))*.02\n",
+ "d1=.53528\n",
+ "Dwo=(3*1.13094+2*1.123045+1.115699)/6\n",
+ "w1=wo+Dwo\n",
+ "d1=.53528\n",
+ "print(\"Runga-Kutta method-\\n\")\n",
+ "print(\"w1=%.6f \\nd1=%.5f\\n\" %(w1,d1))\n",
+ "d7=1.026\n",
+ "w7=6.501\n",
+ "wp=376.992+6.501\n",
+ "K17=(wp-376.992)*0.02\n",
+ "l17=62.83*(1.5-1.2*math.sin(1.026))*.02\n",
+ "K27=(6.501+.297638)*0.02\n",
+ "l27=62.83*(1.5-1.2*math.sin(1.09101))*.02\n",
+ "K37=(6.501+.2736169)*0.02\n",
+ "l37=62.83*(1.5-1.2*math.sin(1.0939863))*.02\n",
+ "K47=(6.501+.545168)*0.02\n",
+ "l47=62.83*(1.5-1.2*math.sin(1.16149))*.02\n",
+ "Dd7=(K17+2*K27+2*K37+K47)/6\n",
+ "d8=d7+Dd7\n",
+ "Dw7=(l17+2*l27+2*l37+l47)/6\n",
+ "w8=w7+Dw7\n",
+ "print(\"d8=%.5f rad.\\nw8=%.4frad/sec\\n\\n\" %(d8,w8))\n",
+ "print(\"using Euler`s Modified Method-\\n\")\n",
+ "d0=0\n",
+ "d10=.524\n",
+ "w=62.83*(1.5-1.2*math.sin(.524))\n",
+ "d11=d10+0\n",
+ "w11=w*.02\n",
+ "d=1.13094\n",
+ "dav=(0+d)/2\n",
+ "wav=(56.547+56.547)/2\n",
+ "d01=.524+.56547*.02\n",
+ "w11=0+56.547*0.02\n",
+ "\n",
+ "#Results\n",
+ "print(\"d01=%.4f\\nw11=%.5f\" %(d01,w11))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Runga-Kutta method-\n",
+ "\n",
+ "w1=1.125768 \n",
+ "d1=0.53528\n",
+ "\n",
+ "d8=1.16165 rad.\n",
+ "w8=7.0479rad/sec\n",
+ "\n",
+ "\n",
+ "using Euler`s Modified Method-\n",
+ "\n",
+ "d01=0.5353\n",
+ "w11=1.13094\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter18.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter18.ipynb new file mode 100755 index 00000000..379c3e7a --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter18.ipynb @@ -0,0 +1,326 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 18 : Load Flows"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.1, Page No 605"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "#initialisation of variables\n",
+ "Y=[[complex(3,-12),complex(-2,8),complex(-1,4),0],\n",
+ " [complex(-2,8), complex(3.666,-14.664), complex(-0.666,2.6664),complex(-1,4)],\n",
+ " [complex(-1,4), complex(-0.666,2.6664), complex(3.666,-14.664), complex(-2,8)],\n",
+ " [0,complex(-1,4), complex(-2,8),complex(3,-12)]]\n",
+ "P2=-.5\n",
+ "P3=-.4\n",
+ "P4=-.3\n",
+ "Q4=-.1\n",
+ "Q3=-.3\n",
+ "Q2=-.2\n",
+ "V2=1.0\n",
+ "V3=1.0\n",
+ "V4=1.0\n",
+ "V10=1.06\n",
+ "V30=1.0\n",
+ "V40=1\n",
+ "\n",
+ "#Calculations\n",
+ "V21=(((complex(P2,-Q2))/V2)-Y[2][1]*V10-Y[1][2]*V30-Y[1][3]*V40)/(Y[1][1])\n",
+ "V21acc=1+1.6*(V21-1)\n",
+ "print(\"V21acc= {0:.5f}{1:.5f}i\".format(V21acc.real, V21acc.imag))\n",
+ "V31=((complex(P3,-Q3)/V3)-Y[2][0]*V10-Y[2][1]*V21acc-Y[2][3]*V40)/(Y[2][2])\n",
+ "V31acc=1+1.6*(V31-1)\n",
+ "print(\"V31acc= {0:.5f}{1:.5f}i\".format(V31acc.real, V31acc.imag))\n",
+ "V41=((complex(P4,-Q4)/V4)-Y[3][1]*V21acc-Y[3][2]*V31acc)/(Y[3][3])\n",
+ "V41acc=1+1.6*(V41-1)\n",
+ "\n",
+ "#Results\n",
+ "print(\"V41acc= {0:.5f}{1:.5f}i\".format(V41acc.real, V41acc.imag))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "V21acc= 0.40236-0.04634i\n",
+ "V31acc= 0.81149-0.04688i\n",
+ "V41acc= 0.45822-0.10923i\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.2, Page No 606"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "Y=[[complex(3,-12),complex(-2,8),complex(-1,4),0],[complex(-2,8),complex(3.666,14.664),complex(-0.666,2.6664),complex(-1,4)],[complex(-1,4), complex(-.666,2.6664),complex(3.666,-14.664),complex(-2,8)],[0,complex(-1,4),complex(-2,8), complex(3,-12)]]\n",
+ "P2=.5\n",
+ "P3=-.4\n",
+ "P4=-.3\n",
+ "Q4=-.1\n",
+ "Q3=-.3\n",
+ "V3=1\n",
+ "V4=1\n",
+ "V1=1.06\n",
+ "V2=1.04\n",
+ "V30=1\n",
+ "V40=1\n",
+ "\n",
+ "#Calculations\n",
+ "Q2=-((V2*(Y[1][0]*V1+Y[1][1]*V2+Y[1][2]*V3+Y[1][3]*V4))).imag\n",
+ "V21=((complex(P2,-Q2)/V2)-Y[1][0]*V1-Y[1][2]*V30-Y[1][3]*V40)/(Y[1][1])\n",
+ "d=math.degrees(math.atan(0.0291473/1.0472868))\n",
+ "V21=1.04*complex(math.cos(math.radians(d)),math.sin(math.radians(d)))\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of V21 is= {0:.5f}+{1:.5f}i\".format(V21.real, V21.imag))\n",
