PK I]Hx x E Basic Electrical Engineering with Numerical Problems/Chapter_01.ipynb{
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"cell_type": "heading",
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"metadata": {},
"source": [
"Introductions"
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}
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}PK I
=2 E Basic Electrical Engineering with Numerical Problems/Chapter_02.ipynb{
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"cells": [
{
"cell_type": "heading",
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"metadata": {},
"source": [
"Wires Cables and General Electrical Accessories"
]
}
],
"metadata": {}
}
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}PK I$h>{9 {9 E Basic Electrical Engineering with Numerical Problems/Chapter_03.ipynb{
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"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
Chapter 3: Electricity and its Fundamental Laws"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.1: Page 47"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"V = 230; # in volts\n",
"I = 10; # in A\n",
"\n",
"# calculations:\n",
"R = V/I;\n",
"\n",
"#Results\n",
"print \"resistance of element,R =\", R,\"ohm\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"resistance of element,R = 23.0 ohm\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.2: Page 47"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"R1 = 0.5; # minimum value of resistance in ohm\n",
"R2 = 20; # maximum value of resistance in ohm\n",
"I = 1.2; # current in A\n",
"\n",
"#Calculation\n",
"V1 = I*R1;\n",
"V2 = I*R2;\n",
"\n",
"#Results\n",
"print \"Voltage drop in Ist case,V1(V)=\",V1,\"volts and voltage drop in IInd case,V2(V)=\", V2,\"volts\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Voltage drop in Ist case,V1(V)= 0.6 volts and voltage drop in IInd case,V2(V)= 24.0 volts\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.3: Page 48"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"L = 1000; # length of wire in cm\n",
"d = 0.14; # diameter of wire in cm\n",
"R1 = 2.5*10**6;# resistance in micro-ohm\n",
"\n",
"# calculations:\n",
"a = (math.pi*d**2)/4; # cross section area\n",
"p = (R1*a)/L;\n",
"\n",
"#Results\n",
"print \"the specific resistance,p =\", round(p,1),\" uOhm-cm\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the specific resistance,p = 38.5 uOhm-cm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.4: Page 49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"Rt1 = 54.3;# resistance in ohm\n",
"alfa = 0.0043;# the resistance temperature of coeficient in per degree celcius\n",
"t1 = 20;# temperature in degree celcius\n",
"t2 = 40;# temperature in degree celcius\n",
"\n",
"# calculations\n",
"Rt2 = (Rt1*(1+(alfa*t2)))/(1+(alfa*t1));\n",
"\n",
"#Results\n",
"print \"resistance at 40 degC ,Rt2=\", Rt2,\" ohms\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"resistance at 40 degC ,Rt2= 58.6 ohms\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.5: Page 50"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"r1=30;# resistance in ohm\n",
"r2=35;# resistance in ohm\n",
"r3=45;# resistance in ohm\n",
"V=220;\n",
"\n",
"# calculations:\n",
"R=r1+r2+r3;\n",
"I=V/R;\n",
"\n",
"#Results\n",
"print \"(a)total resistance,R=\", R,\" ohm\"\n",
"print \"(b)current,I=\", I,\"A\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)total resistance,R= 110 ohm\n",
"(b)current,I= 2.0 A\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.6: Page 50"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"I=75;# current in A\n",
"R=0.15;# resistance in ohm\n",
"v=220;# voltage in volts\n",
"\n",
"#calculations\n",
"V1=I*R;# voltage drop of the feeder in section AB\n",
"V2=I*R;# voltage drop of the feeder in section CD\n",
"V_total=V1+V2;# total voltage drop in the lead and return feeder\n",
"V=v+V_total;\n",
"\n",
"#Results\n",
"print \"voltage at the generating station,V=\", V,\"volts\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"voltage at the generating station,V= 242.5 volts\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.7: Page 52"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"r1=6;# resistance in ohm\n",
"r2=10;# resistance in ohm\n",
"r3=15;# resistance in ohm\n",
"\n",
"#calculations:\n",
"r=(1/r1)+(1/r2)+(1/r3);\n",
"R=1/r;\n",
"\n",
"#Results\n",
"print \"equivalent resistance,R=\", R,\"ohm\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"equivalent resistance,R= 3.0 ohm\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.8: Page 53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"I=5; # current in A\n",
"n=2; # number of resistance in parallel of section BC\n",
"r1=15;# resistance in ohm\n",
"r2=20;# resistance in ohm\n",
"r3=60;# resistance in ohm\n",
"r4=64;# resistance in ohm\n",
"r5=64;# resistance in ohm\n",
"r6=2.5;# resistance in ohm\n",
"\n",
"#calculation\n",
"R1=r4/n;# equivalent resistance of section BC\n",
"R2=(r1*r2*r3)/((r1*r2)+(r2*r3)+(r3*r1));# equivalent resistance of section CD\n",
"R=R1+R2+r6;# equivalent resistance of section AD\n",
"V=I*R;\n",
"\n",
"#Results\n",
"print \"voltage,V= \", V,\"volts\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"voltage,V= 210.0 volts\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.9: Page 53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"V=240;# voltage in volts\n",
"r1=2;# resistance in ohm\n",
"r2=3;# resistance in ohm\n",
"r3=8.8;# resistance in ohm\n",
"r4=10;# resistance in ohm\n",
"r5=3;# resistance in ohm\n",
"\n",
"# calculations:\n",
"R1=(r1*r2)/(r1+r2);# equivalent resistance of parallel branch\n",
"R2=R1+r3;# equivalent resistance of section ABC\n",
"R3=(R2*r4)/(R2+r4);\n",
"R=R3+r5;# total resistance of section AD\n",
"I=V/R;\n",
"V1=I*r5;# voltage drop across r5\n",
"V2=V-V1;# voltage drop across section ABC\n",
"I1=V2/r4;# current flowing through r4 resistance\n",
"I2=I-I1;# current in r3 resistance\n",
"V3=I2*r3;# voltage drop across r3 resistance, section ABC\n",
"V4=V2-V3;# voltage drop between section AB\n",
"I3=V4/r1;# current flowing through r1 resistance\n",
"I4=V4/r2;# current flowing through r2 resistance\n",
"\n",
"#Results\n",
"print \"current flowing through r1 (2 ohms) resistance,I3 =\", I3,\" A\"\n",
"print \"current flowing through r2 (3 ohms)resistance,I4 =\", I4,\" A\"\n",
"print \"total resistance,R = \", R,\" ohm\"\n",
"print \"voltage drop across r5(3 ohms) resistance,V1 =\", V1,\" volts\"\n",
"print \"voltage drop across section ABC,V2 = \", V2,\" volts\"\n",
"print \"voltage drop across r3 resistance(8.8 ohms),V3 = \",V3,\" Volts\"\n",
"print \" voltage drop between section AB,V4 = \", V4,\"volts\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current flowing through r1 (2 ohms) resistance,I3 = 9.0 A\n",
"current flowing through r2 (3 ohms)resistance,I4 = 6.0 A\n",
"total resistance,R = 8.0 ohm\n",
"voltage drop across r5(3 ohms) resistance,V1 = 90.0 volts\n",
"voltage drop across section ABC,V2 = 150.0 volts\n",
"voltage drop across r3 resistance(8.8 ohms),V3 = 132.0 Volts\n",
" voltage drop between section AB,V4 = 18.0 volts\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.10: Page 54"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"I=44;# current in A\n",
"r1=6;# resistance in ohm\n",
"r2=12;# resistance in ohm\n",
"r3=18;# resistance in ohmr1\n",
"\n",
"# calculation:\n",
"a=(1/r1)+(1/r2)+(1/r3);\n",
"R=1/a;\n",
"V=I*R;\n",
"i1=V/r1;\n",
"i2=V/r2;\n",
"i3=V/r3;\n",
"\n",
"#Results\n",
"print \"current in 6 ohm resistance,i1 = \",i1,\"A\" \n",
"print \"current in 12 ohm resistance,i2 = \",i2,\"A\"\n",
"print \"current in 18 ohm resistance,i3 = \",i3,\"A\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current in 6 ohm resistance,i1 = 24.0 A\n",
"current in 12 ohm resistance,i2 = 12.0 A\n",
"current in 18 ohm resistance,i3 = 8.0 A\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.11: Page 55"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given:\n",
