{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 19: Three phase systems
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 299
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the system phase voltage, (b) the phase current and (c) the line current.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"Vl = 415;# in Volts\n",
"Rp = 30;# in ohms\n",
"\n",
"#calculation:\n",
"Vp = Vl/(3**0.5)\n",
"Ip = Vp/Rp\n",
"Il = Ip\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the system phase voltage is \",round(Vp,2),\" V\"\n",
"print \"\\n (b)phase current is \",round(Ip,2),\" A\"\n",
"print \"\\n (c)line current is \",round(Il,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the system phase voltage is 239.6 V\n",
"\n",
" (b)phase current is 7.99 A\n",
"\n",
" (c)line current is 7.99 A"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2, page no. 299
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the line voltage if the supply frequency is 50 Hz\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"R = 30;# in ohms\n",
"L = 0.1273;# in Henry\n",
"Ip = 5.08;# in Amperes\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
"XL = 2*math.pi*f*L\n",
"Zp = (R*R + XL*XL)**0.5\n",
"Il = Ip\n",
"Vp = Ip*Zp\n",
"Vl = Vp*(3**0.5)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)line voltage is \",round(Vl,2),\" V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)line voltage is 439.89 V"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4, page no. 301
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the current in each line and (b) the current in the neutral conductor.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V = 415;# in Volts\n",
"PR = 24000;# in Watt\n",
"Py = 18000;# in Watt\n",
"Pb = 12000;# in Watt\n",
"VR = 240;# in Volts\n",
"Vy = 240;# in Volts\n",
"Vb = 240;# in Volts\n",
"\n",
"#calculation:\n",
" #For a star-connected system VL = Vp*(3**0.5)\n",
"Vp = V/(3**0.5)\n",
"phir = 90*math.pi/180\n",
"phiy = 330*math.pi/180\n",
"phib = 210*math.pi/180\n",
" # I = P/V for a resistive load\n",
"IR = PR/VR\n",
"Iy = Py/Vy\n",
"Ib = Pb/Vb\n",
"Inh = IR*math.cos(phir) + Ib*math.cos(phib) + Iy*math.cos(phiy)\n",
"Inv = IR*math.sin(phir) + Ib*math.sin(phib) + Iy*math.sin(phiy)\n",
"In = (Inh**2 + Inv**2)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)cuurnt in R line is \",round(IR,2),\" A, cuurnt in Y line is \",round(Iy,2),\" A \"\n",
"print \"and cuurnt in B line is \",round(Ib,2),\" A\"\n",
"print \"\\n (b)cuurnt in neutral line is \",round(In,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)cuurnt in R line is 100.0 A, cuurnt in Y line is 75.0 A and cuurnt in B line is 50.0 A\n",
"\n",
" (b)cuurnt in neutral line is 43.3 A"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5, page no. 302
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the phase current, and (b) the line current.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"R = 30;# in ohms\n",
"L = 0.1273;# in Henry\n",
"VL = 440;# in Volts\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
"XL = 2*math.pi*f*L\n",
"Zp = (R*R + XL*XL)**0.5\n",
"Vp = VL\n",
" #Phase current\n",
"Ip = Vp/Zp\n",
" #For a delta connection,\n",
"IL = Ip*(3**0.5)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the phase current \",round(Ip,2),\" A\"\n",
"print \"\\n (b)line current \",round(IL,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the phase current 8.8 A\n",
"\n",
" (b)line current 15.24 A"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6, page no. 302
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine the capacitance of each of the capacitors.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"IL = 15;# in Amperes\n",
"VL = 415;# in Volts\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
" #For a delta connection\n",
"Ip = IL/(3**0.5)#phase current\n",
"Vp = VL\n",
" #Capacitive reactance per phase\n",
"Xc = Vp/Ip\n",