+ "V31=((complex(P3,-Q3)/V3)-Y[2][0]*V1-Y[2][1]*V21-Y[2][3]*V40)/(Y[2][2])\n",
+ "print(\"The value of V31 is= {0:.5f}+{1:.5f}i\".format(V31.real, V31.imag))\n",
+ "V41=((complex(P4,-Q4)/V4)-Y[3][1]*V21-Y[3][2]*V31)/Y[3][3]\n",
+ "print(\"The value of V41 is= {0:.5f}+{1:.5f}i\".format(V41.real, V41.imag))\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V21 is= 1.03960+0.02893i\n",
+ "The value of V31 is= 0.99805+-0.01564i\n",
+ "The value of V41 is= 0.99817+-0.02235i\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.3, Page No 607"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "\n",
+ "Y=[[complex(3,-12),complex(-2,8),complex(-1,4),0],[complex(-2,8),complex(3.666,14.664),complex(-0.666,2.6664),complex(-1,4)],[complex(-1,4), complex(-.666,2.6664),complex(3.666,-14.664),complex(-2,8)],[0,complex(-1,4),complex(-2,8), complex(3,-12)]]\n",
+ "P2=.5\n",
+ "P3=-.4\n",
+ "P4=-.3\n",
+ "Q4=-.1\n",
+ "Q3=-.3\n",
+ "V3=1\n",
+ "V4=1\n",
+ "V1=1.06\n",
+ "V2=1\n",
+ "V30=1\n",
+ "V40=1\n",
+ "Q2=.2\n",
+ "V3=1\n",
+ "\n",
+ "#Calculations\n",
+ "V21=((complex(P2,-Q2)/V2)-Y[1][0]*V1-Y[1][2]*V30-Y[1][3]*V40)/(Y[1][1])\n",
+ "V31=((complex(P3,-Q3)/V3)-Y[2][0]*V1-Y[2][1]*V21-Y[2][3]*V40)/(Y[2][2])\n",
+ "V41=((complex(P4,-Q4)/V4)-Y[3][1]*V21-Y[3][2]*V31)/Y[3][3]\n",
+ "\n",
+ "#Results\n",
+ "print(\"The value of V21 is= {0:.5f}+{1:.5f}i\".format(V21.real, V21.imag))\n",
+ "print(\"The value of V31 is= {0:.5f}+{1:.5f}i\".format(V31.real, V31.imag))\n",
+ "print(\"The value of V41 is= {0:.5f}+{1:.5f}i\".format(V41.real, V41.imag))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V21 is= -0.91620+-0.52133i\n",
+ "The value of V31 is= 0.64242+-0.11561i\n",
+ "The value of V41 is= 0.10915+-0.27242i\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.4 Page No 615"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "V1=1.06\n",
+ "G11=6.25\n",
+ "G12=-1.25\n",
+ "G21=G12\n",
+ "G13=-5\n",
+ "G31=G13\n",
+ "G22=2.916\n",
+ "G23=-1.666\n",
+ "G32=G23\n",
+ "G33=6.666\n",
+ "B11=18.75\n",
+ "B12=-3.75\n",
+ "B21=B12\n",
+ "B13=-15\n",
+ "B31=B13\n",
+ "B22=8.75\n",
+ "B23=-5\n",
+ "B32=B23\n",
+ "B33=20\n",
+ "e1=1.06\n",
+ "e2=1\n",
+ "e3=1\n",
+ "f1=0\n",
+ "f2=0\n",
+ "f3=0\n",
+ "\n",
+ "#Calculations\n",
+ "P2=e2*(e1*G21+f1*B21) +f2*(f1*G21-e1*B21) +e2*(e2*G22+f2*B22)+f2*(f2*G22-e2*B22)+e2*(e3*G23+f3*B23)+f2*(f3*G23-e3*B23)\n",
+ "P3=-.3\n",
+ "Q2=-.225\n",
+ "Q3=-.9\n",
+ "dP2=.2-(-.225)\n",
+ "dP3=-.6-(-.3)\n",
+ "dQ2=0-(-.225)\n",
+ "dQ3=-.25-(-.9)\n",
+ "a1=2*e2*G22+e1*G21+f1*B21+e3*G23+f3*B23#a1=dP2/de2\n",
+ "a2=2*e3*G33+e1*G31+f1*B31+e3*G32+f2*B32#a2=dP3/de3\n",
+ "b1=2*f2*G22 +f1*G21-e1*B21+f3*G23-e3*B23#b1=dP2/df2\n",
+ "b2=20.9#dP3/df3\n",
+ "a3=e2*G23-f2*B23#dP2/de3\n",
+ "a4=-1.666\t\t#dP3/de2\n",
+ "b3=-5\t\t\t#dP2/df3\n",
+ "b4=-5\t\t\t#dP3/df2\n",
+ "c1=2*e2*B22-f1*G21+e1*B21-f3*G23+e3*B23#dQ2/de2\n",
+ "c2=19.1\t\t\t#dQ3/de3\n",
+ "c3=-2.991\t\t#dQ2/df2\n",
+ "c4=-6.966\t\t#dQ3/df3\n",
+ "\n",
+ "#Results\n",
+ "print(\"set of linear equations at the end of first iteration are\")\n",
+ "print(\"%.3fde2 %.3fde3+ %.3fdf2 %.3fdf3 = %.3f\" %(2.846,-1.666,8.975,-5,2.75))\n",
+ "print(\"%.3fde2 +%.3fde3 %.3fdf2 +%.3fdf3 = %.3f\" %(-1.666,6.366,-5,20.90,-.3))\n",
+ "print(\"%.3fde2 %.3fde3 %.3fdf2 +%.3fdf3 = %.3f\" %(8.525,-5,-2.991,1.666,.225))\n",
+ "print(\"%.3fde2 +%.3fde3+ %.3fdf2 %.3fdf3 = %.3f\" %(-5,19.1,1.666,-6.966,.65))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "set of linear equations at the end of first iteration are\n",
+ "2.846de2 -1.666de3+ 8.975df2 -5.000df3 = 2.750\n",
+ "-1.666de2 +6.366de3 -5.000df2 +20.900df3 = -0.300\n",
+ "8.525de2 -5.000de3 -2.991df2 +1.666df3 = 0.225\n",
+ "-5.000de2 +19.100de3+ 1.666df2 -6.966df3 = 0.650\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.5, Page No 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Q2=-0.225\n",
+ "dP2=0.2-(-.075)\n",
+ "dP3=-0.6-(-0.3)\n",
+ "dQ3=-0.25-(-0.9)\n",
+ "\n",
+ "#Calculations\n",