"t=15 #TOTAL CURRENT IN AMPERES\n",
"i1=2 #CURRENT THROUGH UNKNOWN RESISTANCE\n",
"R1=15#in ohms\n",
"R2=50/2#in ohms\n",
"\n",
"# calculations:\n",
"x=(t-i1)*((R1*R2)/(R1+2*R2))#unknown resistance in ohms)\n",
"PD=i1*x#in volts\n",
"RX=((1/R1)+(1/(2*R2))+(1/x))#\n",
"R=1/RX\n",
"i5= PD/(2*R2)#current in 5 ohms resistance\n",
"i15=PD/R1#current in 15 ohms resistance\n",
"\n",
"\n",
"#Results\n",
"print \"(a)unknown resistance in ohms =\", x\n",
"print \"(b)potential drop across the circuit in volts is =\", PD\n",
"print \"(c)current in 5 ohms resistance in ampere =\",i5,\"\\n and current in 15 ohms resistance in ampere =\", i15\n",
"print \"(d)total resistance of the circuit in ohms =\",R"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)unknown resistance in ohms = 75.0\n",
"(b)potential drop across the circuit in volts is = 150.0\n",
"(c)current in 5 ohms resistance in ampere = 3.0 \n",
" and current in 15 ohms resistance in ampere = 10.0\n",
"(d)total resistance of the circuit in ohms = 10.0\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}PK IJ J E Basic Electrical Engineering with Numerical Problems/Chapter_04.ipynb{
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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 4: Work Power and Energy"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.1: Page 61"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"I=11;# current in A\n",
"V1=55;# voltage in V\n",
"V2=220;# voltage in V\n",
"\n",
"#calculations\n",
"V=V2-V1;\n",
"R=V/I; \n",
"P=I**2*R;\n",
"\n",
"#Results\n",
"print \"(a)resistance,R = \", R,\" ohm\" \n",
"print \"(b)power lost,P = \",P,\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)resistance,R = 15.0 ohm\n",
"(b)power lost,P = 1815.0 W\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.2: Page 61"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"V=300;# voltage in volts\n",
"W=360;# power lost in one coil in watt\n",
"I=6; # current in A\n",
"\n",
"#calculations:\n",
"R1=V/I;\n",
"R=V**2/W;\n",
"a=(1/R1)-(1/R);\n",
"r2=1/a;\n",
"\n",
"#Results\n",
"print \"resistance of 360W coil1,R= \",R,\"ohm and \\n resistance of second coil2,r2=\",r2,\"ohm\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"resistance of 360W coil1,R= 250.0 ohm and \n",
" resistance of second coil2,r2= 62.5 ohm\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.3: Page 61"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"W1=100# in watt\n",
"E=110# in volts\n",
"W2=60# in watt\n",
"\n",
"#calculations:\n",
"I1=W1/E# current taken by 100 w lamp\n",
"I2=W2/E# current taken by 60W lamp\n",
"I=I1-I2;\n",
"R=E/I;\n",
"\n",
"#Results\n",
"print \"resistance,R =\", R,\" ohm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"resistance,R = 302.5 ohm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.4: Page 62"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"w=100;# in watt\n",
"V=220;# voltage in volts\n",
"\n",
"#calculations:\n",
"R1=V**2/w;\n",
"Rp=R1/2;# total resistance of the circuit\n",
"Ip=V/Rp;\n",
"Wp=Ip**2*Rp;\n",
"R2=V**2/w;\n",
"Rs=R1+R2;# total resistance of the circuit\n",
"Is=V/Rs;\n",
"Ws=Is**2*Rs;\n",
"\n",
"#Results\n",
"print \"(a)power in case of parallel,W = \",Wp,\"watts\"\n",
"print \"(b)power in case of series, W = \",Ws,\" Watts\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)power in case of parallel,W = 200.0 watts\n",
"(b)power in case of series, W = 50.0 Watts\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.5: Page 63"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"V = 220; #voltage\n",
"l=300;# number of lamps\n",
"w1=60;# in watt\n",
"w2=40;# in watt\n",
"f=100;# number of fan\n",
"\n",
"# Calculations:\n",
"W1=w1*l;# wattage required for 300 lamps, 60 watt each\n",
"W2=w2*f# wattage required for 100 fans, 40 watt each\n",
"W=(W1+W2)*10**-3;\n",
"I=(W*1000)/V;\n",
"\n",
"#Results\n",
"print \"(a)total load,W = \",W,\" kW\" \n",
"print \"(b)current,I = \",I,\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)total load,W = 22.0 kW\n",
"(b)current,I = 100.0 A\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.6: Page 63"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given and Calculations:\n",
"nl=12#no. of lamps\n",
"wl=100#wattage of lamps\n",
"hl=6 #each lamps work 6 hours a days\n",
"nf=6 #no. of fans\n",
"wf=60#wattage of fans\n",
"hf=5 #each fans work 5 hours a days\n",
"nc=2 #no. of electric cookers\n",
"wc=1500#wattage of electric cookers\n",
"hc=4 #each electric cookers work 4 hours a days\n",
"ng=2 #no. of gysers\n",
"wg=1000#wattage of each gyser\n",
"hg=3 #each gyser works 3 hours a day\n",
"Ccg=40#IN PAISA\n",
"Ccg1=35#IN PAISA\n",
"\n",
"# Calculations:\n",
"w12=wl*nl*hl#wattage of 12 lamps in Wh\n",
"w6=wf*nf*hf#wattage of 12 fans in Wh\n",
"w2=wc*nc*hc#wattage of 2 electric cookers in Wh\n",
"w21=wg*hg*ng#total wattage of gysers in Wh\n",
"tcg=(w12+w6)*10**-3#TOTAL WATTAGE OF LAMPS AND FANS\n",
"Ecg= (tcg*Ccg*30)/100#TOTAL ENERGY CHARGES @40 PAISA PER UNIT\n",
"tcg1=(w2+w21)*10**-3#TOTAL WATTAGE OF COOKERS AND GYSERS\n",
"Ecg1= (tcg1*Ccg1*30)/100#TOTAL ENERGY CHARGES @35 PAISA PER UNIT\n",
"tc=Ecg+Ecg1# IN RUPPES\n",
"\n",
"#Results\n",
"print \"total cost of electric charge @40 paisa per unit in rupees\",Ecg\n",
"print \"total cost of electric charge @35 paisa per unit in rupees\",Ecg1\n",
"print \"total charge for ligh and power in rupees\",tc"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"total cost of electric charge @40 paisa per unit in rupees 108.0\n",
"total cost of electric charge @35 paisa per unit in rupees 189.0\n",
"total charge for ligh and power in rupees 297.0\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.7: Page 63"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#given and Calculations\n",
"V = 250;#volts\n",
"r = 0.03; #in ohms\n",
"I = 20; #in Amps\n",
"nl=400#no. of lamps\n",
"wl=100#wattage of lamps\n",
"nf=100#no. of fans\n",
"wf=40#wattage of fans\n",
"nc=200#no. of wall scokets\n",
"wc=60#wattage of wall scckets\n",
"hl=50#heating load in kW\n",
"h=5 # hours\n",
"\n",
"# Calculations:\n",
"w400=wl*nl#wattage of 400 lamps in W\n",
"w6=wf*nf#wattage of 100 fans in W\n",
"w2=wc*nc#wattage of 200 wall sockets in Wh\n",
"tc= (w400+w6+w2)/1000#total consumption in kW\n",
"Ml=V*I/1000#miscellaneous loads in kW\n",
"Mo= ((50*80*746)/(100*1000))#MOTOR AT 80% LOAD IN Kw\n",
"tl=tc+Ml+hl+Mo#total load in kW\n",
"It = tl*1000/V\n",
"Vc=It*r#voltage drop in the cable\n",
"Vs=Vc+V#voltage at the sending end of the feeder in volts\n",
"Pw=It**2*r#power wasted in kW\n",
"ll=tc*h#lightning load in kWh\n",
"te=Ml*2 + ll#TOTAL ENERGY COSNUMED PER DAY\n",
"Nu=te*6#NO. OF UNITS\n",
"Ec=(Nu*30)/100# ENERGY CHARGE @30 PAISA PER UNIT\n",
"eCM=Ec+2+34.80#TOTAL CHARGE AFTER TAX AND RENT IN RUPEES.\n",
"hlh=hl*4#heating load in kWh\n",
"Moh=Mo*8#MOTOR LOAD IN kWh\n",
"TEP=hlh+Moh#total energy per day\n",
"tepl=TEP*6#total energy in 6 days\n",
"tepc=(tepl*35)/100# energy charges @35 paisa per unit in rupees\n",
"tepcl=tepc+50+78.96#total charges in rupess\n",
"\n",
"GTb = eCM + tepcl\n",
"\n",
"\n",
"#Results\n",
"print \"(a)total consumption of factory is \", tl,\"kW\"\n",
"print \"(b)total current taken buy the factory\",It,\" Amp\"\n",
"print \"(c)voltage at the sending end of the feeder is\",round(Vs,1),\"Volts\"\n",
"print \"(d)power wasted is\",round(Pw/1000,2),\"kW\"\n",
"print \"(e)total lightning charges- including meter rent and electricy tax is,(Rs)=\",round(eCM,2)\n",
"print \"total power charges including meter rent and electricy tax is,(Rs)=\",round(tepcl,2)\n",
"print \"grand total of bills is,(Rs)=\",round(GTb,2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)total consumption of factory is 140.84 kW\n",
"(b)total current taken buy the factory 563.36 Amp\n",
"(c)voltage at the sending end of the feeder is 266.9 Volts\n",
