{
"metadata": {
"name": "",
"signature": "sha256:a93d445dad4ffd499630570fa7ced24b4b25ee06fcd5352ae47d0eb721a47db5"
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"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"CHAPTER1 : WHAT MACHINES AND TRANSFORMERS HAVE IN COMMON"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E01 : Pg 16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import sqrt\n",
"horsepower=2.5 # rating of induction motor in horsepower at half load\n",
"Vl=230. # terminal voltage of motor in volts\n",
"Il=7. # load current of motor in amperes\n",
"pf=0.8 # power factor of the machine\n",
"Pin=sqrt(3.)*Vl*Il*pf # input power in watts\n",
"print\"Pin=\",Pin,\"W\"# The answer may vary due to roundoff error\n",
"Whp=746. # watts per hp\n",
"Pout=horsepower*Whp # output power in watts\n",
"print\"Pout=\",Pout,\"W\"\n",
"print\"n=\",Pout/Pin# The answer may vary due to roundoff error # efficiency of the machine\n",
"print\"Losses=Pin-Pout=\",Pin-Pout,\"W\"# The answer may vary due to roundoff error # losses in the machine in watts"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Pin= 2230.88144015 W\n",
"Pout= 1865.0 W\n",
"n= 0.835992431707\n",
"Losses=Pin-Pout= 365.881440149 W\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E02 : Pg 17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# the below exmaple is an extension of Ex1_1.sce\n",
"from math import sqrt \n",
"Vl=230. # terminal voltage of machine in volts\n",
"Il=7. # current drawn by machine in amperes\n",
"pf=0.8 # power factor of machine\n",
"Pin=sqrt(3.)*Vl*Il*pf # from Ex1_1 # input power in watts\n",
"Losses=365. # in watts\n",
"Pout=Pin-Losses # output power in watts\n",
"Whp=746. # watts per hp\n",
"print\"n=1-(Losses/Input)=\",1.-(Losses/Pin) # The answer may vary due to roundoff error # efficiency of the machine\n",
"print\"Pout=\",Pout,\"W\"# The answer may vary due to roundoff error\n",
"print\"Pout=\",Pout/Whp,\"hp\"# The asnwer may vary due to roundoff error # output power in horsepower"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"n=1-(Losses/Input)= 0.836387540175\n",
"Pout= 1865.88144015 W\n",
"Pout= 2.50118155516 hp\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E03 : Pg 17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import sqrt,pi \n",
"f=60. # frequency of voltage source in Hz\n",
"x=1.9 # Steinmetz coefficient\n",
"V=80. # applied sinusoidal voltage in volts\n",
"t=100. # no of turns wound on a coil\n",
"hc=500. # hysteresis coefficient \n",
"w=2.*pi*f # angular frequency in rads/sec\n",
"phimax=(sqrt(2.)*V)/(t*w)# maximum value of flux in the core in webers\n",
"print\"phimax=\",phimax,\"Wb\"# the answer may vary due to roundoff error\n",
"A1=0.0025 # cross-sectional area of core in metre square\n",
"Bmax1=phimax/A1 # flux density in core A in tesla\n",
"print\"Bmax=\",Bmax1,\"T\"# the answer may vary due to roundoff error\n",
"lfe1=0.5 # mean flux path length of core A in meters\n",
"VolA=A1*lfe1 # volume of core A in metre cube\n",
"print\"VolA=\",VolA,\"metre cube\"\n",
"# for core A\n",
"Ph1=VolA*f*hc*(Bmax1**x) # hysteresis loss in core A in watts\n",
"print\"Ph=\",Ph1,\"W\"# the answer may vary due to roundoff error\n",
"# for core B\n",
"A2=A1*3. # cross sectional area of core B in metre square\n",
"lfe2=0.866 # mean flux path length of core B in metres\n",
"Bmax2=phimax/A2 # flux density in core B in tesla\n",
"VolB=A2*lfe2 # volume of core B in metre cubes\n",
"Ph2=VolB*f*hc*(Bmax2**x) # hysteresis loss of core B in watts\n",
"print\"Ph=\",Ph2,\"W\"# the answer may vary due to roundoff error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"phimax= 0.00300105438719 Wb\n",
"Bmax= 1.20042175488 T\n",
"VolA= 0.00125 metre cube\n",
"Ph= 53.0597985532 W\n",
"Ph= 34.1904136606 W\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E04 : Pg 18"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"V1=240. # voltage applied to a winding of transformer(three phase) in volts\n",
"f1=60. # initial applied frequency in Hz\n",
"f2=30. # reduced frequency in Hz\n",
"Phe1=400. # core loss in watts at f1 frequency\n",
"Phe2=169. # core losses in watts at f2 frequency\n",
"print\"V2=\",(f2*V1)/f1,\"V\"# voltage at 30 Hz frequency\n",
"print\"Ph+e/f=Ch+Ce*f\"# equation for claculating hysteresis and eddy current loss coefficients\n",
"#a=[1 f1;1 f2] # left hand side matix for the equation above\n",
"#b=[Phe1/f1;Phe2/f2] # right hand side matrix for the equation above\n",
"#c=inv(a)*b\n",
"Ch=4.6#c(1,:)# hysteresis loss coefficient in W/Hz\n",
"Ce=0.0344#c(2,:)# eddy current loss coefficient in W/(Hz*Hz)\n",
"print\"Ph=\",Ch*f1,\"W\"# ans may vary due to roundoff error # hysteresis loss in watts at 60 Hz\n",
"print\"Pe=\",round(Ce*f1*f1),\"W\"# ans may vary due to roundoff error # eddy current loss at 60 Hz in watts"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"V2= 120.0 V\n",
"Ph+e/f=Ch+Ce*f\n",
"Ph= 276.0 W\n",
"Pe= 124.0 W\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E05 : Pg 20"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import sqrt \n",
"Pk=75. # core loss of transfomer in watts\n",
"R=0.048 # internal resistance in ohms\n",
"V2=240.# secondary voltage in volts\n",
"I2=sqrt(Pk/R)# secondary current in amperes\n",
"print\"I2=\",round(I2),\"A\"# ans may vary due to roundoff error\n",
"print\"|S|=V2*I2=\",round(V2*I2),\"VA\"# The answer in the textbook is wrong # output volt ampere of transformer"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I2= 40.0 A\n",
"|S|=V2*I2= 9487.0 VA\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E06 : Pg 22"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"sfl=1746 # speed at full load in rev/min\n",
"snl=1799.5 # speed at no load in rev/min\n",
"print\"Voltage Regulation=\",round((snl-sfl)/sfl,5) # the ans may vary due to round of error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Voltage Regulation= 0.03064\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E07 : Pg 22"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"Vnl=27.3 # no load voltage in volts\n",
"Vfl1=24. # full load voltage at power factor 1 in volts\n",
"print\"(Vnl-Vfl/Vfl)=\",(Vnl-Vfl1)/Vfl1# ans may vary due to roundoff error\n",
"Vfl2=22.1 # full load voltage at power factor 0.7 in volts\n",
"print\"Voltage Regulation=\",round((Vnl-Vfl2)/Vfl1,4)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(Vnl-Vfl/Vfl)= 0.1375\n",
"Voltage Regulation= 0.2167\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}