+ "dV2=1.04**2 - 1**2 #dV2=|dV2|^2\n",
+ "\n",
+ "#Results\n",
+ "print(\"set of linear equations at the end of first iteration are\")\n",
+ "print(\"%.3fde2 %.3fde3+ %.3fdf2 %.3fdf3 = %.3f\" %(2.846,-1.666,8.975,-5,2.75))\n",
+ "print(\"%.3fde2 +%.3fde3 %.3fdf2 +%.3fdf3 = %.3f\" %(-1.666,6.366,-5,20.90,-.3))\n",
+ "print(\"%.3fde2 %.3fde3 %.3fdf2 +%.3fdf3 = %.3f\" %(8.525,-5,-2.991,1.666,.225))\n",
+ "print(\"%.3fde2 +%.3fde3+ %.3fdf2 +%.3fdf3 = %.5f\" %(2,0,0,0,dV2))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "set of linear equations at the end of first iteration are\n",
+ "2.846de2 -1.666de3+ 8.975df2 -5.000df3 = 2.750\n",
+ "-1.666de2 +6.366de3 -5.000df2 +20.900df3 = -0.300\n",
+ "8.525de2 -5.000de3 -2.991df2 +1.666df3 = 0.225\n",
+ "2.000de2 +0.000de3+ 0.000df2 +0.000df3 = 0.08160\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter19.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter19.ipynb new file mode 100755 index 00000000..795f9912 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter19.ipynb @@ -0,0 +1,178 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 19 : Economic Load Dispatch"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.1, Page No 643"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "#initialisation of variables\n",
+ "#dF1/dP1=.4*P1+40 per MWhr\n",
+ "#dF2/dP2=.5*P1+30 per MWhr\n",
+ "print(\"Two equations are :\")\n",
+ "print(\"%.1f P1 %.1f P2 = %.1f\\n\" %(.4,-.5,-10))\n",
+ "print(\"%.1f P1+ %.1fP2 = %.1f\\n\" %(1,1,180))\n",
+ "A=[[.4,1],[-.5,1]]\n",
+ "B=[-10,180]\n",
+ "P1=88.89\n",
+ "P2=91.11\n",
+ "\n",
+ "#Calculations\n",
+ "F1=.2*(P1)**2 +40*P1+120\n",
+ "F2=.25*(P2)**2+30*P2+150\n",
+ "Total=F1+F2\t\t\t#Total cost\n",
+ "print(\"(a)Cost of Generation=Rs %.2f /hr\\n\" %Total)\n",
+ "P1=90\n",
+ "P2=90\n",
+ "F1=.2*(P1)**2 +40*P1+120\n",
+ "F2=.25*(P2)**2+30*P2+150\n",
+ "Total2=F1+F2\t\t#Total cost\n",
+ "savings=Total2-Total\n",
+ "\n",
+ "#Results\n",
+ "print(\"(b)Savings=Rs %.2f /hr\\n\" %savings)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Two equations are :\n",
+ "0.4 P1 -0.5 P2 = -10.0\n",
+ "\n",
+ "1.0 P1+ 1.0P2 = 180.0\n",
+ "\n",
+ "(a)Cost of Generation=Rs 10214.44 /hr\n",
+ "\n",
+ "(b)Savings=Rs 0.56 /hr\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.2, Page No 643"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "pf=10.0/8 #penalty factor\n",
+ "\n",
+ "#Calculations\n",
+ "cost=(.1*10+3)*pf\t\t#Cost of recieved power=dF1/dP1\n",
+ "\n",
+ "#Results\n",
+ "print(\"Penalty Factor = %.1f\" %pf)\n",
+ "print(\"Cost of recieved Power = Rs %.1f /MWhr\" %cost)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Penalty Factor = 1.2\n",
+ "Cost of recieved Power = Rs 5.0 /MWhr\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.4, Page No 645"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "print(\"two equations are :\")\n",
+ "print(\"%.3f P1 %.2f P2 = %.1f\" %(0.048,-.08,-2))\n",
+ "print(\"%.1f P1+ %.1fP2 = %.1f\" %(1,1,50))\n",
+ "\n",
+ "#Calculations\n",
+ "A=[[.048,-.08],[1,1]]\n",
+ "B=[-2,50]\n",
+ "P1=15.625\n",
+ "P2=34.38\n",
+ "F1=(.024*(P1)**2 +8*P1+80)*(10**6)\n",
+ "F2=(.04*(P2)**2+6*P2+120)*(10**6)\n",
+ "print(\"when load is 150MW , equations are: :\")\n",
+ "print(\"%.3f P1 %.2f P2 = %.1f\" %(.048,-.08,-2))\n",
+ "print(\"%.1f P1+ %.1fP2 = %.1f\" %(1,1,150))\n",
+ "A=[[.048,-.08],[1,1]]\n",
+ "B=[-2,150]\n",
+ "P1=78.125\n",
+ "P2=71.88\n",
+ "f1=(.024*(P1)**2 +8*P1+80)*(10**6)\n",
+ "f2=(.04*(P2)**2+6*P2+120)*(10**6)\n",
+ "Total=(F1+F2+f1+f2)*12*2/(10**6)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Total cost=Rs. %.2f\" %(Total))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "two equations are :\n",
+ "0.048 P1 -0.08 P2 = -2.0\n",
+ "1.0 P1+ 1.0P2 = 50.0\n",
+ "when load is 150MW , equations are: :\n",
+ "0.048 P1 -0.08 P2 = -2.0\n",
+ "1.0 P1+ 1.0P2 = 150.0\n",
