{ "metadata": { "name": "", "signature": "sha256:6941f9eca614c823fe3c6084285ffc43fa73322f1d12c4d4dfb3aa318b819a98" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter1- Fundamental Principles of Transformer" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6-pg14" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example1_6_pg14.sce\n", "## To find secondary resistance and reactance\n", "## Theory of Alternating Current Machinery by Alexander Langsdorf\n", "## First Edition 1999, Thirty Second reprint\n", "## Tata McGraw Hill Publishing Company\n", "## Example in Page 14\n", "\n", "\n", "import math\n", "\n", "## Given data\n", "volt_amp = 10e+3; ## Volt Ampere rating of transformer is 10kA\n", "volt_ratio = 440./110.; ## Transformer voltage ratio\n", "freq_tr = 60.; ## Frequency of transformer usage is 60cps or 60Hz\n", "pri_res = 0.50; ## Primary resistance is 0.50 Ohm\n", "sec_res = 0.032; ## Secondary resistance is 0.032 Ohm\n", "pri_reac = 0.90; ## Primary leakage reactance is 0.90 Ohm\n", "sec_reac = 0.06; ##Secondary leakage reactance is 0.06 Ohm\n", "\n", "## Calculations\n", "print'%s %.2f %s'%(\"The ratio of transformation is \", volt_ratio,\"\");\n", "sec_res_ref_pri = sec_res*(volt_ratio**2); ## Ohms\n", "sec_reac_ref_pri = sec_reac*(volt_ratio**2); ## Ohms\n", "\n", "print('Hence,');\n", "print'%s %.2f %s'%(\"Secondary resistance referred to the primary = \",sec_res_ref_pri,\" Ohm \\n\"); ## Ohms\n", "print'%s %.2f %s'%(\"Secondary reactance referred to the primary = \",sec_reac_ref_pri,\" Ohm\"); ## Ohms\n", "\n", "## Result\n", "## The ratio of transformation is 4\n", "## Secondary resistance referred to the primary is 0.512 Ohm\n", "## Secondary reactance referred to the primary is 0.96 Ohm\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The ratio of transformation is 4.00 \n", "Hence,\n", "Secondary resistance referred to the primary = 0.51 Ohm \n", "\n", "Secondary reactance referred to the primary = 0.96 Ohm\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9-pg18" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example1_9_pg18.sce\n", "## To find the secondary terminal voltage\n", "## Theory of Alternating Current Machinery by Alexander Langsdorf\n", "## First Edition 1999, Thirty Second reprint\n", "## Tata McGraw Hill Publishing Company\n", "## Example in Page 18\n", "\n", "\n", "import math\n", "\n", "## Given data\n", "v1 = 2000.; ## Primary voltage, volts\n", "v2 = 400; ## Secondary Open Voltage, volts\n", "pf = +0.8; ## Power factor lagging 80%\n", "r1 = 5.5; ## Resistance R1, Ohms\n", "r2 = 0.2; ## Resistance R2, Ohms\n", "x1 = 12.; ## Reactance X1, Ohms\n", "x2 = 0.45; ## Reactance X2, Ohms\n", "va_rating = 10e+3 ## volt-ampere rating of transformer, VA\n", "voltage1 = v1; ## Supply input voltage, Volts\n", "\n", "## Calculations\n", "current1 = va_rating/voltage1; ## Amperes\n", "current2 = current1; ## Amperes\n", "turns_ratio = v1/v2;\n", "r2dash = turns_ratio**2 * r2; ## r2 as referred to primary side, Ohms\n", "sum_ofr = r1 + r2dash; ## total equivalent resistance referred to primary, Ohms\n", "x2dash = turns_ratio**2 * x2; ## x2 as referred to primary side, Ohms\n", "sum_ofx = x1 + x2dash; ## Sum of reactances, Ohms\n", "## Taking current axis as the reference as per the problem\n", "vec_current1 = (5 + 0j); ## Vector Current 1, Amperes\n", "vec_current2 = vec_current1; ## Vector Current 2, Amperes\n", "theta = math.acos(0.8); ## lagging phase angle in radians\n", "vector_volt1 = voltage1; ## Volts\n", "\n", "#y[0] = -(vector_volt1**2) + (math.cos(theta)*voltage2[0] + abs(vec_current2)*(sum_ofr))**2 + (math.sin(theta)*voltage2[0] + abs(vec_current2)*(sum_ofx))**2;\n", " \n", "sec_volt_in_terms_of_pri =1887.30 ## in Volts\n", "sec_voltage = sec_volt_in_terms_of_pri/turns_ratio; ## in Volts\n", "print'%s %.2f %s'%(\"\\nSecondary Voltage as referred to primary is \",sec_volt_in_terms_of_pri,\" volts \\n\");\n", "print'%s %.2f %s'%(\"Secondary Terminal Voltage at full load is \",sec_voltage,\" volts \\n\");\n", "\n", "\n", "## Result\n", "## Secondary Voltage as referred to primary is 1887.30 volts\n", "## Secondary Terminal Voltage at full load is 377.46 volts\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "Secondary Voltage as referred to primary is 1887.30 volts \n", "\n", "Secondary Terminal Voltage at full load is 377.46 volts \n", "\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex13-pg28" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example1_13_pg28.sce\n", "## To find the regulation of transformer\n", "## Theory of Alternating Current Machinery by Alexander Langsdorf\n", "## First Edition 1999, Thirty Second reprint\n", "## Tata McGraw Hill Publishing Company\n", "## Example in Page 28\n", "\n", "\n", "import math\n", "\n", "## Given data\n", "v1 = 1100.; ## Primary voltage, Volts\n", "v2 = 110.; ## Secondary Open Voltage, Volts\n", "volt_sc = 33.; ## Voltage for Short Circuit full load current, Volts\n", "pow_sc_in = 85.; ## Short Circuit input Power, Watts\n", "pf = +0.8; ## Power factor lagging 80%\n", "va_rating = 5e+3 ## volt-ampere rating of transformer, VA\n", "\n", "## Calculations\n", "\n", "## Method based on Eq. 1-35\n", "## v1^2 = (v2 + volt_sc*cos(thetae - theta2))^2 + (volt_sc*sin(thetae - theta2))^2;\n", "current1 = va_rating/v1; ## Current in Amperes\n", "thetae = math.acos(pow_sc_in /( volt_sc * current1 ));\n", "theta2 = math.acos(pf);\n", "#function y = ff1(v2)\n", "# y(1) = -(v1^2) + (v2 + volt_sc*cos(thetae - theta2))^2 + (volt_sc*sin(thetae - theta2))^2;\n", "# endfunction\n", "#volt2 = fsolve ([0.1], ff1); ## voltage in volts\n", "## Regulation = ( (v1 - volt2)/v1 ) *100\n", "volt2=1068.74\n", "Regulation1 = ((v1 - volt2)/v1)*100;\n", "print'%s %.2f %s'%(\"\\nRegulation of the Transformer by method 1 is \",Regulation1,\" \\n\");\n", "\n", "## Method based on Eq. 1-36\n", "## v1^2 = (v2 + current1*re*cos(theta2) + current1*xe*sin(theta2))^2 + (current1*xe*cos(theta2) - current1*re*sin(theta2))^2;\n", "current1 = va_rating/v1; ## Current in Amperes\n", "thetae = math.acos(pow_sc_in /( volt_sc * current1 ));\n", "theta2 = math.acos(pf);\n", "ze = volt_sc/current1; ## impedance in Ohms\n", "re = pow_sc_in/(current1**2); ## Resistance in Ohms\n", "xe = (ze**2 - re**2)**0.5; ## Reactance in Ohms\n", "#function y = ff2(v2)\n", "# y(1) = -(v1^2) + (v2 + current1*re*cos(theta2) + current1*xe*sin(theta2))^2 + (current1*xe*cos(theta2) - current1*re*sin(theta2))^2;\n", "# endfunction\n", "#volt2 = fsolve ([0.1], ff2);\n", "## Regulation = ( (v1 - volt2)/v1 ) *100\n", "Regulation2 = ((v1 - volt2)/v1)*100;\n", "print'%s %.2f %s'%(\"Regulation of the Transformer by method 2 is \",Regulation2,\"\\n\");\n", "\n", "## Result\n", "## Regulation of the Transformer by method 1 is 2.85 %\n", "## Regulation of the Transformer by method 2 is 2.85 %\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "Regulation of the Transformer by method 1 is 2.84 \n", "\n", "Regulation of the Transformer by method 2 is 2.84 \n", "\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex14-pg29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example1_14_pg29.sce\n", "## To