{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 8 : Q factor Power and Power Factor"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_1,pg 234"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Q factor of coil\n",
"\n",
"import math\n",
"#Variable declaration\n",
"fr= 400.0*10**3 #resonance frequency\n",
"C = 400.0*10**-12 #tuned capacitance\n",
"R = 10.0 #resistance of coil\n",
"n = 40.0 #Cp=nC\n",
"\n",
"#Calculations\n",
"Cp=n*(100.0/400.0)*10**-12 \n",
"L=(1.0/(4*(math.pi**2)*(fr**2)*(C+Cp)))\n",
"Q=2*math.pi*fr*(L/R)\n",
"\n",
"#Result\n",
"print(\"Inductance:\\nL = %f mH\"%(L*1000))\n",
"print(\"Observed Q-factor:\")\n",
"print(\"Q = %.2f \"%Q)\n",
"# Aanswer do not match with the book"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Inductance:\n",
"L = 0.386133 mH\n",
"Observed Q-factor:\n",
"Q = 97.05 \n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_2,pg 240"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# truncation error\n",
"\n",
"import math\n",
"#Variable declaration\n",
"fs=50*10**3 #sampling rate\n",
"delt=2.0 #summation interval\n",
"f=50.0 #signal frequency\n",
"\n",
"#Calculations\n",
"n=(fs/delt) #value of samples for 2s\n",
"maxer1=100.0/(2*n) #max error for synchronous case\n",
"maxer2=(100.0/(2*fs*delt*math.sin((2*math.pi*f)/fs)))\n",
"\n",
"#Result\n",
"print(\"max error for synchronous case:\")\n",
"print(\"maxer1 = %.3f%% \\n\"%maxer1)\n",
"print(\"max error for asynchronous case:\")\n",
"print(\"maxer2 = %.2f%% \"%maxer2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"max error for synchronous case:\n",
"maxer1 = 0.002% \n",
"\n",
"max error for asynchronous case:\n",
"maxer2 = 0.08% \n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_3,pg 258"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# find ratio errror and phase angle\n",
"\n",
"import math\n",
"#Variable declaration\n",
"#assume no iron loss and magnetizing current=1% of 10A, i.e 0.01A\n",
"Xs=1.884 #reactance of secondary\n",
"Rs=0.5 #resistance of secondary\n",
"Xm=2.0 #reactance of meter\n",
"Rm=0.4 #reactance of meter\n",
"Im=0.01 #magnetizing current\n",
"n2=10\n",
"n1=1\n",
"\n",
"#Calculations\n",
"B=math.atan((Xs+Xm)/(Rs+Rm))\n",
"#nominal ratio (n2/n1)=10/1\n",
"R=n2+((Im*math.sin(B))/n1) #actual impedance\n",
"R1=0.0097 #practical impedance\n",
"perer=(R1/R)*100 #percentage error\n",
"theta=((Im*math.cos(B))/n2)\n",
"\n",
"#Result\n",
"print(\"percentage error = %.3f%% \\n\"%perer)\n",
"print(\"phase angle:\")\n",
"print(\"theta = %.5f rad\"%(math.floor(theta*10**5)/10**5))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"percentage error = 0.097% \n",
"\n",
"phase angle:\n",
"theta = 0.00022 rad\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_4,pg 499"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# inductor Q factor and resistance\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Vc=100.0 #voltage across capacitor\n",
"Vi=12.0 #input voltage\n",
"f=100.0 #frequency of operation\n",
"Vl=100.0 #Vc=Vl at resonance\n",
"Ir=5.0 #current at resonance\n",
"\n",
"#Calculations\n",
"Q=(Vc/Vi) #Q-factor\n",
"Xl=(Vl/Ir) #inductive reactance\n",
"L=(Xl/(2*math.pi*f)) #inductance\n",
"Rl=(Xl/Q) #resistance\n",
"\n",
"#Result\n",
"print(\"Inductance of coil:\")\n",
"print(\"L = %.1f mH\\n\"%(L*1000))\n",
"print(\"Q-factor:\")\n",
"print(\"Q = %.2f\\n\"%Q)\n",
"print(\"Resistance of coil:\")\n",
"print(\"Rl = %.1f ohm\"%Rl)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Inductance of coil:\n",
"L = 31.8 mH\n",
"\n",
"Q-factor:\n",
"Q = 8.33\n",
"\n",
"Resistance of coil:\n",
"Rl = 2.4 ohm\n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_5,pg 499"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# actual Q factor and resistance\n",
"\n",
"import math\n",
"# Variable declaration\n",
"#when switch is open\n",
"C1=0.011*10**-6 #capacitance-1\n",
"Q1=10.0 #Q-factor-1\n",
"#when switch is closed\n",
"C2=0.022*10**-6 #capacitance-2\n",
"Q2=100.0 #Q-factor-2\n",
"\n",
"#Calculations\n",
"Qac=((Q1*Q2)/(Q1-Q2))*((C1-C2)/C1) #actual Q-factor\n",
"Rp=((Q1*Q2)/(Q2-Q1))*(1/(2*math.pi*C2)) #parallel resistance\n",
"\n",
"#Result\n",
"print(\"actual Q-factor:\")\n",
"print(\"Qac = %.2f \\n\"%Qac)\n",
"print(\"parallel resistance:\")\n",
"print(\"Rp = %.f M-ohm\"%(Rp/10**6))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"actual Q-factor:\n",
"Qac = 11.11 \n",
"\n",
"parallel resistance:\n",
"Rp = 80 M-ohm\n"
]
}
],
"prompt_number": 32