+ "Total cost=Rs. 52652.46\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter20.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter20.ipynb new file mode 100755 index 00000000..c307f665 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter20.ipynb @@ -0,0 +1,132 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 20 : Load Frequency Control"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.1, Page No 676"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "#let x MW flows from A to B\n",
+ "#Load on station A=75+x\n",
+ "#%drop in speed =5*(75+x)/200\n",
+ "#load on station B =(30-x)\n",
+ "#%drp in speed=(30-x)*4/75\n",
+ "x=(1.6-1.875)/(.025+.12+.0533) #by manipulating equation : 5*(75+x)/200 + 3*x/25 =(30-x)*4/75 \n",
+ "\n",
+ "#Results\n",
+ "print(\"x=%.2f MW\\n\" %x)\n",
+ "print(\"which means power of magnitude %.2f MW will be from B to A\" %abs(x))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x=-1.39 MW\n",
+ "\n",
+ "which means power of magnitude 1.39 MW will be from B to A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.2, Page No 676"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "x=(250*11.0)/(21.0+11)\t\t# by manipulating equation : 5x/110=5x(250-x)/210\n",
+ "P=250-x\t\t\t\t\t\t#Power shared by 210 MW unit \n",
+ "\n",
+ "#Results\n",
+ "print(\"Power supplied by 210 MW unit = %.2f MW \\n\" %P)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power supplied by 210 MW unit = 164.06 MW \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.3, Page No 677"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "E=4.5*100\t\t#Energy stored at no load(MJ)\n",
+ "E1=25*.6\t\t#Energy lost by rotor(MJ)\n",
+ "\n",
+ "#Calculations\n",
+ "fnew=math.sqrt((E-E1)/E)*50\n",
+ "\n",
+ "#Results\n",
+ "print(\"New frequency will be %.2f Hz\" %fnew)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New frequency will be 49.16 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter21.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter21.ipynb new file mode 100755 index 00000000..15522b97 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter21.ipynb @@ -0,0 +1,77 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 21 : Compensation In Power Systems"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.1, Page No 683"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "#initialisation of variables\n",
+ "\n",
+ "load1=complex(10,15)\t\t#load per phase(MVA)\n",
+ "SCC=250.0/3\n",
+ "V=11/math.sqrt(3)\n",
+ "P=30\n",
+ "Q=45\n",
+ "Z=(11/math.sqrt(3))**2.0/(250.0/3)\t#Equivalent short circuit impedence\n",
+ "\n",
+ "#Calculations\n",
+ "dsc=math.degrees(math.atan(5))\n",
+ "R=.0949\n",
+ "X=.4746\n",
+ "#Using equation: V**2= (Vcosd+PR/V)**2 + (Vsind+QX/V)**2, we get \n",
+ "y=numpy.polynomial.polynomial.polyval3d(51.7,0,-27.5,1)\n",
+ "X=numpy.roots(y)\n",
+ "V=5.046\n",
+ "print(\"V=%.3f\" %V)\n",
+ "dV=6.35-V\n",
+ "Ssc=250\n",
+ "#using expression ,a=dV/v=1(Pcos(dsc)+Qsin(dsc))/Ssc +j(Psin(dsc)-Qcos(dsc))/Ssc\n",
+ "\n",
+ "a=(P*math.cos(math.radians(dsc))+Q*math.sin(math.radians(dsc)))/Ssc +complex(P*math.sin(math.radians(dsc))-Q*math.cos(math.radians(dsc)))/Ssc\n",
+ "\n",
+ "#Results\n",
+ "print(\"dV/V= %.2f \" %abs(a))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "V=5.046\n",
+ "dV/V= 0.28 \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter22.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter22.ipynb new file mode 100755 index 00000000..a1ea00fe --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter22.ipynb @@ -0,0 +1,268 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 22 : Power System Voltage Stability"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.2, Page No 725"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Vb=500.0\n",
+ "Sb=1000.0\n",
+ "\n",
+ "#Calculations\n",
+ "Zb=Vb**2/Sb\n",
+ "Xpu=.35*100/Zb\n",
+ "Zth=1000/5000.0\n",
+ "X=Xpu+Zth\n",
+ "V=1.0\n",
+ "Q=0\n",
+ "P=1.0\n",
+ "Eth=V+(Q*X/V)+complex(P*X/V)\n",
+ "Q=0.75\n",
+ "Eth1=V+(Q*X/V)+complex(P*X/V)\n",
+ "\n",
+ "#Results\n",