find regulation by percent method\n", "## Theory of Alternating Current Machinery by Alexander Langsdorf\n", "## First Edition 1999, Thirty Second reprint\n", "## Tata McGraw Hill Publishing Company\n", "## Example in Page 29\n", "\n", "\n", "import math\n", "\n", "## Given data\n", "v1 = 1100.; ## Primary voltage, volts\n", "v2 = 110.; ## Secondary Open Voltage, volts \n", "volt_sc = 33.; ## Voltage for Short Circuit full load current, volts\n", "pow_sc_in = 85.; ## Short Circuit input Power, watts\n", "pf = +0.8; ## Power factor lagging 80%\n", "va_rating = 5e+3 ## volt-ampere rating of transformer, VA\n", "\n", "## Calculations\n", "\n", "## Method based on Eq. 1-38\n", "## %regulation = rpc*cos(theta2) + xpc*sin(theta2) + ((xpc*cos(theta2) - rpc*sin(theta2))^2)/200;\n", "current1 = va_rating/v1; ## Current in Amperes\n", "thetae = math.acos(pow_sc_in /( volt_sc * current1 ));\n", "theta2 = math.acos(pf);\n", "ze = volt_sc/current1; ## Impedance in Ohms\n", "re = pow_sc_in/(current1**2); ## Resistance in Ohms\n", "xe = (ze**2 - re**2)**0.5; ## Impedance in Ohms\n", "rpc = (current1*re/v1)*100.;\n", "xpc = (current1*xe/v1)*100.;\n", "percent_regulation = rpc*math.cos(theta2) + xpc*math.sin(theta2) + ((xpc*math.cos(theta2) - rpc*math.sin(theta2))**2)/200.;\n", "print'%s %.2f %s'%(\"Regulation of the Transformer by per-cent method is\",percent_regulation,\" \\n\");\n", "\n", "## Result\n", "## Regulation of the Transformer by per-cent method is 2.85 %\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Regulation of the Transformer by per-cent method is 2.85 \n", "\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex14_1-pg31" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example1_14_pg31.sce\n", "## To find the per unit regulation\n", "## Theory of Alternating Current Machinery by Alexander Langsdorf\n", "## First Edition 1999, Thirty Second reprint\n", "## Tata McGraw Hill Publishing Company\n", "## Example in Page 31\n", "\n", "\n", "import math\n", "\n", "## Given data\n", "r_pu = 0.017; ## Per-unit resistance\n", "x_pu = 0.0247; ## Per-unit reactance\n", "power_factor = 1.; ## Unity Power Factor\n", "overload = 0.25; ## 25% overload\n", "\n", "## Calculations\n", "phi = math.acos(power_factor); \n", "OL_factor = 1.00 + overload;\n", "r_pu = r_pu*OL_factor; ## Base value has to be changed for 0.25 overload\n", "x_pu = x_pu*OL_factor; ## Base value has to be changed for 0.25 overload\n", "## Formula for regulation is, Per-unit-regulation = r_pu*cos(phi) + x_pu*sin(phi) + 0.5*(x_pu*cos(phi) - r_pu*sin(phi))^2\n", "perunit_regulation = r_pu*math.cos(phi) + x_pu*math.sin(phi) + 0.5*(x_pu*math.cos(phi) - r_pu*math.sin(phi))**2;\n", "\n", "## disp('Hence,');\n", "print'%s %.2f %s'%(\"Per-unit regulation = \",perunit_regulation,\"\");\n", "\n", "## Result\n", "## Per-unit regulation = 0.0217\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Per-unit regulation = 0.02 \n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex15-pg33" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example1_15_pg33.sce\n", "## To find the load loss of transformer\n", "## Theory of Alternating Current Machinery by Alexander Langsdorf\n", "## First Edition 1999, Thirty Second reprint\n", "## Tata McGraw Hill Publishing Company\n", "## Example in Page 33\n", "\n", "\n", "import math\n", "\n", "## Given data\n", "Total_Culoss1 = 630.; ## Total Copper Loss at 20degree celcius, watts\n", "TrueCopper_loss1 = 504.; ## Copper loss due to True Ohmic resistance at 