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_6,pg 499"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# find Q factor\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Cr=0.01*10**-6 #capacitance at resonance\n",
"Cu=0.014*10**-6 #capacitance at upper half\n",
"Cl=0.008*10**-6 #capacitance at lower half\n",
"\n",
"#Calculations\n",
"Qac=((2*Cr)/(Cu-Cl)) #actual Q-factor\n",
"\n",
"#Result\n",
"print(\"actual Q-factor:\")\n",
"print(\"Qac = %.2f \"%Qac)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"actual Q-factor:\n",
"Qac = 3.33 \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_7,pg 499"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# find lag\n",
"\n",
"import math\n",
"#Variable declaration\n",
"V=10.0 #v=10sin6280t\n",
"I=1.0 #current peak\n",
"P=3.1 #active power\n",
"\n",
"#Calculations\n",
"phi=math.acos((P*2)/V) #phase in radian\n",
"w=6280.0 #v=10sin6280t\n",
"lag=(phi/w) #lag\n",
"\n",
"#Result\n",
"print(\"lag = %.2f ms\"%(lag*10**3))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"lag = 0.14 ms\n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_8,pg 500"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# find truncation error\n",
"\n",
"import math\n",
"#Variable declaration\n",
"V=4.0 #peak voltage\n",
"I=0.4 #peak current\n",
"f=1*10**3 #operating frequency\n",
"fs=40*10**3 #sampling rate\n",
"delt=2.2 #time interval\n",
"\n",
"#Calculations\n",
"phi=((2*math.pi*f)/fs) #phase \n",
"Et=(V*I*phi)/(4*math.pi*f*delt*math.sin(phi))\n",
"\n",
"#Result\n",
"print(\"truncation error:\")\n",
"print(\"Et = %.1f * 10^-6 \"%(Et*10**6))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"truncation error:\n",
"Et = 58.1 * 10^-6 \n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_9,pg 500"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# find frequency of PF meter\n",
"\n",
"import math\n",
"#Variable declaration\n",
"ar=1.0 #gain of rectifier\n",
"nc=40.0 #turns ratio (1:40)\n",
"Vm=4.0 #peak load voltage\n",
"PF=0.85 #power factor\n",
"\n",
"#Calculations\n",
"f=(1/math.pi)*ar*Vm*nc*PF #frequency\n",
"\n",
"#Result\n",
"print(\"frequency of digital power meter:\")\n",
"print(\"f = %.1f Hz\"%f)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"frequency of digital power meter:\n",
"f = 43.3 Hz\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_10,pg 500"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# calculate ratio error and phase angle\n",
"\n",
"import math\n",
"#Variable declaration\n",
"Rp=94.0 #primary resistance\n",
"Xp=64.3 #primary reactance\n",
"Rs=0.85*10**2 #secondary resistance\n",
"Im=31*10**-3 #magnetizing current\n",
"PF=0.4 #power factor\n",
"n=10.0 #PT ratio\n",
"Is=1.0 #load current\n",
"Vs=110.0 #n=(Vp/Vs)\n",
"\n",
"#Calculations\n",
"B=math.acos(PF)\n",
"beta = math.floor(math.sin(B)*10)/10\n",
"R=Rp+Rs #total resistance\n",
"nerr=n+((((Is/n)*((R*PF)+(Xp*beta)))+Im*Xp)/Vs)\n",
"theta=((PF*(Xp/n))-(beta*(R/n))-(Im*Rp))/(Vs*n)\n",
"\n",
"#Result\n",
"print(\"ratio error:\")\n",
"print(\"nerr = %.3f\\n\"%nerr)\n",
"print(\"phase angle:\")\n",
"print(\"theta = %.3f\"%theta)\n",
"#Answer for theta do not match with the book"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ratio error:\n",
"nerr = 10.136\n",
"\n",
"phase angle:\n",
"theta = -0.015\n"
]
}
],
"prompt_number": 52
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example8_11,pg 500"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# calculate ratio error and phase angle\n",
"\n",
"import math\n",
"#Variable declaration\n",
"n=20.0 #(Vs/Is)\n",
"Is=5.0 #n=(Vs/Is)\n",
"Vs=100.0 #n=(Vs/Is)\n",
"N=0.25 #resistance to reactance ratio\n",
"Bur=15.0 #burden of CT=15VA (rating)\n",
"IL=0.13 #iron loss\n",
"Im=1.3 #magnetizing current\n",
"\n",
"#Calculations\n",
"V=(Bur/Is) #voltage rating\n",
"B=math.atan(N) #cos(B)-> power factor\n",
"#B=B*(180/math.pi) #conversion into degree\n",
"I=(Bur/Vs) #current rating\n",
"I1=(IL/I)\n",
"Rac=0.23 #actual value\n",
"R=n+((I1*math.cos(B)+Im*math.sin(B))/Is)\n",
"theta=((Im*math.cos(B)-I1*math.sin(B))/Vs)\n",
"nerr=-(Rac/R)*100 #ratio error\n",
"\n",
"# Result\n",
"print(\"ratio error:\")\n",
"print(\"nerr = %.3f%%\\n \"%nerr)\n",
"print(\"phase angle \\n\")\n",
"print(\"theta = %.4f\u00b0 \"%theta)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ratio error:\n",
"nerr = -1.137%\n",
" \n",
"phase angle \n",
"\n",
"theta = 0.0105\u00b0 \n"
]
}
],
"prompt_number": 56
}
],
"metadata": {}
}
]
}