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| author | Trupti Kini | 2016-07-19 23:30:36 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-07-19 23:30:36 +0600 |
| commit | 10c80b1dd1eb129240f34e72a7404b771a9bbdc2 (patch) | |
| tree | 0562a252e056d175c2b706fe313c15b6ab93545e | |
| parent | 6c4ae200cf807d3837540f33bfbce25acce517e4 (diff) | |
| download | Python-Textbook-Companions-10c80b1dd1eb129240f34e72a7404b771a9bbdc2.tar.gz Python-Textbook-Companions-10c80b1dd1eb129240f34e72a7404b771a9bbdc2.tar.bz2 Python-Textbook-Companions-10c80b1dd1eb129240f34e72a7404b771a9bbdc2.zip | |
Added(A)/Deleted(D) following books
A basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2.ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter4.ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter5.ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter1.ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter11.ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter3_(2).ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter6.ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter7.ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter8.ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter9.ipynb
A basic_electrical_engineering_by_nagsarkar_and_sukhija/screenshots/chap1.png
A basic_electrical_engineering_by_nagsarkar_and_sukhija/screenshots/chapter2.png
A basic_electrical_engineering_by_nagsarkar_and_sukhija/screenshots/chapter6.png
13 files changed, 9208 insertions, 0 deletions
diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2.ipynb new file mode 100644 index 00000000..4f463827 --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter2.ipynb @@ -0,0 +1,1385 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2:Network Analysis And Network Theorems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1:Page number-50" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i= 2.5 A\n", + "voltage across 6 ohm resistor= 6.0 V\n", + "voltage across 4 ohm resistor= 4.0 V\n", + "voltage when 4 ohm resistor is connected= 40.0 V\n", + "voltage when both resistors are in series= 100.0 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "v=10\n", + "r=4\n", + "\n", + "#case a\n", + "\n", + "i=v/float(r)\n", + "\n", + "print \"i=\",format(i,'.1f'),\"A\"\n", + "\n", + "#case b\n", + "\n", + "#6ohm resistor is in series with 4 ohm resistor\n", + "\n", + "i=v/(6+4)\n", + "\n", + "v1=i*6\n", + "v2=i*4\n", + "\n", + "print \"voltage across 6 ohm resistor=\",format(v1,'.1f'),\"V\"\n", + "\n", + "print \"voltage across 4 ohm resistor=\",format(v2,'.1f'),\"V\"\n", + "\n", + "#case c\n", + "\n", + "i=10 #constant in both cases\n", + "\n", + "v4=i*4\n", + "\n", + "print \"voltage when 4 ohm resistor is connected=\",format(v4,'.1f'),\"V\"\n", + "\n", + "v6=i*6\n", + "\n", + "v=v4+v6\n", + "\n", + "print \"voltage when both resistors are in series=\",format(v,'.1f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2:Page number-53" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rs= 0.5 ohm\n", + "the load voltage is expressed as 36rl/(0.5+rl)\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f9cd722d810>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import math\n", + "\n", + "i=72\n", + "v=36\n", + "\n", + "rs=v/float(i)\n", + "\n", + "print \"rs=\",format(rs,'.1f'),\"ohm\"\n", + "\n", + "print \"the load voltage is expressed as 36rl/(0.5+rl)\"\n", + "\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "x=[40,50,60,72]\n", + "y=[36,34,32,30]\n", + "\n", + "plt.plot(x,y)\n", + "plt.xlabel('il(A)')\n", + "plt.ylabel('vl(V)')\n", + "plt.show()\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3:Page number-55" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ir= 32.0 A\n", + "il= 2.23 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "v=24\n", + "r=0.75\n", + "\n", + "ir=v/r\n", + "\n", + "print \"ir=\",format(ir,'.1f'),\"A\"\n", + "\n", + "il=v/(10+r) #since 10 is in series with r\n", + "\n", + "print \"il=\",format(il,'.2f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4:Page number-56" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power= 120.0 W\n", + "power dissipated= 30.0 W\n", + "total power supplied by practical source is= 90.0 W\n", + "current source= 40.0 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "vs=12\n", + "rs=0.3\n", + "il=10\n", + "\n", + "#case a\n", + "\n", + "p=vs*il\n", + "\n", + "print \"power=\",format(p,'.1f'),\"W\"\n", + "\n", + "#case b\n", + "\n", + "power=il**2*rs\n", + "\n", + "print \"power dissipated=\",format(power,'.1f'),\"W\"\n", + "\n", + "#case c\n", + "\n", + "totpow=(vs-il*rs)*il\n", + "\n", + "print \"total power supplied by practical source is=\",format(totpow,'.1f'),\"W\"\n", + "\n", + "i=vs/rs\n", + "\n", + "print \"current source=\",format(i,'.1f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.5:Page number-58" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "r2= 15.0 ohm\n", + "req= 15.0 ohm\n", + "0.0291666666667\n", + "req= 15.0 ohm\n", + "0.230833333333\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "#v0/vs=r2/(r1+r2)=0.4r2=0.6r1\n", + "\n", + "r1=10\n", + "\n", + "r2=(0.6*r1)/float(0.4)\n", + "\n", + "print \"r2=\",format(r2,'.1f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "#when r2 is parallel to r3\n", + "r3=200000\n", + "req=(r2*r3)/(r2+r3)\n", + "\n", + "print \"req=\",format(req,'.1f'),\"ohm\"\n", + "\n", + "#v0/vs=0.5825\n", + "\n", + "change=(0.6-0.5825)/float(0.6)\n", + "\n", + "print change\n", + "\n", + "r3=20000\n", + "\n", + "req=(r2*r3)/(r3+r2)\n", + "\n", + "print \"req=\",format(req,'.1f'),\"ohm\"\n", + "\n", + "#v0/vs=0.4615\n", + "\n", + "change=(0.6-0.4615)/0.6\n", + "\n", + "print change" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6:Page number-60" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "req= 1.09 ohm\n", + "vs= 7.66 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r=2\n", + "i=2\n", + "\n", + "i3=3 #obtained by applying current divider rule to figure\n", + "\n", + "i4=1\n", + "\n", + "req=1/float(0.5+0.25+0.166) #1/2,1/4,1/6 values are converted to decimal form\n", + "\n", + "print \"req=\",format(req,'.2f'),\"ohm\"\n", + "\n", + "i2=(4*i4/float(6))\n", + "\n", + "i1=(6*i2)/float(req)\n", + "\n", + "#tracing circuit cabc via 6 ohm resistor and applying ohms law,\n", + "\n", + "vs=i1*i4+i2*6\n", + "\n", + "print \"vs=\",format(vs,'.2f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7:Page number-61" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of series parallel resistances is 10 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#combining series parallel series\n", + "\n", + "#[(2+2+2)||(6+5+2)||10]+5\n", + "\n", + "#[[6*6/6+6]+7]||10]+5=[10+10/10*10]+5=5+5=10\n", + "\n", + "print \"the value of series parallel resistances is 10 ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.8" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rab= 54.55 ohm\n", + "rab= 54.286 ohm\n", + "rcd= 50.91 ohm\n", + "rab= 50.67 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "#rab=(80+40)||(60+40)\n", + "\n", + "rab=(120*100)/float(120+100)\n", + "\n", + "print \"rab=\",format(rab,'.2f'),\"ohm\"\n", + "\n", + "#rab=(80||60)+(40||40)\n", + "\n", + "rab=(4800/float(140))+(1600/80)\n", + "print \"rab=\",format(rab,'.3f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "#(60+80)||(40+40)\n", + "\n", + "rcd=(140*80)/float(140+80)\n", + "\n", + "print \"rcd=\",format(rcd,'.2f'),\"ohm\"\n", + "\n", + "#(60||40)+(80||40)\n", + "\n", + "rab=float(2400/float(100))+(3200/float(120))\n", + "\n", + "print \"rab=\",format(rab,'.2f'),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 2.9:Page number-65" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ceq= 0.83402836 F\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#simplifying the circuit \n", + "\n", + "ceq=1/float(0.333+0.666+0.2) #converted to decimal form\n", + "\n", + "print \"ceq=\",format(ceq,'.8f'),\"F\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.10:Page number-67" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i= 1.200000 A\n", + "i1= 0.800000 A\n", + "i2= 0.400000 A\n", + "power consumed by 2 ohm resistor= 2.88 W\n", + "power consumed by 12 ohm resistor= 7.68 W\n", + "power consumed by 2 ohm resistor= 3.84 W\n", + "voltage drop= 2.4 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "I=12/(2+((12*24)/float(36))) #values taken from circuit\n", + "\n", + "I1=I*(24/float(36))\n", + "\n", + "I2=I*(12/float(36))\n", + "\n", + "print \"i=\",format(I,'1f'),\"A\"\n", + "\n", + "print \"i1=\",format(I1,'1f'),\"A\"\n", + "\n", + "print \"i2=\",format(I2,'1f'),\"A\"\n", + "\n", + "#case b\n", + "\n", + "power=(I**2)*2\n", + "\n", + "print \"power consumed by 2 ohm resistor=\",format(power,'.2f'),\"W\"\n", + "\n", + "\n", + "power=(I1**2)*12\n", + "\n", + "print \"power consumed by 12 ohm resistor=\",format(power,'.2f'),\"W\"\n", + "\n", + "power=(I2**2)*24\n", + "\n", + "print \"power consumed by 2 ohm resistor=\",format(power,'.2f'),\"W\"\n", + "\n", + "#case c\n", + "\n", + "v=I*2\n", + "print \"voltage drop=\",format(v,'.1f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.11:Page number-69" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rab=3.12ohm\n", + "ran=6 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "#values taken and calculated from figure\n", + "\n", + "r1=6\n", + "r2=12\n", + "r3=18\n", + "\n", + "rab=3.21 #calculating similar to above using parallel in series resistances\n", + "\n", + "print \"rab=3.12ohm\"\n", + "\n", + "#case b\n", + "\n", + "r4=30\n", + "r5=15\n", + "r6=30\n", + "\n", + "ran=6 #similar as above\n", + "\n", + "print \"ran=6 ohm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.12:Page number-73" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "v1=0.0769 V\n", + "v2=-0.3846V\n", + "current in 0.5ohm resistance is 0.154A,0.25ohm resistance is 1.846,0.66ohm resistor is -1.154A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#eqns derived from figure\n", + "\n", + "#6v1-4v2=2-->1\n", + "#-4v1+7v2=-3-->2\n", + "\n", + "#eqn 1 and 2 are written in matrix form and solved using cramers rule\n", + "\n", + "print \"v1=0.0769 V\"\n", + "\n", + "print \"v2=-0.3846V\"\n", + "\n", + "print \"current in 0.5ohm resistance is 0.154A,0.25ohm resistance is 1.846,0.66ohm resistor is -1.154A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.13:Page number-74" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "v1=3.6V\n", + "v2=2.2V\n", + "the current in 0.6 ohm resistor is 10.8A,0.2 ohm resistor is 7A,0.16ohm resistor is 13.2 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#from the figure the eqns are written in matrix form and using cramers rule the value of v1 and v2 can be found\n", + "\n", + "print \"v1=3.6V\"\n", + "\n", + "print \"v2=2.2V\"\n", + "\n", + "print \"the current in 0.6 ohm resistor is 10.8A,0.2 ohm resistor is 7A,0.16ohm resistor is 13.2 A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.14:Page number-76 " + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the voltages of nodes 1 and 3 are 50.29 and 57.71 respectively\n", + "current through 16 ohm resistor is 1.64A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#kcl is applied to the circuit and the eqns obtained are solved using cramer's rule\n", + "\n", + "print \"the voltages of nodes 1 and 3 are 50.29 and 57.71 respectively\"\n", + "\n", + "#i3=v/r\n", + "\n", + "print \"current through 16 ohm resistor is 1.64A\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.15:Page number-78" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage across 3 ohm resistor is= 5.832 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#the eqns obtained are converted to matrix form for solving using cramer's rule values are found\n", + "\n", + "i1=5.224\n", + "i2=0.7463\n", + "i3=3.28\n", + "\n", + "v=(i1-i3)*3\n", + "\n", + "print \"voltage across 3 ohm resistor is=\",format(v,'.3f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.16:page number-79" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "currents obtained are i1=2.013 and i2=1.273\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#kvl eqns are obtained from figure which are solved to obtain currents\n", + "\n", + "print \"currents obtained are i1=2.013 and i2=1.273\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.17:Page number-80" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage at node D= 5.68 v\n", + "current in 4 ohm resistor is= 1.47 A\n", + "power supplied by 18V source is= 27.72 W\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#the currents are obtained by solving the eqns\n", + "\n", + "i1=5.87\n", + "i2=-0.13\n", + "i3=-1.54\n", + "\n", + "v=18-1.54*8\n", + "\n", + "print \"voltage at node D=\",format(v,'.2f'),\"v\"\n", + "\n", + "i=5.86/float(4)\n", + "\n", + "print \"current in 4 ohm resistor is=\",format(i,'.2f'),\"A\"\n", + "\n", + "power=18*1.54\n", + "\n", + "print \"power supplied by 18V source is=\",format(power,'.2f'),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.18:Page number-82" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "va=8.33V and vb=4.17V\n", + "current through 8 ohm resistor is= 1.04 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#node eqns are obtained form the figure\n", + "\n", + "print \"va=8.33V and vb=4.17V\"\n", + "\n", + "i=8.33/float(8)\n", + "\n", + "print \"current through 8 ohm resistor is=\",format(i,'.2f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.19:Page number-83 " + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i1=-1.363A and i2=-3.4A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#eqns obtained are calculated just like above problems and are aolved for i1 and i2\n", + "\n", + "print \"i1=-1.363A and i2=-3.4A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.20:Page number-84" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current supplied by dependent source is= -6.0 A\n", + "power supplied by voltage source is= 41.34 W\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#eqns are obtained from the figure and are solved for currents\n", + "\n", + "i1=6.89\n", + "i2=3.89\n", + "i3=-2.12\n", + "\n", + "i=2*(i2-i1)\n", + "\n", + "print \"current supplied by dependent source is=\",format(i,'.1f'),\"A\"\n", + "\n", + "power=6*i1\n", + "\n", + "print \"power supplied by voltage source is=\",format(power,'.2f'),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.21:Page number-86" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i8= 0.667 A\n", + "i8'= 1.333 A\n", + "total current= 2.0 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#the following problem is based on usage of superposition theorem\n", + "\n", + "i8=12/float(6+4+8) #current for 8 ohm resistor.the resistances are in series with each other.Hence 6+4+8\n", + "\n", + "#next when voltage source is short circuited (8+4) total of resistance is obtained.The 4A is distributed in parallel branches as per current divider rule\n", + "\n", + "i=(4*6)/float(6+12)\n", + "\n", + "print \"i8=\",format(i8,'.3f'),\"A\"\n", + "\n", + "print \"i8'=\",format(i,'.3f'),\"A\"\n", + "\n", + "tot=i8+i\n", + "\n", + "print \"total current=\",format(tot,'.1f'),\"A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exampe 2.22:Page number-88" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.972972972973 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#kvl is applied to circuit\n", + "\n", + "i=1\n", + "\n", + "vth=12-(1*4) #12 is voltage 1 is current and 4 is resistance\n", + "\n", + "rth=(4*5)/float(4+5)\n", + "\n", + "i6=vth/float(rth+6) #since current passes through 6 ohm resistor\n", + "\n", + "print i6,\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.23:Page number-89" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current through 2 ohm resistor is= 2.45 A\n", + "Note that the same problem is again solved using superposition theorem and hence ignored \n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#thevenin's theorem and superposition theorem used here\n", + "\n", + "#applying mesh eqns to the 2 circuits and after getting the eqns they are solved using cramer's rule to obtain i1 and i2\n", + "\n", + "i1=-0.6\n", + "i2=-1.2\n", + "\n", + "#the value of currents indicates that they have assumed to be flowing in directions opposite to the assumed direction\n", + "\n", + "vth=12-1.2*3 #voltage eqn\n", + "\n", + "rth=1.425 #(1+2||12)||3=(1+(2*12)/(2+12))||3=19/7||3=19/7*3/19/7+3=1.425\n", + "\n", + "i2=vth/(rth+2)\n", + "\n", + "print \"current through 2 ohm resistor is=\",format(i2,'.2f'),\"A\"\n", + "\n", + "print \"Note that the same problem is again solved using superposition theorem and hence ignored \" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.24:Page number-91" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current through 5 ohm resistor is= 1.327 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#using thevenin's theorem\n", + "\n", + "#applying kcl at node a va is obtained\n", + "\n", + "va=12\n", + "\n", + "rth=1.33 #2||4\n", + "\n", + "i5=vth/(rth+5)\n", + "\n", + "print \"current through 5 ohm resistor is=\",format(i5,'.3f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.25:Page number-92" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rth= 4.997 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#applying kvl to circuit\n", + "\n", + "i=0.414\n", + "\n", + "vth=12-4*0.414 #using vth formula\n", + "\n", + "#when terminals a and b are short circuited applying kcl to node a gives isc=5*i\n", + "\n", + "isc=2.07\n", + "\n", + "rth=vth/isc\n", + "\n", + "print \"rth=\",format(rth,'.3f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.26:Page number-93" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "iab= 1.5 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#norton's theorem\n", + "\n", + "v=10\n", + "\n", + "#applying kvl to closed circuit \n", + "\n", + "isc=12/float(2+2) \n", + "\n", + "rn=4 #resistance obtained by short circuiting v and opening i\n", + "\n", + "iab=(4*3)/float(4+4) #current through 4 ohm connected across AB\n", + "\n", + "print \"iab=\",format(iab,'.1f'),\"A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.27:Page number-103" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "natural frequency= -0.91668 secinverse\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#natural frequency needs to be determined\n", + "\n", + "#req=[(6+6)||4]+[1||2]=3.6666\n", + "\n", + "req=3.6667\n", + "\n", + "l=4 #inductance\n", + "\n", + "s=-req/float(l)\n", + "\n", + "print \"natural frequency=\",format(s,'.5f'),\"secinverse\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.28" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "natural frequency= -0.15873 secinverse\n", + "time constant= 6.3 sec\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#req=[10+2+(5||15)]=15.75\n", + "\n", + "#case a\n", + "\n", + "c=0.4\n", + "req=15.75\n", + "s=-1/float(c*req)\n", + "\n", + "print \"natural frequency=\",format(s,'.5f'),\"secinverse\"\n", + "\n", + "#case b\n", + "\n", + "tc=req*0.4 #time constant\n", + "\n", + "print \"time constant=\",format(tc,'.1f'),\"sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.30:Page number-109" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage= 1560.0 v\n", + "r=20 ohm\n", + "tc= 0.1667 sec\n", + "balance energy= 2.25 J\n", + "t=0.25 sec\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "v=120\n", + "r=40\n", + "\n", + "i=v/float(r)\n", + "\n", + "#applying kvl to the closed loop\n", + "\n", + "v=3*520\n", + "\n", + "print \"voltage=\",format(v,'.1f'),\"v\"\n", + "\n", + "#when v=120,R can be found by I*(r+20)=120-->r=20\n", + "\n", + "r=20\n", + "\n", + "print \"r=20 ohm\"\n", + "\n", + "#when r=20 total r=20+20+20=60\n", + "\n", + "r=60\n", + "\n", + "l=10\n", + "\n", + "tc=l/float(r) #time constant\n", + "\n", + "print \"tc=\",format(tc,'.4f'),\"sec\"\n", + "\n", + "#i=I0*e^-(t/tc)=3*e^(-6t)\n", + "\n", + "energy=(10*9)/float(2)\n", + "\n", + "benergy=0.05*energy\n", + "\n", + "print \"balance energy=\",format(benergy,'.2f'),\"J\"\n", + "\n", + "#(L*i^2)/2=2.25-->hence i=0.6708\n", + "\n", + "#3*e^-6t=0.6708-->e^-6t=0.2236-->applying log on both sides we get t=0.25\n", + "\n", + "print \"t=0.25 sec\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.34:Page number-116" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R=2.72Mohm\n", + "t=9.16 sec\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "v=120\n", + "\n", + "V=200\n", + "\n", + "#v=V(1-e^-5/2R)\n", + "\n", + "#120=200*(1-e^-5/2R)\n", + "\n", + "#applying log on both sides and solving we get R=2.72 Mohm\n", + "\n", + "print \"R=2.72Mohm\"\n", + "\n", + "R=5 \n", + "tc=10\n", + "\n", + "#applying in the above eqn and solving lograthmically we get t=9.16\n", + "\n", + "print \"t=9.16 sec\"" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter4.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter4.ipynb new file mode 100644 index 00000000..1c1f8fba --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter4.ipynb @@ -0,0 +1,2063 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4:Alternating quantities" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1:Page number-193" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "e(t)=0 V\n", + "e(t)= 362.58 V\n", + "e(t)= 418.67 V\n" + ] + } + ], + "source": [ + "import math \n", + "\n", + "#given\n", + "\n", + "b=0.2\n", + "a=0.04\n", + "n=1000/float(60) #rev/sec\n", + "t=500\n", + "\n", + "#case a\n", + "\n", + "#since coil is at right angles ang=0\n", + "\n", + "print \"e(t)=0 V\"\n", + "\n", + "#case b\n", + "\n", + "#when coil is 30deg to the field ang=60\n", + "\n", + "#p=math.sin(60) \n", + "\n", + "p=0.8660254\n", + "\n", + "e=2*3.14*a*n*b*t*p\n", + "\n", + "\n", + "print \"e(t)=\",format(e,'.2f'),\"V\"\n", + "\n", + "#case c\n", + "\n", + "#when ang=90 that is coil is in the plane of the field\n", + "\n", + "#p=math.sin(90)\n", + "\n", + "p=1\n", + "e=2*3.14*b*a*n*p*t\n", + "\n", + "print \"e(t)=\",format(e,'.2f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.2:Page number-202" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "t= 0.0167 sec\n", + "f= 60.0 Hz\n", + "t= -0.0014 sec\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "vm=155\n", + "omega=377\n", + "\n", + "#case a\n", + "\n", + "t=(2*3.14)/float(omega)\n", + "\n", + "print \"t=\",format(t,'.4f'),\"sec\"\n", + "\n", + "#case b\n", + "\n", + "f=1/float(t)\n", + "\n", + "print \"f=\",format(f,'.1f'),\"Hz\"\n", + "\n", + "#case c\n", + "\n", + "v=109.60 #rms value\n", + "\n", + "#at t=0 -77.5=155*sin(ang)\n", + "\n", + "#therefore, ang=-0.5236 rad\n", + "\n", + "ang=-0.5236\n", + "\n", + "t=ang/omega\n", + "\n", + "print \"t=\",format(t,'.4f'),\"sec\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i= 10.0 A\n", + "f= 50.0 A\n", + "i= 0.15 A\n", + "NOTE:Answer calculated wrongly in textbook for i obtained here\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "#i=14.14*sin(314t)-->i=im*sin(omega*t)\n", + "\n", + "#case a\n", + "\n", + "im=14.14\n", + "i=14.14/1.414 #1.414 is the value of root 2\n", + "\n", + "print \"i=\",format(i,'.1f'),\"A\"\n", + "\n", + "#case b\n", + "\n", + "#omega=314=2*3.14*f\n", + "\n", + "f=314/float(2*3.14)\n", + "\n", + "print \"f=\",format(f,'.1f'),\"A\"\n", + "\n", + "#case c\n", + "\n", + "t=0.002\n", + "\n", + "#i=im*sin(omega*t)\n", + "\n", + "p=0.01096 #value of sin(omega*t)\n", + "i=im*p\n", + "\n", + "print \"i=\",format(i,'.2f'),\"A\" \n", + "\n", + "print \"NOTE:Answer calculated wrongly in textbook for i obtained here\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4:Page number-203" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I= 24.496 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "i=20\n", + "im=i/float(1.414) #that is i*root 2\n", + "\n", + "#the heat produced by i is the sum of heat produced by dc and ac current\n", + "p=i**2\n", + "q=im**2\n", + "r=p+q\n", + "I=(r**0.5)\n", + "\n", + "print \"I=\",format(I,'.3f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i= 0.2 A\n", + "i= 1.4 A\n", + "NOTE:The answer given in text is printed wrongly\n", + "t= 0.00333 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "f=50\n", + "irms=10\n", + "\n", + "im=irms/float(0.707)\n", + "\n", + "#omega*t=2*3.14*f*t here the value for t can be substituted and value for i can be found from i=im*sin(omega*t)\n", + "\n", + "t=0.0025\n", + "p=0.0137 #value of sin(314*0.0025)\n", + "i=(10*p)/float(0.707)\n", + "\n", + "print \"i=\",format(i,'.1f'),\"A\"\n", + "\n", + "#maximum value is when 314*t=pi/2 (in radians)-->t=0.005\n", + "\n", + "#hence at t=0.005+0.0125=0.0175 the value of i nedds to be found\n", + "p=0.0957\n", + "i=(10*p)/float(0.707)\n", + "\n", + "print \"i=\",format(i,'.1f'),\"A\"\n", + "print \"NOTE:The answer given in text is printed wrongly\"\n", + "\n", + "i=7.07\n", + "\n", + "#7.07=(10*sin314t)/0.707-->t=0.00833 sec\n", + "\n", + "t=0.00833-0.005 #the time at which the instaneous value is 7.07A after positive maximum value is at this time\n", + "\n", + "print \"t=\",format(t,'.5f'),\"A\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.6:Page number-204" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "v= 25.79 V\n", + "vavg= 20.0 v\n", + "1.28937969582\n", + "1.93891683582\n", + "rms value for a sin wave with the same peak value is= 35.35 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#from graph \n", + "a=0\n", + "b=5**2\n", + "c=10**2\n", + "c=20**2\n", + "d=40**2\n", + "e=50**2\n", + "f=40**2\n", + "g=20**2\n", + "h=10**2\n", + "i=5**2\n", + "v=(0.1*(a+b+c+d+e+f+g+h+i))**0.5 #pi and omega values get cancelled\n", + "\n", + "print \"v=\",format(v,'.2f'),\"V\"\n", + "vavg=0.1*(0+5+10+20+40+50+40+20+10+5)\n", + "print \"vavg=\",format(vavg,'.1f'),\"v\"\n", + "ff=v/float(vavg)\n", + "print ff\n", + "\n", + "pf=50/float(v) #50 is the maximum value\n", + "print pf\n", + "\n", + "v=0.707*50 \n", + "\n", + "print \"rms value for a sin wave with the same peak value is=\",format(v,'.2f'),\"V\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8:Page number-210" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "vac= 130.77 v\n", + "phase position with respect to vbc=60-36.59=23.41\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#from phasor diagram vac=vab+vbc\n", + "\n", + "hcab=60\n", + "vcab=60\n", + "hcbc=45\n", + "vcbc=77.94 #vbc=60*sin(60)\n", + "\n", + "p=(vcab+hcbc)**2\n", + "q=vcbc**2\n", + "vac=((p+q)**0.5)\n", + "\n", + "print \"vac=\",format(vac,'.2f'),\"v\"\n", + "\n", + "#the angle is given by ang=taninverse(vcbc/(vcab+hcbc))=36.59\n", + "\n", + "print \"phase position with respect to vbc=60-36.59=23.41\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9:Page number-210" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "E1**2+2*E1*E2*cos(alpha)+E2**2=5836.96\n", + "E1**2+2*E1*E2*cos(alpha)+E2**2=712.89\n", + "E1=46.12V,E2=33.88V\n", + "alpha=34.93\n" + ] + } + ], + "source": [ + "Example 4.9:Page number-210" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.10:Page number-215" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a+b=22.72+j2.12\n", + "a/b=-0.13+j0.74\n", + "Thus (a+b)/(a-b) gives -0.24-j0.81\n", + "(a+b*b/(a-b)*a)=-1.01+j0.5\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#a=6.34+j*13.59\n", + "#b=20angle(35)\n", + "\n", + "#case a-->(a+b)\n", + "\n", + "#in polar form a=15 at angle 65\n", + "#in rectangular form b=16.38-j*11.47\n", + "\n", + "#a+b=6.34+j13.59+16.38-j11.47=22.72+j2.12\n", + "\n", + "print \"a+b=22.72+j2.12\"\n", + "\n", + "#a/b=15angle(65)/20angle(-35)=0.75angle(100)=-0.13+j0.74\n", + "\n", + "print \"a/b=-0.13+j0.74\"\n", + "\n", + "#a-b=-10.04+j25.06\n", + "\n", + "print \"Thus (a+b)/(a-b) gives -0.24-j0.81\"\n", + "\n", + "#(a+b)*b/(a-b)*a\n", + "\n", + "print \"(a+b*b/(a-b)*a)=-1.01+j0.5\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.11" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I=7.5+j4.75. Its value in polar form is obtained as 8.8776 at angle 32.34\n", + "instantaneous value of resultant i is 12.5548*sin(314t+32.34)\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#i1=20*sin(314t+60),i2=-10*sin(314t),i3=15*sin(314t-45)-->angles are in degrees\n", + "\n", + "#I1=(7.7072+j12.25),I2=(-7.072),I3=7.5-j7.5\n", + "\n", + "#adding phasor currents I1,I2 and I3\n", + "\n", + "#I=7.702+j12.25-7.702+7.5-j7.5=7.5+j4.75\n", + "\n", + "print \"I=7.5+j4.75. Its value in polar form is obtained as 8.8776 at angle 32.34\"\n", + "\n", + "#i=2**0.5*8.8776*sin(314t+32.34)-->instantaneous value of resultant i\n", + "\n", + "print \"instantaneous value of resultant i is 12.5548*sin(314t+32.34)\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.12:Page number-226" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i= 12.35 A\n", + "phase angle of current=57.52 lag\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "v=230\n", + "f=50\n", + "L=50*10**-3\n", + "r=10\n", + "\n", + "#case a\n", + "xl=2*3.14*f*L\n", + "\n", + "z=complex(r,xl)\n", + "\n", + "#the value of z in polar form is 18.62 ohm\n", + "\n", + "z=18.62\n", + "\n", + "i=v/float(z)\n", + "\n", + "print \"i=\",format(i,'.2f'),\"A\"\n", + "\n", + "#case b\n", + "\n", + "#phy=taninverse(xl/r)=57.52 lag\n", + "\n", + "print \"phase angle of current=57.52 lag\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.13" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "v= 279.21 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "vr=150\n", + "r=50\n", + "l=250*10**-3\n", + "f=50\n", + "\n", + "i=vr/r\n", + "\n", + "xl=2*3.14*f*l\n", + "\n", + "vl=i*xl\n", + "\n", + "v=(((vr**2)+(vl**2))**0.5)\n", + "\n", + "print \"v=\",format(v,'.2f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.14" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.6875\n", + "power consumed= 500.0 w\n", + "power consumed in choke oil= 187.5 W\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "v=200\n", + "f=50\n", + "r=20\n", + "vr=100\n", + "vc=144\n", + "vl=150\n", + "\n", + "#case a\n", + "\n", + "#from eqn ((vr**2+vl*cos(angle))**2)+((vl*sin(angle))**2)=v**2\n", + "\n", + "#on substituting values in the above eqn the value of angle can be found by isolating cos\n", + "\n", + "#angle=75.52\n", + "\n", + "cos=0.25\n", + "\n", + "pf=(vr+vl*cos)/float(v)\n", + "\n", + "print pf\n", + "\n", + "#case b\n", + "\n", + "i=vr/r\n", + "power=i**2*r\n", + "\n", + "print \"power consumed=\",format(power,'.1f'),\"w\"\n", + "\n", + "#case c\n", + "\n", + "power=vl*i*cos\n", + "\n", + "print \"power consumed in choke oil=\",format(power,'.1f'),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.15:Page number-230" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " xc= 31.85 ohm\n", + "i= 6.89 A\n", + "0.299580587178\n", + "phase angle=72.6\n", + "v= 68.9 v\n", + "v= 219.4 v\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r=10\n", + "c=10**-4\n", + "v=230\n", + "f=50\n", + "omega=314\n", + "\n", + "#case a\n", + "xc=1/float(omega*c)\n", + "\n", + "print \"xc=\",format(xc,'.2f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "zc=33.38 #zc=10-j31.85 into polar form is 33.38\n", + "\n", + "i=v/zc\n", + "\n", + "print \"i=\",format(i,'.2f'),\"A\"\n", + "\n", + "#case c\n", + "\n", + "pf=r/zc\n", + "\n", + "print pf\n", + "\n", + "#case d\n", + "\n", + "#phase angle=cosinverse(0.3)=72.6\n", + "\n", + "print \"phase angle=72.6\"\n", + "\n", + "#case e\n", + "\n", + "v=r*i\n", + "\n", + "print \"v=\",format(v,'.1f'),\"v\"\n", + "\n", + "v=xc*i\n", + "\n", + "print \"v=\",format(v,'.1f'),\"v\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.16:Page number-230" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "vc= 207.12 v\n", + "c= 0.00007688 F\n", + "maximum voltage across c= 292.92 V\n", + "phase angle=64.2\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "v=230\n", + "f=50\n", + "\n", + "#voltage vr across r is in phase with the current i while voltage vc across c lage i by 90\n", + "\n", + "#from phasor diagram v**2=vr**2+vc**2\n", + "\n", + "vr=100\n", + "\n", + "vc=((v**2)-(vr**2))**0.5\n", + "\n", + "print \"vc=\",format(vc,'.2f'),\"v\"\n", + "p=500 #power\n", + "\n", + "i=p/vr\n", + "\n", + "c=i/float(2*3.14*f*vc)\n", + "\n", + "print \"c=\",format(c,'.8f'),\"F\"\n", + "\n", + "#case b\n", + "\n", + "v=(2**0.5)*vc\n", + "\n", + "print \"maximum voltage across c=\",format(v,'.2f'),\"V\"\n", + "\n", + "#case c\n", + "\n", + "#phase angle=cosinverse(vr/v)=cosinverse(0.4348)=64.2\n", + "\n", + "print \"phase angle=64.2\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.17:Page number-234" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "xl= 47.1 ohm\n", + "xc= 25.48 ohm\n", + "complex impedance=8+j21.62 at an impedance angle of 69.7\n", + "current= 9.98 A\n", + "voltage across coil=446.8 at 10.66 degrees\n", + "voltage across capacitor=-254.29 at -159.7 degrees\n", + "phase difference between supply and current i is 69.7 lag\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r=8\n", + "l=0.15\n", + "f=50\n", + "v=230\n", + "c=125*10**-6\n", + "\n", + "#case a inductive reactance\n", + "\n", + "xl=2*3.14*f*l\n", + "\n", + "print \"xl=\",format(xl,'.1f'),\"ohm\"\n", + "\n", + "#case b capacitance reactance\n", + "\n", + "xc=1/float(2*3.14*f*c)\n", + "\n", + "print 'xc=',format(xc,'.2f'),\"ohm\"\n", + "\n", + "#case c complex impedance\n", + "\n", + "#z=r+j(xl-xc)-->on substituting valuees we get z=8+j21.62\n", + "\n", + "#z=((8**2)+(21.62**2))**0.5\n", + "\n", + "print \"complex impedance=8+j21.62 at an impedance angle of 69.7\"\n", + "\n", + "#impedance angle=taninverse(xl-xr)/r\n", + "\n", + "#case d\n", + "\n", + "v=230\n", + "z=23.05\n", + "i=v/z\n", + "\n", + "print \"current=\",format(i,'.2f'),\"A\"\n", + "\n", + "#case e\n", + "\n", + "#(r+jxl)*i=446.8 at 10.66 degrees\n", + "\n", + "print \"voltage across coil=446.8 at 10.66 degrees\"\n", + "\n", + "#-j*xc*i=25.48*9.98\n", + "print \"voltage across capacitor=-254.29 at -159.7 degrees\"\n", + "\n", + "#case e\n", + "\n", + "print 'phase difference between supply and current i is 69.7 lag'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.18:Page number-235 " + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "capacitive reactance= 63.7 ohm\n", + "f= 50.0 cycles/sec\n", + "power loss in iron cored choke is= 53.69 w\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "c=50*10**-6\n", + "i=2.355\n", + "\n", + "#case a\n", + "\n", + "vl=120\n", + "vr=70\n", + "vac=150\n", + "\n", + "#the phasor sum of vr and vl is OC;the applied voltage v is the phasor sum of vc and OC and is represented by OV\n", + "\n", + "#the theta be the impedance angle of RL combination\n", + "\n", + "#from right angled triangle OCD,theta can be determined as follows:\n", + "#(vr+vl*costheta)**2+(vl*costheta)**2=vac**2\n", + "#substitute the values then value of costheta can be found\n", + "\n", + "zl=vl/i #impedance of the coil\n", + "\n", + "p=0.981 #value of sin(79)\n", + "xl=zl*p\n", + "\n", + "q=0.19 #value of cos(79)\n", + "r=zl*q\n", + "\n", + "dc=i*xl\n", + "bd=i*r\n", + "#from right angled triangle ODB in fig.