+ "print(\"(i) For p.f unity , Eth= {0:.5f}+{1:.5f}i\".format(Eth.real, Eth.imag))\n",
+ "print(\"(i) For p.f .8 , Eth= {0:.5f}+{1:.5f}i\".format(Eth1.real, Eth1.imag))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) For p.f unity , Eth= 1.34000+0.00000i\n",
+ "(i) For p.f .8 , Eth= 1.59500+0.00000i\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.3, Page No 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "X=.625\n",
+ "P=1.0\n",
+ "Q=0.6\n",
+ "V=1.0\n",
+ "\n",
+ "#Calculations\n",
+ "Eth=V+(Q*X/V)+complex(P*X/V)\n",
+ "Phase_Eth=math.degrees(math.atan(Eth.imag/Eth.real))\n",
+ "\n",
+ "#Results\n",
+ "print(\"Eth=%.2f at an angle %.0f degrees\" %(abs(Eth),Phase_Eth))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Eth=2.00 at an angle 0 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.4, Page No 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "P=0.5\n",
+ "toff=4.0\n",
+ "\n",
+ "#Calculations\n",
+ "ton=(P*toff-0*toff)/(0.8-P)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Toff= 4min .\")\n",
+ "print(\"ton(min.)=%.3f min.\" %ton)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Toff= 4min .\n",
+ "ton(min.)=6.667 min.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.6 Page No 739"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "V=1.0\n",
+ "Qload=1.0*V\n",
+ "Qcap=-0.75*V**2\n",
+ "\n",
+ "#Calculations\n",
+ "Qnet=Qload+Qcap\n",
+ "VS=1-0.75*2*V # voltage sensitivity\n",
+ "\n",
+ "#Results\n",
+ "print(\"Voltage sensitivity=%.3f\" %VS)\n",
+ "print(\"since the voltage sensitivity is negative,\\nvoltage regulation by tap changing will reduce net reactive load and improive voltage stability \")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage sensitivity=-0.500\n",
+ "since the voltage sensitivity is negative,\n",
+ "voltage regulation by tap changing will reduce net reactive load and improive voltage stability \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.7, Page No 740"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Y=complex(-10)\n",
+ "n=1+0.1\n",
+ "\n",
+ "#Calculations\n",
+ "Y1=n*(n-1)*Y\n",
+ "Y2=(1-n)*Y\n",
+ "\n",
+ "#Results\n",
+ "print(\"Y1= {0:.2f}+{1:.2f}i\".format(Y1.real, Y1.imag))\n",
+ "print(\"Y2= {0:.2f}+{1:.2f}i\".format(Y2.real, Y2.imag))\n",
+ "print(\"The shunt elements equal to a reactor of 1.1V1^2 size oin the primary side and a capacitive of sixe 1V2**2 on the secondary side\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Y1= -1.10+0.00i\n",
+ "Y2= 1.00+-0.00i\n",
+ "The shunt elements equal to a reactor of 1.1V1^2 size oin the primary side and a capacitive of sixe 1V2**2 on the secondary side\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.8, Page No 745"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "P=1.0\t\t#assuming\n",
+ "S1=P/.95\t#For pf .95\n",
+ "S2=P/.8\t\t#For pf .8\n",
+ "\n",
+ "#Calculations\n",
+ "dMVA=(S2-S1)*100.0/P\t\t#Increase in MVA rating \n",
+ "Q1=P*math.tan(math.radians(math.degrees(math.acos(0.95))))\t\t#Q for pf .95\n",
+ "Q2=P*math.tan(math.radians(math.degrees(math.acos(0.8))))\t\t#Q for pf .8\n",
+ "dPc=(Q2-Q1)*100.0/Q1\t\t#Percent additional Reactive Power Capability \n",
+ "\n",
+ "#Results\n",
+ "print(\"Percent additional Reactive Power Capability is %.2f\" %dPc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percent additional Reactive Power Capability is 128.18\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter23.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter23.ipynb new file mode 100755 index 00000000..7d2d8122 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter23.ipynb @@ -0,0 +1,298 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 23 : State Estimation In Power Systems"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.1, Page No 148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "C1=0.02*100\n",
+ "C2=0.05\n",
+ "Fs=100\n",
+ "S1=complex(0.41,-0.11)\n",
+ "S2=complex(-0.4, 0.10)\n",
