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"source": [
"# Chapter 2 Electric Fields"
]
},
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"metadata": {},
"source": [
"## Example 2_5 pgno:65"
]
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"text": [
"Maximum field = V/m per volt 42064315640.1\n"
]
}
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"source": [
"#Chapter 2, Example 5, page 65\n",
"#Calculate the maximum field at the sphere surface\n",
"#Calulating Field at surface E based on figure 2.31 and table 2.3\n",
"from math import pi\n",
"Q1 = 0.25\n",
"e0 = 8.85418*10**-12 #Epselon nought\n",
"RV1= ((1/0.25**2)+(0.067/(0.25-0.067)**2)+(0.0048/(0.25-0.067)**2))\n",
"RV2= ((0.25+0.01795+0.00128)/(0.75-0.067)**2)\n",
"RV= RV1+RV2\n",
"E = (Q1*RV)/(4*pi*e0)\n",
"print\"Maximum field = V/m per volt\",E\n",
"\n",
"#Answers vary due to round off error\n"
]
},
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"metadata": {},
"source": [
"## Example 2_6 pgno:66"
]
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"#Chapter 2, Exmaple 6, page 66\n",
"#calculation based on figure 2.32\n",
"\n",
"#(a)Charge on each bundle\n",
"print\"Part a\\t\"\n",
"req = (0.0175*0.45)**0.5\n",
"print\"Equivalent radius = m \", req\n",
"from math import log\n",
"from math import pi\n",
"V = 400*10**3 #Voltage\n",
"H = 12. #bundle height in m\n",
"d = 9. #pole to pole spacing in m\n",
"e0 = 8.85418*10**-12 #Epselon nought\n",
"Hd = ((2*H)**2+d**2)**0.5#2*H**2 + d**2\n",
"Q = V*2*pi*e0/(log((2*H/req))-log((Hd/d)))\n",
"q = Q/2\n",
"print\"Charge per bundle = uC/m \",Q #micro C/m\n",
"print\"Charge per sunconducter = uC/m \",q #micro C/m\n",
"\n",
"#(b part i)Maximim & average surface feild\n",
"print\"\\tPart b\"\n",
"print\"\\tSub part 1\\t\"\n",
"r = 0.0175 #subconductor radius\n",
"R = 0.45 #conductor to subconductor spacing\n",
"MF = (q/(2*pi*e0))*((1/r)+(1/R)) # maximum feild\n",
"print\"Maximum feild = kV/m \\t\",MF\n",
"MSF = (q/(2*pi*e0))*((1/r)-(1/R)) # maximum surface feild\n",
"print\"Maximum feild = kV/m \\t\",MSF\n",
"ASF = (q/(2*pi*e0))*(1/r) # Average surface feild\n",
"print\"Maximum feild = kV/m \\t\",ASF\n",
"\n",
"#(b part ii) Considering the two sunconductors on the left\n",
"print\"\\tSub part 2\\t\"\n",
"#field at the outer point of subconductor #1 \n",
"drO1 = 1/(d+r)\n",
"dRrO1 = 1/(d+R+r)\n",
"EO1 = MF -((q/(2*pi*e0))*(drO1+dRrO1))\n",
"print\"EO1 = kV/m \\t\",EO1\n",
"#field at the outer point of subconductor #2 \n",
"drO2 = 1/(d-r)\n",
"dRrO2 = 1/(d-R-r)\n",
"EO2 = MF -((q/(2*pi*e0))*(dRrO2+drO2))\n",
"print\"EO2 = kV/m \\t\",EO2\n",
"\n",
"#field at the inner point of subconductor #1 \n",
"drI1 = 1/(d-r)\n",
"dRrI1 = 1/(d+R-r)\n",
"EI1 = MSF -((q/(2*pi*e0))*(drI1+dRrI1))\n",
"print\"EI1 = kV/m \\t\",EI1\n",
"#field at the inner point of subconductor #2 \n",
"drI2 = 1/(d+r)\n",
"dRrI2 = 1/(d-R+r)\n",
"EI2 = MSF -((q/(2*pi*e0))*(dRrI2+drI2)) \n",
"print\"EI2 = kV/m \\t\",EI2\n",
"\n",
"#(part c)Average of the maximim gradient\n",
"print\"\\tPart c\\t\"\n",
"Eavg = (EO1+EO2)/2\n",
"print\"The average of the maximum gradient = kV/m \\t\",Eavg\n",
"\n",
"\n",
"#Answers might vary due to round off error\n"
]
},
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"metadata": {},
"source": [
"## Example 2_7 pgno:69"
]
},
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"execution_count": 4,
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{
"name": "stdout",
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"text": [
"Electric Feild = V/m \t30015596280.4\n"
]
}
],
"source": [
"#Chapter 2, Exmaple 7, page 69\n",
"#Electric feild induced at x\n",
"from math import pi\n",
"e0 = 8.85418*10**-12 #Epselon nought\n",
"q = 1 # C/m\n",
"C = (q/(2*pi*e0))\n",
"#Based on figure 2.33\n",
"E = C-(C*(1./3.+1./7.))+(C*(1+1./5.+1./9.))+(C*(1./5.+1./9.))-(C*(1./3.+1./7.))\n",
"print\"Electric Feild = V/m \\t\",E\n",
"\n",
"#Answers might vary due to round off error\n"
]
},
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"metadata": {},
"source": [
"## Example 2_8 pgno:70"
]
},
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"execution_count": 5,
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"text": [
"\n",
"Thickness of graded design= cm 4.24264068712\n",
"Curve = cm**2 62.4264068712\n",
"V1 = cm**3 47402.906725\n",
"Thickness of regular design = cm 14.684289433\n",
"V2 = cm**3 861.944682812\n"
]
}
],
"source": [
"#Chapter 2, Exmaple 8, page 70\n",
"#Calculate the volume of the insulator\n",
"#Thinkness of graded design\n",
"from math import e\n",
"from math import pi\n",
"V = 150*(2)**0.5\n",
"Ebd = 50\n",
"T = V/Ebd\n",
"print\"\\nThickness of graded design= cm \",T\n",
"#Based on figure 2.24\n",
"r = 2 # radius of the conductor\n",
"l = 10 #length of graded cylinder; The textbook uses 10 instead of 20\n",
"zr = l*(T+r)\n",
"print\"Curve = cm**2 \",zr\n",
"#Volume of graded design V1\n",
"V1 = 4*pi*zr*(zr-r)\n",
"print\"V1 = cm**3 \",V1 #Unit is wrong in the textbook\n",
"#Thickness of regular design as obtained form Eq.2.77\n",
"pow = V/(2*Ebd)\n",
"t = 2*(e**pow-1)\n",
"print\"Thickness of regular design = cm \",t\n",
"#Volume of regular design V2\n",
"V2 = pi*((2+t)**2-4)\n",
"print\"V2 = cm**3 \",V2#unit not mentioned in textbook\n",
" \n",
"#Answers may vary due to round off error\n"
]
},
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"metadata": {},
"source": [
"## Example 2_11 pgno:75"
]
},
{
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"execution_count": 6,
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{
"name": "stdout",
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"text": [
"The values of Phi2 and Phi4 are: [[ -3.6568 326.5 ]\n",
" [ 261.92857143 -4.37537287]]\n"
]
}
],
"source": [
"#Chapter 2, Exmaple 11, page 75\n",
"#Calculate the potential within the mesh\n",
"#Based on figure 2.38(b)\n",
"#equations are obtained using Eq.2.46\n",
"import numpy\n",
"from numpy import linalg\n",
"A1 = 1/2*(0.54+0.16)\n",
"A2 = 1/2*(0.91+0.14)\n",
"S = numpy.matrix([[0.5571, -0.4571, -0.1],[-0.4751, 0.828, 0.3667],[-0.1, 0.667, 0.4667]])\n",
"#By obtaining the elements of the global stiffness matrix(Sadiku,1994)\n",
"#and by emplying the Eq.2.49(a)\n",
"S1 = numpy.matrix([[1.25, -0.014],[-0.014, 0.8381]])\n",
"S2 = numpy.matrix([[-0.7786, -0.4571],[-0.4571, -0.3667]])\n",
"Phi13 = numpy.matrix([[0], [10]])\n",
"val1 = S2*Phi13\n",
"Phi24 = val1/S1\n",
"print\"The values of Phi2 and Phi4 are:\",Phi24\n",
"\n",
"#Answers may vary due to round of error \n",
"\n"
]
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