\n", + "\n", + "v=98.3\n", + "\n", + "xc=vac/i\n", + "\n", + "print \"capacitive reactance=\",format(xc,'.1f'),\"ohm\"\n", + "\n", + "f=1/float(xc*2*3.14*c)\n", + "\n", + "print \"f=\",format(f,'.1f'),\"cycles/sec\"\n", + "\n", + "ploss=i**2*r\n", + "\n", + "\n", + "print \"power loss in iron cored choke is=\",format(ploss,'.2f'),\"w\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.19:Page number-238" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i= 12.07 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r=20\n", + "l=200*10**-3\n", + "v=230\n", + "f=50\n", + "\n", + "xl=314*l #314 is omega\n", + "\n", + "ir=v/float(r)\n", + "\n", + "il=v/float(xl)\n", + "\n", + "i=((ir**2)+(il**2))**0.5\n", + "\n", + "print \"i=\",format(i,'.2f'),\"A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.20:Page number-240 " + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current with a lead of 57.5 is obtained as= 4.13 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r=100\n", + "c=50*10**-6\n", + "f=50\n", + "v=230\n", + "\n", + "#case a\n", + "\n", + "xc=-1/float(314*c) #314 is omega\n", + "\n", + "ir=v/r #with angle 0\n", + "\n", + "ic=230/float(xc) #with angle of 90 deg\n", + "\n", + "i=((ir**2)+(ic**2))**0.5\n", + "\n", + "print \"current with a lead of 57.5 is obtained as=\",format(i,'.2f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.21:Page number-242" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current at 56.76 lead= 4.196 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r=100\n", + "l=0.1\n", + "c=150*10**-6\n", + "v=230\n", + "f=50\n", + "\n", + "#case a\n", + "\n", + "xl=314*l #at 90 deg\n", + "\n", + "xc=1/float(314*c) #at lag -90 deg\n", + "\n", + "ir=v/r #at 0 deg\n", + "il=v/xl\n", + "ic=v/xc\n", + "\n", + "#i=ir+ic+il-->2.3+j3.51\n", + "\n", + "i=((2.3**2)+(3.51**2))**0.5\n", + "\n", + "print \"current at 56.76 lead=\",format(i,'.3f'),\"A\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.22:Page number-244" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of zbc is 8.159-j9.553\n", + "zac=18.159+j5.447(in rectangular form)\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "z1=18.03 #z1=10+j15 converted to polar form also it is at angle 56.31\n", + "z2=32.02\n", + "z3=10.77\n", + "\n", + "#ybc=1/zbc=(1/z2+1/z3)=1/32.02+1/10.77\n", + "\n", + "#on performing the add operation we get the value of zbc as 8.159-j9.553 that is in rectangular form\n", + "\n", + "print \"The value of zbc is 8.159-j9.553\"\n", + "\n", + "#thus total impedance between terminals A and C is given by zac=z1+zbc\n", + "\n", + "print \"zac=18.159+j5.447(in rectangular form)\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.23:Page number-246" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I= 5.76 A\n", + "z= 39.91 ohm\n", + "R= 36.97 ohm\n", + "x= -15.03 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r1=25\n", + "l1=0.159\n", + "r2=60\n", + "c=125*10**-6\n", + "v=230\n", + "f=50\n", + "\n", + "#case a\n", + "\n", + "xl=2*3.14*f*l1\n", + "\n", + "z1=((r1**2)+(xl**2))**0.5\n", + "\n", + "i1=v/z1\n", + "\n", + "#phy1=cosinverse(r1/z1)=63.43 lag\n", + "\n", + "xc=1/float(2*3.14**c)\n", + "\n", + "z2=((r2**2)+(xc**2))**0.5\n", + "\n", + "i2=v/z2\n", + "\n", + "#i2 has 23 deg lead calculated similar to i1\n", + "#p=cosphy1\n", + "#q=cosphy2\n", + "\n", + "p=0.44\n", + "q=0.92\n", + "I1=i1*p+i2*q\n", + "a=-0.89\n", + "b=0.39\n", + "I2=i1*a+i2*b\n", + "\n", + "I=((I1**2)+(I2**2))**0.5\n", + "\n", + "print \"I=\",format(I,'.2f'),\"A\"\n", + "\n", + "#case b\n", + "\n", + "z=v/I\n", + "\n", + "print \"z=\",format(z,'.2f'),\"ohm\"\n", + "\n", + "R=(z*I1)/I #note the value of I in text is printed wrongly so the result may vary\n", + "\n", + "print \"R=\",format(R,'.2f'),\"ohm\"\n", + "\n", + "x=(z*I2)/I #same note applicable here as well\n", + "\n", + "print \"x=\",format(x,'.2f'),\"ohm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.24:Page number-247" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I1=6.78A\n", + "I2=13.22A\n", + "power loss in z1= 689.53 W\n", + "power loss in z2= 1398.15 W\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "#z1=15+j20\n", + "#z2=8-j10\n", + "I=20\n", + "z1=25 #in polar form at angle 53.13\n", + "z2=12.81 #at angle -51.34\n", + "\n", + "#v=I1z1=I2z2\n", + "#I2=1.95I1\n", + "\n", + "#from diagram I**2=(I1cosang1+I2cosang2)**2+(I2sinang2-I1sinang1)**2\n", + "#on substituting values in the above eqn and simplifying\n", + "I1=6.78\n", + "print \"I1=6.78A\"\n", + "I2=13.22\n", + "#substitute this in I2=1.95I1\n", + "\n", + "print \"I2=13.22A\"\n", + "\n", + "pow1=I1**2*15\n", + "pow2=I2**2*8\n", + "\n", + "print \"power loss in z1=\",format(pow1,'.2f'),\"W\"\n", + "print \"power loss in z2=\",format(pow2,'.2f'),\"W\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.25:Page number-248" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i1= 7.19 A\n", + "current lags by voltage 38.66\n", + "c= 0.00006218 F\n", + "ir= 5.606 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r=25\n", + "f=50\n", + "xl=20\n", + "v=230\n", + "\n", + "#case a\n", + "\n", + "#z1=r+jxl\n", + "\n", + "z1=32 #in polar form\n", + "i1=v/float(z1)\n", + "\n", + "print \"i1=\",format(i1,'.2f'),'A'\n", + "\n", + "#case b\n", + "\n", + "print \"current lags by voltage 38.66\"\n", + "\n", + "#case c\n", + "\n", + "p=0.78 #cos value\n", + "q=-0.62 #sin value\n", + "\n", + "ir=i1*p\n", + "il=i1*q\n", + "\n", + "#from phasor diagram current c is equal to il\n", + "\n", + "ic=il=4.491\n", + "\n", + "c=ic/float(v*2*3.14*50)\n", + "\n", + "print \"c=\",format(c,'.8f'),\"F\"\n", + "\n", + "#case d\n", + "\n", + "print \"ir=\",format(ir,'.3f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.26:Page number-249" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(12.9596827495-2.78255122274j)\n", + "the phase angle is -12.11\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "z1=complex(6,-10)\n", + "z2=complex(10,15)\n", + "z3=complex(18,12)\n", + "\n", + "#z1+z2 is parallel to z3\n", + "\n", + "zab=z1+(z2*z3)/(z2+z3)\n", + "\n", + "print zab\n", + "\n", + "print \"the phase angle is -12.11\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.27:Page number-258" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current at -63 lag is= 13.05 A\n", + "phase angle between supply voltage and current is -63\n", + "power= 3002.3 VA\n", + "active power= 1351.0 W\n", + "reactive power= 2672.0 VAR\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r=8\n", + "l=0.05\n", + "v=230\n", + "f=50\n", + "\n", + "#case a\n", + "\n", + "xl=2*3.14*f*l\n", + "\n", + "zl=complex(r,xl)\n", + "\n", + "zl=17.62\n", + "\n", + "i=v/zl #since v=230 at angle 0 and zl in polar form has 63 deg i has a lag of 63\n", + "\n", + "print \"current at -63 lag is=\",format(i,'.2f'),'A'\n", + "\n", + "#case b\n", + "\n", + "print \"phase angle between supply voltage and current is -63\"\n", + "\n", + "#case c\n", + "\n", + "power=v*i\n", + "\n", + "print \"power=\",format(power,'.1f'),\"VA\"\n", + "\n", + "#case d\n", + "p=0.45 #cos63\n", + "actpow=v*i*p\n", + "print \"active power=\",format(actpow,'.1f'),\"W\"\n", + "\n", + "#case e\n", + "\n", + "q=0.89 #sin63\n", + "\n", + "reapow=v*i*q\n", + "\n", + "print 'reactive power=',format(reapow,'.1f'),\"VAR\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.28:Page number-259" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input= 13392.86 VA\n", + "active component= 40.76 A\n", + "reactive component= 41.34 A\n", + "reactive power= 9508.9 VAR\n", + "c= 0.00006218 F\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "v=230\n", + "f=50\n", + "pf=0.7\n", + "n=0.8\n", + "op=7500\n", + "\n", + "#case a\n", + "\n", + "ip=op/float(0.7*0.8)\n", + "\n", + "print \"input=\",format(ip,'.2f'),\"VA\"\n", + "\n", + "#case b\n", + "\n", + "im=ip/v\n", + "\n", + "p=0.71 #sin\n", + "\n", + "activecompo=im*pf\n", + "\n", + "print \"active component=\",format(activecompo,'.2f'),\"A\"\n", + "\n", + "reacompo=p*im\n", + "\n", + "print \"reactive component=\",format(reacompo,'.2f'),\"A\"\n", + "\n", + "#case c\n", + "\n", + "reacpow=p*ip\n", + "\n", + "print \"reactive power=\",format(reacpow,'.1f'),\"VAR\"\n", + "\n", + "#case d\n", + "\n", + "cos=0.95\n", + "\n", + "i=activecompo/cos\n", + "\n", + "isin=13.40 #i*sinang=i*(1-cos**2)**0.5ic=28.18 #since i=ic+im\n", + "\n", + "c=ic/float(2*3.14*f*v)\n", + "\n", + "print \"c=\",format(c,'.8f'),\"F\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.29:Page number-266" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "c= 0.00004057\n", + "i= 115.0 A\n", + "39.25\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "l=0.25\n", + "f=50\n", + "v=230\n", + "r=2\n", + "\n", + "c=1/float(((2*3.14*f)**2)*l)\n", + "\n", + "print \"c=\",format(c,'.8f')\n", + "\n", + "#case b\n", + "\n", + "i=v/r\n", + "\n", + "print \"i=\",format(i,'.1f'),\"A\"\n", + "\n", + "#case c\n", + "\n", + "vl=2*3.14*f*l*i\n", + "vc=i/float(c*2*3.14*f)\n", + "\n", + "q=(2*3.14*f*l)/float(r)\n", + "\n", + "print q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.30:Page number-266" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "f0= 102.0860 Hz\n", + "f1= 101.2898 Hz\n", + "f2= 102.8822 Hz\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "l=10\n", + "r=100\n", + "i=1\n", + "f=100\n", + "i1=0.5\n", + "\n", + "c=1/float(4*(3.14**2)*(r**2)*l)\n", + "\n", + "v=i*r\n", + "z=v/i1\n", + "\n", + "#z=100+jX\n", + "\n", + "x=((200**2)-(100**2))**0.5\n", + "\n", + "omega=641.1 #angular frequency in rad/sec\n", + "\n", + "f0=omega/float(2*3.14)\n", + "\n", + "f1=f0-(r/float(4*3.14*l))\n", + "\n", + "f2=f0+(r/float(4*3.14*l))\n", + "\n", + "print \"f0=\",format(f0,'.4f'),\"Hz\"\n", + "\n", + "print \"f1=\",format(f1,'.4f'),\"Hz\"\n", + "\n", + "print \"f2=\",format(f2,'.4f'),\"Hz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.31:Page number-271 " + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "l= 0.00507 H\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "v=3*10**8\n", + "lamb=3000\n", + "c=0.0005*10**-6\n", + "f=v/lamb\n", + "\n", + "l=1/float(4*3.14*3.14*f**2*c)\n", + "\n", + "print \"l=\",format(l,'.5f'),\"H\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.32:Page number-272" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "c= 0.0000000005599 F\n", + "z= 238130.4 ohm\n", + "i= 0.0000063 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "r=1500\n", + "l=0.2\n", + "v=1.5\n", + "f=15000\n", + "\n", + "#case a\n", + "\n", + "#p=1/0.2c\n", + "\n", + "p=(4*3.14*3.14*f**2)+(r**2)/float(l**2)\n", + "\n", + "c=1/float(0.2*p)\n", + "\n", + "print \"c=\",format(c,'.13f'),\"F\"\n", + "\n", + "#case b\n", + "\n", + "z=l/float(c*r)\n", + "\n", + "print \"z=\",format(z,'.1f'),\"ohm\"\n", + "\n", + "#case c\n", + "\n", + "i=v/float(z)\n", + "\n", + "print \"i=\",format(i,'.7f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.33:Page number-274" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "v1 at -47.63 is= 18.80 V\n", + "v2 at -42.30 is= 21.55 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#the eqns are formed using the given diagram\n", + "#the derivations from the eqns are obtained as below using matrices for their construction\n", + "#the below eqns are in polar form\n", + "delta=0.3165\n", + "delta1=5.95\n", + "delta2=6.82\n", + "\n", + "v1=delta1/delta\n", + "\n", + "print \"v1 at -47.63 is=\",format(v1,'.2f'),\"V\"\n", + "\n", + "v2=delta2/delta\n", + "\n", + "print \"v2 at -42.30 is=\",format(v2,'.2f'),\"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.34:Page number-275 " + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i at -84.21 is= -1.32 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#in polar form\n", + "\n", + "z1=10\n", + "z2=12.806\n", + "z3=13.416\n", + "\n", + "#the mesh currents are written in matrix form\n", + "\n", + "delta=329.31 #in polar form\n", + "\n", + "delta1=360\n", + "delta2=793.22\n", + "\n", + "i1=delta1/delta\n", + "i2=delta2/delta\n", + "\n", + "i=i1-i2 #answer obtained in text is wrongly printed\n", + "\n", + "print \"i at -84.21 is=\",format(i,'.2f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.35:Page number-276 " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(1.638+4.839j)\n", + "(0.732-5.144j)\n", + "(2.37-0.305j)\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#superposition theorem\n", + "\n", + "r=4\n", + "\n", + "#z=4+(8+6j)*(0-j10)/8+j6+0-j10\n", + "\n", + "#z=14-j5\n", + "\n", + "z=14.87\n", + "l=40\n", + "#I1a=z/l=2.69 in polar form\n", + "I1a=complex(2.533,0.904)\n", + "\n", + "I2a=complex(-0.324,-2.67)\n", + "\n", + "\n", + "#from fig c\n", + "\n", + "z=complex(2.93,-9.47)\n", + "\n", + "I1b=complex(-0.895,3.935)\n", + "\n", + "I2b=complex(1.056,-2.474)\n", + "\n", + "I1=I1a+I1b\n", + "\n", + "print I1\n", + "\n", + "I2=I2a+I2b\n", + "\n", + "print I2\n", + "\n", + "I=I1+I2\n", + "\n", + "print I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.36:Page number-278" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(3.07692307692+5.38461538462j)\n", + "(8.81695846645+9.55403833866j)\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#thevenin's theorem\n", + "#all the values are derived from the figures\n", + "z1=complex(8,-6)\n", + "z2=complex(0,5)\n", + "\n", + "zth=(z1*z2)/(z1+z2)\n", + "\n", + "print zth\n", + "\n", + "vth=complex(-17.71,141.54)\n", + "\n", + "zload=complex(4,3)\n", + "\n", + "I=vth/(zth+zload)\n", + "\n", + "print I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.37:Page number-279" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(8.8178913738+9.55271565495j)\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#norton's theorem\n", + "\n", + "#values derived and calculated from figure\n", + "\n", + "v=complex(230,0)\n", + "xl=complex(8,-6)\n", + "\n", + "isc=v/xl\n", + "\n", + "IN=isc\n", + "\n", + "rl=complex(0,5)\n", + "zn=(rl*xl)/(rl+xl)\n", + "zload=complex(4,3)\n", + "\n", + "I=(IN*zn)/(zn+zload)\n", + "\n", + "print I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.38:Page number-281" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I= 4.51 A\n", + "pl= 18.75 w\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#all values derived from figure\n", + "\n", + "\n", + "#zth=complex(0.923,2.615)\n", + "\n", + "#vth=complex(-4.615,-6.923) #derived using formula\n", + "\n", + "#zl=complex(0.923,-2.615)\n", + "\n", + "#z=zl+zth\n", + "vth=8.32 #polar form\n", + "z=1.846\n", + "I=vth/z\n", + "\n", + "print \"I=\",format(I,'.2f'),\"A\"\n", + "\n", + "rl=0.923\n", + "pl=(I**2)*rl\n", + "\n", + "print \"pl=\",format(pl,'.2f'),\"w\"" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter5.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter5.ipynb new file mode 100644 index 00000000..dfdf53ef --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter5.ipynb @@ -0,0 +1,505 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: Three Phase Systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.1: Page number-317" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ia= 51.962 A\n", + "ib= 43.30129 A\n", + "ic= 34.64103 A\n", + "IN= 15.0 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "vl=400 #line voltage\n", + "\n", + "va=vl/math.sqrt(3)\n", + "vb=230.94 #angle(-120)\n", + "vc=230.94 #angle(-240)\n", + "\n", + "#case a\n", + "\n", + "#the line currents are given by\n", + "\n", + "ia=12000/230.94 #with angle 0\n", + "\n", + "ib=10000/230.94 #with angle 120\n", + "\n", + "ic=8000/230.94 #with angle 240\n", + "\n", + "print\"ia=\",round(ia,3),\"A\"\n", + "print \"ib=\",round(ib,5),\"A\"\n", + "print \"ic=\",round(ic,5),\"A\"\n", + "\n", + "#case b\n", + "\n", + "#IN=ia+ib+ic\n", + "\n", + "#ia,ib and ic are phase currents hence contain with angles they are in the form sin(angle)+icos(angle)\n", + "\n", + "#IN=51.96*(sin(0)+i*cos(0))+43.3*(sin(120)+i*cos(120))+34.64*(sin(240)+i*cos(240))\n", + "\n", + "#IN=51.96+(-21.65+i*37.5)+34.64*(-0.5-i*0.866)\n", + "\n", + "#12.99+i*7.5 on which the sin+icos=sin**2+cos**2 operation is performed\n", + "#therefore \n", + "\n", + "IN=15 #at angle 30\n", + "print \"IN=\",round(IN,10),\"A\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2:Page number-320 " + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "iab= 2.0 A\n", + "ibc=5.4414-j3.1416 A\n", + "ica=3.1463+j4.2056 A\n", + "ia=4.2328 with an angle of -96.51 A\n", + "ib=4.1915 with angle of -48.55 A\n", + "ic=7.6973 with an angle of 107.35 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "vab=400 #phase angle of 0\n", + "vbc=400 #phase angle of 120\n", + "vca=400 #phase angle of 240\n", + "\n", + "#the phase currents are given by iab,ibc,ica\n", + "\n", + "iab=400/150 #from the diagram \n", + "\n", + "print \"iab=\",round(iab,5),\"A\"\n", + "#ibc=(400*314*50)/10**6 numerator with an angle of -120 and denominator angle of -90 which amounts to -30 in numerator\n", + "#this leads to simplifying with the formula as the value obtained for ibc after simplification from above mutiplied by values of cos(-30)+jsin(-30)\n", + "#therefore print as below\n", + "\n", + "print\"ibc=5.4414-j3.1416\",\"A\"\n", + "\n", + "#same method for ica\n", + "\n", + "\n", + "print \"ica=3.1463+j4.2056\",\"A\"\n", + "\n", + "#case b\n", + "\n", + "#ia=iab-ica\n", + "\n", + "#ia=2.667-(3.1463+j4.2056)\n", + "\n", + "#leads to 4.2328 with an angle of -96.51\n", + "#angle calculated using tan formula\n", + "print \"ia=4.2328 with an angle of -96.51\",\"A\"\n", + "\n", + "#same for ib and ic\n", + "\n", + "print \"ib=4.1915 with angle of -48.55\",\"A\"\n", + "\n", + "print \"ic=7.6973 with an angle of 107.35\",\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3:Page number:321" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power factor =0.8\n", + "p= 25601.1 KW\n", + "q= 19200.82 Kvar\n", + "t= 32001.0 KVA\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "#given\n", + "zl=5 #load impedanc with an angle of 36.87 degrees\n", + "vl=400 #line voltage\n", + "il=46.19\n", + "va=400/3**0.5 #phase voltage\n", + "\n", + "ia=va/zl #line current with an angle of -36.87 degrees\n", + "\n", + "#ib and ic are also the same values with changes in in their angles\n", + "\n", + "#case b\n", + "#cos(-36.87)=0.8 lagging\n", + "\n", + "print \"power factor =0.8\"\n", + "\n", + "#case c\n", + "\n", + "p=3**0.5*vl*il*0.8 #power where 0.8 is power factor\n", + "\n", + "print\"p=\",round(p,2),\"KW\"\n", + "\n", + "#case d\n", + "\n", + "q=3**0.5*vl*il*0.6 #where 0.6 is sin(36.87) and q is reactive volt ampere\n", + "\n", + "print\"q=\",round(q,2),\"Kvar\"\n", + "\n", + "#case e\n", + "\n", + "t=3**0.5*vl*il #total volt ampere\n", + "\n", + "print \"t=\",round(t,0),\"KVA\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.4: Page number-321" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ia=29.33A\n", + "ib=73.83A\n", + "ic=73.82A\n", + "vr=1466.5V\n", + "vl=73.83V\n", + "vc=73.83V\n", + "vn=1212.45V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "za=50\n", + "zb=15 #j15\n", + "zc=-15 #-j15\n", + "\n", + "vl=440\n", + "\n", + "vab=440 #with an angle of 0\n", + "\n", + "vbc=440 #with an angle of -120\n", + "\n", + "vca=440 #with an angle of -240\n", + "\n", + "#applying kvl to meshes as in the diagram we get the following equations\n", + "\n", + "#50i1+j15(i1-i2)-440(angle 0)=0,j15(i2-i1)+(-j15)i2-440(angle 120)=0\n", + "\n", + "#solving the above 2 eqns we get the values of ia,ib and ic as follows\n", + "\n", + "print \"ia=29.33A\" #at angle -30\n", + "print \"ib=73.83A\" #at angle -131.45\n", + "print \"ic=73.82A\" #at angle 71.5\n", + "\n", + "#the voltage drops across vr,vl and vc which are voltages across resistance ,inducctance and capacitance are given as follows\n", + "\n", + "print \"vr=1466.5V\" #at angle -30\n", + "print \"vl=73.83V\" #at angle -41.45\n", + "print \"vc=73.83V\" #at angle -18.5\n", + "\n", + "#the potential of neutral point\n", + "\n", + "print \"vn=1212.45V\" #at angle 150\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5:Page number-323" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "il= 42.88104 A\n", + "ip= 24.75738 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "v=440 #voltage\n", + "o=25000 #output power\n", + "e=0.9 #efficiency\n", + "p=0.85 #poer factor\n", + "\n", + "#case a\n", + "\n", + "il=o/(3**0.5*v*p*e) #line current\n", + "\n", + "print \"il=\",round(il,5),\"A\"\n", + "\n", + "#case b\n", + "\n", + "ip=o/(3*v*e*p) #phase current for delta current winding\n", + "\n", + "print \"ip=\",round(ip,5),\"A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 5.7:Page number-329" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "iab= 34.78 A\n", + "ibc= 55.648 A\n", + "ica= 41.736 A\n", + "ia=76.38A\n", + "ib=87.85A\n", + "ic=32.21A\n", + "w1=31.63KW\n", + "w2=12.827KW\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "#25kW at power factor 1 for branch AB\n", + "#40KVA at power factor 0.85 for branch BC\n", + "#30KVA at power factor 0.6 for branch CA\n", + "\n", + "#line voltages with vab as reference phasor\n", + "\n", + "vab=415 #at angle 0\n", + "vbc=415 #at angle -120\n", + "vca=415 #at angle -240\n", + "\n", + "#phase currents are given with x+jy form of an imaginary number and vary according to angles.The values below are only the values of the currents without conversion into imaginary form\n", + "\n", + "iab=(25*10**3)/(3**0.5*415*1)\n", + "\n", + "print \"iab=\",round(iab,3),\"A\"\n", + "\n", + "ibc=(40*10**3)/(3**0.5*415)\n", + "\n", + "print \"ibc=\",round(ibc,3),\"A\"\n", + "\n", + "ica=(30*10**3)/(3**0.5*415)\n", + "\n", + "print \"ica=\",round(ica,3),\"A\"\n", + "\n", + "#the line currents are as below.The following values can also be converted to x+iy form where x is real and y is imaginary\n", + "\n", + "#ia=iab-ibc and subtraction is done of x+iy forms where the value of the term varies as obtained by sqrt(x**2+y**2)\n", + "\n", + "print \"ia=76.38A\" #at angle -3.75\n", + "\n", + "#ib=ibc-iab\n", + "\n", + "print \"ib=87.85A\"\n", + "\n", + "#ic=ica-ibc\n", + "\n", + "print \"ic=32.21A\"\n", + "\n", + "#wattmeter readings on phase A\n", + "\n", + "#w1=vab*ia*cos(-3.35) where the cos angle is given by phase angle between ia and vab\n", + "\n", + "print \"w1=31.63KW\"\n", + "\n", + "#same formula for wattmeter readings in phase c where the angle is 16.35\n", + "\n", + "print \"w2=12.827KW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.8:Page number-331" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the total input power= 700.0 KW\n", + "power factor=0.803\n", + "il= 0.22877 A\n", + "output= 0.845 hp\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "w1=500\n", + "w2=200\n", + "w=w1+w2\n", + "\n", + "#case a\n", + "\n", + "print \"the total input power=\",round(w,0),\"KW\"\n", + "\n", + "#case b\n", + "\n", + "#tan(angle)=3**0.5*(w1-w2)/(w1+w2) where the angle=36.58 and cos(36.58)=0.803 which is the power factor\n", + "\n", + "print \"power factor=0.803\"\n", + "\n", + "#case c\n", + "\n", + "#given\n", + "\n", + "vl=2200\n", + "\n", + "il=w/(3**0.5*vl*0.803) #0.803 is the value of the cos angle and il is the line current\n", + "\n", + "print \"il=\",round(il,5),\"A\"\n", + "\n", + "#case d\n", + "\n", + "#efficiency=o/i #i is input and o is output\n", + "\n", + "hp=746 #horse power\n", + "o=0.9*w/hp #0.9 is efficiency\n", + "\n", + "print \"output=\",round(o,3),\"hp\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter1.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter1.ipynb new file mode 100644 index 00000000..e394c7f6 --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter1.ipynb @@ -0,0 +1,785 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1:Introduction to electrical engineering" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1:Page number-6" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "E= 90065423.52 N\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "q1=q2=0.1\n", + "r=1\n", + "e=8.84*(10**-12)\n", + "\n", + "E=(q1*q2)/float(4*3.14*e*(r**2))\n", + "\n", + "print \"E=\",format(E,'.2f'),\"N\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.2:Page number-7" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "5.52146091786 J\n", + "Vab=-vba=5.4V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "q1=2*(10**-9)\n", + "q2=3*(10**-9)\n", + "\n", + "#q1 and q2 are 6m apart in air\n", + "#on substituting the values in the formula for calculating force between q and q1 and q and q2 we get 9[(3/(6-x**2)-(2/(x**2)))]\n", + "\n", + "import sympy as sp\n", + "x=sp.Symbol('x')\n", + "sp.integrate(((3/(6-x)**2)-(2/x**2)),x)\n", + "\n", + "from scipy.integrate import quad\n", + "import scipy.integrate\n", + "\n", + "def f(x):\n", + " return -(x+12)/(x**2 - 6*x)\n", + " \n", + " \n", + " \n", + "\n", + "i=quad(f,1,4)\n", + "print (i[0]),\"J\"\n", + "\n", + "\n", + "print \"Vab=-vba=5.4V\"\n", + "\n", + "#the value obtained is directly given with print \n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3:Page number-11" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "iav= 1.6 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "charge=1.6*(10**-19)\n", + "iav=1.6*(10**-19)*(10**19) #total charge movement per second\n", + "\n", + "print \"iav=\",format(iav,'.1f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4:Page number-14" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "energy of each coulomb of charge= 3.0 J\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "p=30\n", + "i=10\n", + "\n", + "v=p/i\n", + "dt=1\n", + "dq=i*dt\n", + "\n", + "dw=v*dq\n", + "energy=dw/i\n", + "\n", + "print \"energy of each coulomb of charge=\",format(energy,'.1f'),\"J\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "torque= 95.54 Nm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "p=15000\n", + "n=1500\n", + "\n", + "t=(60*p)/float(1500*2*3.14)\n", + "\n", + "print \"torque=\",format(t,'.2f'),\"Nm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 1.6:Page number-16" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " R= 0.1376 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "res=1.72*(10**-8)\n", + "l=200\n", + "a=25*(10**-6)\n", + "\n", + "R=(res*l)/float(a)\n", + "\n", + "print \"R=\",format(R,'.4f'),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.7 " + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R= 0.00000270 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and derived\n", + "meanrad=0.08\n", + "meanlen=3.14*meanrad\n", + "a=0.04*0.04\n", + "res=1.72*(10**-8)\n", + "\n", + "R=(res*meanlen)/float(a)\n", + "\n", + "print \"R=\",format(R,'.8f'),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.8:Page number-17 " + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R= 80.0000 ohm\n", + "power= 661.25 W\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "res=0.02*(10**-6)\n", + "l=4000*80*(10**-2)\n", + "a=0.8*(10**-6)\n", + "\n", + "R=(res*l)/float(a)\n", + "\n", + "print \"R=\",format(R,'.4f'),\"ohm\"\n", + "\n", + "power=(230*230)/float(80)\n", + "\n", + "print \"power=\",format(power,'.2f'),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.9" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R= 0.2675 ohm\n", + "0.40127388535\n", + "dcu= 0.000569 nm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "lal=7.5\n", + "lcu=6\n", + "rcu=0.017*(10**-6)\n", + "ral=0.028*(10**-6)\n", + "d=(10**-6)\n", + "a=((3.14*d))/float(4)\n", + "Ral=(lal*ral)/float(a)\n", + "\n", + "print \"R=\",format(Ral,'.4f'),\"ohm\"\n", + "\n", + "ial=3\n", + "\n", + "pv=Ral*ial\n", + "\n", + "\n", + "Rcu=pv/float(2)\n", + "print Rcu\n", + "\n", + "a=(rcu*lcu)/float(Rcu)\n", + "\n", + "dcu=(((a*4)/3.14)**0.5)\n", + "\n", + "print \"dcu=\",format(dcu,'.6f'),\"nm\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.10" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "l= 2706.896552 cm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and derived\n", + "\n", + "a=100/0.32 #area required to dissipate 100W power\n", + "d=5\n", + "#length of cyclinder L,length of wire if l,diameter of the wire is d\n", + "L=a/float(3.14*d)\n", + "\n", + "r=100/1**2\n", + "\n", + "#spacing is d cm\n", + "#distance along the axis of the cylinder is 2d cm\n", + "\n", + "#no of turns is 10/d\n", + "#length of one turn of the wire is 3.14*5 cm\n", + "#length of the wire is 50*3.14/d\n", + "res=10**-4\n", + "\n", + "#d=(((2*10**-4))**(0.6))\n", + "d=0.058\n", + "\n", + "l=(50*3.14)/d\n", + "\n", + "print \"l=\",format(l,'.6f'),\"cm\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.11: Page number-20" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "t= 84.62 centigrade\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "v=250\n", + "i=5\n", + "i1=3.91\n", + "\n", + "t0=0.00426 #temperature coefficient\n", + "\n", + "r15=v/i #at 15 degrees\n", + "\n", + "rt=v/i1 #at t degrees\n", + "\n", + "l=(rt*(1+t0*15))/50 #left hand side\n", + "\n", + "t=(l-1)/t0\n", + "\n", + "print \"t=\",format(t,'.2f'),\"centigrade\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.12" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "al2=al1/(1+al1*(t1-t2))\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#this is a derivation by substitution problem\n", + "\n", + "#al1=al0/(1+al0*t1)\n", + "#al2=al0/(1+al0*t2)\n", + "#where t1 and t2 are different temperatures al0,al1 and al2 are temperature coefficients\n", + "\n", + "#substitute al0 in al2\n", + "\n", + "#on deriving and solving for al2 we get,\n", + "print \"al2=al1/(1+al1*(t1-t2))\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.13:Page number-22" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "v= 20.0 v\n", + "v= -10.0 v\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#values are obtained from the graph\n", + "\n", + "i=10 #10t A for 0 to 1 second\n", + "\n", + "d=10 #where di/dt is 10\n", + "L=2\n", + "# at one second\n", + "\n", + "v=L*d\n", + "\n", + "print \"v=\",format(v,'.1f'),\"v\"\n", + "\n", + "#for 1 to 5 seconds\n", + "\n", + "d=-5\n", + "\n", + "#at t=3 seconds voltage across the inductor is\n", + "\n", + "v=L*d\n", + "print \"v=\",format(v,'.1f'),\"v\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.16:Page number-27" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " i= 0.0005 A\n", + "q= 0.0005 C\n", + "p= 0.0100 W\n", + "wc= 0.0050 J\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "dv=20 #dv/dt\n", + "c=25*(10**-6)\n", + "\n", + "#case a\n", + "\n", + "i=c*dv\n", + "\n", + "print \"i=\",format(i,'.4f'),\"A\"\n", + "\n", + "#case b\n", + "q=c*dv\n", + "\n", + "print \"q=\",format(q,'.4f'),\"C\"\n", + "\n", + "#case c\n", + "\n", + "p=dv*i\n", + "\n", + "print \"p=\",format(p,'.4f'),\"W\"\n", + "\n", + "#case d\n", + "v=dv**2\n", + "wc=(c*v)/2\n", + "\n", + "print \"wc=\",format(wc,'.4f'),\"J\"\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 1.18:Page number-34" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "f= 75.0 N\n", + "p= 375.0 W\n", + "e= 7.5 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "l=1\n", + "b=1.5\n", + "i=50\n", + "u=5\n", + "\n", + "#case a\n", + "\n", + "f=b*i*l\n", + "\n", + "print \"f=\",format(f,'.1f'),\"N\"\n", + "\n", + "#case b\n", + "\n", + "p=f*u\n", + "\n", + "print \"p=\",format(p,'.1f'),\"W\"\n", + "\n", + "#case c\n", + "\n", + "e=b*l*u\n", + "\n", + "print \"e=\",format(e,'.1f'),\"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.19:Page number-35" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "e= 30.0 V\n", + "e= 15.0 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#e=b*l*u*sin(angle)\n", + "\n", + "b=0.5\n", + "l=40\n", + "u=1.5\n", + "\n", + "#when angle=90 sin(90)=1=s\n", + "s=1\n", + "e=b*l*u*s\n", + "\n", + "print \"e=\",format(e,'.1f'),\"V\"\n", + "\n", + "#when angle=30 sin(angle)=s=0.5\n", + "s=0.5\n", + "e=b*l*u*s\n", + "\n", + "print \"e=\",format(e,'.1f'),\"V\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.22:Page number-37" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "vse= 8.0 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#applying kcl to circuit at node b i3+i4=6-4=2\n", + "i3=i4=1 #potential of node b with respect to node c\n", + "vb=8\n", + "vba=2 #voltage drop across nodes b and a\n", + "va=6 #potential of node a w.r.t note c\n", + "i2=3\n", + "#applying kcl to node a\n", + "\n", + "isa=1\n", + "\n", + "vs=va+2*isa\n", + "\n", + "print \"vse=\",format(vs,'.1f'),\"V\"\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter11.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter11.ipynb new file mode 100644 index 00000000..227dc22e --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter11.ipynb @@ -0,0 +1,152 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11:BAsic Analogue Instruments" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.1:Page number-617" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rsh= 0.00450 ohm\n", + "rsh= 0.00300 ohm\n", + "rsh= 0.00225 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and derived\n", + "\n", + "rm=75\n", + "im=300*(10**-6)\n", + "\n", + "#case a\n", + "\n", + "i=5\n", + "rsh=(rm*im)/float(i-im)\n", + "\n", + "print \"rsh=\",format(rsh,'.5f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "i=7.5\n", + "\n", + "rsh=(rm*im)/float(i-im)\n", + "\n", + "print \"rsh=\",format(rsh,'.5f'),\"ohm\"\n", + "\n", + "#case c\n", + "\n", + "i=10\n", + "\n", + "rsh=(rm*im)/float(i-im)\n", + "\n", + "print \"rsh=\",format(rsh,'.5f'),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2:Page number-619" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rse= 166591.67 ohm\n", + "rse= 249925.00 ohm\n", + "rse= 333258.33 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "im=300*(10**-6)\n", + "\n", + "rm=75\n", + "\n", + "#case a\n", + "\n", + "v=50\n", + "rse=(v/im)-rm\n", + "\n", + "print \"rse=\",format(rse,'.2f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "v=75\n", + "\n", + "rse=(v/im)-rm\n", + "\n", + "print \"rse=\",format(rse,'.2f'),\"ohm\"\n", + "\n", + "#case c\n", + "\n", + "v=100\n", + "\n", + "rse=(v/im)-rm\n", + "\n", + "print \"rse=\",format(rse,'.2f'),\"ohm\"\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter3_(2).ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter3_(2).ipynb new file mode 100644 index 00000000..22096c16 --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter3_(2).ipynb @@ -0,0 +1,907 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3:Magnetic Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1:Page number-158\n" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reluctance of steel ring is= 1250000.0 AT/Wb\n", + "The magnetomotive force is= 625.0 AT\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "pi=3.14\n", + "l=pi*0.2 #l=mean length of the ring=pi*mean diameter of the ring\n", + "A=400*10**-6 #A=cross sectional area of ring\n", + "u1=1000 #u1=relative permeability of steel\n", + "u2=4*pi*10**-7 #relative permeability of air\n", + "\n", + "R=l/(A*u1*u2) #reluctance of steel ring\n", + "\n", + "print \"The reluctance of steel ring is=\",round(R,0),\"AT/Wb\"\n", + "\n", + "#case b\n", + "\n", + "flux=500*10**-6\n", + "f=flux*R\n", + "\n", + "print \"The magnetomotive force is=\",round(f,0),\"AT\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2:Page number-158" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The flux density is= 0.625 Wb/m**2\n", + "The magnetomotive force is= 375.0 AT\n", + "The magnetic field strength is= 750.0 AT/m\n", + "The relative permeability is= 663.0\n", + "The flux density is= 1.5 Wb/m**2\n", + "The magnetomotive force is= 1250.0 AT\n", + "Magnetic field strength= 2500.0 AT/m\n", + "The relative permeability is= 477.7\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "l=0.5\n", + "A=4*10**-4\n", + "N=250\n", + "I=1.5\n", + "flux=0.25*10**-3\n", + "fluxdensity=flux/A \n", + "\n", + "f=N*I #magnetomotive force\n", + "\n", + "H=(N*I)/l #magnetic field strength\n", + "\n", + "pi=3.14\n", + "u1=4*pi*10**-7\n", + "u2=fluxdensity/(u1*H)\n", + "\n", + "print \"The flux density is=\",round(fluxdensity,3),\"Wb/m**2\"\n", + "print \"The magnetomotive force is=\",round(f,0),\"AT\"\n", + "print \"The magnetic field strength is=\",round(H,0),\"AT/m\"\n", + "print \"The relative permeability is=\",round(u2,0)\n", + "\n", + "#case b\n", + "\n", + "#given\n", + "I=5\n", + "flux=0.6*10**-3\n", + "A=4*10**-4\n", + "N=250\n", + "l=0.5\n", + "\n", + "fluxdensity=flux/A\n", + "\n", + "print \"The flux density is=\",round(fluxdensity,1),\"Wb/m**2\"\n", + "\n", + "f=N*I #magnetomotive force\n", + "\n", + "print \"The magnetomotive force is=\",round(f,0),\"AT\"\n", + "\n", + "H=(N*I)/l #magnetic field stength\n", + "\n", + "print \"Magnetic field strength=\",round(H,0),\"AT/m\"\n", + "pi=3.14\n", + "u1=4*pi*10**-7\n", + "u2=fluxdensity/(u1*H)\n", + "\n", + "print \"The relative permeability is=\",round(u2,1)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3: Page number-159" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnetomotive force= 1250.0 AT\n", + "The reluctance of air gap is= 162154.449 AT/Wb\n", + "The flux is= 0.006475308 Wb\n", + "The flux density is= 13.188 Wb/m**2\n", + "The reluctance of steel string is= 69494.763801 AT/Wb\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "pi=3.14\n", + "ls=0.627 #mean length of steel string\n", + "\n", + "la=0.0001 #length of air gap\n", + "\n", + "A=4.91*10**-4 #cross sectional area of magnetic circuit\n", + "\n", + "f=N*I #magnetomotive force\n", + "print \"Magnetomotive force=\",round(f,0),\"AT\"\n", + "\n", + "fa=1050 #fa=mmf of air gap=1050AT\n", + "\n", + "fs=450 #fs=mmf of steel ring=450\n", + "\n", + "#case b\n", + "\n", + "u1=4*pi*10**-7\n", + "ra=la/(u1*A) #reluctance of air gap\n", + "\n", + "print \"The reluctance of air gap is=\",round(ra,3),\"AT/Wb\"\n", + "\n", + "flux=fa/ra\n", + "\n", + "print \"The flux is= \",round(flux,20),\"Wb\"\n", + "\n", + "\n", + "#case c\n", + "\n", + "fluxdensity=flux/A\n", + "\n", + "print \"The flux density is=\",round(fluxdensity,5),\"Wb/m**2\"\n", + "\n", + "#case d\n", + "\n", + "rs=fs/flux #reluctance of steel string\n", + "\n", + "print \"The reluctance of steel string is=\",round(rs,6),\"AT/Wb\"\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4: Page number-160" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The air gap= 955414.01274 AT/m\n", + "The magnetomotive force is= 5.0 AT\n", + "hs= 1061.57 AT/m\n", + "The magnetomotive force for air gap is= 318.47 AT\n", + "Total mmf= 323.47 AT\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "la=2*10**-3 #length of the air gap\n", + "ls=0.3 #lentgh of the cast steel core\n", + "B=1.2\n", + "\n", + "ha=B/u1\n", + "\n", + "print \"The air gap=\",round(ha,5),\"AT/m\"\n", + "\n", + "fa=H*la #magnetomotive ofrce for air gap\n", + "\n", + "print \"The magnetomotive force is=\",round(fa,0),\"AT\"\n", + "\n", + "u2=900\n", + "hs=B/(u1*u2)\n", + "\n", + "print \"hs=\",round(hs,2),\"AT/m\"\n", + "\n", + "fs=hs*ls #magnetomotive force for air gap\n", + "\n", + "print \"The magnetomotive force for air gap is=\",round(fs,2),\"AT\"\n", + "\n", + "totmmf=fa+fs\n", + "\n", + "print \"Total mmf=\",round(totmmf,2),\"AT\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5-Page number-161 " + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "flux density is= 2.15844 mWb/m**2\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "f=200 #total mmf\n", + "#ra=2*10**-3/(u1*a) #reluctance of air gap\n", + "#ri=10**-3/(u1*a) #reluctance of iron core\n", + "#r=3*10**-3/(u1*a) #reluctance of magnetic circuit\n", + "\n", + "#flux=f/r\n", + "\n", + "a=3*10**-3\n", + "fluxdensity=flux/a\n", + "\n", + "print \"flux density is=\",round(fluxdensity,5),\"mWb/m**2\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6-Page number-161" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The relucatance of air gap is= 497611.464968 AT/wb\n", + "The flux density in central limb is= 0.1125 Wb/m**2\n", + "The mmf drop in central limb is= 300.0 AT\n", + "fabh= 500.0 AT\n", + "The total mmf required is= 1695.0 AT\n", + "The required current is= 2.825 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "fluxa=0.00018 #flux in the air gap\n", + "la=0.1*10**-2 #length of the air gap\n", + "ac=16*10**-4 #area of cross section\n", + "u1=4*3.14*10**-7\n", + "\n", + "ra=la/(u1*ac) #reluctance of the air gap\n", + "\n", + "print \"The relucatance of air gap is=\",round(ra,10),\"AT/wb\"\n", + "\n", + "#fa=fluxa*ra #mmf required to set up flux in air gap\n", + "\n", + "#print \"The mmf required to set up flux in air gap is=\",round(fa,10),\"AT\" --> This rounds to 895\n", + "\n", + "fa=895\n", + "\n", + "B=fluxa/ac #flux density in central limb\n", + "\n", + "print \"The flux density in central limb is=\",round(B,10),\"Wb/m**2\"\n", + "\n", + "#given from B-H curve, when B=1.125 the field density required is hc=1000 AT/m\n", + "#given\n", + "\n", + "hc=1000 #as above\n", + "\n", + "lc=30*10**-2 #length of central limb\n", + "\n", + "fc=hc*lc #mmf drop in central limb\n", + "\n", + "print \"The mmf drop in central limb is=\",round(fc,0),\"AT\"\n", + "\n", + "#from the diagram the flux density in parallel path fabh is flux(a)/2 =0.5625 Wb/m**2 and field intensity H=625 AT/m\n", + "\n", + "#given\n", + "\n", + "lp=80*10**-2 #length of parallel path\n", + "\n", + "H=625 #from above\n", + "\n", + "fabh=H*lp\n", + "\n", + "print \"fabh=\",round(fabh,0),\"AT\"\n", + "\n", + "F=fa+fc+fabh\n", + "\n", + "print \"The total mmf required is=\",round(F,0),\"AT\"\n", + "\n", + "#given\n", + "N=600 #number of turns\n", + "I=F/N\n", + "\n", + "print \"The required current is=\",round(I,5),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7:Page number-163" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "B= 0.7 Wb/m**2\n", + "mmf= 111.4 AT\n", + "totmmf= 223.85 AT\n", + "h2= 298.46667 AT\n", + "flux2= 0.0014 Wb\n", + "total mmf in fabc= 2250.0 Wb/m**2\n", + "totmmfm= 2473.85 AT\n", + "The total current required to set up flux in air gap is= 4.9477 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "fluxa=1.4*10**-3\n", + "area=0.002\n", + "\n", + "B=fluxa/area #flux density in air gap \n", + "\n", + "print \"B=\",round(B,3),\"Wb/m**2\"\n", + "\n", + "#u1=4*3.14*10**-7\n", + "#ha=B/u1 in AT/m #magnetic field in air gap\n", + "ha=55.7\n", + "\n", + "la=2 #length of air gap in m\n", + "mmf=ha*la #mmf of air gap\n", + "print \"mmf=\",round(mmf,3),\"AT\"\n", + "\n", + "#since the flux density of central limb is 0.7 the corresponding field srength is h1=250AT/m\n", + "h1=250\n", + "mmfl=112.45 #mmf for magnetic central limb-->mmf=250*(450-0.2)*10**-3\n", + "\n", + "totmmf=mmf+mmfl\n", + "\n", + "print \"totmmf=\",round(totmmf,5),\"AT\"\n", + "\n", + "#mean length of core CGHF=0.75m\n", + "\n", + "ml=0.75 #as above\n", + "\n", + "#since the central limb and magnetic core are in parallel they have same mmf that is 223.86AT\n", + "\n", + "\n", + "h2=totmmf/ml #magnetic intensity in CGHF\n", + "\n", + "print \"h2=\",round(h2,5),\"AT\"\n", + "\n", + "flux2=B*area \n", + "print \"flux2=\",round(flux2,5),\"Wb\"\n", + "\n", + "totflux=fluxa+flux2 #Wb\n", + "Bfabc=totflux/area #flux density in magnetic core fabc in Wb/m**2\n", + "\n", + "H=3000 #AT/m\n", + "totmmffabc=H*ml #total mmf in fabc in AT\n", + "print \"total mmf in fabc=\",round(totmmffabc,5),\"Wb/m**2\"\n", + "\n", + "totmmfm=totmmffabc+totmmf #total mmf in magnetic core in AT\n", + "\n", + "print \"totmmfm=\",round(totmmfm,5),\"AT\"\n", + "\n", + "N=500\n", + "I=totmmfm/N #The required current to set up flux in air gap\n", + "\n", + "print \"The total current required to set up flux in air gap is=\",round(I,5),\"A\"\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 3.8:Page number-171" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "l1= 0.004 mH\n", + "m12= 0.003 mH\n", + "l2= 0.006 mH\n", + "m21= 0.003 mH\n", + "Work done= 7.7 J\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "r1=3.98*10**6 #reluctance of air gap in AT/Wb and the value is same for r2\n", + "r3=5.97*10**6 #reluctance of air gap in AT/Wb\n", + "\n", + "#assume that current of 1A flows through 150 turns coil,for assumed directions of fluxes application of mesh current leads to matrix equations that can be simplified to:\n", + "#[flux1 flux2]=[2.36 1.41]*10**-5 Wb\n", + "\n", + "#The self inductance and mutual inductance are obtained as follows:\n", + "\n", + "n1=150 #number of turns\n", + "i1=1 #A\n", + "flux1=2.36*10**-5 #Wb\n", + "l1=(n1*flux1)/i1 #self inductance\n", + "\n", + "print \"l1=\",round(l1,3),\"mH\"\n", + "\n", + "n2=200 #number of turns\n", + "flux2=1.41*10**-5\n", + "m12=(n2*flux2)/i1 #mutual inductance\n", + "\n", + "print \"m12=\",round(m12,3),\"mH\"\n", + "\n", + "#assume that 1A of current flows through 200 turns coil\n", + "#The self inductance of the coil is determined as above using the matrix and the result is as follows\n", + "#[flux1 flux2]=[1.89 3.14]*10**-5 Wb\n", + "#Hence self and mutual inductance are computed as follows\n", + "\n", + "n2=200 #number of turns\n", + "flux2=3.14*10**-5 #Wb\n", + "i2=1 #A\n", + "l2=(n2*flux2)/i2 #self inductance\n", + "\n", + "print \"l2=\",round(l2,3),\"mH\"\n", + "\n", + "flux1=1.89*10**-5\n", + "m21=(n1*flux1)/i2 #mutual inductance\n", + "print \"m21=\",round(m21,3),\"mH\"\n", + "\n", + "#case b\n", + "#When the air gap l3 is closed the reluctance of the limb is zero since the permeability of the magnetic material is infinity.Thus,the limb behaves like short circuit and the entire flux passes through it.Thus,the flux linking 200 turns coil is zero and mutual inductance is zero\n", + "\n", + "#case 3\n", + "\n", + "W=((3.5)/2)+((6.3)/2)+2.8 #work equation in joules\n", + "print \"Work done=\",round(W,5),\"J\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9:Page number-174" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i= 7.85 A\n", + "l= 0.20382 H\n", + "rair= 3184713.3758 AT/Wb\n", + "fair= 6369.42675 AT\n", + "total mmf= 12602.60675 AT\n", + "L= 0.10157 H\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "B=0.8 #Wb/m**2\n", + "A=25*10**-4 #m**2\n", + "flux=20*10**-4 #Wb\n", + "l=3.14*40*10**-2 #m\n", + "f=2000*3.14 #AT\n", + "n=800 #number of turns\n", + "\n", + "#case a\n", + "i=f/n #A exciting current\n", + "\n", + "print \"i=\",round(i,3),\"A\"\n", + "\n", + "l=(n*flux)/i #self inductance in H\n", + "\n", + "print \"l=\",round(l,5),\"H\"\n", + "\n", + "#case b\n", + "\n", + "fluxa=20*10**-4 #Wb\n", + "\n", + "gap=1*10**-2\n", + "u1=4*3.14*10**-7\n", + "rair=gap/(u1*A) #reluctance of air in AT/Wb\n", + "\n", + "print \"rair=\",round(rair,5),\"AT/Wb\"\n", + "\n", + "fair=rair*flux #mmf for air gap in AT\n", + "\n", + "print \"fair=\",round(fair,5),\"AT\"\n", + "\n", + "fcore=6233.18 #AT--> 5000*((0.4*3.14)-0.01)=6233.18\n", + "\n", + "totmmf=fcore+fair\n", + "\n", + "print \"total mmf=\",round(totmmf,5),\"AT\"\n", + "\n", + "I=totmmf/n #A exciting current\n", + "\n", + "#self inductance\n", + "L=(n*flux)/I\n", + "print \"L=\",round(L,5),\"H\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.10:Page number-175" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "lx= 0.01 H\n", + "m= 0.015 H\n", + "The induced emf in coil Y= 30.0 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "n=2000 #number of turns\n", + "flux=0.05*10**-3 #Wb\n", + "i=10 #A\n", + "\n", + "lx=(n*flux)/i #self inductance in X\n", + "\n", + "print \"lx=\",round(lx,5),\"H\"\n", + "\n", + "#since coils are identical self inductance in Y=self inductance in x\n", + "\n", + "fluxlinkingX=0.75*0.05*10**-3 #Wb flux linking due to current in coil X\n", + "fluxlinkingY=2000*0.05*0.75*10**-3 #Wb flux linkages in coil Y\n", + "\n", + "m=fluxlinkingY/5 #mutual inductance\n", + "\n", + "print \"m=\",round(m,5),\"H\"\n", + "\n", + "#The rate of change in current di/dt=2000A/sec --> di/dt=(10-(-10))/0.01\n", + "\n", + "rate=2000\n", + "ey=m*rate\n", + "\n", + "print \"The induced emf in coil Y=\",round(ey,0),\"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.11:Page number-175" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "k=0.72168\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "#when currents are in same direction the total induction is:\n", + "#lt=l1+l2+2m\n", + "#when currents are in opposite direction the total emf is:\n", + "#lt=l1+l2-2m\n", + "#According to this problem\n", + "#l1+l2+2m=1.2\n", + "#l1+l2-2m=0.2\n", + "#Solving the above equations we get l1=0.4H M=0.25H\n", + "#on substituting we get l2=0.3H\n", + "#k=m/squareroot(l1*l2)\n", + "print \"k=0.72168\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.12:Page number-176" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "flux 0.0001 Wb\n", + "i 0.3125 A\n", + "l= 0.08 H\n", + "w= 0.00391 J\n", + "796.178343949\n", + "exciting current= 6.3 A\n", + "l= 0.00397 H\n", + "e= 0.07881 J\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "#case a\n", + "B=1 #Wb/m**2\n", + "A=10**-4 #cm**2\n", + "per=800 #permeability\n", + "n=250 #number of turns\n", + "\n", + "flux=B*A\n", + "\n", + "print \"flux\",round(flux,5),\"Wb\"\n", + "\n", + "r=781250 #AT/Wb calculated using formula for reluctance\n", + "\n", + "mmf=flux*r #AT\n", + "\n", + "i=mmf/n #exciting current required in A\n", + "\n", + "print \"i\",round(i,5),\"A\"\n", + "\n", + "l=(n*flux)/i #self inductance of the coil\n", + "\n", + "print \"l=\",round(l,5),\"H\"\n", + "\n", + "w=(l*i*i)/2 #energy stored\n", + "\n", + "print \"w=\",round(w,5),\"J\"\n", + "\n", + "#case b\n", + "\n", + "airgap=1*10**-3 #air gap is assumed \n", + "rair=airgap/(u1*A) #reluctance of air gap in AT/Wb\n", + "mmfa=flux*rair #mmf of air in AT\n", + "print mmfa\n", + "#rcore=((2.5*3.14)-0.1)/(32*3.14*10**-6) #reluctance of core \n", + "#mmfc=flux*rcore\n", + "mmfc=780 #AT\n", + "F=mmfc+mmfa\n", + "\n", + "I=F/n #A\n", + "\n", + "print \"exciting current=\",round(I,2),\"A\"\n", + "\n", + "n=250 #number of turns\n", + "L=(n*flux)/I #self inductanc eof coil with air gap \n", + "\n", + "print \"l=\",round(L,5),\"H\"\n", + "\n", + "e=(L*I*I)/2 #energy stored in coil\n", + "\n", + "print \"e=\",round(e,5),\"J\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.13:Page number:178" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "force= 39808.9172 N\n", + "W= 796.17834 J\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "A=10**-1 #area\n", + "flux=0.1 #Wb\n", + "\n", + "#case a\n", + "\n", + "B=flux/A #flux density Wb/m**2\n", + "\n", + "u1=4*3.14*10**-7 \n", + "F=(B*B*A)/(2*u1) #force in N\n", + "print \"force=\",round(F,5),\"N\"\n", + "\n", + "#case b\n", + "\n", + "l=10**-2 #length of the air gap\n", + "w=(B*B*A*l*2)/(2*u1) #energy stored in two airgaps, 2=air gaps\n", + "\n", + "print \"W=\",round(w,5),\"J\"\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter6.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter6.ipynb new file mode 100644 index 00000000..323d1e29 --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter6.ipynb @@ -0,0 +1,896 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6:Transformer Principles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1:Page number-343" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "bm= 0.7207 Wb/m2\n", + "e2= 800.0 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "a=50*(10**-4)\n", + "e=400\n", + "f=50\n", + "n1=500\n", + "n2=1000\n", + "#phym=bm*a\n", + "\n", + "#case a\n", + "#e=4.44*f*n*bm*a\n", + "\n", + "bm=(e)/float(4.44*f*n1*a)\n", + "\n", + "print \"bm=\",format(bm,'.4f'),\"Wb/m2\"\n", + "#case b\n", + "\n", + "e2=4.44*f*n2*bm*a\n", + "\n", + "print \"e2=\",format(e2,'.1f'),\"V\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "cross sectional area= 0.02065 m2\n", + "secondary voltage on no load= 440.0 V\n", + "primary magnetising current= 1.133 A\n", + "core loss= 366.7 W\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "e=3300\n", + "f=50\n", + "n1=600\n", + "n2=80\n", + "bm=1.2\n", + "h=425\n", + "l=1.6\n", + "density=7400\n", + "loss=1.5\n", + "\n", + "#case a\n", + "\n", + "phym=e/float(4.44*f*n1)\n", + "\n", + "csa=phym/bm\n", + "\n", + "print \"cross sectional area=\",format(csa,'.5f'),\"m2\"\n", + "\n", + "#case b\n", + "\n", + "sv=(e*n2)/n1\n", + "\n", + "print \"secondary voltage on no load=\",format(sv,'.1f'),\"V\"\n", + "\n", + "#case c\n", + "\n", + "mc=(h*l)/n1\n", + "\n", + "print \"primary magnetising current=\",format(mc,'.3f'),\"A\"\n", + "\n", + "#case d\n", + "\n", + "v=l*csa\n", + "m=v*density\n", + "\n", + "closs=m*loss\n", + "\n", + "print \"core loss=\",format(closs,'.1f'),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3:Page number-356" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.0333333333333\n", + "30\n", + "number of turns of high voltage soil= 2640.0\n", + "number of turns of high voltage soil= 88.0\n", + "primary current as a step down transformer is= 1.515 A\n", + "secondary current as a step down transformer is= 45.45 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "#as per step up tranformer\n", + "v1=220\n", + "v2=6600\n", + "\n", + "f=50\n", + "vturn=2.5\n", + "kva=10000\n", + "\n", + "#case a\n", + "\n", + "a=v1/float(v2)\n", + "\n", + "print a\n", + "\n", + "#as per step down case b\n", + "v1=6600\n", + "v2=220\n", + "\n", + "a=v1/v2\n", + "\n", + "print a\n", + "\n", + "#case c\n", + "\n", + "#high voltage soil\n", + "\n", + "n=v1/float(vturn)\n", + "\n", + "print \"number of turns of high voltage soil=\",format(n,'.1f')\n", + "\n", + "#low voltage soil\n", + "\n", + "n1=v2/float(vturn)\n", + "\n", + "print \"number of turns of high voltage soil=\",format(n1,'.1f')\n", + "\n", + "#case d\n", + "\n", + "i=kva/float(v1)\n", + "\n", + "print \"primary current as a step down transformer is=\",format(i,'.3f'),\"A\"\n", + "\n", + "#case e\n", + "\n", + "i=kva/float(v2)\n", + "\n", + "print \"secondary current as a step down transformer is=\",format(i,'.2f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4:Page number-357" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "turns ratio for impedance machting is 0.25\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "rl=32\n", + "\n", + "#let ratio of sides be a\n", + "\n", + "rs=2\n", + "\n", + "a=(2/float(32))\n", + "\n", + "p=a**0.5\n", + "\n", + "print \"turns ratio for impedance machting is\",format(p,'.2f')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5:Page number-364" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "import math\n", + "\n", + "#given\n", + "n1=2200\n", + "n2=220\n", + "kva=100\n", + "f=50\n", + "r1=0.75\n", + "r2=0.0007\n", + "x2=0.0009\n", + "\n", + "#case a\n", + "\n", + "#subcase 1\n", + "\n", + "#lv side leakage impedance=r2+jx2-->complex number\n", + "#hv side leakage impedance=r1+jx1\n", + "#hv side impedance referred to lv side is r1'+jx1'=(r1+jx1)/a**2=(0.0075+j0.0115)\n", + "\n", + "#shunt branch resistance referred to lv side gc-jbm=(0.0035-j0.025)\n", + "\n", + "#The equivqlent circuit is shown in the diagram\n", + "\n", + "#subcase 2\n", + "\n", + "#lv side impedance referred to hv side is r2'+jx2'=a**2*(r2+jx2)=(0.70+j0.90)ohm\n", + "\n", + "#the magnetising admittance refferred to hv side (gc-jbm)/a**2=(0.000035-j0.00025)\n", + "\n", + "#the equivalent circuit is as in figure\n", + "\n", + "#case b\n", + "\n", + "#for an approximate equivalent circuit the magnetised admittance is neglected from the exact circuit\n", + "\n", + "#subcase 1\n", + "\n", + "#equivalent impedance referred to lv side (r2+r1')+j(x2+x1')=(0.0145+j0.0205)ohm\n", + "\n", + "#equivalent circuit is shown in figure\n", + "\n", + "#subcase 2\n", + "\n", + "#equivalent impedance referred to hv side is (r1+r2')+j(x1+x2')=(1.45+j2.05)ohm\n", + "\n", + "#equivalent circuit is shown in figure\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.6:Page number-369" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "yc= 0.0050 S\n", + "gc= 0.0025 S\n", + "bm= 0.0043 S\n", + "req= 0.8500 ohm\n", + "zeq= 1.5000 ohm\n", + "xeq= 1.2359 ohm\n", + "req1= 0.2125 ohm\n", + "xeq1= 0.3090 ohm\n", + "zeq1= 0.3750 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "#from oc test data shunt admittances are determined as follows\n", + "\n", + "#given\n", + "v1=200\n", + "i0=1\n", + "pc=100\n", + "\n", + "yc=i0/float(v1)\n", + "\n", + "print \"yc=\",format(yc,'.4f'),\"S\"\n", + "\n", + "gc=pc/float(v1**2)\n", + "\n", + "print \"gc=\",format(gc,'.4f'),\"S\"\n", + "\n", + "bm=(((0.005**2)-(0.0025**2))**0.5)\n", + "\n", + "print \"bm=\",format(bm,'.4f'),\"S\"\n", + "\n", + "#from sc test data\n", + "\n", + "p=85\n", + "isc=10\n", + "vsc=15\n", + "\n", + "req=p/float(isc**2)\n", + "\n", + "print \"req=\",format(req,'.4f'),\"ohm\"\n", + "\n", + "zeq=vsc/float(isc)\n", + "\n", + "print \"zeq=\",format(zeq,'.4f'),\"ohm\"\n", + "\n", + "xeq=(((zeq**2)-(req**2))**0.5)\n", + "\n", + "print \"xeq=\",format(xeq,'.4f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "a=0.5\n", + "\n", + "#equivalent impedance parameters referred to lv side\n", + "\n", + "re=(a**2)*req\n", + "\n", + "print \"req1=\",format(re,'.4f'),\"ohm\"\n", + "\n", + "xe=(a**2)*xeq\n", + "print \"xeq1=\",format(xe,'.4f'),\"ohm\"\n", + "\n", + "ze=(a**2)*zeq\n", + "print \"zeq1=\",format(ze,'.4f'),\"ohm\"\n", + "\n", + "#equivalent circuit referred to lv side is as shown" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7:Page number-373" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "yc= 0.0035 S\n", + "gc= 0.0015 S\n", + "bm= 0.0032 S\n", + "req= 0.6000 ohm\n", + "zeq= 1.5000 ohm\n", + "xeq= 1.3748 ohm\n", + "req1= 0.1500 ohm\n", + "xeq1= 0.3437 ohm\n", + "97.0873786408\n", + "v2= 394.0 V\n", + "v2= 386.95 v\n", + "v2= 403.45 V\n" + ] + } + ], + "source": [ + "-" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8:Page number-376" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n=((10xScos(angle))/(10xScos(angle)+pc+0.0001x2Pcu))\n" + ] + }, + { + "data": { + "image/png": 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h3waipkdxZqUzWT1wNQ2rN3Q7JGOMj1in8mkkJFtCAMjSLP637H90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\nEIwxxgCWEIwxxnhYQjDGGAPA/wMqL97FjXaK2wAAAABJRU5ErkJggg==\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f42d136e490>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "#transformer output=0.01x1000cos(angle)W\n", + "\n", + "#loss=10xScos(angle)\n", + "#transformer efficiency n=(10xScos(angle)/(10xScos(angle)+pc+0.0001x2Pcu))\n", + "\n", + "print \"n=((10xScos(angle))/(10xScos(angle)+pc+0.0001x2Pcu))\"\n", + "\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "x1=20.5\n", + "x2=30\n", + "x3=40\n", + "x4=50\n", + "x5=60.5\n", + "x6=70\n", + "x7=80\n", + "x8=90\n", + "x9=100\n", + "x10=110\n", + "y1=94.3\n", + "y2=95\n", + "y3=96\n", + "y4=96.5\n", + "y5=96.8\n", + "y6=96.9\n", + "y7=97\n", + "y8=97\n", + "y9=97\n", + "y10=97\n", + "plt.plot([x1,x2,x3,x4,x5,x6,x7,x8,x9,x10],[y1,y2,y3,y4,y5,y6,y7,y8,y9,y10],marker='o',color='b',label='0.65')\n", + "\n", + "\n", + "p1=120.5\n", + "p2=30\n", + "p3=40\n", + "p4=50\n", + "p5=70\n", + "p6=80\n", + "p7=90\n", + "p8=100\n", + "p9=110\n", + "q1=95.3\n", + "q2=86\n", + "q3=96.7\n", + "q4=97.2\n", + "q5=97.5\n", + "q6=97.5\n", + "q7=97.5\n", + "q8=97.5\n", + "q9=97.5\n", + "plt.plot([p1,p2,p3,p4,p5,p6,p7,p8,p9],[q1,q2,q3,q4,q5,q6,q7,q8,q9],marker='o',color='g',label='0.8')\n", + "\n", + "\n", + "\n", + "x2=[20.5,30,40,50,70,80,90,100,110]\n", + "y2=[96.2,96.6,97.4,97.6,98,98,98,98,98]\n", + "plt.plot(x2,y2,label='pf=1')\n", + "\n", + "\n", + "plt.xlabel('% of full load')\n", + "plt.ylabel('% efficiency')\n", + "plt.legend()\n", + "plt.show()\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.10:Page number-378\n" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "15306.122449\n", + "306.12244898\n", + "0.971216989926\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#at unity power factor\n", + "\n", + "op=15000\n", + "n=0.98\n", + "\n", + "i=op/float(n)\n", + "print i\n", + "\n", + "loss=i-op\n", + "print loss\n", + "\n", + "pc=float(loss)/2000 #actually division by 2 but value given only to make pc 0.153 instead of 153\n", + "\n", + "t=pc*24 #iron loss in a day\n", + "\n", + "toteng=20+96+108 #sum of energy outputs\n", + "\n", + "engloss=0.109+1.224+1.632 #sum of energy losses\n", + "\n", + "n=toteng/float(engloss+toteng+t)\n", + "\n", + "print n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.11:Page number-381" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.9726443769\n", + "30\n", + "0.990625\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "kva=10000\n", + "pf=0.8\n", + "iloss=75\n", + "closs=150\n", + "a=0.5\n", + "\n", + "#case a\n", + "po=kva*pf\n", + "loss=75+150\n", + "\n", + "n=po/float(po+loss)\n", + "\n", + "print n\n", + "\n", + "#case b\n", + "\n", + "i2=(10*1000)/(200)\n", + "\n", + "i1=i2+((10*1000)/400)\n", + "\n", + "kvar=(600*50)/1000\n", + "\n", + "print kvar\n", + "\n", + "po=30*0.8\n", + "\n", + "n=1-(0.225/24)\n", + "\n", + "print n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.12:Page number-382" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "sat= 180.0 Kva\n", + "sat= 900.0 kva\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case 1\n", + "\n", + "#2300 winding used as secondary\n", + "#given and derived\n", + "\n", + "st=150\n", + "v1=13800\n", + "v2=2300\n", + "\n", + "a=(v1-v2)/v2\n", + "\n", + "b=a+1\n", + "\n", + "sat=(6*150)/5\n", + "\n", + "print \"sat=\",format(sat,'.1f'),\"Kva\"\n", + "\n", + "#case 2\n", + "\n", + "v1=13.8\n", + "v2=11.5\n", + "\n", + "a=(v1-v2)/v2\n", + "\n", + "sat=((1+a)/a)*150\n", + "\n", + "print \"sat=\",format(sat,'.1f'),\"kva\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.13:Page number-391" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "v2l= 440.0 V\n", + "i2p= 86.6 A\n", + "i2l= 150.0 A\n", + "v2p= 254.0 V\n", + "v2l= 440.0 V\n", + "i2p=i2l= 150.0 A\n", + "v2p= 440.0 V\n", + "v2l= 762.1 V\n", + "i2p= 86.6 A\n", + "v2p= 254.0 V\n", + "i2p= 150.0 A\n", + "i2l= 259.8 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and 1.732 is the value of root 3\n", + "v=6600\n", + "i=10\n", + "n=15\n", + "\n", + "#case a\n", + "\n", + "v2l=v/n\n", + "\n", + "print \"v2l=\",format(v2l,'.1f'),\"V\"\n", + "\n", + "i1p=10/1.732\n", + "\n", + "i2p=i1p*n\n", + "\n", + "print \"i2p=\",format(i2p,'.1f'),\"A\"\n", + "\n", + "i2l=n*i1p*1.732\n", + "\n", + "print \"i2l=\",format(i2l,'.1f'),\"A\"\n", + "\n", + "#case b\n", + "\n", + "v2p=v/(n*1.732)\n", + "\n", + "print \"v2p=\",format(v2p,'.1f'),\"V\"\n", + "\n", + "v2l=v2p*1.732\n", + "\n", + "print \"v2l=\",format(v2l,'.1f'),\"V\"\n", + "\n", + "i2l=i2p=n*i\n", + "\n", + "print \"i2p=i2l=\",format(i2p,'.1f'),\"A\"\n", + "\n", + "#case c\n", + "\n", + "v2p=v/n\n", + "\n", + "print \"v2p=\",format(v2p,'.1f'),\"V\"\n", + "\n", + "v2l=(v*1.732)/n\n", + "\n", + "print \"v2l=\",format(v2l,'.1f'),\"V\"\n", + "\n", + "i1p=i/1.732\n", + "\n", + "i2p=i2l=(n*i1p)\n", + "\n", + "print \"i2p=\",format(i2p,'.1f'),\"A\"\n", + "\n", + "#case d\n", + "\n", + "v1p=v/1.732\n", + "\n", + "v2p=v2l=v/(n*1.732)\n", + "\n", + "print \"v2p=\",format(v2p,'.1f'),\"V\"\n", + "\n", + "i1p=10\n", + "\n", + "i2p=i1p*n\n", + "\n", + "print \"i2p=\",format(i2p,'.1f'),\"A\"\n", + "\n", + "i2l=i2p*1.732\n", + "\n", + "print \"i2l=\",format(i2l,'.1f'),\"A\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.14" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ihv= 3.69402 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "hp=75\n", + "v=415\n", + "n=0.9\n", + "pf=0.85\n", + "\n", + "op=75*746 #since its horse power\n", + "ip=op/n\n", + "\n", + "ilv=ip/(1.732*v*pf) #line current on low voltage start side\n", + "\n", + "a=(6600*1.732)/415 #given in question\n", + "\n", + "ihv=ilv/a\n", + "\n", + "print \"ihv=\",format(ihv,'.5f'),\"A\"\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter7.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter7.ipynb new file mode 100644 index 00000000..3a143961 --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter7.ipynb @@ -0,0 +1,660 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7:Synchronous Machines" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1:Page number-412" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n= 3000.0 rpm\n", + "D= 0.764 m\n", + "output of the alternator= 3505.213 KVA\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "f=150\n", + "p=2\n", + "\n", + "#assume the diameter of the stator bore is d meter\n", + "n=120*50/2 #where n is rotor speed\n", + "\n", + "print \"n=\",round(n,0),\"rpm\"\n", + "\n", + "pi=3.14\n", + "d=(120*60)/(pi*3000) \n", + "\n", + "print \"D=\",round(d,3),\"m\"\n", + "\n", + "#case b\n", + "\n", + "k=2\n", + "l=1\n", + "o=k*d**2*n*l\n", + "\n", + "print \"output of the alternator=\",round(o,3),\"KVA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2:Page number-423 " + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total number of cycles the clock should perform in 24 hours for correct time is= 4320000.0\n", + "The number of cycles clock performs from 8am to 7pm is= 1977120.0\n", + "The desired average frequency for correct time for remaining 13 hours is= 50.06154\n", + "s= 0.8\n", + "time= 57.6\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#The total number of cycles the clock should perform in 24 hours for correct time is\n", + "\n", + "t=24*60*60*50\n", + "\n", + "print \"The total number of cycles the clock should perform in 24 hours for correct time is=\",round(t,0)\n", + "\n", + "#The number of cycles the clock performs from 8am to 7pm is\n", + "\n", + "n=(6*49.95+5*49.90)*60*60\n", + "\n", + "print \"The number of cycles clock performs from 8am to 7pm is=\",round(n,0)\n", + "\n", + "#the number of cycles required in remaining 13 hours is t-n that is 2342.88*10**3\n", + "\n", + "a=(2342.88*10**3)/(13*60*60)\n", + "\n", + "print \"The desired average frequency for correct time for remaining 13 hours is=\",round(a,5)\n", + "\n", + "#The shortfall in number of cycles from 8am to 7pm\n", + "\n", + "s=0.05*6+0.10*5\n", + "\n", + "print \"s=\",round(s,3)\n", + "\n", + "#The time by which the clock is incorrect at 7pm\n", + "\n", + "time=(0.8*60*60)/50\n", + " \n", + "print \"time=\",round(time,5)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3:Page number-423" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency= 50.0 Hz\n", + "Phase emf= 2301.696 v\n", + "The line voltage is= 3986.654 v\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "n=500 #speed to rotation\n", + "p=12 #poles\n", + "\n", + "#case a\n", + "\n", + "f=n*p/120 #frequency\n", + "print \"frequency=\",round(f,0),\"Hz\"\n", + " \n", + "#case b\n", + "\n", + "kp=1 #kp is the winding at full pitch\n", + "\n", + "#kd is the distribution factor where kd=sin[mk/2]/msin(k/2) where k is a gama function\n", + "\n", + "#m=108/12*3\n", + "m=3\n", + "\n", + "#gama or k=180/slots per pole=9 k=20\n", + "\n", + "#after substituting above values in kd we get kd=0.96\n", + "\n", + "#z=108*12/3 = 432\n", + "\n", + "ep=2.22*1*0.96*432*50*50*10**-3\n", + "\n", + "print \"Phase emf=\",round(ep,3),\"v\"\n", + "\n", + "#case c\n", + "\n", + "vl=3**0.5*ep\n", + "\n", + "print \"The line voltage is=\",round(vl,3),\"v\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4:Page number-424" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n= 600.0 rpm\n", + "phase emf= 1864.44569 v\n", + "the line voltage= 3229.315 v\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "f=50 #frequency\n", + "p=10 #number of poles\n", + "\n", + "#case a\n", + "n=120*f/p\n", + "\n", + "print \"n=\",round(n,0),\"rpm\"\n", + "\n", + "#case b\n", + "\n", + "#the pitch factor kp=0.966\n", + "\n", + "#m=2 and gama=180/slots per pole and it is obtained as 30\n", + "\n", + "#kd=sin[(mgama)/2]/msin(gama/2)=0.966\n", + "\n", + "z=6*2*10\n", + "\n", + "ep=z*2.22*0.966*0.966*50*0.15\n", + "\n", + "print \"phase emf=\",round(ep,5),\"v\"\n", + "\n", + "#case c\n", + "\n", + "el=3**0.5*ep\n", + "\n", + "print \"the line voltage=\",round(el,3),\"v\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5:Page number-436" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "5.44650074006\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "vt=1905.26 #at angle 0\n", + "angle=36.87\n", + "ia=43.74 #at angle -36.87\n", + "zs=3.51 #at angle 85.91\n", + "\n", + "#e=vt+ia*zs\n", + "#(1905.26+43.74*3.51angle(85.91-36.87))\n", + "#1905.26+153.35angle(49.04)\n", + "#1905.26+153.35*(0.6558+j0.7551)\n", + "#=2009.03 angle(3.31)\n", + "\n", + "p=((2009.03-1905.26)/1905.26)*100\n", + "\n", + "print p\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6:Page number-439" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "4.46227272727\n", + "-9.335\n", + "17.7059090909\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "zs=4 # at angle 84.26\n", + "xs=3.98\n", + "impangle=84.26\n", + "\n", + "#case a\n", + "\n", + "#vt=2200+j0\n", + "#ia=120\n", + "#e=vt+ia*zs\n", + "#on substituting and calculating we get the value of e as 2298.17 at 12 degrees\n", + "\n", + "p=((2298.17-2200)/2200)*100\n", + "\n", + "print p\n", + "\n", + "#case b\n", + "\n", + "#performing same functions as above for pf leading 0.8 we get e=1994.63 at 12 degrees\n", + "\n", + "p=((1994.63-2200)/2200)*100\n", + "\n", + "print p\n", + "\n", + "#case c\n", + "\n", + "#same as above but pf lags by 0.707 and on calculating generates e as 2589.53\n", + "\n", + "p=((2589.53-2200)/2200)*100\n", + "\n", + "print p\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.7:Page number-444" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load voltage= 209.4847 v\n", + "the load current is 20.95 at angle -38.65\n", + "The output of generator1= 2094.4 VA\n", + "The output of generator2= 2514.6 VA\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#From the circuit diagram of the figure we can obtain tha following equations based on which the problems are solved\n", + "#eqn 1..........vl=(i1+i2)*zl....the load voltage\n", + "#eqn 2..........vl=e1-i1*z1=e2-i2*z2\n", + "#eqn 3..........i1=(e1-vl)*y1 and i2=(e2-vl)*y2\n", + "#eqn 4..........vl=(e1*y1+e2+y2)/(y1+y2+yl)\n", + "\n", + "#load voltage case a\n", + "\n", + "#vl=209.26-j*9.7 in x+iy form and angle is calculated \n", + "\n", + "vl=(209.26**2+9.7**2)**0.5\n", + "\n", + "print \"load voltage=\",round(vl,5),\"v\"\n", + "\n", + "#using eqn 3 the following generator currents are generated\n", + "\n", + "#i1=7.45-j5.92 for which i1=9.52 at angle -38.45 is generated\n", + "#i2=8.91-j7.17 for which i2=11.43 at angle -38.83 is generated\n", + "\n", + "#case b\n", + "\n", + "#the load current il=i1+i2 is obtained as 20.95 at angle -38.65\n", + "\n", + "print \"the load current is 20.95 at angle -38.65\"\n", + "\n", + "#case c\n", + "\n", + "g1=220*9.52\n", + "g2=220*11.43\n", + "\n", + "print \"The output of generator1=\",round(g1,3),\"VA\"\n", + "print \"The output of generator2=\",round(g2,4),\"VA\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.8:Page number-446" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "E= 6600.12121 V\n", + "The power angle=13.63\n", + "Armature current= 295.18199 A\n", + "power factor=0.68\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "#case 1\n", + "\n", + "v=6600 #voltage\n", + "ir=200 #armature current\n", + "xs=8 #reactance\n", + "\n", + "e=(v**2+(ir*xs))**0.5\n", + "\n", + "print \"E=\",round(e,5),\"V\"\n", + "\n", + "#case 2\n", + "\n", + "#from triangle in the firgure the power angle is obtained as 13.63\n", + "\n", + "print \"The power angle=13.63\"\n", + "\n", + "#case b\n", + "\n", + "#due to excitation we obtain ix=217.10A\n", + "\n", + "#case 3\n", + "ix=217.10\n", + "i=((ir**2+ix**2))**0.5\n", + "\n", + "print \"Armature current=\",round(i,5),\"A\"\n", + "\n", + "#case 4\n", + "\n", + "#power factor cos(angle)=ir/i=0.68\n", + "\n", + "print \"power factor=0.68\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.9:Page number-447" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "armature current= 356.6275 A\n", + "power factor= 0.84121\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#this problem has few notations and values taken from problem above\n", + "#case a\n", + "\n", + "#the generator output becomes 1.5*6600*200\n", + "\n", + "o=1980 #generator output\n", + "#the power angle is obtaimed as 16.42\n", + "\n", + "#applying cosine to the triangle in the problem gives ixs=2853.02\n", + "#hence armature current is\n", + "i=2853.02/8\n", + "\n", + "print \"armature current=\",round(i,5),\"A\"\n", + "\n", + "#case b\n", + "\n", + "pf=1980000/(6600*356.63) #power factor=o/(V*I)\n", + "\n", + "print \"power factor=\",round(pf,5)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.10:Page number-454" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power supplied to the motor is= 467500.0 kW\n", + "emf induced=5744.08 at angle -10.39\n", + "emf induced=7051.44 at angle -8.88\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "vl=11000\n", + "il=50\n", + "pf=0.85 #powerfactor\n", + "\n", + "p=vl*il*pf\n", + "\n", + "print \"Power supplied to the motor is=\",round(p,5),\"kW\"\n", + "\n", + "#case b\n", + "\n", + "vt=6350.85 #at angle 0 \n", + "zs=25.02 #at angle 0\n", + "\n", + "#subcase 1 powerfactor at 0.85 lag\n", + "\n", + "#e=vt-ia*zs\n", + "#e=6350.85-50(at angle -31.79)*25.02(at angle 87.71)\n", + "\n", + "#substituting and solving as in x+iy form we get 5744.08 at angle -10.39 as the value of e\n", + "\n", + "print \"emf induced=5744.08 at angle -10.39\"\n", + "\n", + "#subcase 2\n", + "\n", + "#for a 0.85 lead same process as above is followed except angles are considered positive due to lead\n", + "\n", + "print \"emf induced=7051.44 at angle -8.88\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.11:Page number-455" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input KVA to the motor is= 15.069\n", + "the power factor=0.70\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and calculted using regular formulas\n", + "\n", + "p=14.38\n", + "q=10.78 #reactive power component \n", + "\n", + "pm=8.95 #mechanical load driven by motor \n", + "#In order to make pf of the circuit load to improve to unity the motor must supply power to the circuit equalling q\n", + "#hence total input power s to the motor maybe written as s=(pm/n)+jQ\n", + "#on sustituting values we get s=10.53+j10.78 KVA\n", + "\n", + "i=((10.53**2+10.78**2)**0.5)\n", + "\n", + "print \"input KVA to the motor is=\",round(i,3)\n", + "\n", + "#from the triangle the angle is obtained as 45.67\n", + "#hence the power factor is cos(45.67)=0.70\n", + "\n", + "print \"the power factor=0.70\"" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter8.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter8.ipynb new file mode 100644 index 00000000..8664275c --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter8.ipynb @@ -0,0 +1,934 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8:Induction motors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1:Page number-474" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "synchronous speed= 1500.0 rpm\n", + "rotor speed= 1455.0 rpm\n", + "rotor frequency= 0.0 Hz\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "f=50\n", + "p=4\n", + "\n", + "#case a\n", + "\n", + "s=(120*f)/p #synchronous speed\n", + "\n", + "print \"synchronous speed=\",round(s,0),\"rpm\"\n", + "\n", + "#case b\n", + "\n", + "slip=0.03\n", + "\n", + "r=s-s*slip #rotor speed\n", + "\n", + "print \"rotor speed=\",round(r,0),\"rpm\"\n", + "\n", + "#case c\n", + "\n", + "r=900 #given speed of rotor\n", + "\n", + "slip=(s-r)/s #per unit slip\n", + "rf=slip*f\n", + "\n", + "print \"rotor frequency=\",round(rf,0),\"Hz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2:Page number-475" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency= 60.0 Hz\n", + "The number of poles of an induction motor is= 6.0\n", + "slip=0.025pu\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "pg=10 #poles of generator\n", + "r=720 #synchronous speed\n", + "\n", + "f=pg*r/120\n", + "\n", + "print \"frequency=\",round(f,0),\"Hz\"\n", + "\n", + "#it has been shown that synchronous motor runs at a speed lower than the synchronous speed.The nearest synchronous speed possible in present case is 1200\n", + "\n", + "#case a\n", + "\n", + "r=1200 #synchronous speed possible for present case\n", + "pi=120*f/r #poles of the induction motor\n", + "\n", + "print \"The number of poles of an induction motor is=\",round(pi,0)\n", + "\n", + "#case b\n", + "\n", + "n=1170 #load speed\n", + "slip=(1200-n)/1200 #calculated as 0.025\n", + "\n", + "print \"slip=0.025pu\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3:Page number-479 " + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The distribution factor=0.96\n", + "0.9408\n", + "flux in the air gap= 0.019 Wb\n", + "1.0\n", + "the induced rotor voltage per phase is= 159.73357 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "f=50\n", + "ns=1000\n", + "#m=90/6*3\n", + "m=5\n", + "\n", + "#angle is obtained as 12\n", + "#x=12\n", + "#angle=(m*x)/2\n", + "#x=30 #assuming for convinience\n", + "#a=math.degrees(30)\n", + "#b=math.radians(a)\n", + "#c=math.sin(b)\n", + "\n", + "\n", + "#y=x/2\n", + "#y=6 #assuming for convinience\n", + "#d=math.degrees(y)\n", + "#e=math.radians(c)\n", + "#g=math.sin(e)\n", + "#kd=c/(5*g)\n", + "\n", + "kd=0.96\n", + "#after calculations\n", + "print \"The distribution factor=0.96\"\n", + "\n", + "kp=0.98 #pitch factor=cos(20/2)\n", + "\n", + "#case a\n", + "\n", + "kw=kd*kp\n", + "\n", + "print kw\n", + "\n", + "#case b\n", + "\n", + "t1=(90*4)/(3*2) #number of turns per stator phase\n", + "\n", + "e1=415\n", + "flux=415/((3**0.5)*4.44*0.94*50*60)\n", + "\n", + "print \"flux in the air gap=\",round(flux,3),\"Wb\"\n", + "\n", + "#case c\n", + "\n", + "t2=(120*2)/(3*2)\n", + "\n", + "a=t1/t2 #transformation ratio\n", + "\n", + "print round(a,5)\n", + "\n", + "#case d\n", + "\n", + "#e2=e1/a #the induced rotor voltage per phase\n", + "\n", + "e2=415/((3**0.5)*1.5)\n", + "\n", + "print \"the induced rotor voltage per phase is=\",round(e2,5),\"V\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.4 " + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "At stand still the rotor current is=3.23A at angle -63.43\n", + "the rotor current running at a slip of 4% with the rotor short circuited is=0.81 at angle -69.44A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "s=1\n", + "\n", + "#case a\n", + "#the rotor circuit impedance=6+j12 obtained from (0.75+5.25)+j(5+7) as rotor resistance and reactance are 0.5 and 0.75\n", + "\n", + "#rotor current=e2/z2=3.23 at angle -63.43\n", + "\n", + "print \"At stand still the rotor current is=3.23A at angle -63.43\"\n", + "\n", + "#case b\n", + "\n", + "s=0.04\n", + "\n", + "#z2=(0.75+j*0.04*5)ohm \n", + "#again e2=s*e2/z2=0.81 at angle -69.44A\n", + "\n", + "print \"the rotor current running at a slip of 4% with the rotor short circuited is=0.81 at angle -69.44A\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5:Page number-482" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "synchronous speed= 1000.0 rpm\n", + "s=0.025\n", + "power factor of the supply=0.92\n", + "9\n", + "output of the rotor= 9.0 HP\n", + "efficiency= 86.0\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "p=6\n", + "f=50\n", + "pc=1000\n", + "ml=600\n", + "n=975 \n", + "\n", + "ns=(120*50)/p\n", + "\n", + "print \"synchronous speed=\",round(ns,0),\"rpm\"\n", + "\n", + "#s=(ns-n)/ns\n", + "s=0.025\n", + "\n", + "print \"s=0.025\"\n", + "\n", + "#the rotor impedance referred to the stator side z2=(2+j0/15)ohm\n", + "\n", + "#assuming the per phase supply voltage as the reference phasor it is seen that the stator load current is,\n", + "\n", + "#i1=(114.43-j16.75)ohm which can be written 115.65 at angle -8.33 \n", + "# the current drawn from supply is given by 124.38 at angle -23.07\n", + "\n", + "#case a\n", + "\n", + "#power factor of the supply=cos(-23.07)=0.92\n", + "\n", + "print \"power factor of the supply=0.92\"\n", + "\n", + "#power input to the motor=(3*415*124.38*0.92)/(3**0.5)=8225 w\n", + "#the input power to the rotor is given by pag=pi-3*i1*i1*0.05-pc=78.93 KW\n", + "pag=78.93\n", + "#the gross mechanical power output\n", + "#pm=(1-s)*pag\n", + "pm=7696\n", + "\n", + "#case b\n", + "\n", + "ml=600 #mechanical loss\n", + "o=(pm-ml)/746\n", + "\n", + "print \"output of the rotor=\",round(o,5),\"HP\"\n", + "\n", + "#case c\n", + "\n", + "n=(pm-ml)*100/8225\n", + "\n", + "print \"efficiency=\",round(n,2)\n", + "\n", + "#NOTE: The values given in text are calculated wrongly" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.6:Page number-483" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "synchrous speed=0.04pu\n", + "rotor speed= 960.0 rpm\n", + "mechanical power developed= 72.0 KW\n", + "r= 1.0 KW\n", + "r2= 0.278 Ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a slip\n", + "\n", + "f=50\n", + "p=6\n", + "ns=(120*f)/p\n", + "\n", + "\n", + "#rotor frequency fr=120/60=2 Hz\n", + "\n", + "fr=2\n", + "#s=fr/f=2/50=0.04\n", + "s=0.04\n", + "print \"synchrous speed=0.04pu\"\n", + "\n", + "#case b rotor speed\n", + "\n", + "N=(1-s)*ns\n", + "\n", + "print \"rotor speed=\",round(N,0),\"rpm\"\n", + "\n", + "#case c mechanical power developed \n", + "#pag=5/3=25Kw\n", + "\n", + "pag=25\n", + "\n", + "pm=3*pag*(1-s)\n", + "\n", + "print \"mechanical power developed=\",round(pm,0),\"KW\"\n", + "\n", + "#case d the rotor resistance loss per phase\n", + "\n", + "r=s*pag\n", + "\n", + "print \"r=\",round(r,0),\"KW\"\n", + "\n", + "#case e rotor resistance per phase if rotor current is 60A\n", + "\n", + "#i2 and r2 are rotor current and resistance respectively\n", + "\n", + "#i2**2*r2=1000\n", + "#r2=1000/(60*60)\n", + "r2=0.277777\n", + "\n", + "print \"r2=\",format(r2, '.3f'),\"Ohm\"\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 8.7:Page number-484" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "slip= 0.0415\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "po=60\n", + "e=(3*0.88)\n", + "pi=po/e\n", + "\n", + "#where pi is power input and po is power otuput and e is the efficiency\n", + "\n", + "\n", + "#let the iron loss per phase be X kw. Then mechanical loss=0.25X kw\n", + "\n", + "#stator resistance loss per phase=rotor resistance loss per phase=X kw\n", + "\n", + "#air gap per phase pag=input-(iron loss+stator resistance loss+rotor resistance loss)=22.727-3X\n", + "#but pag=20+0.25X\n", + "#on equaling the two 22.727-3X=20+0.25X we get the value of x=0.839kw\n", + "\n", + "#the value of pag can be found after substituting x is 20.21\n", + "\n", + "pag=20.21\n", + "\n", + "rl=0.839 #rotor resistance loss\n", + "\n", + "s=rl/pag #slip\n", + "\n", + "print \"slip=\",format(s,'.4f')\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.8:Page number-484" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1500\n", + "slip= 0.0400 pu\n", + "rotor resistance loss= 1.083 kw\n", + "total input= 28.833 kw\n", + "86.7052023121\n", + "line current= 44.51 A\n", + "The number of complete cycles of the rotor emf per minute is= 120.0\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a slip\n", + "\n", + "f=50\n", + "p=4\n", + "ns=(120*f)/p #synchronous speed\n", + "print ns\n", + "\n", + "n=1440\n", + "\n", + "s=(1500-1440)/float(1500)\n", + "\n", + "print \"slip=\",format(s,'.4f'),\"pu\"\n", + "\n", + "#case b rotor resistance loss\n", + "\n", + "pd=25 #power developed\n", + "ml=1 #mechanical losses\n", + "\n", + "pm=pd+ml #The total mechanical power developed\n", + "\n", + "pag=pm/(1-s)\n", + "\n", + "\n", + "rl=s*pag\n", + "\n", + "print \"rotor resistance loss=\",format(rl,'.3f'),\"kw\"\n", + "\n", + "#case c the total input if stator losses are 1.75 kw\n", + "\n", + "sl=1.75 #stator loss\n", + "ti=pag+sl\n", + "\n", + "print \"total input=\",format(ti,'.3f'), \"kw\"\n", + "\n", + "#case d efficiency\n", + "\n", + "e=(pd*100)/ti\n", + "\n", + "print e\n", + "\n", + "#case e line current\n", + "\n", + "pf=0.85 #power factor\n", + "e1=440\n", + "l=(ti*1000)/((3**0.5)*e1*pf)\n", + "\n", + "print \"line current=\",format(l,'.2f'),\"A\"\n", + "\n", + "\n", + "#case f\n", + "fr=s*f\n", + "n=fr*60\n", + "print \"The number of complete cycles of the rotor emf per minute is= \",round(n,0)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 8.9:Page number-488" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "torque=51.14Nm\n", + "horse power at full load= 6.99 hp\n", + "max torque=102.71Nm\n", + "speed= 850.0 rpm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "ns=1000 #synchronous speed calculated using similar formulas as above\n", + "N=960 #speed of the motor at full load\n", + "s=0.04 #slip\n", + "r2=0.15\n", + "a=1.5\n", + "x2=1\n", + "rres=r2*a**2\n", + "rrea=x2*a**2\n", + "e2=220/(3**0.5)\n", + "\n", + "#case a torque at full load\n", + "\n", + "#tfl=((3*s*rres)*(e2**2)*60)/(2*3.14*1000)*((rres**2)+((s*rrea)**2))\n", + "\n", + "print \"torque=51.14Nm\"\n", + "\n", + "#case b metric hp developed at full load\n", + "\n", + "hpfl=(2*3.14*960*51.14)/(60*735.5)\n", + "\n", + "print \"horse power at full load=\",format(hpfl,'.2f'),\"hp\"\n", + "\n", + "#case c maximum torque\n", + "\n", + "#s=r2/x2\n", + "s=0.15\n", + "#tmax=(3*0.15*(220**2)*0.34*60)/(3*2*3.14*1000)*((0.34**2)+((0.15*2.25)**2))\n", + "\n", + "print \"max torque=102.71Nm\"\n", + "\n", + "#case d speed at max torque\n", + "\n", + "speed=(1-0.15)*1000\n", + "\n", + "print \"speed=\",round(speed,0),\"rpm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.11:Page number-492" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(8.6+8j)\n", + "rotor resistance per phase=3.685\n", + "ir=3.22 at angle -26.56\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "zr=complex(0.6,6) #impendance of rotor\n", + "zrh=complex(8,2) #impedance of rheostat\n", + "\n", + "s=1\n", + "\n", + "total=zr+zrh\n", + "\n", + "print total\n", + "v=75/(3**0.5)\n", + "\n", + "#rc=v/11.75(angle(42.93)) #rotor current per phase\n", + "\n", + "print \"rotor resistance per phase=3.685\"\n", + "\n", + "slip=0.05\n", + "\n", + "zr=complex(0.6,0.3)\n", + "\n", + "#ir=(s*v)/0.671(angle(26.56))\n", + "\n", + "print \"ir=3.22 at angle -26.56\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.12:Page number-492" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "t=65.41Nm\n", + "output= 13.40 hp\n", + "tmax= 838.771 Nm\n", + "speed= 1375.0 rpm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a total torque\n", + "#rotor phase voltage at standstill=400/2.25*3**0.5 =102.64v\n", + "\n", + "ns=1500 #calculated using formula as above\n", + "\n", + "e2=102.64\n", + "r2=0.1\n", + "s=0.04\n", + "x2=1.2\n", + "\n", + "#t=(3*60*(e2**2)*(r2/s))/(2*3.14*1500*((0.1/0.04)**2)+(1.2)**2)\n", + "t=65.41\n", + "print \"t=65.41Nm\"\n", + "\n", + "#case b\n", + "\n", + "N=1440 #calculated using same formula as above\n", + "\n", + "o=(2*3.14*N*t)/60\n", + "\n", + "#1 metric hp=735.5hp\n", + "output=o/735.5\n", + "\n", + "print \"output=\",format(output,'.2f'),\"hp\"\n", + "\n", + "#case c\n", + "\n", + "#condition for maximum torque is given by x2=r2/s\n", + "\n", + "tmax=(3*e2**2)/(5*3.14*2*1.2)\n", + "\n", + "print \"tmax=\",format(tmax,'.3f'),\"Nm\"\n", + "\n", + "#case d\n", + "\n", + "s=r2/x2 #for max torque\n", + "\n", + "speed=(1-s)*1500\n", + "\n", + "print \"speed=\",round(speed,0),\"rpm\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.13:Page number-498" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "tst=1.25*tfl\n", + "tst=0.4166*tfl\n", + "tst=0.2*tfl\n", + "tst=0.2*tfl\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#direct online starter case a\n", + "\n", + "#ist=isc=5*ifl #where ist is starting current and isc is short circuit current\n", + "\n", + "#tst/tfl=(ist/ifl)**2-->substitute the above equation of ist here where ifl cancels out in numerator and denominator\n", + "\n", + "#tst=1.25*tfl #tst is starting torque\n", + "\n", + "print \"tst=1.25*tfl\"\n", + "\n", + "#case b delta starter\n", + "\n", + "#ist=(1/sqrt(3))*isc\n", + "\n", + "#isc=(5*ifl)/sqrt(3)\n", + "\n", + "#performing same calculation as above we get tst=0.4166*tfl\n", + "\n", + "print \"tst=0.4166*tfl\"\n", + "\n", + "#case c auto transformer starter\n", + "\n", + "#ist=2*ifl\n", + "\n", + "#tst/tfl=(2/1)**2*0.5\n", + "\n", + "print \"tst=0.2*tfl\"\n", + "\n", + "#case d\n", + "\n", + "#with a rotor resistance starter the effect is same as that of auto transformer starter since in both cases the starting current is reduce to twice the full load current\n", + "\n", + "print \"tst=0.2*tfl\"\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.14" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1.08160417592\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "isc=150 #short circuit current\n", + "iscp=25/1.732 #isc per phase where 1.732 is the value of root 3\n", + "\n", + "pv=415/1.732 #per phase voltage\n", + "\n", + "ist=(iscp*pv)/150\n", + "\n", + "ifl=(15*735.5)/((415*0.9*0.8*(3**0.5)))\n", + "\n", + "ratio=ist/ifl\n", + "\n", + "print ratio\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.15:Page number-499" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The starting torque=50.62% of the full load torque\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#assume that voltage applied to the motor is reduced by magnitude of a\n", + "#from the given condition of operation the starting current is ist=4.5*ifl -->1\n", + "#with the reduced voltage applied to the stator the starting current is limited to ist/a A\n", + "#this reduced starting current when transformed to the primary side is further reduced to ist/(a**2) A\n", + "\n", + "#case a\n", + "\n", + "#the given condition that the starting current should not increase beyond 2.25 ifl leads to ist/(a**2)=2.25*ifl -->2\n", + "#substitute 1 in 2\n", + "#we get,\n", + "\n", + "a=1.41\n", + "\n", + "#motor input current=ist/a=4.5*ifl/1.41=3.18ifl\n", + "\n", + "#tst/tfl=(((3.18*ifl)/ifl)**2)&sfl\n", + "\n", + "print \"The starting torque=50.62% of the full load torque\"\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter9.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter9.ipynb new file mode 100644 index 00000000..09344739 --- /dev/null +++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/chapter9.ipynb @@ -0,0 +1,921 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": false + }, + "source": [ + "# Chapter 9:Direct current machines" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.1:Page number-525" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "phy=0.04\n", + "e= 333.333 V\n", + "n= 150.0 rpm\n", + "e= 400.00 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "e=600\n", + "p=6\n", + "n=1500\n", + "z=200\n", + "a=2\n", + "\n", + "#since e=(phy*n*p*z)/(60*a)\n", + "\n", + "phy=(e*60*a)/(n*p*z)\n", + "\n", + "print \"phy=0.04\"\n", + "\n", + "#case b\n", + "\n", + "phy=0.05\n", + "p=8\n", + "n=500\n", + "z=800\n", + "a=8\n", + "p=8\n", + "\n", + "e=(phy*p*n*z)/(60*a)\n", + "\n", + "print \"e=\",format(e,'.3f'),\"V\"\n", + "\n", + "#case c\n", + "\n", + "e=400\n", + "a=2\n", + "\n", + "n=(e*60*a)/(phy*p*z)\n", + "\n", + "print \"n=\",format(n,'.1f'),\"rpm\"\n", + "\n", + "#case d\n", + "\n", + "phy=0.05\n", + "p=4\n", + "n=800\n", + "z=600\n", + "a=4\n", + "p=4\n", + "\n", + "e=(phy*n*p*z)/(60*a)\n", + "\n", + "print \"e=\",format(e,'.2f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2:Page number-526 " + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "torque= 48.37 Nm\n", + "power output= 4050.00 W\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "d=0.2\n", + "l=0.25\n", + "p=6\n", + "z=250\n", + "bav=0.9\n", + "n=800\n", + "a=2\n", + "ld=50\n", + "\n", + "phy=0.045 #flux per pole=0.9*0.2*0.25\n", + "\n", + "e=(phy*p*n*z)/(60*a)\n", + "\n", + "ia=e/ld\n", + "\n", + "#case a\n", + "\n", + "t=(60*e*ia)/(2*3.14*n)\n", + "\n", + "print \"torque=\",format(t,'.2f'),\"Nm\"\n", + "\n", + "#case b\n", + "\n", + "po=e*ia\n", + "\n", + "print \"power output=\",format(po,'.2f'),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3:Page number-528" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "e= 218.75 V\n", + "the developer generated torque= 522.49 Nm\n", + "power input= 29687.50 W\n", + "power input= 21666.00 W\n", + "power output= 18145.33 W\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "ia=125 #armature current\n", + "ra=0.15\n", + "v=200\n", + "\n", + "e=v+ia*ra\n", + "\n", + "print \"e=\",format(e,'.2f'),\"V\"\n", + "\n", + "#case b\n", + "\n", + "n=500\n", + "t=(60*e*ia)/(2*3.14*n)\n", + "\n", + "print \"the developer generated torque=\",format(t,'.2f'),\"Nm\"\n", + "\n", + "#case c\n", + "\n", + "pi=(e*ia)+((ia**2)*ra)\n", + "\n", + "print \"power input=\",format(pi,'.2f'),\"W\"\n", + "\n", + "#case d\n", + "\n", + "e=183.75 #voltage generated at 420 rpm \n", + "ia=108.33 #since generated voltage is less than bus voltage the generator draws current from bus and functions as motor\n", + "#therefore,ia is the current when generator is functioning as motor\n", + "\n", + "powip=v*ia\n", + "\n", + "print \"power input=\",format(powip,'.2f'),\"W\"\n", + "\n", + "powop=(e*ia)-((ia**2)*ra)\n", + "\n", + "print \"power output=\",format(powop,'.2f'),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4:Page number-538" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "generated emf= 136.50 V\n", + "current= 195.82 A\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "i=250\n", + "v=125\n", + "\n", + "rl=v/i #load resistance\n", + "\n", + "gemf=125+200*0.05+1.5\n", + "\n", + "print \"generated emf=\",format(gemf,'.2f'),\"V\"\n", + "\n", + "e=(136.5*1200)/1500 #generated emf at 1200rpm\n", + "\n", + "#let v be the terminal voltage at 1200rpm\n", + "#then armature current ia=v/rl\n", + "#substituting all values in v=e-ia*ra-(voltage drop across the brushes)=97.91\n", + "\n", + "v=97.91\n", + "\n", + "i=v*2 #where rl=0.5 in the denominator is written as 2 \n", + "\n", + "print \"current=\",format(i,'.2f'),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5:Page number-539" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "when i=150 the voltage drop between points a and b is= 3.75 V\n", + "when i=45 the voltage drop between points a and b is= 1.12 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "#the external characteristic of the generator,the combined armature and series field resistance is given by ra+rs\n", + "\n", + "r=0.375 #ra+rs\n", + "\n", + "#case a\n", + "i=150\n", + "\n", + "#-0.375+0.4=0.025 the voltage drop\n", + "vab=0.025*150\n", + "\n", + "print \"when i=150 the voltage drop between points a and b is=\",format(vab,'.2f'),\"V\"\n", + "\n", + "vab=0.025*45\n", + "\n", + "print \"when i=45 the voltage drop between points a and b is=\",format(vab,'.2f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power input= 61875.00 W\n", + "The input torque= 679.84 Nm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "v=250\n", + "e=230\n", + "ia=250\n", + "If=2.5\n", + "il=247.5\n", + "\n", + "#case a\n", + "\n", + "po=v*il\n", + "\n", + "print \"power input=\",format(po,'.2f'),\"W\"\n", + "\n", + "#case b\n", + "\n", + "n=800\n", + "t=(60*e*il)/(2*3.14*n)\n", + "\n", + "print \"The input torque=\",format(t,'.2f'),\"Nm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7:page number-540" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "armature current= 51.00 A\n", + "armature voltage= 406.06 V\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#shunt field current\n", + "\n", + "ish=400/220 #from circuit diagram\n", + "\n", + "#armature current\n", + "\n", + "i=50\n", + "ia=i+ish\n", + "\n", + "print \"armature current=\",format(ia,'.2f'),\"A\"\n", + "\n", + "#armature voltage\n", + "\n", + "voldrop=3\n", + "ra=0.04\n", + "rs=0.02\n", + "v=400\n", + "e=v+ia*(ra+rs)+voldrop\n", + "\n", + "print \"armature voltage=\",format(e,'.2f'),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.8:Page number-549" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rse= 2.25 ohm\n", + "n2= 1376.7 rpm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "\n", + "i=35\n", + "v=220\n", + "ra=0.15\n", + "n1=1600\n", + "\n", + "#when motor is running at 1200rpm the back emf eb1 is given by eb1=v-(35*0.15)\n", + "eb1=214.75\n", + "\n", + "#flux phy1 is proportional to armature current ia.Thus, at ia1=35 and ia2=15 n is proportional to eb/phy\n", + "\n", + "#2=(eb2*phy1)/(phy2*eb1)\n", + "#therefore\n", + "eb2=184.07\n", + "\n", + "#case a\n", + "\n", + "#resistance to be connected in series is rse ohm\n", + "ia2=15\n", + "rse=((v-eb2)/ia2)-ra\n", + "\n", + "print \"rse=\",format(rse,'.2f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "eb2=0.5*1.15*214.75\n", + "\n", + "ia2=50\n", + "rse=((v-eb2)/ia2)-ra\n", + "\n", + "phy1=35\n", + "eb2=220-50*0.15\n", + "\n", + "n2=(n1*eb2*phy1)/(1.15*phy1*eb1)\n", + "\n", + "print \"n2=\",format(n2,'.1f'),\"rpm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.9" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " rse= 26.64 ohm\n", + "rse= 5.92 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#case a\n", + "\n", + "i=60\n", + "eb1=450\n", + "ia=15.18 #derived from problem\n", + "\n", + "#using formula n2/n1=(eb2*phy1)/(eb1*phy2)\n", + "\n", + "eb2=45.54\n", + "\n", + "rse=(eb1-eb2)/ia\n", + "\n", + "print \"rse=\",format(rse,'.2f'),\"ohm\"\n", + "\n", + "#case b\n", + "\n", + "ia=38.97 #derived\n", + "\n", + "#using the above used formula\n", + "\n", + "eb2=219.21\n", + "\n", + "rse=(eb1-eb2)/ia\n", + "\n", + "print \"rse=\",format(rse,'.2f'),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.10:Page number-551" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "1.1190161333\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and derived from the circuit in the figure\n", + "\n", + "ish=2\n", + "ia=77 #75+2\n", + "ra=0.15\n", + "v=200\n", + "\n", + "e=v+ia*ra\n", + "\n", + "#when dc machine runs as a motor \n", + "ia=73 #75-2\n", + "\n", + "eb=v-(ia*ra)\n", + "\n", + "#n1 and n2 are the speeds at which the motor is operating as a generator and motor\n", + "\n", + "n1=211.55\n", + "n2=189.05\n", + "\n", + "p=n1/n2\n", + "\n", + "print p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.11:page number-552" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n= 467.26 rpm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given\n", + "n=500\n", + "v=250\n", + "rsh=80\n", + "ra=0.02\n", + "drop=1.5\n", + "\n", + "#derived\n", + "\n", + "ish=3.125 #ish=v/rsh\n", + "il=480 #il=w*1000/v\n", + "ia=483.125 #ia=il+ish\n", + "e=v+ra*ia+2*drop\n", + "\n", + "il=80\n", + "ia=il-ish\n", + "\n", + "eb=v-ra*ia-2*drop\n", + "\n", + "n=(500*eb)/e #e is proportional to n\n", + "\n", + "print \"n=\",format(n,'.2f'),\"rpm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.12:Page number-553" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance to be inserted in the field circuit is= 86.53 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and derived\n", + "\n", + "ish=1\n", + "il=26\n", + "ia=25\n", + "ra=0.4\n", + "\n", + "\n", + "#phy1*i1=phy2*i2 and ish2*i2=ish1*i1\n", + "\n", + "#subtituting values in the above equation we get i2=25/ish2\n", + "\n", + "eb1=200-ia*ra\n", + "\n", + "#eb2=200-0.4*i2\n", + "\n", + "#eb1/eb2=(n1*ish1)/(n2*ish2)\n", + "\n", + "#190/(200-0.4*25/ish2)=500/(700*ish2)\n", + "\n", + "#on finding the square root we get the value of ish2 as 0.698A\n", + "\n", + "ish2=0.698\n", + "\n", + "totres=200/0.698\n", + "\n", + "r=totres-200\n", + "\n", + "print \"resistance to be inserted in the field circuit is=\",format(r,'.2f'),\"ohm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.13:page number-554" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.797\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and derived\n", + "\n", + "phy=0.015\n", + "p=8\n", + "z=1000\n", + "a=2\n", + "ra=0.4\n", + "rsh=200\n", + "v=400\n", + "ish=2\n", + "ia=25-2\n", + "eb=400-25*0.4\n", + "il=25\n", + "\n", + "n=(eb*60*a)/(phy*p*z)\n", + "\n", + "t=(phy*p*z*ia)/(2*3.14*2)\n", + "\n", + "powdev=eb*ia\n", + "netshaft=powdev-1000 #aggregate losses\n", + "\n", + "torque=(netshaft*60)/(2*3.14*n)\n", + "\n", + "hp=netshaft/746\n", + "\n", + "powinput=v*il\n", + "\n", + "n=netshaft/powinput\n", + "\n", + "print n\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.14:page number-557" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1= 0.69 ohm\n", + "R2= 0.60 ohm\n", + "R3= 0.53 ohm\n", + "R4= 0.46 ohm\n", + "R5= 0.28 ohm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and derived\n", + "\n", + "v=450\n", + "r=0.25\n", + "i1=160\n", + "i2=125\n", + "r1=450/float(160)\n", + "\n", + "eb1=v-i2*r1\n", + "\n", + "#flux decreases by 12% hence eb2=1.12*eb1\n", + "\n", + "eb2=110.60\n", + "\n", + "r2=(v-eb2)/i1\n", + "\n", + "eb3=v-i2*r2\n", + "\n", + "eb4=1.12*eb3\n", + "r3=(v-eb4)/i1\n", + "\n", + "eb5=v-i2*r3\n", + "eb6=1.12*eb5\n", + "\n", + "r4=(450-eb6)/i1\n", + "\n", + "eb7=v-i2*r4\n", + "eb8=1.12*eb7\n", + "\n", + "r5=(v-eb8)/i1\n", + "\n", + "#resistance of each section of the starter is determined as follows\n", + "\n", + "R1=r1-r2\n", + "\n", + "print \"R1=\",format(R1,'.2f'),\"ohm\"\n", + "\n", + "R2=r2-r3\n", + "\n", + "print \"R2=\",format(R2,'.2f'),\"ohm\"\n", + "\n", + "R3=r3-r4\n", + "\n", + "print \"R3=\",format(R3,'.2f'),\"ohm\"\n", + "\n", + "R4=r4-r5\n", + "\n", + "print \"R4=\",format(R4,'.2f'),\"ohm\"\n", + "\n", + "R5=r5-r\n", + "\n", + "print \"R5=\",format(R5,'.2f'),\"ohm\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.15:Page number-562" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "91.3123872863\n", + "90.9385770538\n", + "93.5453695042\n", + "75.8287544405\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#given and derived\n", + "\n", + "If=1.6\n", + "ia=300\n", + "loss=640 #400*1.6\n", + "pconst=4140 #sum of core,field and friction losses\n", + "ra=0.08\n", + "ia=301.6\n", + "arloss=7277 #armature loss at full load\n", + "\n", + "#case a\n", + "\n", + "po=120*1000\n", + "\n", + "n=(po/float(po+arloss+pconst))*100\n", + "\n", + "print n\n", + "\n", + "arlosshalfload=150+1.6 #il/2+if\n", + "arlossfullload=1838.6 #ia**2*ra\n", + "\n", + "#case b\n", + "\n", + "n=((60*1000)/((60*1000)+1838.6+4140))*100\n", + "\n", + "print n\n", + "\n", + "#for maximum n ia=il\n", + "\n", + "ia=(pconst/ra)**0.5\n", + "\n", + "nmax=((120*1000)/float((120*1000)+2*4140))*100\n", + "\n", + "print nmax\n", + "\n", + "maxn=(ia*100)/300\n", + "\n", + "print maxn\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/screenshots/chap1.png 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