"C = 1/(2*math.pi*f*Xc)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n capacitance is \",round(C*1E6,2),\"uF\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" capacitance is 66.43 uF"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 7, page no. 303
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"R = 3;# in ohms\n",
"XL = 4;# in ohms\n",
"VL = 415;# in Volts\n",
"\n",
"#calculation:\n",
" #For a star connection:\n",
" #IL = Ip\n",
" #VL = Vp*(3**0.5)\n",
"VLs = VL\n",
"Vps = VLs/(3**0.5)\n",
" #Impedance per phase,\n",
"Zp = (R*R + XL*XL)**0.5\n",
"Ips = Vps/Zp\n",
"ILs = Ips\n",
" #For a delta connection:\n",
" #VL = Vp\n",
" #IL = Ip*(3**0.5)\n",
"VLd = VL\n",
"Vpd = VLd\n",
"Ipd = Vpd/Zp\n",
"ILd = Ipd*(3**0.5)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the line voltage for star connection is \",round(VLs,2),\" V \"\n",
"print \"and the phase voltage for star connection is \",round(Vps,2),\" V \"\n",
"print \"and the line voltage for delta connection is \",round(VLd,2),\" V \"\n",
"print \"and the phase voltage for delta connection is \",round(Vpd,2),\" V\"\n",
"print \"\\n (b)the line current for star connection is \",round(ILs,2),\" A \"\n",
"print \"and the phase current for star connection is \",round(Ips,2),\" A \"\n",
"print \"and the line current for delta connection is \",round(ILd,2),\" A \"\n",
"print \"and the phase current for delta connection is \",round(Ipd,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the line voltage for star connection is 415.0 V \n",
"and the phase voltage for star connection is 239.6 V \n",
"and the line voltage for delta connection is 415.0 V \n",
"and the phase voltage for delta connection is 415.0 V\n",
"\n",
" (b)the line current for star connection is 47.92 A \n",
"and the phase current for star connection is 47.92 A \n",
"and the line current for delta connection is 143.76 A \n",
"and the phase current for delta connection is 83.0 A\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 8, page no. 304
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the total power dissipated by the resistors.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"Rp = 12;# in ohms\n",
"VL = 415;# in Volts\n",
"\n",
"#calculation:\n",
" #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
"Vp = VL/(3**0.5)# since the resistors are star-connected\n",
" #Phase current, Ip\n",
"Zp = Rp\n",
"Ip = Vp/Zp\n",
" #For a star connection\n",
"IL = Ip\n",
" # For a purely resistive load, the power factor cos(phi) = 1\n",
"pf = 1\n",
"P = VL*IL*(3**0.5)*pf\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)total power dissipated by the resistors is \",round(P,2),\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)total power dissipated by the resistors is 14352.08 W"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 9, page no. 304
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine the power factor of the system.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"P = 5000;# in Watts\n",
"IL = 8.6;# in amperes\n",
"VL = 400;# in Volts\n",
"\n",
"#calculation:\n",
" #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
"pf = P/(VL*IL*(3**0.5))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n power factor is \",round(pf,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" power factor is 0.839"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 10, page no. 304
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the total power dissipated in each case.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"R = 10;# in ohms\n",
"L = 0.042;# in Henry\n",
"VL = 415;# in Volts\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
" #For a star connection:\n",
" #IL = Ip\n",
" #VL = Vp*(3**0.5)\n",
"XL = 2*math.pi*f*L\n",
"Zp = (R*R + XL*XL)**0.5\n",
"VLs = VL\n",
"Vps = VLs/(3**0.5)\n",
" #Impedance per phase,\n",
"Ips = Vps/Zp\n",
"ILs = Ips\n",
" #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
"pfs = R/Zp\n",
"Ps = VLs*ILs*(3**0.5)*pfs\n",
"\n",
" #For a delta connection:\n",
" #VL = Vp\n",
" #IL = Ip*(3**0.5)\n",
"VLd = VL\n",
"Vpd = VLd\n",
"Ipd = Vpd/Zp\n",
"ILd = Ipd*(3**0.5)\n",
" #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
"pfd = R/Zp\n",
"Pd = VLd*ILd*(3**0.5)*pfd\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n total power dissipated in star is \",round(Ps,2),\" W and in delta is \",round(Pd,2),\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" total power dissipated in star is 6283.29 W and in delta is 18849.88 W"
]
}
],
"prompt_number": 9
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 11, page no. 305
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine (a) the power input, (b) the line current and (c) the phase current\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"Po = 12750;# in Watts\n",
"pf = 0.77;# power factor\n",
"eff = 0.85;\n",
"VL = 415;# in Volts\n",
"\n",
"#calculation:\n",
" #eff = power_out/power_in\n",
"Pi = Po/eff\n",
" #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
"IL = Pi/(VL*(3**0.5)*pf)# line current\n",
" #For a delta connection:\n",
" #IL = Ip*(3**0.5)\n",
"Ip = IL/(3**0.5)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)power input is \",round(Pi,2),\" W\"\n",
"print \"\\n (b)line current is \",round(IL,2),\" A\"\n",
"print \"\\n (c)phase current is \",round(Ip,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)power input is 15000.0 W\n",
"\n",
" (b)line current is 27.1 A\n",
"\n",
" (c)phase current is 15.65 A"
]
}
],
"prompt_number": 10
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 13, page no. 308
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"R = 30;# in ohms\n",
"XL = 40;# in ohms\n",
"VL = 400;# in Volts\n",
"\n",
"#calculation:\n",
"Zp = (R*R + XL*XL)**0.5\n",
" #a delta-connected load\n",
"Vp = VL\n",
" #Phase current\n",
"Ip = Vp/Zp\n",
"IL = Ip*(3**0.5)\n",
" #Alternator output power is equal to the power dissipated by the load.\n",
" #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
"pf = R/Zp\n",
"P = VL*IL*(3**0.5)*pf\n",
" #Alternator output kVA,\n",
"S = VL*IL*(3**0.5)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the current supplied by the alternator is \",round(IL,2),\" A\"\n",
"print \"\\n (b)output power is \",round(P/1000,2),\"KW and kVA of the alternator is \",round(S/1000,2),\"kVA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the current supplied by the alternator is 13.86 A\n",
"\n",
" (b)output power is 5.76 KW and kVA of the alternator is 9.6 kVA"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14, page no. 308
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate (a) the phase current, (b) the line current, (c) the total power dissipated and \n",
"#(d) the kVA rating of the load. Draw the complete phasor diagram for the load.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"R = 30;# in ohms\n",
"C = 80E-6;# in Farads\n",
"f = 50;# in Hz\n",
"VL = 400;# in Volts\n",
"\n",
"#calculation:\n",
" #Capacitive reactance\n",
"Xc = 1/(2*math.pi*f*C)\n",
"Zp = (R*R + Xc*Xc)**0.5\n",
"pf = R/Zp\n",
" #a delta-connected load\n",
"Vp = VL\n",
" #Phase current\n",
"Ip = Vp/Zp\n",
"IL = Ip*(3**0.5)\n",
" #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
"P = VL*IL*(3**0.5)*pf\n",
" #Alternator output kVA,\n",
"S = VL*IL*(3**0.5)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the phase current is \",round(Ip,2),\" A\"\n",
"print \"\\n (b)the line current is \",round(IL,2),\" A\"\n",
"print \"\\n (c) power is \",round(P/1000,2),\"kW\"\n",
"print \"\\n (d)kVA of the alternator is \", round(S/1000,2),\"kVA\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the phase current is 8.03 A\n",
"\n",
" (b)the line current is 13.9 A\n",
"\n",
" (c) power is 5.8 kW\n",
"\n",
" (d)kVA of the alternator is 9.63 kVA"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 15, page no. 309
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine (a) the total power input and (b) the load power factor.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"Pi1 = 8000;# in Watts\n",
"Pi2 = 4000;# in Watts\n",
"\n",
"#calculation:\n",
" #Total input power\n",
"Pi = Pi1 + Pi2\n",
"phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n",
" #Power factor\n",
"pf = math.cos(phi)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)power input is \",round(Pi,2),\"W\"\n",
"print \"\\n (b)power factor is \",round(pf,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)power input is 12000.0 W\n",
"\n",
" (b)power factor is 0.87"
]
}
],
"prompt_number": 14
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16, page no. 310
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the readings of each wattmeter.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"Pi = 12000;# in Watts\n",
"pf = 0.6;# power factor\n",
"\n",
"#calculation:\n",
" #If the two wattmeters indicate Pi1 and Pi2 respectively\n",
" # Pit = Pi1 + Pi2\n",
"Pit = Pi\n",
" # Pid = Pi1 - Pi2\n",
" #power factor = 0.6 = cos(phi)\n",
"phi = math.acos(pf)\n",
"Pid = Pit*math.tan(phi)/(3**0.5)\n",
" #Hence wattmeter 1 reads\n",
"Pi1 = (Pid + Pit)/2\n",
" #wattmeter 2 reads\n",
"Pi2 = Pit - Pi1\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n reading in each wattameter are \",round(Pi1,2),\"W and \",round(Pi2,2),\"W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" reading in each wattameter are 10618.8 W and 1381.2 W"
]
}
],
"prompt_number": 15
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 17, page no. 310
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the input power and (b) the load power factor.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"Pi1 = 10000;# in Watts\n",
"Pi2 = -3000;# in Watts\n",
"\n",
"#calculation:\n",
" #Total input power\n",
"Pi = Pi1 + Pi2\n",
"phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n",
" #Power factor\n",
"pf = math.cos(phi)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)power input is \",round(Pi,2),\"W\"\n",
"print \"\\n (b)power factor is \",round(pf,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)power input is 7000.0 W\n",
"\n",
" (b)power factor is 0.3"
]
}
],
"prompt_number": 16
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 18, page no. 311
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"R = 8; # in ohms\n",
"XL = 8; # in ohms\n",
"VL = 415; # in Volts\n",
"\n",
"#calculation:\n",
"#For a star connection:\n",
"#IL = Ip\n",
"#VL = Vp*(3**0.5)\n",
"VLs = VL\n",
"Vps = VLs/(3**0.5)\n",
"#Impedance per phase,\n",
"Zp = (R*R + XL*XL)**0.5\n",
"Ips = Vps/Zp\n",
"ILs = Ips\n",
"#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
"pf = R/Zp\n",
"Ps = VLs*ILs*(3**0.5)*pf\n",
"#If wattmeter readings are P1 and P2 then P1 + P2 = Pst\n",
"Pst = Ps\n",
"# Pid = Pi1 - Pi2\n",
"phi = math.acos(pf)\n",
"Psd = Pst*math.tan(phi)/(3**0.5)\n",
"#Hence wattmeter 1 reads\n",
"Ps1 = (Psd + Pst)/2\n",
"#wattmeter 2 reads\n",
"Ps2 = Pst - Ps1\n",
"\n",
"#For a delta connection:\n",
"#VL = Vp\n",
"#IL = Ip*(3**0.5)\n",
"VLd = VL\n",
"Vpd = VLd\n",
"Ipd = Vpd/Zp\n",
"ILd = Ipd*(3**0.5)\n",
"#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n",
"Pd = VLd*ILd*(3**0.5)*pf\n",
"#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt\n",
"Pdt = Pd\n",
"# Pid = Pi1 - Pi2\n",
"Pdd = Pdt*math.tan(phi)/(3**0.5)\n",
"#Hence wattmeter 1 reads\n",
"Pd1 = (Pdd + Pdt)/2\n",
"#wattmeter 2 reads\n",
"Pd2 = Pdt - Pd1\n",
"\n",
"#results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"(a)When the coils are star-connected the wattmeter readings are\", round(Ps1,2),\"W and \",round(Ps2,2),\"W\"\n",
"print \"(b)When the coils are delta-connected the wattmeter readings are are\", round(Pd1,2),\"W and\", round(Pd2,2),\"W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"(a)When the coils are star-connected the wattmeter readings are 8489.35 W and 2274.71 W\n",
"(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W"
]
}
],
"prompt_number": 17
}
],
"metadata": {}
}
]
}