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diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb new file mode 100755 index 00000000..8285878f --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb @@ -0,0 +1,974 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 19: Three phase systems</h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1, page no. 299</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 2, page no. 299</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 4, page no. 301</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 5, page no. 302</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 6, page no. 302</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 7, page no. 303</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 8, page no. 304</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 9, page no. 304</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 10, page no. 304</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 11, page no. 305</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 13, page no. 308</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 14, page no. 308</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 15, page no. 309</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 16, page no. 310</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 17, page no. 310</h3>"
+ ]
+ },
+ {
+ "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": [
+ "<h3>Example 18, page no. 311</h3>"
+ ]
+ },
+ {
+ "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": {}
+ }
+ ]
+}
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