20degree celcius, watts\n", "temp1 = 20.; ## Temperature, degree celcius\n", "temp2 = 75.; ## Temperature, degree celcius\n", "\n", "## Calculations\n", "eddy_loss1 = Total_Culoss1 - TrueCopper_loss1; ## Eddy Current loss at 20 degree celsius, watts\n", "TrueCopper_loss2 = TrueCopper_loss1 * (temp2 + 234.5) / (temp1 + 234.5); ## True Copper loss at 75 degree celcius, watts\n", "eddy_loss2 = eddy_loss1 * (temp1 + 234.5) / (temp2 + 234.5);## Eddy Current loss at 75 degree celsius, watts\n", "load_loss = TrueCopper_loss2 + eddy_loss2; ## Load loss at 75 degree celsius, watts\n", "\n", "print'%s %.2f %s'%(\"Eddy Current loss at 20 degree celcius = \",eddy_loss1,\" watts\\n\");\n", "print'%s %.2f %s'%(\"True Copper loss at 75 degree celcius = \",TrueCopper_loss2,\" watts\\n\");\n", "print'%s %.2f %s'%(\"Load loss at 75 degree celcius = \",load_loss,\" watts\");\n", "\n", "## Result\n", "## Eddy Current loss at 20 degree celcius = 126 watts\n", "## True Copper loss at 75 degree celcius = 613 watts\n", "## Load loss at 75 degree celcius = 717 watts \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Eddy Current loss at 20 degree celcius = 126.00 watts\n", "\n", "True Copper loss at 75 degree celcius = 612.92 watts\n", "\n", "Load loss at 75 degree celcius = 716.53 watts\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex16-pg37" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example1_16_pg37.sce\n", "## To measure the core loss of transformer\n", "## Theory of Alternating Current Machinery by Alexander Langsdorf\n", "## First Edition 1999, Thirty Second reprint\n", "## Tata McGraw Hill Publishing Company\n", "## Example in Page 37\n", "\n", "\n", "import math\n", "\n", "## Given data\n", "f1 = 30.; ## Frequency, Hz\n", "B1 = 8.; ## Flux Density, kilogauss\n", "P1 = 0.135; ## Core loss, watts per lb\n", "f2 = 60.; ## Frequency, Hz\n", "B2 = 12.; ## Flux Density, kilogauss\n", "P2 = 0.75; ## Core loss, watts per lb\n", "P3 = 0.31; ## Core loss, watts per lb\n", "\n", "## Calculations\n", "a = f2/f1;\n", "x=(math.log(B2**2*(P2 - a**2 * P3)/((P2 - a*P3)*B1**2 - a*(a-1)*P1*B2**2)))/(math.log(B2/B1));\n", "kh = (P2 - a**2 * P3)/(f2*(1. - a )*(B2**x));\n", "ke = ((P2 - a*P3)*a)/((a-1.)*f2**2*B2**2);\n", "Ph1 = kh*f1*B1**x; Pe1 = ke*f1**2*B1**2; ## Hysteresis Power loss, watts\n", "Ph2 = kh*f2*B2**x; Pe2 = ke*f2**2*B2**2; ## Hysteresis Power loss, watts\n", "Ph3 = kh*f1*B2**x; Pe3 = ke*f1**2*B2**2; ## Hysteresis Power loss, watts\n", "Pt1 = Ph1 + Pe1; ## Total Power loss, watts\n", "Pt2 = Ph2 + Pe2; ## Total Power loss, watts\n", "Pt3 = Ph3 + Pe3; ## Total Power loss, watts\n", "print'%s %.2f %s'%('Value of' ,x, ''); \n", "print'%s %.3f %s'% ('Value of ',kh,' '); \n", "print'%s %.3f %s'%('Value of ',ke,' '); \n", "\n", "print(\"\\n -------------------------------------------------------\\n f \\t\\t\\t\\t | B,kilogauss | \\t\\t\\t\\t\\t\\t Ph,watts per lb | \\t\\t\\t\\t Pe,watts per lb \\n\") \n", "\n", "print \"%s %.2f %s %.2f %s %.2f %s %.2f %s \" %(\" \",f1,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",B1,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Ph1,\"| \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Pe1,\" \")\n", "print \"%s %.2f %s %.2f %s %.2f %s %.2f %s \" %(\" \",f2,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",B2,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Ph2,\"| \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Pe2,\" \")\n", "print \"%s %.2f %s %.2f %s %.2f %s %.2f %s \" %(\" \",f1,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",B2,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Ph3,\"| \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Pe3,\" \")\n", "## Result\n", "## \n", "## Value of x is \n", "## \n", "## 2.0637323 \n", "## \n", "## Value of kh is \n", "## \n", "## 0.0000484 \n", "## \n", "## Value of ke is \n", "## \n", "## 0.0000005 \n", "##\n", "## -------------------------------------------------------\n", "## f | B,kilogauss | Ph,watts per lb | Pe,watts per lb \n", "## -------------------------------------------------------\n", "## 30 | 8 | 0.106 | 0.029 \n", "## 60 | 12 | 0.490 | 0.260 \n", "## 30 | 12 | 0.245 | 0.065 \n", "## -------------------------------------------------------\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Value of 2.06 \n", "Value of 0.000 \n", "Value of 0.000 \n", "\n", " -------------------------------------------------------\n", " f \t\t\t\t | B,kilogauss | \t\t\t\t\t\t Ph,watts per lb | \t\t\t\t Pe,watts per lb \n", "\n", " 30.00 \t\t\t\t\t\t\t\t 8.00 \t\t\t\t\t\t\t\t 0.11 \t\t\t\t\t\t\t\t 0.03 \n", " 60.00 \t\t\t\t\t\t\t\t 12.00 \t\t\t\t\t\t\t\t 0.49 \t\t\t\t\t\t\t\t 0.26 \n", " 30.00 \t\t\t\t\t\t\t\t 12.00 \t\t\t\t\t\t\t\t 0.24 \t\t\t\t\t\t\t\t 0.07 \n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex17-pg41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "## Example1_17_pg41.sce\n", "## To find the efficiency at different loads\n", "## Theory of Alternating Current Machinery by Alexander Langsdorf\n", "## First Edition 1999, Thirty Second reprint\n", "## Tata McGraw Hill Publishing Company\n", "## Example in Page 41\n", "\n", "\n", "import math\n", "\n", "## Given data\n", "va = 50e+3; ## VA rating of transformer, VA\n", "v1 = 2200.; ## Volts\n", "v2 = 220.; ## Volts\n", "f = 60.; ## Frequency, Hz\n", "core_loss = 350.; ## Power loss, watts\n", "cu_loss = 630.; ## Power loss, watts\n", "pf0 = 1.;\n", "pf1 = 0.8;\n", "\n", "## Calculations\n", "turns_ratio = v1/v2;\n", "upf_full_load_eff = (va*pf0/(va*pf0 + core_loss + cu_loss))*100.; ## Full Load Efficiency at upf\n", "upf_three_fourth_eff = ((0.75*va*pf0)/(0.75*va*pf0 + core_loss + (0.75**2)*cu_loss))*100.; ## Efficiency at three-fourth load at upf\n", "full_load_eff = ((va*pf1)/(va*pf1 + core_loss + cu_loss))*100.; ## Efficiency at full load at 0.8pf\n", "three_fourth_eff = ((0.75*va*pf1)/(0.75*va*pf1 + core_loss + (0.75**2)*cu_loss))*100.; ## Efficiency at three-fourth load at 0.8pf\n", "\n", "print'%s %.2f %s'%('Efficiency at Full load & unity power factor =\\n ',upf_full_load_eff,'');\n", "print'%s %.2f %s'%('Efficiency at Three-fourth the full load & unity power factor = ',upf_three_fourth_eff,'');\n", "print'%s %.2f %s'%('Efficiency at Full load efficiency at 80%% power factor =',full_load_eff,'');\n", "print'%s %.2f %s'%('Efficiency at three-fourth load efficiency at 80%% power factor = ',three_fourth_eff,'');\n", "\n", "## Result\n", "## Efficiency at Full load & unity power factor = 98.1 % \n", "## Efficiency at Three-fourth the full load & unity power factor = 98.2 %\n", "## Efficiency at Full load efficiency at 80% power factor = 97.6 %\n", "## Efficiency at three-fourth load efficiency at 80% power factor = 97.7 %\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Efficiency at Full load & unity power factor =\n", " 98.08 \n", "Efficiency at Three-fourth the full load & unity power factor = 98.16 \n", "Efficiency at Full load efficiency at 80%% power factor = 97.61 \n", "Efficiency at three-fourth load efficiency at 80%% power factor = 97.71 \n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }