{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 22: Three-phase induction motors
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 389
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the synchronous speed of the motor in rev/min.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"f = 50;# in Hz\n",
"p = 2/2;# number of pairs of poles\n",
"\n",
"#calculation:\n",
" #ns is the synchronous speed, \n",
" #f is the frequency in hertz of the supply to the stator and \n",
" #p is the number of pairs of poles.\n",
"ns = f/p\n",
"nsrpm = ns*60\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\nsynchronous speed of the motor is \",nsrpm,\" rev/min\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"synchronous speed of the motor is 3000.0 rev/min"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2, page no. 389
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the number of poles.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"f = 60;# in Hz\n",
"ns = 900/60;# in rev/sec\n",
"\n",
"#calculation:\n",
" #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
" #p is the number of pairs of poles.\n",
"p = f/ns\n",
"np = p*2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\nnumber of poles is \", round(np,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"number of poles is 8.0"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3, page no. 390
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the frequency of the supply voltage.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"p = 2/2;# number of pairs of poles\n",
"ns = 6000/60;# in rev/sec\n",
"\n",
"#calculation:\n",
" #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
" #\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"frequency is \",f,\" Hz\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"frequency is 100.0 Hz"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4, page no. 391
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the synchronous speed and (b) the slip at full load.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"p = 4/2;# number of pairs of poles\n",
"f = 50;# in Hz\n",
"nr = 1455/60;# in rev/sec\n",
"\n",
"#calculation:\n",
" #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
" #p is the number of pairs of poles.\n",
"ns = f/p\n",
" #The slip, s\n",
"s = ((ns - nr)/ns)*100# in percent\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
"print \"\\n(b) slip is \",s,\" percent\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a) synchronous speed is 25.0 rev/sec\n",
"\n",
"(b) slip is 3.0 percent"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5, page no. 392
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine (a) the synchronous speed,\n",
"#(b) the speed of the rotor and (c) the frequency of the induced e.m.f.\u2019s in the rotor\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"p = 2/2;# number of pairs of poles\n",
"f = 60;# in Hz\n",
"s = 0.02;# slip\n",
"\n",
"#calculation:\n",
" #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
" #p is the number of pairs of poles.\n",
"ns = f/p\n",
" #The the rotor runs at\n",
"nr = ns*(1 - s)\n",
" #frequency of the e.m.f. induced in the rotor bars is\n",
"fr = ns - nr\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
"print \"\\n(b) rotor speed is \",nr,\" rev/sec\"\n",
"print \"\\n(c) frequency of the e.m.f. induced in the rotor bars is is \",fr,\" Hz\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a) synchronous speed is 60.0 rev/sec\n",
"\n",
"(b) rotor speed is 58.8 rev/sec\n",
"\n",
"(c) frequency of the e.m.f. induced in the rotor bars is is 1.2 Hz"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6, page no. 392
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the synchronous speed.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"f = 50;# in Hz\n",
"nr = 1200/60;# in rev/min\n",
"s = 0.04;# slip\n",
"\n",
"#calculation:\n",
" #the synchronous speed.\n",
"ns = nr/(1 - s)\n",
"nsrpm = ns*60\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n synchronous speed is \",nsrpm,\" rev/min\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" synchronous speed is 1250.0 rev/min"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 7, page no. 394
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the slip, and (b) the rotor speed.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"p = 8/2;# number of pairs of poles\n",
"f = 50;# in Hz\n",
"fr = 3;# in Hz\n",
"\n",
"#calculation:\n",
" #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
" #p is the number of pairs of poles.\n",
"ns = f/p\n",
" #fr = s*f\n",
"s = (fr/f)\n",
" #the rotor speed.\n",
"nr = ns*(1 - s)\n",
"nrrpm = nr*60\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a) slip is \",s*100,\" percent\"\n",
"print \"\\n (b) rotor speed is \",nrrpm,\" rev/min\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a) slip is 6.0 percent\n",
"\n",
" (b) rotor speed is 705.0 rev/min"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 8, page no. 396
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine (a) the rotor copper loss, \n",
"#(b) the total mechanical power developed by the rotor,\n",
"#(c) the output power of the motor if friction and windage losses are 750 W, and \n",
"#(d) the efficiency of the motor, neglecting rotor iron loss.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"Psi = 32000;# in Watts\n",
"Psl = 1200;# in Watts\n",
"s = 0.05;# slip\n",
"Pfl = 750;# in Watts\n",
"\n",
"#calculation:\n",
" #Input power to rotor = stator input power -\u0006 stator losses\n",
"Pi = Psi - Psl\n",
" #slip = rotor copper loss/rotor input\n",
"Pl = s*Pi\n",
" #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
"Pr = Pi - Pl\n",
" #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
"Po = Pr - Pfl\n",
" #Efficiency of induction motor = (output power/input power)*100\n",
"eff = (Po/Psi)*100# in percent\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
"print \"\\n(b) Total mechanical power developed by the rotor is \",Pr,\" W\"\n",
"print \"\\n(c) Output power of motor is \",Po,\" Watt\"\n",
"print \"\\n(d) efficiency of induction motor is \",round(eff,2),\" percent\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a) rotor copper loss is 1540.0 Watt\n",
"\n",
"(b) Total mechanical power developed by the rotor is 29260.0 W\n",
"\n",
"(c) Output power of motor is 28510.0 Watt\n",
"\n",
"(d) efficiency of induction motor is 89.09 percent"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 9, page no. 397
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine (a) the rotor copper loss, and (b) the efficiency of the motor.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"Psi = 32000;# in Watts\n",
"Psl = 1200;# in Watts\n",
"Pfl = 750;# in Watts\n",
"x = 0.35;\n",
"\n",
"#calculation:\n",
" #The slip, s\n",
"s = 1-x\n",
" #Input power to rotor = stator input power - stator losses\n",
"Pi = Psi - Psl\n",
" #slip = rotor copper loss/rotor input\n",
"Pl = s*Pi\n",
" #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
"Pr = Pi - Pl\n",
" #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
"Po = Pr - Pfl\n",
" #Efficiency of induction motor = (output power/input power)*100\n",
"eff = (Po/Psi)*100# in percent\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
"print \"\\n(b) efficiency of induction motor is \",round(eff,2),\" percent\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a) rotor copper loss is 20020.0 Watt\n",
"\n",
"(b) efficiency of induction motor is 31.34 percent"
]
}
],
"prompt_number": 9
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 10, page no. 398
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate (a) the synchronous speed, (b) the slip, \n",
"#(c) the full load torque, (d) the power output if mechanical losses amount to 770 W, \n",
"#(e) the maximum torque, (f) the speed at which maximum torque occurs,\n",
"#and (g) the starting torque.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V = 415;# in Volts\n",
"f = 50 ;# in Hz\n",
"nr = 24;# in rev/sec\n",
"p = 4/2;# no. of pole pairs\n",
"R2 = 0.35;# in Ohms\n",
"X2 = 3.5;# in Ohms\n",
"tr = 0.85;# turn ratio N2/N1\n",
"Pl = 770;# in Watt\n",
"m = 3;# no. of phases\n",
"\n",
"#calculation:\n",
" #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
" #p is the number of pairs of poles.\n",
"ns = f/p\n",
" #The slip, s\n",
"s = ((ns - nr)/ns)*100# in percent\n",
" #Phase voltage, E1 = V/(3**0.5)\n",
"E1 = V/(3**0.5)\n",
" #Full load torque\n",
"T = (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2 + (X2*(s/100))**2))\n",
" #Output power, including friction losses\n",
"Pm = 2*math.pi*nr*T\n",
" #power output\n",
"Po = Pm - Pl\n",
" #Maximum torque occurs when R2 = Xr = 0.35 ohm\n",
" #Slip \n",
"sm = R2/X2\n",
" #maximum torque, Tm \n",
"Tm = (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2 + (X2*sm)**2))\n",
" #speed at which maximum torque occurs\n",
"nrm = ns*(1 - sm)\n",
"nrmrpm = nrm*60\n",