"(d)power wasted is 9.52 kW\n",
"(e)total lightning charges- including meter rent and electricy tax is,(Rs)= 558.8\n",
"total power charges including meter rent and electricy tax is,(Rs)= 1050.27\n",
"grand total of bills is,(Rs)= 1609.07\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.8: Page 64"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"V=250;# voltage in volts\n",
"L=5*746;# 1 hp=746 watt\n",
"eta=80# eficiency of motor in %\n",
"\n",
"# Calculations:\n",
"Input=(L*100)/80; \n",
"I=Input/V;\n",
"\n",
"#Results\n",
"print \"cureent,I(A) = \", I"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"cureent,I(A) = 18.65\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.9: Page 64"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given:\n",
"p=30 #horse power of motor\n",
"r=24 # rupees per kWh\n",
"ec=35#paisa per unit\n",
"n=80 #percentage of load\n",
"t=8 # in hours\n",
"d=25 # total days\n",
"ne=96#efficiency of motor in percentage\n",
"\n",
"#calculations:\n",
"mo=(n*p)/100#output of motor at 80% of load\n",
"mi=(mo*100*746)/(ne)#input of motor in watts\n",
"ecm=mi*10**-3*t*d#energy consumed in a month\n",
"ecu=(ecm*35)/100#energy charges\n",
"mid=(30*100*746)/(ne*1000)#input of motor in kW at demanded\n",
"ecud=(mid*24)# demanded connection in rupees\n",
"ta=ecu+ecud#total bill in rupees\n",
"\n",
"#Results\n",
"print \"total bill in rupees is\",ta"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"total bill in rupees is 1865.0\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.10: Page 65"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given:\n",
"V=400#three phase voltage\n",
"lp=50#no. of light points\n",
"lw=60#wattage of light points\n",
"fp=20#no. of fan points\n",
"fw=100#wattage of fan points\n",
"wpp=10#no. of wall plug points\n",
"wppw=60#wattage of wall plug points\n",
"bp=5 #no. of bell points\n",
"bpw=40#wattage of bell points\n",
"ppp=8#power plug points\n",
"pppw=500#wattage of power plug points\n",
"\n",
"#calculations:\n",
"lpw=lp*lw#wattage of 50 lamps\n",
"fpw=fp*fw#wattage of 20 fans\n",
"wpppw=wpp*wppw#wattage of wall plug points\n",
"bpww=bp*bpw#wattage of bell points \n",
"tl=lpw+fpw+wpppw+bpww#total wattage\n",
"ppppw=ppp*pppw#wattage of power plug points\n",
"tw=tl+ppppw#total wattage\n",
"Il=(tl/V)# CURRENT THROUGH LIGHTNING LOAD\n",
"Ip=ppppw/V# current through power load\n",
"ttl=Il+Ip#total load curent\n",
"\n",
"#Results\n",
"print \"total wattage of lightning load is in watts = \", tl\n",
"print \"total load current in amperes = \",ttl"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"total wattage of lightning load is in watts = 5800\n",
"total load current in amperes = 24.5\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.11: Page 66"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data and calculations:\n",
"h=30;# in m\n",
"Fl=10# friction loss in %\n",
"eta=90;# eficiency of pump\n",
"w=1000;# water weight in kg\n",
"flow_rate=243;# in per hour\n",
"\n",
"# Calculations:\n",
"Hl=(Fl/100)*h;\n",
"total_H=h+Hl;\n",
"W_done=(flow_rate*w*total_H)/60;# in kg-m/min\n",
"output=W_done/4500;#output of pump in hp\n",
"In=(output*100)/eta;\n",
"O=In;\n",
"\n",
"#Results\n",
"print \"output of the motor,O(hp) = \",O"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"output of the motor,O(hp) = 33.0\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 4.12: Page 66"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"l=7.5#load in tonnes\n",
"h=135#height in meters\n",
"c=0.5#cge weight in tonnes\n",
"b=3 #balance weight in tonnes\n",
"td=90#time in seconds\n",
"onet=1000# in kg\n",
"onehp=746#watt\n",
"\n",
"#calculations:\n",
"wl=l+c-b#weight lifted during upward journey in tonnes\n",
"wld=b-c#weight lifted during downward journey in tonnes\n",
"wdu=(wl*10**3*h*60)/td#work done by the lift per minute during upward journey\n",
"wdd=(wld*10**3*h*60)/td#work done by the lift per minute during downward journey\n",
"mou=wdu/4500# in hp\n",
"miu=(mou*100*746)/(n*1000)# input of motor in kW\n",
"mod=wdd/4500# in hp\n",
"mid=(mod*100*746)/(n*1000)# input of motor in kW\n",
"tc=miu+mid#total energy consumption in kW\n",
"Eh=tc*10#total energy consuption per hour\n",
"rate=40#rate in paisa\n",
"ce=Eh*(rate/100)#cost of energy in rupees\n",
"\n",
"#Results\n",
"print \"(a1)BHP of the motor in upward journey in hp\",mou \n",
"print \"(a2)BHP of the motor in downward journey in hp\",mod\n",
"print \"(b)cost of energy in rupees is\" ,ce"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a1)BHP of the motor in upward journey in hp 100.0\n",
"(a2)BHP of the motor in downward journey in hp 50.0\n",
"(b)cost of energy in rupees is 559.5\n"
]
}
],
"prompt_number": 12
}
],
"metadata": {}
}
]
}PK I=
E Basic Electrical Engineering with Numerical Problems/Chapter_05.ipynb{
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"name": "",
"signature": "sha256:179d1aa791f84982595aec77701ac00e5f4718cb4b0fb539bad5702fddb2a8a1"
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5:Chemical and Heating Effects of Electric Current"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.1: Page 74:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"t=200;# time in sec\n",
"M=111.83;# silver in mg\n",
"I=0.5;# current in A\n",
"\n",
"#calculations:\n",
"Z=(M/(I*t*1000))*1000# electro-chemical-equivalent\n",
"\n",
"#Results\n",
"print \"E.C.E,Z(mg/C) = \",Z"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"E.C.E,Z(mg/C) = 1.1183\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.2: page 74:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"Z=0.329*10**-3# IN g/C\n",
"I=1 # in amperes\n",
"t=90*60# in seconds\n",
"\n",
"#calculation:\n",
"M=Z*I*t# in grams \n",
"A=200#area in centimete square\n",
"S=8.9#density in g/cc\n",
"T=(M)/(2*A*S)#thickness in cm\n",
"\n",
"#Results\n",
"print \"thickness of copper in cm is\", round(T,6)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"thickness of copper in cm is 0.000499\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.3: Page 76:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"w=15 # in kg\n",
"t1=15# in degree celsius\n",
"t2=100#in degree celsius\n",
"t=25 # time in minutes\n",
"I=10 # in ampere\n",
"n=85 #efficiency of conversion in percentage\n",
"\n",
"#calculations:\n",
"ho=w*(t2-t1)#output heat required in kcal\n",
"R=((ho*4187*100)/(I**2*t*60*n))# resistance in ohms\n",
"\n",
"#Results\n",
"print \"resistance in ohms\",R"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"resistance in ohms 41.87\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.4: page 76:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"w=20 # in kg\n",
"t1=10# in degree celsius\n",
"t2=90#in degree celsius\n",
"t=2*3600+19*60+34# time in seconds\n",
"I=4 # in ampere\n",
"n=80 #efficiency of conversion in percentage\n",
"\n",
"#calculations:\n",
"ho=w*(t2-t1)#output heat required in kcal\n",
"V=((ho*4187*100)/(I*t*n))# POTENTIAL DROP IN VOLTS\n",
"\n",
"#Results\n",
"print \"potential drop across heater element in volts is\", V"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"potential drop across heater element in volts is 250.0\n"
]
}
],
"prompt_number": 4
}
],
"metadata": {}
}
]
}PK Iݦ) ) E Basic Electrical Engineering with Numerical Problems/Chapter_06.ipynb{
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"name": "",
"signature": "sha256:574b119cc7251d26efe27fb23031d2a66fe48006b301b495653100a11e0e2a3e"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 6: Magnetism, Electromagnetism and Electromagnetic Induction"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.1: page 99:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"I=1.5# in amperes\n",
"n=50 #turns\n",
"l=0.25#length of coil in meter\n",
"\n",
"#calculations:\n",
"H=(I*n)/l#field strength in ampere-turns/m\n",
"\n",
"#Results\n",
"print \"field strength in ampere-turns/m is\",H"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"field strength in ampere-turns/m is 300.0\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.2: page 100:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"I=70 # in amperes\n",
"B=0.4#flus density in Wb/m**2\n",
"n=1 #turns\n",
"\n",
"#calculations:\n",
"F=B*n*I# in newton\n",
"\n",
"#Results\n",
"print \"force in newtons is\", F"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"force in newtons is 28.0\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.3: page 104:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"b=2 #in Wb/m**2\n",
"l=6 #in cm\n",
"s=0.75#in m's\n",
"alpha=90#\n",
"\n",
"#calculations:\n",
"emf=b*l*s*(math.sin(alpha*math.pi/180))#\n",
"\n",
"#Results\n",
"print \"emf induced in volts is\",emf"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"emf induced in volts is 9.0\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}PK IRHA# # E Basic Electrical Engineering with Numerical Problems/Chapter_07.ipynb{
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"name": "",
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7: DC Generators"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.1: Page 114:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"# given data:\n",
"p=8; # number of poles\n",
"a1=p; # in lap winding\n",
"a2=2; # in wave winding\n",
"fi=15*10**-3;# in wb\n",
"N=500;# rev/min\n",
"Z=800;# number of conductors on armature\n",
"\n",
"#calculations:\n",
"emf1=(fi*Z*N*p)/(60*a1)# when the armature is lap wound\n",
"emf2=(fi*Z*N*p)/(60*a2)# when the armature is wave wound\n",
"\n",
"#Results\n",
"print \"when the armature is lap wound, emf(V) = \",emf1\n",
"print \"when the armature is wave wound, emf(V) = \",emf2"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"when the armature is lap wound, emf(V) = 100.0\n",
"when the armature is wave wound, emf(V) = 400.0\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.2: Page 119:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"Vt=200# terminal voltage in volts\n",
"Rsh=100;#shunt fieldresistance in ohm\n",
"Ra=0.1;# armature resistance in ohm\n",
"l=60;# number of lamps\n",
"w=40 # in watt\n",
"N=4; # number of poles\n",
"\n",
"#calculations:\n",
"total_l=l*w# in watt\n",
"Il=total_l/Vt# load current\n",
"Ish=Vt/Rsh# shunt field current\n",
"Ia=Il+Ish;\n",
"I=Ia/N;\n",
"Va=Ia*Ra#armature voltage drop \n",
"Vb=1+1;# brush contact drop for 2 pair of poles\n",
"E=Vt+Va+Vb;\n",
"\n",
"#Results\n",
"print \"(a)armature current,Ia(A) = \",Ia\n",
"print \"(b)current per path in a armature,I(A) =\",I\n",
"print \"(c)emf,E(Volts) = \",E"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)armature current,Ia(A) = 14.0\n",
"(b)current per path in a armature,I(A) = 3.5\n",
"(c)emf,E(Volts) = 203.4\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.3: Page 119:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"# given data:\n",
"W=10 # output of the generator in k-w\n",
"V=250;# voltage in volts\n",
"R=0.07;# in ohm\n",
"Rsh=63.2;# shunt resistance in ohm\n",
"Ra=0.05;# armature resistance in ohm\n",
"Vb=2;# brush contact drop\n",
"\n",
"#calculations:\n",
"Il=(W*1000)/V# load current in A\n",
"Vf=Il*R# voltage drop in feeder\n",
"Vt=V+Vf;\n",
"Ish=Vt/Rsh;\n",
"Ia=Il+Ish;\n",
"Vd=Ia*Ra# voltage drop in the armature\n",
"E=Vt+Vd+Vb;\n",
"#Results\n",
"print \"(a)terminal voltage,Vt(V) = \",Vt \n",
"print \"(b)emf,E(V) = \", E"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)terminal voltage,Vt(V) = 252.8\n",
"(b)emf,E(V) = 257.0\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.4: page 129:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"# given data:\n",
"W=20000# in watt\n",
"V=200;# in volts\n",
"R=0.08;# in ohm\n",
"Rs=0.02;# series field resistance in ohm\n",
"Rsh=42;# shunt ield resistance in ohm\n",
"Ra=0.04;# armature resistance in ohm\n",
"iron_losses=309.5;# iron and friction losses\n",
"\n",
"#calculations:\n",
"I=W/V;# in A\n",
"Vf=I*R;\n",
"Vs=I*Rs;\n",
"V1=Vf+Vs;# voltage drop of feeder and series field\n",
"Vg=V+V1;\n",
"Ish=Vg/Rsh# shunt field current\n",
"Ia=I+Ish;\n",
"Vd=Ia*Ra;\n",
"emf=Vg+Vd;\n",
"Ed=emf*Ia# in watt\n",
"copper_losses=Ed-W;\n",
"mech_in=W+copper_losses+iron_losses;\n",
"Bhp=mech_in/735.5;\n",
"efficiency=(W/mech_in)*100;\n",
"\n",
"#Results\n",
"print \"(a)terminal voltage,Vg(V) = \",Vg\n",
"print \"(b)emf(V) =\",emf\n",
"print \"(c)copper losses(Watt) = \",copper_losses\n",
"print \"(d)bhp metric of the primemover,Bhp = \",Bhp \n",
"print \"(e)efficiency(%) = \",round(efficiency,1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)terminal voltage,Vg(V) = 210.0\n",
"(b)emf(V) = 214.2\n",
"(c)copper losses(Watt) = 2491.0\n",
"(d)bhp metric of the primemover,Bhp = 31.0\n",
"(e)efficiency(%) = 87.7\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.5: page 129:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"# given data:\n",
"n=3 # number of motors\n",
"n1=4 # number of parallel path in winding\n",
"i=30;#current in A\n",
"Bhp=65# in hp\n",
"Rsh=44;# shunt field resistance\n",
"Ra=0.08;# armature resistance in ohm\n",
"V=440;# voltage in V\n",
"Vb=2 # we know , brush contact drops\n",
"\n",
"#calculations:\n",
"I=i*n# current taken by three motors\n",
"Ish=V/Rsh# shunt field current\n",
"Ia=I+Ish;\n",
"I1=Ia/n1# current in each path\n",
"Va=Ia*Ra;# armature drop\n",
"E=V+Va+Vb;\n",
"E_power=E*Ia;\n",
"W=V*I# in watt\n",
"M_power=Bhp*746# assume Bhp=746 W\n",
"Copper_losses=E_power-W;\n",
"S_loses=M_power-E_power;\n",
"eta_e=(W/E_power)*100;\n",
"eta_c=(W/M_power)*100;\n",
"eta_m=(E_power/M_power)*100;\n",
"\n",
"#Results\n",
"print \"(a)total armature current,Ia(A) =\",Ia\n",
"print \"(b)current in each path,I1(A) = \",I1\n",
"print \"(c)emf,E(V) = \",E # answer is wrong in a book \n",
"print \"(d)electrical power developed in watt = \",E_power # answer is wrong in a book \n",
"print \"(e)copper losses (W) = \",Copper_losses\n",
"print \"(f)stray losses(W) = \",S_loses\n",
"print \"(g1)electrical efficiency,eta_e(%) = \",eta_e\n",
"print \"(g2)commercial efficiency,eta_c(%) = \",round(eta_c,2)\n",
"print \"(g3)mechanical efficiency,eta_m(%) = \",round(eta_m,1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)total armature current,Ia(A) = 100.0\n",
"(b)current in each path,I1(A) = 25.0\n",
"(c)emf,E(V) = 450.0\n",
"(d)electrical power developed in watt = 45000.0\n",
"(e)copper losses (W) = 5400.0\n",
"(f)stray losses(W) = 3490.0\n",
"(g1)electrical efficiency,eta_e(%) = 88.0\n",
"(g2)commercial efficiency,eta_c(%) = 81.67\n",
"(g3)mechanical efficiency,eta_m(%) = 92.8\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}PK I{46* * E Basic Electrical Engineering with Numerical Problems/Chapter_08.ipynb{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 8: DC Motors"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.1: page 137:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"s=22;# shaft of the motor in hp\n",
"Tsh=210;# torue in hp\n",
"\n",
"#calculations:\n",
"N=(s*60*746)/(2*math.pi*Tsh);\n",
"\n",
"#Results\n",
"print \"speed,N(rpm) = \",N "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"speed,N(rpm) = 746.300264578\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.2: Page 143:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"N=955;# in r.p.m\n",
"V=230;# voltage in volts\n",
"I=72;# current in A\n",
"s=968;# stray losses\n",
"Rsh=115;# shunt field resistance in ohm\n",
"Ra=0.5# armature resistance in ohm\n",
"\n",
"#calculations:\n",
"W=V*I;\n",
"Ish=V/Rsh# shunt field resistance\n",
"Ia=I-Ish;\n",
"Eb=V-(Ia*Ra)# back emf in volts\n",
"Dpd=Eb*Ia# driving power developed\n",
"Mpo=Dpd-s;\n",
"bhp=Mpo/746;\n",
"c_losses=W-Dpd;\n",
"Ta=(9.55*Eb*Ia)/N;\n",
"Tsh=(bhp*60*746)/(2*math.pi*N);\n",
"Tl=Ta-Tsh;\n",
"eta=(Mpo/W)*100;\n",
"\n",
"#Results\n",
"print \"(a)bhp = \",bhp\n",
"print \"(b)copper losses(W) = \",c_losses\n",
"print \"(c)torque armature,Ta(N-m) = \",Ta\n",
"print \"(d)shaft torque,Tsh(N-m) = \",round(Tsh,2)\n",
"print \"(e)lost torque,Tl(N-m) = \",round(Tl,2)\n",
"print \"(f)commercial efficioency,eta(%) = \",round(eta,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)bhp = 17.0\n",
"(b)copper losses(W) = 2910.0\n",
"(c)torque armature,Ta(N-m) = 136.5\n",
"(d)shaft torque,Tsh(N-m) = 126.81\n",
"(e)lost torque,Tl(N-m) = 9.69\n",
"(f)commercial efficioency,eta(%) = 76.58\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.3: page 144:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"V=230# in volts\n",
"I=5 # in amperes\n",
"rpm=914#turns\n",
"ra=0.5#resistance of armature in ihms\n",
"rsh=115#shunt field in ohms\n",
"Il=30# in amperes\n",
"ar=10# in percent\n",
"\n",
"#calculations:\n",
"Ish=V/rsh# in amperes\n",
"anl=I-Ish#armature current in amperes at no load\n",
"al=Il-Ish#armature currentin amperes at load\n",
"Eb1=(V-anl*ra)#back emf at no load\n",
"Eb2=(V-al*ra)#back emf at load\n",
"ph1=100#\n",
"ph2=90#\n",
"Ns=(rpm*Eb2*ph1)/(Eb1*ph2)#speed when loaded in rpm\n",
"\n",
"#Results\n",
"print \"speed when loaded in rpm is \",Ns"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"speed when loaded in rpm is 960.0\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.4: page 144:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"Il=83# WHEN LOADED IN AMPERES\n",
"V=110# in volts\n",
"I=5 # in amperes without load\n",
"ra=0.5#armature resistance in ohms\n",
"rsh=110#shunt field in ohms\n",
"\n",
"#calculations:\n",
"Ish=V/rsh# in ampere\n",
"anl=I-Ish#armature current in amperes at no load\n",
"al=Il-Ish#armature currentin amperes at load\n",
"Eb1=(V-anl*ra)#back emf at no load\n",
"Eb2=(V-al*ra)#back emf at load\n",
"Dp=Eb1*anl#driving power at no load in watt\n",
"Dpl=Eb2*al#driving power at load in watt\n",
"mo=Dpl-Dp#out of motor in watt\n",
"bhp=mo/746#horse power\n",
"mi=V*Il#input power in watt\n",
"n=(mo/mi)*100#efficiency in percentage\n",
"\n",
"#Results\n",
"print \"(a)stray losses in watt is\",Dp\n",
"print \"(b)horse power in ampere is\",round(bhp,1)\n",
"print \"(c)efficiency of motor when it is work on full ,load in percentage is\",round(n,2)\n",
"#answer(c) is wrong in the textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)stray losses in watt is 432.0\n",
"(b)horse power in ampere is 7.0\n",
"(c)efficiency of motor when it is work on full ,load in percentage is 57.24\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.5: page 146"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"V=230# in volts\n",
"I=60# in amperes\n",
"rpm=955#turns\n",
"ra=0.2#resistance of armature in ihms\n",
"rsh=0.15#shunt field in ohms\n",
"sl=604#stray losses in watts\n",
"\n",
"#calculations:\n",
"Rm=ra+rsh# in ohms\n",
"Eb=(V-I*Rm)# back emf in volts\n",
"Dp=Eb*I#driving power in watts\n",
"mi=V*I#input power in watts\n",
"Cl=mi-Dp# copper losses in watts\n",
"mo=Dp-sl#output of motor\n",
"bhp=mo/746# horse power in bhp\n",
"Ta=(9.55*Eb*I)/rpm#total torque in N-m\n",
"Ts=(bhp*60*746)/(2*math.pi*rpm)#shaft torque in N-m\n",
"Tl=Ta-Ts#lost torque in N-m\n",
"nc=(mo/mi)*100#commercial efficiency in percentge\n",
"\n",
"#Results\n",
"print \"(a)back emf in volts is\",Eb\n",
"print \"(b)copper losses in watts is \",Cl\n",
"print \"(c)horse power is\", bhp\n",
"print \"(d)total torque in N-m is\",Ta\n",
"print \"(e)shaft torque in N-m is\",round(Ts,1)\n",
"print \"(f)lost torque in N-m is\",round(Tl,1)\n",
"print \"(g)commercial efficiency in percentge is\",round(nc,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)back emf in volts is 209.0\n",
"(b)copper losses in watts is 1260.0\n",
"(c)horse power is 16.0\n",
"(d)total torque in N-m is 125.4\n",
"(e)shaft torque in N-m is 119.4\n",
"(f)lost torque in N-m is 6.0\n",
"(g)commercial efficiency in percentge is 86.49\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.6: page 146:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"V=220# in volts\n",
"I=60# in amperes\n",
"rpm=728#turns\n",
"Ts=150#shaft torque in N-m\n",
"nc=80#commercial efficiency in percentge\n",
"\n",
"#calculations:\n",
"I=((Ts*2*math.pi*rpm*746)/(60*746*(nc/100)*V))# CURRENT TAKEN IN AMPERES\n",
"\n",
"#Results\n",
"print \"current taken in amperes is\",round(I,1) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current taken in amperes is 65.0\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.7: page 147:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"V=220# in volts\n",
"rpm=2100#turns\n",
"ra=0.5#resistance of armature in ihms\n",
"rsh=220#shunt field in ohms\n",
"Il=21# in amperes\n",
"R1=220# in ohms\n",
"ph1=50#\n",
"ph2=100#\n",
"\n",
"#calculations:\n",
"Ish=V/rsh# in amperes\n",
"Ifs=V/(rsh+R1)#shunt field current in second case in ampere\n",
"n2=(rpm*ph2)/ph1#speed in rpm\n",
"\n",
"#Results\n",
"print \"speed in rpm is\",n2"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"speed in rpm is 4200.0\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}PK IZ8 8 E Basic Electrical Engineering with Numerical Problems/Chapter_09.ipynb{
"metadata": {
"name": "",
"signature": "sha256:31450cb4fbdeff508f73b1f734eeeccc850933334f8ee10004b73433a5bc86e4"
},
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 9: Cells and Batteries"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.1: page 166:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"n=20;# dry cells of emf\n",
"E=1.5;# emf in volts\n",
"R=5; # external resistance in ohm\n",
"r=0.5;# internal resistance in ohm\n",
"\n",
"#calculations:\n",
"I=(n*E)/(R+(n*r));\n",
"\n",
"#Results\n",
"print \"current flowing,I(A) = \",I"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current flowing,I(A) = 2.0\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.2: Page 167:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"n=10;# dry cells of emf\n",
"E=1.5;# emf in volts\n",
"R=4.9;# resistance in ohm\n",
"r=1; # internal resistance in ohm\n",
"\n",
"#calculations:\n",
"I=(n*E)/((n*R)+(r));\n",
"\n",
"#Results\n",
"print \"current flowing,I(A) = \",I"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current flowing,I(A) = 0.3\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.3: page 167:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"m=3; \n",
"n=10;# dry cells of emf\n",
"E=1.5;# emf in volts\n",
"R=2.5;# resistance in ohm\n",
"r=0.5;# internal resistance in ohm\n",
"\n",
"#colculations:\n",
"I=(m*n*E)/((m*R)+(n*r));\n",
"\n",
"#Results\n",
"print \"current flowing,I(A) = \",I"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current flowing,I(A) = 3.6\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.4: page 172:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"n=10 # no. of cells\n",
"Rl=4 # LOAD RESISTANCE\n",
"V=12 # in volts\n",
"Va=18# IN VOLTS\n",
"\n",
"#calculations:\n",
"r=((Va-V)*Rl)/(n*V)# internal resistance in ohms\n",
"Il=V/Rl# IN AMPERES\n",
"\n",
"#Results\n",
"print \"(a)internal resistance in ohms is\",r\n",
"print \"(b)load current in amperes is\", Il"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)internal resistance in ohms is 0.2\n",
"(b)load current in amperes is 3.0\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.5: page 173:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"n=6 # no. of cells\n",
"Rl=3 # LOAD RESISTANCE\n",
"I=2.5# IN AMPERES\n",
"r1=9 # in ohms\n",
"I2=1.25# om amperes\n",
"\n",
"#calculations:\n",
"r=((r1*I2)-(Rl*I))/(n*(I-I2))# internal resistance in ohms\n",
"E=((I*(Rl+n*r))/n)# emf of each cell in volts\n",
"\n",
"#Results\n",
"print \"emf of each cell in volts is\", E \n",
"print \"internal resistance of each cell in ohms is\",r "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"emf of each cell in volts is 2.5\n",
"internal resistance of each cell in ohms is 0.5\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.6: page 173:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"I=20 # in amperes\n",
"t=15 # in hours\n",
"\n",
"#calculations\n",
"Ah=I*t# ampere hour capacity of the battery\n",
"\n",
"#Results\n",
"print \"ampere hour capacity of the battery in A-h is\",Ah "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ampere hour capacity of the battery in A-h is 300\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.7: page 174:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"I=30 # in amperes\n",
"t=6 # in hours\n",
"Vt=2 # terminal voltage\n",
"Ic=40# in amperes\n",
"tc=5 # in hours\n",
"Vc=2.5# in volts\n",
"\n",
"#calculations:\n",
"Aho=I*t# ampere hour output of the battery\n",
"Ahi=Ic*tc# ampere hour input of the battery\n",
"nAh=(Aho/Ahi)*100# ampere hour efficiency\n",
"Who=I*t*Vt# watt hour output of the battery\n",
"Whi=Ic*tc*Vc# watt hour input of the battery\n",
"nWh=(Who/Whi)*100# ampere hour efficiency\n",
"\n",
"#Results\n",
"print \"(a)ampere hour efficiency of the battery in percentage is\",nAh\n",
"print \"(b)watt hour efficiency of the battery in percentage is\",nWh"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)ampere hour efficiency of the battery in percentage is 90.0\n",
"(b)watt hour efficiency of the battery in percentage is 72.0\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.8: page 176:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"n=50 # no. of cells\n",
"Vc=250# in volts\n",
"Vd=1.8#in volts\n",
"Vcs=2.2#in volts\n",
"r=0.01#internal resistance of each cell in ohms\n",
"rl=0.1#lead resistance in ohms\n",
"Re=19.4#external resitance in ohms\n",
"\n",
"#calculations:\n",
"Ib=n*r# internal resistnce of battery\n",
"Tb=rl+Ib#total resistance of battery\n",
"Eb=Vd*n#total rmf of battery\n",
"I=(Vc-Eb)/(Re+Tb)# initial charging current in amperes\n",
"Ebf=Vcs*n#emf of the battery at the end of charging\n",
"If=(Vc-Ebf)/(Re+Tb)# initial charging current in amperes\n",
"#Results\n",
"print \"(a)initial charging current in amperes is\",I\n",
"print \"(b)final charging current in amperes is\",If"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)initial charging current in amperes is 8.0\n",
"(b)final charging current in amperes is 7.0\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.9: page 182:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import numpy as np\n",
"\n",
"#given data:\n",
"V=230# in volts\n",
"emf1=122#in volts\n",
"r=0.4#internal resistance in ohms\n",
"emf2=130#in volts\n",
"r1=0.5#in ohms\n",
"R = 5; #in ohm\n",
"\n",
"# calculations:\n",
" #apllying kirchoff's low\n",
" # x ampere is the total current taken by battery\n",
" # x1 ampere is the total current taken by battery A\n",
" # x-x1 ampere is the total current taken by battery B\n",
" # 5*x+0.4*y=180 is the equation in mesh ABEF\n",
" # 5.5*x+0.5*y=100 equation in the mesh CDEF\n",
" # equation 1 is 25*x+2*y=540 and equation 2 is 22*x-2*y=400\n",
"A=np.array([[25, 2],[22,-2]])# EQUATIONS \n",
"B=np.array([540,400])# VALUES\n",
"X=np.linalg.solve(A, B)# UNKNOW VALUES\n",
"I=X[0]#TOTAL CURRENT IN AMPERES\n",
"x1=X[1]#current taken by battery A\n",
"x2=I-x1#\n",
"p=(I**2)*R# in watt\n",
"\n",
"\n",
"#Results\n",
"print \"(a1)current in battery A in amperes (discharging) is\",x1\n",
"print \"(a2)current in bettery B in amperes is\",round(x2,1)\n",
"print \"(b)total current in battery A and B in amperes (charging)\",I\n",
"print \"(c)power dissipated in watts is\",p\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a1)current in battery A in amperes (discharging) is 20.0\n",
"(a2)current in bettery B in amperes is 0.0\n",
"(b)total current in battery A and B in amperes (charging) 20.0\n",
"(c)power dissipated in watts is 2000.0\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 9.10: page 183:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"import numpy as np\n",
"\n",
"#given data:\n",
"V=34 # in volts\n",
"emf1=2#in volts\n",
"r1=6 #in ohms\n",
"r2=1 #in ohms\n",
"r3=2 #in ohms\n",
"r4=4 # in ohms\n",
"\n",
"# calculations:\n",
" #apllying kirchoff's low\n",
" # x ampere is the current in branch AB\n",
" # x1 ampere is the current in branch AC\n",
" #x2 ampere is the current in the Branch BD\n",
" # x-x2 ampere is the current in the branch BC\n",
" # x1+x2 ampere is the current in the branch DC\n",
" # x-6*x1+8*x2=2 in mesh ABD\n",
" # 2*x-4*x1-14*x2=-2 in mesh BCD\n",
" # 10*x1+4*x2=34;//in mesh ADCEF\n",
"A=np.array([[1,-6,8],[2,-4,-14],[0,10,4]])# EQUATIONS \n",
"B=np.array([2,-2,34])# VALUES\n",
"X=np.linalg.solve(A, B)# UNKNOW VALUES\n",
"x=X[0]#TOTAL CURRENT IN AMPERES\n",
"x1=X[1]#current taken by battery A\n",
"x2=X[2]#\n",
"b1=x-x2# in amperes\n",
"b2=x1+x2#in amperes\n",
"R=((r1*x1+r4*(x2+x1))/(x+x1))#total resistance in ohms\n",
"\n",
"#Results\n",
"print \"current in 1 ohms resistance from A to B in amperes is\",x\n",
"print \"current in 6 ohms resistance from A to D in amperes is\",x1\n",
"print \"current in 8 ohms resistance from B to D in amperes is\",x2\n",
"print \"current in 2 ohm resistance from B to C in amperes is\",b1\n",
"print \"current in 4 ohm resistance from D to C in amperes is\",b2\n",
"print \"total reistance in ohms is\",round(R,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current in 1 ohms resistance from A to B in amperes is 12.0\n",
"current in 6 ohms resistance from A to D in amperes is 3.0\n",
"current in 8 ohms resistance from B to D in amperes is 1.0\n",
"current in 2 ohm resistance from B to C in amperes is 11.0\n",
"current in 4 ohm resistance from D to C in amperes is 4.0\n",
"total reistance in ohms is 2.27\n"
]
}
],
"prompt_number": 10
}
],
"metadata": {}
}
]
}PK I/Āy y E Basic Electrical Engineering with Numerical Problems/Chapter_10.ipynb{
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"name": "",
"signature": "sha256:40665d77e6c03cce80f80ccba51c958481fca359588ebaf3851ecf724eb38bbb"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Wiring Systems"
]
}
],
"metadata": {}
}
]
}PK Is2I I E Basic Electrical Engineering with Numerical Problems/Chapter_11.ipynb{
"metadata": {
"name": "",
"signature": "sha256:e41e7a899079e095312b80bb51a6a04abde629cd10b5f06d1688520f00c1a4b7"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 11: Single Phase A-C Circuits"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.1: page 211:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"E_max=500;# emf in volts\n",
"thita=30;# in degree\n",
"\n",
"#calculations:\n",
"e=E_max*math.sin(thita*math.pi/180);\n",
"\n",
"#Results\n",
"print \"instantaneous value,e(v) = \",e "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"instantaneous value,e(v) = 250.0\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.2: page 212:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"I_max=1.414;# maximum value of current in A\n",
"\n",
"#calculations:\n",
"I_rms=I_max*0.707;\n",
"\n",
"#Results\n",
"print \"rms value of current,I_rms(A) = \",round(I_rms)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"rms value of current,I_rms(A) = 1.0\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.3: page 220:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"f=50;# frequency in Hz\n",
"L=0.2;# inductance in H\n",
"V=220;# voltage in volts\n",
"\n",
"#calculations:\n",
"XL=2*math.pi*f*L# in ohm\n",
"Z=XL;\n",
"I=V/Z;\n",
"\n",
"#Results\n",
"print \"current drawn,I(A) = \",round(I,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current drawn,I(A) = 3.5\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.4: page 222:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"f=50;# frequency in Hz\n",
"C=100*10**-6# capacitor in Farad\n",
"V=210;# voltage in volts\n",
"\n",
"#calculations:\n",
"XC=(1/(2*math.pi*f*C));\n",
"Z=XC;\n",
"I=V/Z;\n",
"\n",
"#Results\n",
"print \"current flowing,I(A) = \",round(I,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current flowing,I(A) = 6.6\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.5: page 222:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"f=50;# frequency in Hz\n",
"L=0.4;# inductance in H\n",
"V=220;# voltage in volts\n",
"f1=25;# frequency is halved\n",
"f2=100;# frequency is doubled\n",
"\n",
"#calculations:\n",
"XL=2*math.pi*f*L;\n",
"I=V/XL;\n",
"\n",
"XL1=2*math.pi*f1*L;\n",
"I1=V/XL1;\n",
"\n",
"XL2=2*math.pi*f2*L;\n",
"I2=V/XL2;\n",
"\n",
"#Results\n",
"print \"current flowing,I(A) = \",round(I,2) \n",
"print \"(a)current when frequency is halved,I(A) = \",round(I1,2)\n",
"print \"current when frequency is doubled,I(A) = \",round(I2,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current flowing,I(A) = 1.75\n",
"(a)current when frequency is halved,I(A) = 3.5\n",
"current when frequency is doubled,I(A) = 0.875\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.6: page 223:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"f=50;# frequency in Hz\n",
"C=28*10**-6# capacitor in Farad\n",
"V=250;# voltage in volts\n",
"f1=25# when frequency is halved\n",
"f2=100# when frequency is doubled\n",
"\n",
"#calculations:\n",
"XC=1/(2*math.pi*f*C);\n",
"I=V/XC;\n",
"\n",
"XC1=1/(2*math.pi*f1*C);\n",
"I1=V/XC1;\n",
"\n",
"XC2=1/(2*math.pi*f2*C);\n",
"I2=V/XC2;\n",
"\n",
"#Results\n",
"print \"current flowing,I(A) = \",round(I,1)\n",
"print \"current flowing when frequency is halved,I(A) = \",round(I1,1)\n",
"print \"current flowing when frequency is doubled ,I(A) =\",round(I2,1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"current flowing,I(A) = 2.2\n",
"current flowing when frequency is halved,I(A) = 1.1\n",
"current flowing when frequency is doubled ,I(A) = 4.4\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.7: Page 224:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"R=40 #in ohms\n",
"L=0.07#IN HENRY\n",
"V=223#IN VOLTS\n",
"F=50 # IN HERTS\n",
"\n",
"#calculations:\n",
"Xl=2*math.pi*F*L# inductive reactance in ohms\n",
"Z=(R**2+Xl**2)**0.5#IMPEDENCE IN OHMS\n",
"I=V/Z;#in amperes\n",
"csp=R/Z#pf\n",
"phi=math.acos(csp)#angle of phase differnce in degree\n",
"\n",
"def decdeg2dms(dd):\n",
" mnt,sec = divmod(dd*3600,60)\n",
" deg,mnt = divmod(mnt,60)\n",
" return deg,mnt,sec\n",
"phiAct = decdeg2dms(phi*180/math.pi)\n",
"\n",
"#Results\n",
"print \"inductive reactance in ohms is\",round(Xl)\n",
"print \"impedence in ohms is\",round(Z,2) \n",
"print \"current in amperes is\",round(I,1)\n",
"print \"angle of phase difference is \",phiAct[0],\"Degrees and \",phiAct[1],\"minutes\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"inductive reactance in ohms is 22.0\n",
"impedence in ohms is 45.65\n",
"current in amperes is 4.9\n",
"angle of phase difference is 28.0 Degrees and 48.0 minutes\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.8: Page 226:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"V=200#in volts\n",
"I=2.5# in amperes\n",
"Vo=250# in volts\n",
"f=50 # in hertz\n",
"\n",
"#calculations:\n",
"R=V/I# in ohms\n",
"Z=Vo/I# in ohms\n",
"Xl=(Z**2-R**2)**0.5#inductive reactance in ohms\n",
"L=(Xl/(2*math.pi*f))#inductance in henry\n",
"pf=R/Z#power factor\n",
"phi=math.acos(pf)#angle of phase differnce in degree\n",
"def decdeg2dms(dd):\n",
" mnt,sec = divmod(dd*3600,60)\n",
" deg,mnt = divmod(mnt,60)\n",
" return deg,mnt,sec\n",
"phiAct = decdeg2dms(phi*180/math.pi)\n",
"\n",
"#Results\n",
"print \"inductance in henry is\",round(L,4)\n",
"print \"angle of phase difference is \",phiAct[0],\"Degrees and \",phiAct[1],\"minutes\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"inductance in henry is 0.191\n",
"angle of phase difference is 36.0 Degrees and 52.0 minutes\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.9: page 226:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"W=100#in watts\n",
"V=110#in volts\n",
"Vc=220#in volts\n",
"f=50 #in hertz\n",
"\n",
"#calculations:\n",
"I=W/V# in amperes\n",
"R=V/I#in ohms\n",
"Z=Vc/I# in ohms\n",
"Xc=math.sqrt(Z**2-R**2)# IN OHMS\n",
"C=(1/(2*math.pi*f*Xc))# in farads\n",
"\n",
"#Results\n",
"print \"capacitance in micro farads is\",round(C*10**6,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"capacitance in micro farads is 15.19\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.10: page 227:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"R=5.94#in ohms\n",
"L=0.35#IN HENRY\n",
"C=35 # in micro farads\n",
"V=220#IN VOLTS\n",
"F=50 # IN HERTS\n",
"\n",
"#calculations:\n",
"Xc=(1/(2*math.pi*F*C*10**-6))# capacitive reactance in ohms\n",
"Xl=2*math.pi*F*L# inductive reactance in ohms\n",
"Z=math.sqrt(R**2+(Xl-Xc)**2)# impedence in ohms\n",
"I=V/round(Z)# in amperes\n",
"pf=R/round(Z)# power factor\n",
"Zc=math.sqrt(R**2+Xl**2)#impedence of the coil\n",
"Vl=I*Zc#voltage drop across the coil\n",
"Vc=I*Xc#voltage drop across the capacitor\n",
"W=I**2*R#total power taken in watts\n",
"\n",
"#Results\n",
"print \"(a)impedence in ohms is\",round(Z)\n",
"print \"(b)current in amperes is\",I\n",
"print \"(c)angle of phase diffence between voltage and current is\",pf\n",
"print \"(d)voltage across the coil in volts is\",round(Vl,1)\n",
"print \"(e)voltage across capacitor in volts is\",round(Vc,1)\n",
"print \"(f)total power taken in watts is\",round(W,1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)impedence in ohms is 20.0\n",
"(b)current in amperes is 11.0\n",
"(c)angle of phase diffence between voltage and current is 0.297\n",
"(d)voltage across the coil in volts is 1211.3\n",
"(e)voltage across capacitor in volts is 1000.4\n",
"(f)total power taken in watts is 718.7\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.11: page 233:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"r1=6 #in ohms\n",
"r2=3.95#in ohms\n",
"R=r1+r2#in ohms\n",
"L1=0.21#IN HENRY\n",
"L2=0.14#in henry\n",
"C1=30# in micro farads\n",
"C2=60#in micro farads\n",
"V=220#IN VOLTS\n",
"F=50 # IN HERTS\n",
"\n",
"#calculations:\n",
"Xc1=(1/(2*math.pi*F*C1*10**-6))# capacitive reactance in ohms\n",
"Xc2=(1/(2*math.pi*F*C2*10**-6))# capacitive reactance in ohms\n",
"Xc=Xc1+Xc2#IN OHMS\n",
"Xl1=2*math.pi*F*L1# inductive reactance in ohms\n",
"Xl2=2*math.pi*F*L2# inductive reactance in ohms\n",
"Xl=Xl1+Xl2#in ohms\n",
"Z=math.sqrt(R**2+(Xl-Xc)**2)# impedence in ohms\n",
"I=V/Z#\n",
"pf=R/Z# leading power factor\n",
"\n",
"#Results\n",
"print \"(a)impedence in ohms is\",round(Z)\n",
"print \"(b)current in amperes is\",round(I)\n",
"print \"(c)power factor (leading) is\",round(pf,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)impedence in ohms is 50.0\n",
"(b)current in amperes is 4.0\n",
"(c)power factor (leading) is 0.198\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.12: Page 233:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"V=200# in volts\n",
"L=0.04# in henry\n",
"C=100#in micro fards\n",
"f=50 # hertz\n",
"Z1=10#ohms\n",
"R1=10# in ohms\n",
"X1=0 # in ohms\n",
"R2=5 # in ohms\n",
"R3=15# in ohms\n",
"\n",
"#calculations:\n",
"Xl=2*math.pi*f*L#inductive reactance in ohms\n",
"Xc=(1/(2*math.pi*f*C*10**-6))#CAPACITIVE REACTANCE IN OHMS\n",
"Z2=math.sqrt(R2**2+Xl**2)#in ohms\n",
"X2=Xl#\n",
"Z3=math.sqrt(R3**2+Xc**2)# IN OHMS\n",
"X3=Xc#\n",
"g1=R1/(Z1)**2# conductance of branch 1 in mho\n",
"b1=X1/(Z1)**2#susceptance in mho in branch 1\n",
"g2=R2/(Z2)**2# conductance of branch 2 in mho\n",
"b2=X2/(Z2)**2#susceptance in mho in branch 2\n",
"g3=R3/(Z3)**2# conductance of branch 3 in mho\n",
"b3=X3/(Z3)**2#susceptance in mho in branch 3\n",
"G=g1+g2+g3# total conductance in mho\n",
"B=b1+b2-b3# total susceptance in mho\n",
"Y=math.sqrt(G**2+B**2)#in ohms\n",
"I0=V*Y#curent in ampere\n",
"theta=math.acos(G/Y)#\n",
"\n",
"def decdeg2dms(dd):\n",
" mnt,sec = divmod(dd*3600,60)\n",
" deg,mnt = divmod(mnt,60)\n",
" return deg,mnt,sec\n",
"phiAct = decdeg2dms(theta*180/math.pi)\n",
"\n",
"I=V/Z3#curent in amperes\n",
"pf3=R3/Z3#power factor\n",
"phi=math.acos(pf3)#angle of phase differnce in degree\n",
"\n",
"\n",
"tc3=pf3#\n",
"ts3=math.sin(phi)\n",
"pf1=R1/R1#\n",
"tc1=pf1#\n",
"ts1=math.sin(math.acos(pf1))#\n",
"I1=V/Z1#\n",
"E1=I1*tc1# energy component in branch 1\n",
"EL1=I1*ts1# idel current component in branch 1\n",
"I2=V/Z2#\n",
"pf2=R2/Z2#\n",
"tc2=pf2#\n",
"ts2=math.sin(math.acos(pf2))#\n",
"E2=I2*tc2#ENERGY COMPONENT IN BRANCH2\n",
"EL2=I2*ts2#idele current component in branch 2\n",
"E3=I*tc3#energy component in branch3\n",
"EL3=I*ts3#idle component of current in branch 3\n",
"E=E1+E2+E3#sum of energy component of current\n",
"EL=EL1+EL2-EL3#sum of idel component of current\n",
"It=math.sqrt(E**2+EL**2)# total current\n",
"pft=E/It#power factor of the complete circuit\n",
"phi=math.acos(0.95)#angle of phase differnce in degree\n",
"\n",
"Zt=V/It#in ohms\n",
"R=Zt*pft#equivalent series resistance\n",
"X=Zt*(math.sin(phi))#equivalent series reactance\n",
"\n",
"\n",
"#Results\n",
"print \"(a)current in amperes is\",round(I0)\n",
"print \"Phase angle is \",phiAct[0],\"Degrees and \",phiAct[1],\"minutes\"\n",
"print \"(c)equivalent series resistance in ohms is\",round(R,2)\n",
"print \"euivalent series reactance in ohms is\",round(X,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)current in amperes is 29.0\n",
"Phase angle is 17.0 Degrees and 8.0 minutes\n",
"(c)equivalent series resistance in ohms is 6.55\n",
"euivalent series reactance in ohms is 2.14\n"
]
}
],
"prompt_number": 12
}
],
"metadata": {}
}
]
}PK Ih^N$ N$ E Basic Electrical Engineering with Numerical Problems/Chapter_12.ipynb{
"metadata": {
"name": "",
"signature": "sha256:a058a43eb8f17e03501b9b96349ed2d08d5db7b72234021ac10c6272f4188cf4"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 12: Polyphase System"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.1: page 248:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"L=30 #load in kW\n",
"pf=0.8#power factor\n",
"Vl=250#line voltage in volts\n",
"\n",
"#calculations:\n",
"I=((L*10**3)/(Vl*pf*math.sqrt(3)))#line current in ampers\n",
"Ip1=I # in star connection\n",
"Ip2=I/(math.sqrt(3))#phase current\n",
"Il=math.sqrt(3)*Ip2#line current in amperes\n",
"\n",
"#Results\n",
"print \"(a)line current (star connection) in amperes is\",round(I,2)\n",
"print \"phase current (start connection) in amperes is\",round(Ip1,2)\n",
"print \"(b)phase current in ampere is\",round(Ip2,2)\n",
"print \"line current (delta connection ) in amperes is\",round(Il,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)line current (star connection) in amperes is 86.6\n",
"phase current (start connection) in amperes is 86.6\n",
"(b)phase current in ampere is 50.0\n",
"line current (delta connection ) in amperes is 86.6\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.2: page 248:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"R=11.88#coil resistance in ohms\n",
"L=0.07#inductance in henry\n",
"f=50 # in hertz\n",
"pf=0.48#power factor\n",
"Vl=433#line voltage in volts\n",
"\n",
"#calculations:\n",
"Vp1= Vl/(math.sqrt(3))#phase voltage\n",
"Xl1=(2*math.pi*f*L)#in ohms\n",
"Zb1=math.sqrt(R**2+Xl1**2)# in ohms\n",
"Ie1=Vp1/Zb1#current in each winding in amperes\n",
"Il1=Ie1#line current in amperes\n",
"W1=math.sqrt(3)*Vl*Il1*pf#power in watts\n",
"\n",
"Vp2= Vl#phase voltage\n",
"Xl2=(2*math.pi*f*L)#in ohms\n",
"Zb2=math.sqrt(R**2+Xl2**2)# in ohms\n",
"Ie2=Vp2/Zb2#current in each winding in amperes\n",
"Il2=math.sqrt(3)*Ie2#line current in amperes\n",
"W2=math.sqrt(3)*Vl*Il2*pf#power in watts\n",
"\n",
"#Results\n",
"print \"(a)line current in ampere is\",round(Il1)\n",
"print \"power taken in connection in kW is\",round(W1*10**-3,1)\n",
"print \"(b)line current in ampere is\",round(Il2)\n",
"print \"power taken in connection in kW is\",round(W2*10**-3,1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)line current in ampere is 10.0\n",
"power taken in connection in kW is 3.6\n",
"(b)line current in ampere is 30.0\n",
"power taken in connection in kW is 10.8\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.3: page 250:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"\n",
"#given data:\n",
"Vl=1100#line voltage in volts\n",
"n=99 #motor efficiency in percentage\n",
"pf= 0.8#power factor\n",
"\n",
"#calculations:\n",
"Mo=n*735.5#output of the motor\n",
"Mi=(Mo*100)/75# INPUT OF THE MOTOR IN WATTS\n",
"Il=(Mi)/(math.sqrt(3)*Vl*pf)#line current in amperes\n",
"Ip=Il/(math.sqrt(3))#phase current in amperes\n",
"Ipm=Il#phase curent of the motor\n",
"Ac1=Ip*pf#active component of phase current in the motor\n",
"Rc1=Ip*(math.sqrt(1-pf**2))#reactive component of phase current of motor\n",
"Ac2=Ipm*pf#active component of phase current in the generator\n",
"Rc2=Ipm*(math.sqrt(1-pf**2))#reactive component of phase current of generator\n",
"#Results\n",
"print \"(a)phase current of motor in amperes is\",round(Ip,2)\n",
"print \"active component of phase current in the motor in amperes\",round(Ac1,2)\n",
"print \"reactive component of phase current in the motor in amperes\",round(Rc1,2)\n",
"print \"(b)phase current of generator in amperes is\",round(Ipm,2)\n",
"print \"active component of phase current in the generator in amperes\",round(Ac2,3)\n",
"print \"reactive component of phase current in the generator in amperes\",round(Rc2,3)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)phase current of motor in amperes is 36.77\n",
"active component of phase current in the motor in amperes 29.42\n",
"reactive component of phase current in the motor in amperes 22.06\n",
"(b)phase current of generator in amperes is 63.7\n",
"active component of phase current in the generator in amperes 50.957\n",
"reactive component of phase current in the generator in amperes 38.218\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.4: Page 253:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"ni=74.6#efficiency\n",
"Mo=40#HP OF MOTOR\n",
"tw=40#total in kW\n",
"pf=0.8#power factor\n",
"\n",
"#calculations:\n",
"mo=Mo*ni#output of motor in watts\n",
"mi=(mo*100)/(ni*1000)#input of motor in kW\n",
"theta=math.acos(pf)#in degree\n",
"v=math.tan(theta)#\n",
"dw=(v*tw)/(3**0.5)#\n",
"w1=(tw+dw)/2#FIRST READING IN kW\n",
"w2=tw-w1#second reading in kW\n",
"\n",
"#Results\n",
"print \"first reading in kW is\",round(w1,2)\n",
"print \"second reading in kW is\",round(w2,2) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"first reading in kW is 28.66\n",
"second reading in kW is 11.34\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.5: page 253:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"w1=4.5#first reading in kW\n",
"w2=3 #second reading in kW , this value is given wrong in question\n",
"\n",
"#calculations:\n",
"tw1=w1+w2#in kW\n",
"dw1=w1-w2#in kW\n",
"pfa1=math.atan(math.sqrt(3)*(dw1/tw1));\n",
"pf1=math.cos(pfa1)#//power factor when both the eadings are positive\n",
"\n",
"tw2=w1-w2#in kW\n",
"dw2=w1+w2#in kW\n",
"pfa2=math.atan(math.sqrt(3)*(dw2/tw2));\n",
"pf2=math.cos(pfa2)#//power factor when second reading is obtained by reversing the connection\n",
"#Results\n",
"print \"(a)power factor when both the readings are positive\", round(pf1,3)\n",
"print \"(b)power factor when second reading is obtained by reversing the connections \",round(pf2,3) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)power factor when both the readings are positive 0.945\n",
"(b)power factor when second reading is obtained by reversing the connections 0.115\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}PK I]Hx x E Basic Electrical Engineering with Numerical Problems/Chapter_01.ipynbPK I
=2 E Basic Electrical Engineering with Numerical Problems/Chapter_02.ipynbPK I$h>{9 {9 E Basic Electrical Engineering with Numerical Problems/Chapter_03.ipynbPK IJ J E = Basic Electrical Engineering with Numerical Problems/Chapter_04.ipynbPK I=
E Basic Electrical Engineering with Numerical Problems/Chapter_05.ipynbPK Iݦ) ) E U Basic Electrical Engineering with Numerical Problems/Chapter_06.ipynbPK IRHA# # E Basic Electrical Engineering with Numerical Problems/Chapter_07.ipynbPK I{46* * E <