{
"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": {}
}
]
}