+ "S3=complex(-.105,0.11)\n",
+ "S4=complex(-.105,.11)\n",
+ "S5=complex(0.14,-.14)\n",
+ "S6=complex(-0.7,.35)\n",
+ "Z12=complex(0.08,.24)\n",
+ "Z23=complex(0.06,.18)\n",
+ "Z31=complex(0.02,.06)\n",
+ "Z21=Z12\n",
+ "Z32=Z23\n",
+ "Z13=Z31\n",
+ "\n",
+ "#Calculations\n",
+ "W1=(50*10**(-6))/((C1*abs(S1)+(C2*(Fs)))**2)\n",
+ "W2=(50*10**(-6))/((C1*abs(S2)+C2*(Fs))**2)\n",
+ "W3=(50*10**(-6))/((C1*abs(S3)+C2*(Fs))**2)\n",
+ "W4=(50*10**(-6))/((C1*abs(S4)+C2*(Fs))**2)\n",
+ "W5=(50*10**(-6))/((C1*abs(S5)+C2*(Fs))**2)\n",
+ "W6=(50*10**(-6))/((C1*abs(S6)+C2*(Fs))**2)\n",
+ "print(\"W1= %.2f\" %W1)\t\t\t#Answers for W1,W2,W3,W4,W5,W6 in the book is wrongly Calculated\n",
+ "print(\"W2= %.2f\" %W2)\t\n",
+ "print(\"W3= %.2f\" %W3)\t\n",
+ "print(\"W4= %.2f\" %W4)\t\n",
+ "print(\"W5= %.2f\" %W5)\t\n",
+ "print(\"W6= %.2f\" %W6)\t\n",
+ "a1=W1/(abs(13)**2)\n",
+ "[D]=diag([W1/(abs(Z13)**2)W2/(abs(Z31)**2)W3/(abs(Z12)**2)W4/(abs(Z21)**2)W5/(abs(Z23)**2)W6/(abs(Z32)**2)])\n",
+ "A=[-1 0 11 0 -11 -1 0-1 1 00 1 -10 -1 1]\n",
+ "B=[-1 01 01 -1-1 10 10 -1]\n",
+ "b=[1-100-11]\n",
+ "C=(B')*D#Assuming Transpose(B)D=C\n",
+ "F=(B')*D*B#Assuming Transpose(B)*D*B=F\n",
+ "G=(inv(F))*C#Assuming(BTDB)-1*(BT)*D=F\n",
+ "E1=1.05\n",
+ "E2=E1\n",
+ "E3=E1\n",
+ "invH=diag([Z31/E3Z13/E1Z12/E1Z21/E2Z23/E2Z32/E2])\n",
+ "Sm=[.41+%i*.11-.4-%i*.1-.105-%i*.11.14+%i*.14.72+%i*.37-.7+%i*.35]\n",
+ "EMo=invH*Sm\n",
+ "a=EMo-b*E1\n",
+ "E=G*a\n",
+ "\n",
+ "#Results\n",
+ "print(E,\"E=\") #Answers differs due to wrong calculation of W1,W2,W3,W4,W5,W6"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) For p.f unity , Eth= 1.34000+0.00000i\n",
+ "(i) For p.f .8 , Eth= 1.59500+0.00000i\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.3, Page No 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "X=.625\n",
+ "P=1.0\n",
+ "Q=0.6\n",
+ "V=1.0\n",
+ "\n",
+ "#Calculations\n",
+ "Eth=V+(Q*X/V)+complex(P*X/V)\n",
+ "Phase_Eth=math.degrees(math.atan(Eth.imag/Eth.real))\n",
+ "\n",
+ "#Results\n",
+ "print(\"Eth=%.2f at an angle %.0f degrees\" %(abs(Eth),Phase_Eth))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Eth=2.00 at an angle 0 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.4, Page No 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "P=0.5\n",
+ "toff=4.0\n",
+ "\n",
+ "#Calculations\n",
+ "ton=(P*toff-0*toff)/(0.8-P)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Toff= 4min .\")\n",
+ "print(\"ton(min.)=%.3f min.\" %ton)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Toff= 4min .\n",
+ "ton(min.)=6.667 min.\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.6 Page No 150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "V=1.0\n",
+ "Qload=1.0*V\n",
+ "Qcap=-0.75*V**2\n",
+ "\n",
+ "#Calculations\n",
+ "Qnet=Qload+Qcap\n",
+ "VS=1-0.75*2*V # voltage sensitivity\n",
+ "\n",
+ "#Results\n",
+ "print(\"Voltage sensitivity=%.3f\" %VS)\n",
+ "print(\"since the voltage sensitivity is negative,\\nvoltage regulation by tap changing will reduce net reactive load and improive voltage stability \")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage sensitivity=-0.500\n",
+ "since the voltage sensitivity is negative,\n",
+ "voltage regulation by tap changing will reduce net reactive load and improive voltage stability \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.7, Page No 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Y=complex(-10)\n",
+ "n=1+0.1\n",
+ "\n",
+ "#Calculations\n",
+ "Y1=n*(n-1)*Y\n",
+ "Y2=(1-n)*Y\n",
+ "\n",
+ "#Results\n",
+ "print(\"Y1= {0:.2f}+{1:.2f}i\".format(Y1.real, Y1.imag))\n",
+ "print(\"Y2= {0:.2f}+{1:.2f}i\".format(Y2.real, Y2.imag))\n",
+ "print(\"The shunt elements equal to a reactor of 1.1V1^2 size oin the primary side and a capacitive of sixe 1V2**2 on the secondary side\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Y1= -1.10+0.00i\n",
+ "Y2= 1.00+-0.00i\n",