" #At the start, i.e., at standstill, slip, s=1\n",
"ss = 1\n",
" #starting torque\n",
"Ts = (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2 + (X2*ss)**2))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a)Synchronous speed is \",round(ns,2),\" rev/sec\"\n",
"print \"\\n(b)Slip is \",round(s,2),\" percent\"\n",
"print \"\\n(c)Full load torque is \",round(T,2),\" Nm\"\n",
"print \"\\n(d)power output is \",round(Po,2),\"W\"\n",
"print \"\\n(e)maximum torque is \",round(Tm,2),\" Nm\"\n",
"print \"\\n(f)speed at which maximum torque occurs is \",round(nrmrpm,2),\"rev/min\"\n",
"print \"\\n(g)starting torque is \",round(Ts,2),\" Nm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a)Synchronous speed is 25.0 rev/sec\n",
"\n",
"(b)Slip is 4.0 percent\n",
"\n",
"(c)Full load torque is 78.05 Nm\n",
"\n",
"(d)power output is 10998.99 W\n",
"\n",
"(e)maximum torque is 113.17 Nm\n",
"\n",
"(f)speed at which maximum torque occurs is 1350.0 rev/min\n",
"\n",
"(g)starting torque is 22.41 Nm"
]
}
],
"prompt_number": 10
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 11, page no. 400
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine for the induction motor in problem 10 at full load, \n",
"#(a) the rotor current, (b) the rotor copper loss, and (c) the starting current.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V = 415;# in Volts\n",
"f = 50 ;# in Hz\n",
"nr = 24;# in rev/sec\n",
"p = 4/2;# no. of pole pairs\n",
"R2 = 0.35;# in Ohms\n",
"X2 = 3.5;# in Ohms\n",
"tr = 0.85;# turn ratio N2/N1\n",
"m = 3;# no. of phases\n",
"\n",
"#calculation:\n",
" #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
" #p is the number of pairs of poles.\n",
"ns = f/p\n",
" #The slip, s\n",
"s = ((ns - nr)/ns)*100# in percent\n",
" #Phase voltage, E1 = V/(3**0.5)\n",
"E1 = V/(3**0.5)\n",
" #rotor current,\n",
"Ir = (s/100)*E1*tr/((R2**2 + (X2*(s/100))**2)**0.5)\n",
" #Rotor copper loss \n",
"Pcl = m*R2*(Ir**2)\n",
" #starting current,\n",
"ss =1\n",
"I2 = ss*tr*E1/((R2**2 + (X2*ss)**2)**0.5)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(a)rotor current is \",round(Ir,2),\" A\"\n",
"print \"\\n(b)Total copper loss is \",round(Pcl,2),\" W\"\n",
"print \"\\n(c)starting current is \",round(I2,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(a)rotor current is 21.61 A\n",
"\n",
"(b)Total copper loss is 490.37 W\n",
"\n",
"(c)starting current is 57.9 A"
]
}
],
"prompt_number": 11
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 12, page no. 401
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine (a) the power input at full load, \n",
"#(b) the efficiency of the motor at full load and \n",
"#(c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V = 415;# in Volts\n",
"Psl = 650;# in Watt\n",
"pf = 0.87;# power factor\n",
"\n",
"#calculation:\n",
"Pm = 11770;# watts from part (d), Problem 22.10\n",
"Pcl = 490.35;# watts, Rotor copper loss, from part (b), Problem 22.11\n",
" #Stator input power\n",
"P1 = Pm + Pcl + Psl\n",
"Po = 11000# watts, Net power output, from part (d), Problem 22.10\n",
" #efficiency = (output/input) *100\n",
"eff = (Po/P1)*100# in percent\n",
" #Power input, P1 = (3**0.5)*VL*IL*cos(phi)\u000e\n",
" # pf = cos(phi)\n",
" #supply current, IL\n",
"I = P1/((3**0.5)*V*pf)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n(aStator input power is \",round(P1,2),\" W\"\n",
"print \"\\n(b)efficiency is \",round(eff,2),\" percent\"\n",
"print \"\\n(c)supply current is \",round(I,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"(aStator input power is 12910.35 W\n",
"\n",
"(b)efficiency is 85.2 percent\n",
"\n",
"(c)supply current is 20.64 A"
]
}
],
"prompt_number": 12
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 13, page no. 401
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine the resistance of the rotor winding required for maximum starting torque.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V = 415;# in Volts\n",
"f = 50 ;# in Hz\n",
"nr = 24;# in rev/sec\n",
"p = 4/2;# no. of pole pairs\n",
"R2 = 0.35;# in Ohms\n",
"X2 = 3.5;# in Ohms\n",
"\n",
"#calculation:\n",
" #At the moment of starting, slip, \n",
"s = 1\n",
" #Maximum torque occurs when rotor reactance equals rotor resistance\n",
" #for maximum torque\n",
"R2 = s*X2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\nresistance of the rotor is \",R2,\" Ohm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"resistance of the rotor is 3.5 Ohm"
]
}
],
"prompt_number": 13
}
],
"metadata": {}
}
]
}