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{
"metadata": {
"name": "",
"signature": "sha256:7879febd57e90b9a18589786b4cd69033cefdde9f9003a10376af967492a8b6b"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"CHAPTER15 : MAGNETIC CIRCUIT COMPUTATIONS"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E01 : Pg 634"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# a\n",
"import math \n",
"phi = 6.*10.**-4.; # given magnetic flux (in Wb)\n",
"A = 0.001; # cross sectional area (in meter square)\n",
"B = phi/A ; # \n",
"Ha = 10.; # magnetic field intensity of material a needed to establish the given magnetic flux \n",
"Hb = 77.; # magnetic field intensity of material b\n",
"Hc = 270.; # magnetic field intensity of material c\n",
"La = 0.3; # arc length of material a (in meters)\n",
"Lb = 0.2; # arc length of material b (in meters) \n",
"Lc = 0.1; # arc length of material c (in meters)\n",
"\n",
"F = Ha*La + Hb*Lb + Hc*Lc; # magnetomotive force\n",
"print '%s' %(\"a\")\n",
"print '%s %.2f' %(\"magnetomotive force needed to establish a flux of 6*10**-4(in At) = \",F)\n",
"\n",
"# b\n",
"N = 100.; # no. of turns \n",
"I = F/N; # current in amps\n",
"print '%s' %(\"b\")\n",
"print '%s %.2f' %(\"current that must be made to flow through the coil(in amps) = \",I)\n",
"\n",
"# c\n",
"MU0 = 4.*math.pi*10.**-7.; \n",
"MUa = B/Ha; # permeability of material a\n",
"MUb = B/Hb; # permeability of material b\n",
"MUc = B/Hc; # permeability of material c\n",
"\n",
"MUra = MUa/MU0; # relative permeability of material a\n",
"MUrb = MUb/MU0; # relative permeability of material b\n",
"MUrc = MUc/MU0; # relative permeability of material c\n",
"\n",
"Ra = Ha*La/phi; # reluctance of material a \n",
"Rb = Hb*Lb/phi; # reluctance of material b\n",
"Rc = Hc*Lc/phi; # reluctance of material c\n",
"\n",
"print '%s' %(\"c\")\n",
"print '%s %.2f' %(\"relative permeability of material a = \",MUra)\n",
"print '%s %.2f' %(\"relative permeability of material b = \",MUrb)\n",
"print '%s %.2f' %(\"relative permeability of material c = \",MUrc)\n",
"print '%s %.2f' %(\"reluctance of material a = \",Ra)\n",
"print '%s %.2f' %(\"reluctance of material b = \",Rb)\n",
"print '%s %.2f' %(\"reluctance of material c = \",Rc)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a\n",
"magnetomotive force needed to establish a flux of 6*10**-4(in At) = 45.40\n",
"b\n",
"current that must be made to flow through the coil(in amps) = 0.45\n",
"c\n",
"relative permeability of material a = 47746.48\n",
"relative permeability of material b = 6200.84\n",
"relative permeability of material c = 1768.39\n",
"reluctance of material a = 5000.00\n",
"reluctance of material b = 25666.67\n",
"reluctance of material c = 45000.00\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E03 : Pg 637"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"mu0 = 4.*math.pi*10.**-7.;\n",
"A = 0.0025; # cross sectional area of the coil\n",
"# dimensions of the coil (in meters)\n",
"Lg = 0.002; # air gap length (in meters)\n",
"Lbd = 0.025; \n",
"Lde = 0.1;\n",
"Lef = 0.025;\n",
"Lfk = 0.2;\n",
"Lbc = 0.175;\n",
"Lcab = 0.5;\n",
"\n",
"Lbghc = 2.*(Lbd + Lde + Lef + (Lfk/2.)) - Lg;# length of the ferromagnetic material involved here\n",
"\n",
"phig = 4.*10.**-4.; # air gap flux (in Wb)\n",
"Bg = phig/A ; # air gap flux density (in tesla)\n",
"Hg = Bg/mu0 ; # feild intensity of the air gap \n",
"mmfg = Hg*Lg ; # mmf produced in the air gap (in At)\n",
"\n",
"Bbc = 1.38 ; # flux density corresponding to cast steel\n",
"\n",
"Hbghc = 125.; # field intensity corresponding to flux density of 0.16T in the steel\n",
"mmfbghc = Hbghc*Lbghc ; # mmf corresponding to bghc\n",
"\n",
"mmfbc = mmfg + mmfbghc ; # mmf across path bc\n",
"Hbc = mmfbc/Lbc;\n",
"phibc = Bbc*A ; # flux produced in bc \n",
"\n",
"phicab = phig + phibc; # total fiux existing in leg cab \n",
"Bcab = phicab/0.00375; # flux density\n",
"Hcab = 690.; \n",
"mmfcab = Hcab*Lcab; # mmf in leg cab\n",
"\n",
"mmf = mmfbc + mmfcab ; # mmf produced by the coil\n",
"\n",
"print '%s %.2f' %(\"mmf produced by the coil(in At) = \",mmf)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"mmf produced by the coil(in At) = 661.90\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E05 : Pg 646"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# b\n",
"import math \n",
"mu0 = 4.*math.pi*10.**-7. ;\n",
"# plunger magnet dimensions (in meters)\n",
"x = 0.025; \n",
"h = 0.05;\n",
"a = 0.025;\n",
"g = 0.00125; \n",
"\n",
"mmf = 1414.; # (in At)\n",
"\n",
"F = math.pi*a*mu0*(mmf**2.)*(h**2.)*(1./(x + h)**2.)/g; # magnitude of the force\n",
"print '%s %.2f' %(\"magnitude of the force (in Newtons) = \",F)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"magnitude of the force (in Newtons) = 70.16\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