+ "The shunt elements equal to a reactor of 1.1V1^2 size oin the primary side and a capacitive of sixe 1V2**2 on the secondary side\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.8, Page No 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "P=1.0\t\t#assuming\n",
+ "S1=P/.95\t#For pf .95\n",
+ "S2=P/.8\t\t#For pf .8\n",
+ "\n",
+ "#Calculations\n",
+ "dMVA=(S2-S1)*100.0/P\t\t#Increase in MVA rating \n",
+ "Q1=P*math.tan(math.radians(math.degrees(math.acos(0.95))))\t\t#Q for pf .95\n",
+ "Q2=P*math.tan(math.radians(math.degrees(math.acos(0.8))))\t\t#Q for pf .8\n",
+ "dPc=(Q2-Q1)*100.0/Q1\t\t#Percent additional Reactive Power Capability \n",
+ "\n",
+ "#Results\n",
+ "print(\"Percent additional Reactive Power Capability is %.2f\" %dPc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percent additional Reactive Power Capability is 128.18\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter24.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter24.ipynb new file mode 100755 index 00000000..07826bb5 --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter24.ipynb @@ -0,0 +1,171 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 24 : Unit Commitment"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.3, Page No 803"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "Fc1=1.1\t\t\t#Fuel cost(1)=Rs 1.1/MBtu\n",
+ "Fc2=1\t\t\t#Fuel cost(2)=1/MBtu\n",
+ "Fc3=1.2\t\t\t#Fuel cost(3)=1.2/MBtu\n",
+ "P1max=600.0\n",
+ "P1=P1max\n",
+ "\n",
+ "#Calculations\n",
+ "F1=600+7.1*P1+0.00141*(P1**2)\t#For P1= Pm1ax\n",
+ "Favg1=F1*Fc1/600.0\t\t\t\t#Full load average production cost\n",
+ "P2max=450.0\n",
+ "P2=P2max\n",
+ "F2=350+7.8*P2+0.00195*(P2**2)\t#For P2= P2max\n",
+ "Favg2=F2*Fc2/450.0\t\t\t\t#Full load average production cost\n",
+ "P3max=250.0\n",
+ "P3=P3max\n",
+ "F3=80+8*P3+0.0049*(P3**2)\t\t#For P3= P3max\n",
+ "Favg3=F3*Fc3/250.0\t\t\t\t#Full load average production cost\n",
+ "\n",
+ "#Results\n",
+ "print(\"Priority List is as follows\")\n",
+ "print(\"Unit Rs/MWhr MinMW Max MW\")\n",
+ "print(\" 2 %.3f 100 %.0f \" %(Favg2,P2max))\n",
+ "print(\" 1 %.4f 60 %.0f \" %(Favg1,P1max))\n",
+ "print(\" 3 %.2f 50 %.0f \" %(Favg3,P3max))\n",
+ "Fmax1=P1max+P2max+P3max\n",
+ "Fmax2=P2max+P1max\n",
+ "Fmax3=P2max\n",
+ "print(\"Unit Commitment Scheme is follows\")\n",
+ "print(\"Combination Min.MW from Combination Max.MW from Combination\")\n",
+ "print(\"2+1+3 310 %.0f \" %Fmax1)\n",
+ "print(\"2+1 260 %.0f \" %Fmax2)\n",
+ "print(\"2 100 %.0f \" %Fmax3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Priority List is as follows\n",
+ "Unit Rs/MWhr MinMW Max MW\n",
+ " 2 9.455 100 450 \n",
+ " 1 9.8406 60 600 \n",
+ " 3 11.45 50 250 \n",
+ "Unit Commitment Scheme is follows\n",
+ "Combination Min.MW from Combination Max.MW from Combination\n",
+ "2+1+3 310 1300 \n",
+ "2+1 260 1050 \n",
+ "2 100 450 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.4, Page No 805"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "#Calculations\n",
+ "\n",
+ "def F1(P1):\n",
+ " F1=7.1*P1+.00141*(P1^2)\n",
+ " print(\"F1(%.0f)=%.1f\" %(P1,F1))\n",
+ "\n",
+ "def F2(P2):\n",
+ " f2=7.8*P2+.00195*(P2^2)\n",
+ " print(\"f2(%.0f)=%.0f\" %(P2,f2))\n",
+ "\n",
+ "def F(P1,P2):\n",
+ " F1=7.1*P1+.00141*(P1**2)\n",
+ " F2=7.8*P2+.00195*(P2**2)\n",
+ " F=F1+F2\n",
+ " print(\"F1(%.0f)+f2(%.0f)=%.0f\" %(P1,P2,F))\n",
+ "\n",
+ "\n",
+ " \n",
+ "#Results\n",
+ "P1max=600\n",
+ "P2max=450\n",
+ "print(\"Unit Commitment using Load 500MW\")\n",
+ "F1(500)\n",
+ "print(\"\\n Since min. Power of second unit is 100MW , we find\")\n",
+ "F(400,100)\n",
+ "F(380,120)\n",
+ "F(360,140)\n",
+ "print(\"\\n Therefore for load 500 MW , the load commitment on unit 1 is 400 MW and that on 2 is 100 MW which gives min. cost\")\n",
+ "print(\"Next we increase the load by 50 MW and loading unit 1 we get, \\n\")\n",
+ "F1(550)\n",
+ "print(\"Also if we distribute a part of load to unit 2 we get ,\")\n",
+ "F(450,100)\n",
+ "F(400,150)\n",
+ "F(350,200)\n",
+ "print(\"\\n Therefore for load 550 MW , the load commitment on unit 1 is 400 MW and that on 2 is 150 MW which gives min. cost\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Unit Commitment using Load 500MW\n",
+ "F1(500)=3550.7\n",
+ "\n",
+ " Since min. Power of second unit is 100MW , we find\n",
+ "F1(400)+f2(100)=3865\n",
+ "F1(380)+f2(120)=3866\n",
+ "F1(360)+f2(140)=3869\n",
+ "\n",
+ " Therefore for load 500 MW , the load commitment on unit 1 is 400 MW and that on 2 is 100 MW which gives min. cost\n",
+ "Next we increase the load by 50 MW and loading unit 1 we get, \n",
+ "\n",
+ "F1(550)=3905.8\n",
+ "Also if we distribute a part of load to unit 2 we get ,\n",
+ "F1(450)+f2(100)=4280\n",
+ "F1(400)+f2(150)=4279\n",
+ "F1(350)+f2(200)=4296\n",
+ "\n",
+ " Therefore for load 550 MW , the load commitment on unit 1 is 400 MW and that on 2 is 150 MW which gives min. cost\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Power_System_by_C.L._Wadhwa/Chapter25.ipynb b/Electrical_Power_System_by_C.L._Wadhwa/Chapter25.ipynb new file mode 100755 index 00000000..ed59ab6b --- /dev/null +++ b/Electrical_Power_System_by_C.L._Wadhwa/Chapter25.ipynb @@ -0,0 +1,116 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 25 : Economic Scheduling of Hydrothermal Plants and Optimal Power Flows"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.1, Page No 817"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#initialisation of variables\n",
+ "q2=25.0\n",
+ "q3=25.0\n",
+ "q1=70-(q2+q3)\n",
+ "Wo=120.0\n",
+ "W3=50.0\n",
+ "Wi1=0\n",
+ "Wi2=0\n",
+ "W1=Wo+Wi1-q1\n",
+ "W2=W1+Wi2-q2\n",
+ "\n",
+ "#Calculations\n",
+ "PH1=9.81*(10**-3)*20*(1+(.5*.006*(120+100)))*(20-2)\n",
+ "PH2=9.81*(10**-3)*20*(1+(.5*.006*(100+75)))*(23)\t\t# Answer in the book is not Correct due to wrong calculation\n",
+ "PH3=9.81*(10**-3)*20*(1+(.5*.006*(75+50)))*(23)\n",
+ "PT1=8-PH1\n",
+ "PT2=12-PH2\n",
+ "PT3=7-PH3\n",
+ "L11=20+PT1\t\t\t#dFT/dPT=PT+20\n",
+ "L12=20+PT2\t\t\t#dF/dp=PT+20\n",
+ "L13=20+PT3\t\t\t#dF/dp=PT+20\n",
+ "#dPL/dPH=0\n",
+ "L31=L11\n",
+ "L32=L12\n",
+ "L33=L13\n",
+ "e=0.006\n",
+ "ho=0.1962\n",
+ "Rho=2\n",
+ "L21=L31*ho*(1+(.5*e*(2*Wo+Wi1-2*q1+Rho)))\n",
+ "L22=L21-L31*(.5*ho*e*(q1-Rho))-L32*(.5*ho*e*(q2-Rho))#for m=1\n",
+ "L23=L22-L32*(.5*ho*e*(q2-Rho))-L33*(.5*ho*e*(q3-Rho))#for m=2\n",
+ "G1=L22-L32*ho*(1+.5*.006*(2*100-2*25+2))\t\t#G1=dF/dq2 Answer doent match due to wrong calculation of PH2 in a book\n",
+ "G2=L23-L33*ho*(1+.5*.006*(2*W2+0-2*q3+Rho))\t\t#G1=dF/dq3\n",
+ "a=0.4\n",
+ "qnew2=q2-a*G1# Answer differs due to wrong calculation of PH2 in the book\n",
+ "qnew3=q3-a*G2\n",
+ "q1=120-50-(qnew2+qnew3)\n",
+ "\n",
+ "#Results\n",
+ "print(\"Let q2=%.0f q3=%.0f q1=%.0f\" %(q2,q3,q1))\n",
+ "print(\"W1=%.0f W2=%.0f\" %(W1,W2))\n",
+ "print(\"PH1=%.2f PH2=%.3f PH3=%.1f\" %(PH1,PH2,PH3))\n",
+ "print(\"Thermal generation during Three Intervals \\n PT1=%.2f PT2=%.2f PT3=%.1f\" %(PT1,PT2,PT3))\n",
+ "print(\"Value of L1 for the three intervals, \\n L11=%.2f L12=%.2f L13=%.1f\" %(L11,L12,L13))\n",
+ "print(\"Neglecting transmission losses we get\\n L11=L31 L12=L32 L13=L33\")\n",
+ "print(\"L21=%.3f\" %(L21))\n",
+ "print(\"For m=1 and 2 we get \\n L22=%.1f \\n L23=%.1f\" %(L22,L23))\n",
+ "print(\"Gradient Vectors \\n dF/dq2=%.2f\\n dF/dq3=%.1f\" %(G1,G2))\n",
+ "print(\"q2new=%.3f \\n q3new=%.1f\\n q1=%.0f\" %(qnew2,qnew3,q1))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Let q2=25 q3=25 q1=20\n",
+ "W1=100 W2=75\n",
+ "PH1=5.86 PH2=6.882 PH3=6.2\n",
+ "Thermal generation during Three Intervals \n",
+ " PT1=2.14 PT2=5.12 PT3=0.8\n",
+ "Value of L1 for the three intervals, \n",
+ " L11=22.14 L12=25.12 L13=20.8\n",
+ "Neglecting transmission losses we get\n",
+ " L11=L31 L12=L32 L13=L33\n",
+ "L21=6.975\n",
+ "For m=1 and 2 we get \n",
+ " L22=6.4 \n",
+ " L23=5.8\n",
+ "Gradient Vectors \n",
+ " dF/dq2=-0.77\n",
+ " dF/dq3=0.5\n",
+ "q2new=25.310 \n",
+ " q3new=24.8\n",
+ " q1=20\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |