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diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter13.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter13.ipynb new file mode 100755 index 00000000..163431b5 --- /dev/null +++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter13.ipynb @@ -0,0 +1,1498 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13: TRANSFORMERS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.1,Page number: 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the voltage across the secondary of the transformer.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "E_p=6400 #Primary voltage(in Volts) \n",
+ "f=50 #Frequency of primary supply(in Hertz)\n",
+ "N1=480 #Number of turns in the primary of the transformer\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "flux_m=E_p/(4.44*f*N1)\n",
+ "N2=20.0\n",
+ "Es=4.44*f*N2*flux_m\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The peak value of the flux produced in the core is %.2f Wb.\" %(flux_m)\n",
+ "print \"(b)The voltage across the secondary winding if it has 20 turns is %.2f V.\" %(Es)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The peak value of the flux produced in the core is 0.06 Wb.\n",
+ "(b)The voltage across the secondary winding if it has 20 turns is 266.67 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.2,Page number: 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the peak value of the flux density in the core.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "f=50.0 #Operating frequency of the transformer(in Hertz)\n",
+ "N1=30.0 #Number of turns in the primary of transformer\n",
+ "N2=350.0 #Number of turns in the secondary of transformer\n",
+ "A=250e-04 #Cross-sectional area of the core(in square-metres)\n",
+ "E1=230.0 #Voltage of the supply(in Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "flux_m=E1/(4.44*f*N1)\n",
+ "B_m=flux_m/A\n",
+ "E2=E1*(N2/N1)\n",
+ "I2=100.0\n",
+ "I1=I2*(N2/N1)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The peak value of flux density in the core is %.2f T.\" %(B_m)\n",
+ "print \"(b)The voltage induced in the secondary winding is %e V.\" %(E2)\n",
+ "print \"(c)The primary current when the secondary current is 100 A is %e A.\" %(I1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The peak value of flux density in the core is 1.38 T.\n",
+ "(b)The voltage induced in the secondary winding is 2.683333e+03 V.\n",
+ "(c)The primary current when the secondary current is 100 A is 1.166667e+03 A.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.3,Page number: 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the turns-ratio of the transformer.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Req=50.0 #Output resistance of the source(in Ohms) \n",
+ "R_L=800.0 #Load resistance(in Ohms) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "K=sqrt(R_L/Req)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The turns-ratio of the transformer to be used for maximising the load power is %d.\" %(K)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The turns-ratio of the transformer to be used for maximising the load power is 4.\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.4,Page number: 378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the load current in the ac circuit.\"\"\"\n",
+ "\n",
+ "from cmath import rect,phase\n",
+ "from math import degrees\n",
+ "\n",
+ "#Calculations:\n",
+ "V=rect(30,0)\n",
+ "Ip=V/(20+20*1j+(pow(2,2)*(2-10*1j)))\n",
+ "I_L=2.0*Ip\n",
+ "\n",
+ "#Result:\n",
+ "print \"The load current is %.3f A at a phase angle of %.3f degrees.\" %(abs(I_L),degrees(phase(I_L)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The load current is 1.744 A at a phase angle of 35.538 degrees.\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.5,Page number: 378 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the output of the transformer in kVA.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "B_m=1.1 #Maximum magnetic flux density(in Weber per square-metre)\n",
+ "A=150e-04 #Cross-sectional area of the core(in square-metres)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "flux_m=B_m*A\n",
+ "N2=66\n",
+ "f=50\n",
+ "Z_L=4.0\n",
+ "E2=4.44*N2*f*flux_m\n",
+ "V2=E2\n",
+ "I2=V2/Z_L\n",
+ "output=(I2*V2)/1000.0\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The output when connected to a load of 4 Ohms impedance is %.3f kVA.\" %(output)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The output when connected to a load of 4 Ohms impedance is 14.612 kVA.\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.6,Page number: 378 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the number of turns in each winding of the transformer.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "A=9e-04 #Cross-sectional area of the core(in square-metre) \n",
+ "E1=230.0 #Primary Voltage(in Volts)\n",
+ "E2=110.0 #Secondary Voltage(in Volts)\n",
+ "E3=6.0 #Tertiary Voltage(in Volts)\n",
+ "f=50.0 #Operating frequency of the transformer(in Hertz)\n",
+ "Bm=1.0 #Maximum magnetic flux density(in Tesla) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "flux_m=Bm*A\n",
+ "N3_half=E3/(4.44*f*flux_m)\n",
+ "N3=2*N3_half\n",
+ "N1=N3_half*(E1/E3)\n",
+ "N2=N3_half*(E2/E3)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The total number of turns on the primary winding is %d turns.\" %(N1)\n",
+ "print \"The total number of turns on the secondary winding is %d turns.\" %(N2)\n",
+ "print \"The total number of turns on the tertiary winding is %d turns.\" %(N3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The total number of turns on the primary winding is 1151 turns.\n",
+ "The total number of turns on the secondary winding is 550 turns.\n",
+ "The total number of turns on the tertiary winding is 60 turns.\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.7,Page number: 380"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the no-load power factor of the transformer.\"\"\"\n",
+ "\n",
+ "from math import sqrt,pow\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=350.0 #Input at no-load(in Volt-Amperes) \n",
+ "V1=230.0 #Primary voltage(in Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Io=VA/V1\n",
+ "Pi=110.0\n",
+ "pf=Pi/(V1*Io)\n",
+ "Iw=Io*pf\n",
+ "Im=sqrt(pow(Io,2)-pow(Iw,2))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The loss component of no-load current is given as %.4f A.\" %(Iw)\n",
+ "print \"The magnetising component of no-load current is %.4f A.\" %(Im)\n",
+ "print \"The no-load power factor is %.4f.\" %(pf)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The loss component of no-load current is given as 0.4783 A.\n",
+ "The magnetising component of no-load current is 1.4446 A.\n",
+ "The no-load power factor is 0.3143.\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.8,Page number: 382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the hysterisis and eddy-current losses of the transformer.\"\"\"\n",
+ "\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "f=50.0 #Operating frequency of the transformer(in Hertz)\n",
+ "eq1=100.0 #Iron loss at 60 Hz(in Watts)\n",
+ "eq2=60.0 #Iron loss at 40 Hz(in Watts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "\"\"\" P_h=A*f ; P_e=B*f*f ; \n",
+ " \n",
+ " P_i=P_h+P_e=(A*f)+(B*f*f); \"\"\"\n",
+ "\n",
+ "A=((eq2*36)-(eq1*16))/((40*36)-(60*16))\n",
+ "B=((eq1*4)-(eq2*6))/((3600*4)-(1600*6))\n",
+ "P_h=A*f\n",
+ "P_e=B*f*f\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The Hysteresis loss at 50 Hz is %.2f W.\" %(P_h) \n",
+ "print \"The Eddy-current loss at 50 Hz is %.2f W.\" %(P_e) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Hysteresis loss at 50 Hz is 58.33 W.\n",
+ "The Eddy-current loss at 50 Hz is 20.83 W.\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.9,Page number: 385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the primary current snd the primary power factor.\"\"\"\n",
+ "\n",
+ "from math import acos,degrees,cos\n",
+ "from cmath import phase,rect\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V1=440.0 #Primary voltage(in Volts)\n",
+ "V2=110.0 #Secondary voltage(in Volts)\n",
+ "I_0_mod=5.0 #No-load current(in Amperes) \n",
+ "I2=120.0 #Secondary current(in Amperes) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "phi_0=acos(0.2)\n",
+ "phi_2=acos(0.8)\n",
+ "K=V2/V1\n",
+ "I1_load_mod=K*I2\n",
+ "I1_load=rect(I1_load_mod,-phi_2)\n",
+ "I0=rect(I_0_mod,-phi_0)\n",
+ "I1_total=I0+I1_load\n",
+ "pf_primary=cos(phase(I1_total))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The primary current is %.2f A at a phase angle of %.2f degrees.\" %(abs(I1_total),degrees(phase(I1_total)))\n",
+ "print \"The primary power factor is %.3f lagging.\" %(pf_primary)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The primary current is 33.90 A at a phase angle of -42.49 degrees.\n",
+ "The primary power factor is 0.737 lagging.\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.10,Page number: 390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the equivalent resistance as referred to the primary and secondary.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=50e03 #Power rating of the transformer(in Volt-Amperes) \n",
+ "V1=4400.0 #Primary voltage(in Volts) \n",
+ "V2=220.0 #Secondary voltage(in Volts)\n",
+ "R1=3.45 #Primary resistance(in Ohms)\n",
+ "R2=0.009 #Secondary resistance(in Ohms)\n",
+ "X1=5.2 #Leakage reactance of primary(in Ohms)\n",
+ "X2=0.015 #Leakage reactance of secondary(in Ohms)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "I1=VA/V1\n",
+ "I2=VA/V2\n",
+ "K=V2/V1\n",
+ "Re1=R1+(R2/(K*K))\n",
+ "Re2=(K*K*R1)+R2\n",
+ "Xe1=X1+(X2/(K*K))\n",
+ "Xe2=(K*K*X1)+X2\n",
+ "Ze1=sqrt((Re1*Re1)+(Xe1*Xe1))\n",
+ "Ze2=sqrt((Re2*Re2)+(Xe2*Xe2))\n",
+ "tot_copp_loss=(I1*I1*R1)+(I2*I2*R2)\n",
+ "tot_copp_eq_p=I1*I1*Re1\n",
+ "tot_copp_eq_s=I2*I2*Re2\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The equivalent resistance as referred to the primary is %.2f Ohms.\" %(Re1)\n",
+ "print \"(b)The equivalent resistance as referred to the secondary is %.4f Ohms.\" %(Re2)\n",
+ "print \"(c)The equivalent reactance as referred to the primary is %.2f Ohms.\" %(Xe1)\n",
+ "print \"(d)The equivalent reactance as referred to the secondary is %.3f Ohms.\" %(Xe2)\n",
+ "print \"(e)The equivalent impedance as referred to the primary is %.2f Ohms.\" %(Ze1)\n",
+ "print \"(f)The equivalent impedance as referred to the secondary is %.4f Ohms.\" %(Ze2)\n",
+ "print \"(g)The total copper loss by using the individual resistances of the two windings is %.2f W.\" %(tot_copp_loss)\n",
+ "print \" By considering equivalent resistances,\"\n",
+ "print \" Total copper loss(referred to primary equivalent resistance)=%.2f W.\" %(tot_copp_eq_p) \n",
+ "print \" Total copper loss(referred to secondary equivalent resistance)=%.2f W\" %(tot_copp_eq_s)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The equivalent resistance as referred to the primary is 7.05 Ohms.\n",
+ "(b)The equivalent resistance as referred to the secondary is 0.0176 Ohms.\n",
+ "(c)The equivalent reactance as referred to the primary is 11.20 Ohms.\n",
+ "(d)The equivalent reactance as referred to the secondary is 0.028 Ohms.\n",
+ "(e)The equivalent impedance as referred to the primary is 13.23 Ohms.\n",
+ "(f)The equivalent impedance as referred to the secondary is 0.0331 Ohms.\n",
+ "(g)The total copper loss by using the individual resistances of the two windings is 910.38 W.\n",
+ " By considering equivalent resistances,\n",
+ " Total copper loss(referred to primary equivalent resistance)=910.38 W.\n",
+ " Total copper loss(referred to secondary equivalent resistance)=910.38 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.11,Page number: 393"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the full-load regulation for different load power factors.\"\"\"\n",
+ "\n",
+ "from math import cos,acos,sin\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=40e03 #Power rating of the transformer(in Volt-Amperes) \n",
+ "V1=6600.0 #Primary voltage(in Volts) \n",
+ "V2=250.0 #Secondary voltage(in Volts)\n",
+ "R1=10.0 #Primary resistance(in Ohms)\n",
+ "R2=0.02 #Secondary resistance(in Ohms)\n",
+ "Xe1=35.0 #Equivalent leakage reactance as referred to the primary(in Ohms)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "K=V2/V1\n",
+ "I2=VA/V2\n",
+ "Re2=(K*K*R1)+R2\n",
+ "Xe2=(K*K*Xe1)\n",
+ "pf=1\n",
+ "phi=acos(pf)\n",
+ "per_reg_a=(((I2*Re2*pf)+(I2*Xe2*sin(phi)))/V2)*100\n",
+ "pf=0.8\n",
+ "phi=acos(pf)\n",
+ "per_reg_b=(((I2*Re2*pf)+(I2*Xe2*sin(phi)))/V2)*100\n",
+ "pf=0.8\n",
+ "phi=acos(pf)\n",
+ "per_reg_c=(((I2*Re2*pf)-(I2*Xe2*sin(phi)))/V2)*100\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)For unity power factor of the load:\"\n",
+ "print \"The full-load percentage regulation is %.3f percent.\" %(per_reg_a)\n",
+ "print \"\\n(b)For power factor of the load=0.8 lagging:\"\n",
+ "print \"The full-load percentage regulation is %.3f percent.\" %(per_reg_b)\n",
+ "print \"\\n(c)For power factor of the load=0.8 leading:\"\n",
+ "print \"The full-load percentage regulation is %.3f percent.\" %(per_reg_c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)For unity power factor of the load:\n",
+ "The full-load percentage regulation is 2.198 percent.\n",
+ "\n",
+ "(b)For power factor of the load=0.8 lagging:\n",
+ "The full-load percentage regulation is 3.687 percent.\n",
+ "\n",
+ "(c)For power factor of the load=0.8 leading:\n",
+ "The full-load percentage regulation is -0.170 percent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.12,Page number: 396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the number of turns in each winding.\"\"\"\n",
+ "\n",
+ "from math import sqrt \n",
+ "\n",
+ "#Variable Declaration:\n",
+ "f=50.0 #Operating frequency of the transformer(in Hertz)\n",
+ "E1=5000.0 #Primary voltage at no-load(in Volts)\n",
+ "E2=250.0 #Secondary voltage at no-load(in Volts)\n",
+ "VA_full=150e03 #Power rating of the transformer(in Volt-Ampere) \n",
+ "flux=0.06 #Maximum core flux(in Weber)\n",
+ "Pi=1500.0 #Core losses(in Watts)\n",
+ "Pc_FL=1800.0 #Full-load copper losses(in Watts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "N2=E2/(4.44*f*flux)\n",
+ "N1=(E1/E2)*round(N2,0)\n",
+ "pf=1\n",
+ "Po=0.5*VA_full*pf\n",
+ "Pc=0.5*0.5*Pc_FL\n",
+ "effi_b=(Po/(Po+Pi+Pc))*100\n",
+ "pf=0.8\n",
+ "Po=VA_full*pf\n",
+ "Pc=Pc_FL\n",
+ "effi_c=(Po/(Po+Pi+Pc))*100\n",
+ "x=sqrt(Pi/Pc_FL)\n",
+ "VA_load=VA_full*x\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The number of turns in the primary winding is %d.\" %(round(N1,0)) \n",
+ "print \" The number of turns in the secondary winding is %d.\" %(round(N2,0))\n",
+ "print \"(b)The efficiency at half rated kVA and unity power factor is %.3f percent\" %(effi_b)\n",
+ "print \"(c)The efficiency at full load and 0.8 power factor lagging is %.3f percent.\" %(effi_c)\n",
+ "print \"(d)The kVA load for maximum efficiency is %d kVA.\" %(round((VA_load/1000),0))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The number of turns in the primary winding is 380.\n",
+ " The number of turns in the secondary winding is 19.\n",
+ "(b)The efficiency at half rated kVA and unity power factor is 97.466 percent\n",
+ "(c)The efficiency at full load and 0.8 power factor lagging is 97.324 percent.\n",
+ "(d)The kVA load for maximum efficiency is 137 kVA.\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.13,Page number: 397"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the all-day efficiency of a distribution transformer.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "kVA_FL=200.0 #Power rating of transformer at full-load(in Volt-Ampere) \n",
+ "Pc_FL=3.02 #Full-load copper losses(in kilo-Watts) \n",
+ "Pi=1.6 #Iron-losses(in kilo-Watts) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Pi_24_hrs=24.0*Pi\n",
+ "\"\"\" For 80 kW load at unity power factor in 6 hrs:\"\"\"\n",
+ "Po_1=80\n",
+ "t1=6\n",
+ "pf_1=1\n",
+ "output_ene_1=Po_1*t1\n",
+ "kVA_1=Po_1/pf_1\n",
+ "Pc_1=pow((kVA_1/kVA_FL),2)*Pc_FL*t1\n",
+ "\"\"\" For 160 kW load at 0.8 power factor in 8 hrs:\"\"\"\n",
+ "Po_2=160\n",
+ "t2=8\n",
+ "pf_2=0.8\n",
+ "output_ene_2=Po_2*t2\n",
+ "kVA_2=Po_2/pf_2\n",
+ "Pc_2=pow((kVA_2/kVA_FL),2)*Pc_FL*t2\n",
+ "\"\"\" For no load period of 10 hrs:\"\"\"\n",
+ "Po_3=0\n",
+ "t3=10\n",
+ "output_ene_3=0\n",
+ "Pc_3=0\n",
+ "Po_total=output_ene_1+output_ene_2+output_ene_3\n",
+ "Pc_total=Pc_1+Pc_2+Pc_3\n",
+ "all_day_effi=Po_total/(Po_total+Pc_total+Pi_24_hrs)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The all day efficiency of the distribution transformer is %.3f percent.\" %(all_day_effi*100)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The all day efficiency of the distribution transformer is 96.414 percent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.14,Page number: 398"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the apparent power rating of an autotransformer.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_rated=120.0 #Voltage rating of the transformer(in Volts)\n",
+ "VA=12e03 #Power rating of the transformer(in Volt-Ampere)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "I1=VA/V_rated\n",
+ "I2=I1\n",
+ "input_app_pow=240*I1\n",
+ "output_app_pow=120*2*I1\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"In auto-tranformer mode, the input apparent power is %d kVA and the output apparent power is %d kVA.\" %(round((input_app_pow/1000),0),round((output_app_pow/1000),0))\n",
+ "print \"Thus, the apparent power capacity of the 12-kVA transformer is doubled by the auto-transformer connection.\"\n",
+ "print \"In effect,half the apparent power is transformed and half is conducted directly to the secondary side.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "In auto-tranformer mode, the input apparent power is 24 kVA and the output apparent power is 24 kVA.\n",
+ "Thus, the apparent power capacity of the 12-kVA transformer is doubled by the auto-transformer connection.\n",
+ "In effect,half the apparent power is transformed and half is conducted directly to the secondary side.\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.15,Page number: 401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the secondary line voltage on no load when the windings are connected(a)star/delta,(b)delta/star.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_L1=3300.0 #Supply voltage(in Volts)\n",
+ "Np=840.0 #Number of turns in the primary\n",
+ "Ns=72.0 #Number of turns in the secondary\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Vph_1a=V_L1/sqrt(3)\n",
+ "Vph_2a=Vph_1a*(Ns/Np)\n",
+ "V_L2a=Vph_2a\n",
+ "Vph_1b=V_L1\n",
+ "Vph_2b=Vph_1b*(Ns/Np)\n",
+ "V_L2b=Vph_2b*sqrt(3)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print(\"(a)For star/delta connection: \")\n",
+ "print \"The secondary line voltage on no load is %.2f V.\" %(V_L2a)\n",
+ "print(\"(b)For delta/star connection: \")\n",
+ "print \"The secondary line voltage on no load is %.2f V.\" %(V_L2b)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)For star/delta connection: \n",
+ "The secondary line voltage on no load is 163.31 V.\n",
+ "(b)For delta/star connection: \n",
+ "The secondary line voltage on no load is 489.92 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.16,Page number: 403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the magnetising current and the core-loss current in a single-phase transformer.\"\"\" \n",
+ " \n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=12e03 #Power rating of the transformer(in Volt-Ampere) \n",
+ "Vp=400.0 #Primary voltage(in Volts)\n",
+ "Vs=200.0 #Secondary voltage(in Volts)\n",
+ "Wo=120.0 #Power in open-circuit test(in Watts) \n",
+ "V1=200.0 #Voltage in open-circuit test(in Volts)\n",
+ "I_0=1.3 #Current in open-circuit test(in Amperes)\n",
+ "Isc=30.0 #Current in short-circuit test(in Amperes)\n",
+ "Wsc=200.0 #Power in short-circuit test(in Watts)\n",
+ "Vsc=22.0 #Voltage in short-circuit test(in Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Iw=Wo/V1\n",
+ "Im=sqrt((I_0*I_0)-(Iw*Iw))\n",
+ "R_0=V1/Iw\n",
+ "X_0=V1/Im\n",
+ "K=Vs/Vp\n",
+ "I_FL=VA/Vp\n",
+ "Re1=Wsc/(Isc*Isc)\n",
+ "Ze1=Vsc/Isc\n",
+ "Xe1=sqrt((Ze1*Ze1)-(Re1*Re1))\n",
+ "Re2=K*K*Re1\n",
+ "Xe2=K*K*Xe1\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The magnetising current is %.2f A and the core-loss current is %.2f A\" %(Im,Iw)\n",
+ "print \"(b)The parameters of equivalent circuit as referred to the low voltage winding(secondary winding) are: \\n Re2=%.4f ohm \\n Xe2=%.4f ohm\" %(Re2,Xe2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The magnetising current is 1.15 A and the core-loss current is 0.60 A\n",
+ "(b)The parameters of equivalent circuit as referred to the low voltage winding(secondary winding) are: \n",
+ " Re2=0.0556 ohm \n",
+ " Xe2=0.1747 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.17,Page number: 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the secondary emf in a transformer.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=25.0e03 #Power rating of the transformer(in VA)\n",
+ "N1=500.0 #Number of turns in the primary winding \n",
+ "N2=40.0 #Number of turns in the secondary winding\n",
+ "V1=3e03 #Voltage of the supply connected to primary(in Volts)\n",
+ "f=50.0 #Frequency of the supply(in Hertz)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "K=N2/N1\n",
+ "E2=K*V1\n",
+ "I1=VA/V1\n",
+ "I2=I1/K\n",
+ "flux=V1/(4.44*f*N1)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The secondary emf is %.2f V.\" %(E2)\n",
+ "print \"(b)The primary current on full-load is %.2f A and the secondary current on full-load is %.2f A.\" %(I1,I2)\n",
+ "print \"(c)The maximum flux in the core is %.4f Wb.\" %(flux)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The secondary emf is 240.00 V.\n",
+ "(b)The primary current on full-load is 8.33 A and the secondary current on full-load is 104.17 A.\n",
+ "(c)The maximum flux in the core is 0.0270 Wb.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.18,Page number: 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the active cross-sectional area of the core.\"\"\"\n",
+ "\n",
+ "from math import sqrt\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N1=50.0 #Number of turns in the primary winding\n",
+ "B=1.0 #Maximum flux density(in Tesla)\n",
+ "f=50.0 #Frequency rating of the transformer(in Hertz)\n",
+ "V=230.0 #Voltage rating of the transformer(in Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "E1=V\n",
+ "flux=E1/(4.44*f*N1)\n",
+ "A=flux/B\n",
+ "\"\"\"Due to the insulation of laminations from each other,the gross area is about 10% greater than the active area.\"\"\" \n",
+ "gross=1.1*A\n",
+ "a=sqrt(gross)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The active cross sectional area of the core is %.5f square m.\" %(A)\n",
+ "print \"(b)The side of a square core is %.2f m.\" %(a)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The active cross sectional area of the core is 0.02072 square m.\n",
+ "(b)The side of a square core is 0.15 m.\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.19,Page number: 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the output of the transformer in kVA.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "A=150e-04 #Cross-sectional area of the core(in square metres)\n",
+ "Bm=1.1 #Maximum flux density(in Tesla)\n",
+ "f=50.0 #Frequency of the supply(in Hertz)\n",
+ "N2=66.0 #Number of turns in the secondary winding\n",
+ "Z_L=4.0 #Load impedance(in Ohms)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "flux=Bm*A\n",
+ "E2=4.44*flux*f*N2\n",
+ "I2=E2/Z_L\n",
+ "kVA=(E2*I2)/1000.0\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The output in kVA when connected to a 4 Ohms load impedance is %.2f kVA.\" %(kVA)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The output in kVA when connected to a 4 Ohms load impedance is 14.61 kVA.\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.20,Page number: 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the magnetising current and the iron loss.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I0=1.0 #No-load primary current(in Amperes)\n",
+ "pf=0.24 #Power factor\n",
+ "V1=11e03 #Primary voltage(in Volts)\n",
+ "V2=400 #Secondary voltage(in Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Iw=I0*pf\n",
+ "Im=sqrt((I0*I0)-(Iw*Iw))\n",
+ "Pi=V1*I0*pf\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The core-loss current is %.2f A.\" %(Iw) \n",
+ "print \"(b)The magnetising current is %.3f A.\" %(Im)\n",
+ "print \"(c)The iron loss is %.2f W.\" %(Pi)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The core-loss current is 0.24 A.\n",
+ "(b)The magnetising current is 0.971 A.\n",
+ "(c)The iron loss is 2640.00 W.\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.21,Page number: 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the supply voltage and the power factor.\"\"\"\n",
+ "\n",
+ "from cmath import rect,phase\n",
+ "from math import degrees,radians\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "K=0.5 #Turns ratio of the step-down transformer\n",
+ "R1=2.5 #Resistance of the primary winding(in Ohms)\n",
+ "X1=6.0 #Reactance of the primary winding(in Ohms)\n",
+ "R2=0.25 #Resistance of the secondary winding(in Ohms)\n",
+ "X2=1 #Reactance of the secondary winding(in Ohms)\n",
+ "Im=51.5e-03 #Magnetising current(in Amperes)\n",
+ "Iw=20.6e-03 #Core-loss current(in Amperes)\n",
+ "Z_L=rect(25,radians(30)) #Load impedance(in Ohms)\n",
+ "Vo=50.0 #Output voltage(in Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Z1=R1+1j*X1\n",
+ "Z2=R2+1j*X2\n",
+ "V2=rect(Vo,0)\n",
+ "I2=V2/Z_L\n",
+ "E2=V2+(I2*Z2)\n",
+ "E1=E2/K\n",
+ "E1_minus=-E1\n",
+ "I1_a=-I2*K\n",
+ "\"\"\"Im lags -E1 by 90 degrees and Iw is in phase with -E1.\"\"\"\n",
+ "Im_com=rect(Im,(phase(E1_minus)-radians(90)))\n",
+ "Iw_com=rect(Iw,phase(E1_minus))\n",
+ "I1=I1_a+Im_com+Iw_com\n",
+ "V1=E1_minus+(I1*Z1)\n",
+ "pf_ang=phase(V1)-phase(I1)\n",
+ "pf=cos(pf_ang)\n",
+ "\n",
+ "\n",
+ "#Result:I\n",
+ "print \"The supply voltage is %.4f V at a phase angle of %.2f degrees.\" %(abs(V1),degrees(phase(V1)))\n",
+ "print \"The current drawn from the supply is %.4f A at a phase angle of %.2f degrees.\" %(abs(I1),degrees(phase(I1)))\n",
+ "print \"The power factor is %.3f lagging.\" %(pf)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The supply voltage is 108.6120 V at a phase angle of -176.35 degrees.\n",
+ "The current drawn from the supply is 1.0451 A at a phase angle of 148.19 degrees.\n",
+ "The power factor is 0.815 lagging.\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.22,Page number: 406"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the copper loss in the transformer.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "K=0.25 #Turns ratio of the step-down transformer\n",
+ "R1=1.4 #Resistance of the primary(in Ohms)\n",
+ "X1=5.5 #Reactance of the primary(in Ohms)\n",
+ "R2=0.06 #Resistance of the secondary(in Ohms)\n",
+ "X2=0.04 #Reactance of the secondary(in Ohms)\n",
+ "Vsc=24.0 #Voltage of the HV winding(in Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Re1=R1+(R2/(K*K))\n",
+ "Xe1=X1+(X2/(K*K))\n",
+ "Ze1=sqrt((Re1*Re1)+(Xe1*Xe1))\n",
+ "Isc=Vsc/Ze1\n",
+ "I1=Isc\n",
+ "I2=I1/K\n",
+ "P=I1*I1*Re1\n",
+ "pf=P/(Vsc*I1)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The current in the LV winding is %.3f A.\" %(I2)\n",
+ "print \"(b)The copper loss in the transformer is %.2f W.\" %(P)\n",
+ "print \"(c)The power factor is %.4f.\" %(pf)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The current in the LV winding is 14.594 A.\n",
+ "(b)The copper loss in the transformer is 31.42 W.\n",
+ "(c)The power factor is 0.3588.\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.23,Page number: 406"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the regulation and efficiency.\"\"\"\n",
+ "\n",
+ "from math import acos\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=20e03 #Power rating of the transformer(in Volt-Amperes)\n",
+ "V1=2200.0 #Voltage of the primary winding(in Volts)\n",
+ "V2=220.0 #Voltage of the secondary winding(in Volts)\n",
+ "f=50.0 #Frequency rating of the transformer(in Hertz)\n",
+ "Vsc=86.0 #Voltage measured during short-circuit test(in Volts)\n",
+ "Isc=10.5 #Current measured during short-circuit test(in Amperes)\n",
+ "Psc=360.0 #Power measured during short-circuit test(in Watts)\n",
+ "Voc=220.0 #Voltage measured during open-circuit test(in Volts)\n",
+ "Ioc=4.2 #Current measured during open-circuit test(in Amperes)\n",
+ "Poc=148.0 #Power measured during open-circuit test(in Watts)\n",
+ "pf=0.8 #Lagging power factor\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Ze1=Vsc/Isc\n",
+ "Re1=Psc/(Isc*Isc)\n",
+ "Xe1=sqrt((Ze1*Ze1)-(Re1*Re1))\n",
+ "I1=VA/V1\n",
+ "reg=(I1*((Re1*pf)+(Xe1*sin(acos(pf)))))/V1\n",
+ "Pc=(I1/Isc)*(I1/Isc)*Psc\n",
+ "Pi=Poc\n",
+ "Po=VA*pf\n",
+ "effi=Po/(Po+Pc+Pi)\n",
+ "pf_sc=Re1/Ze1\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The regulation at 0.8 pf lagging at full load is %.2f per cent.\" %(reg*100) \n",
+ "print \" The efficiency is %.2f per cent.\" %(effi*100)\n",
+ "print \"(b)The power factor on short circuit is %.3f lagging.\" %(pf_sc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The regulation at 0.8 pf lagging at full load is 2.94 per cent.\n",
+ " The efficiency is 97.45 per cent.\n",
+ "(b)The power factor on short circuit is 0.399 lagging.\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.24,Page number: 407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the efficiency at half of full-load current.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=200e03 #Power rating of the transformer(in VA)\n",
+ "effi_FL=0.98 #Full-load efficiency of the transformer\n",
+ "pf=0.8 #Lagging power factor\n",
+ "x=0.75 #Fraction of load at which maximum efficiency occurs\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Po=VA*pf\n",
+ "Pin=Po/effi_FL\n",
+ "tot_loss=Pin-Po\n",
+ "Pc=tot_loss/(1+(x*x))\n",
+ "Pi=tot_loss-Pc\n",
+ "x_new=0.5\n",
+ "P1=(x_new*x_new*Pc)+Pi\n",
+ "effi_half=(Po/2.0)/((Po/2.0)+P1)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The efficiency at half of full-load current is %.3f per cent.\" %(effi_half*100)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The efficiency at half of full-load current is 97.922 per cent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.25,Page number: 408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the efficiency at different rated kVAs.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=150e03 #Power rating of the transformer(in Volt-Amperes)\n",
+ "V1=5000.0 #Voltage of the primary winding(in Volts)\n",
+ "V2=250.0 #Voltage of the secondary winding(in Volts)\n",
+ "f=50.0 #Frequency rating of the transformer(in Hertz)\n",
+ "Pc=1.8e03 #Full-load copper losses(in Watts)\n",
+ "Pi=1.5e03 #Core losses(in Watts)\n",
+ "flux=60e-03 #Maximum core flux(in Webers)\n",
+ "pf=0.8 #Lagging power factor\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "N2=V2/(4.44*f*flux)\n",
+ "N1=round(N2,0)*(V1/V2)\n",
+ "\"\"\"Case 1:\"\"\"\n",
+ "Po=(VA*pf)\n",
+ "effi_a=Po/(Po+Pi+Pc)\n",
+ "\"\"\"Case 2:\"\"\"\n",
+ "pf=1.0\n",
+ "Po=0.5*VA*pf\n",
+ "Pc_new=0.5*0.5*Pc\n",
+ "effi_b=Po/(Po+Pi+Pc_new)\n",
+ "\n",
+ "x=sqrt(Pi/Pc)\n",
+ "VA_max_effi=x*VA\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The number of turns in the primary winding is %d turns.\" %(round(N1,0))\n",
+ "print \" The number of turns in the secondary winding is %d turns.\" %(round(N2,0))\n",
+ "print \"(b)The efficiency at full rated kVA with 0.8 pf lagging is %.2f percent.\" %(effi_a*100) \n",
+ "print \"(c)The efficiency at half rated kVA with unity pf is %.2f percent.\" %(effi_b*100)\n",
+ "print \"(d)The kVA load for maximum efficieny is %d kVA.\" %(round((VA_max_effi/1000),0)) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The number of turns in the primary winding is 380 turns.\n",
+ " The number of turns in the secondary winding is 19 turns.\n",
+ "(b)The efficiency at full rated kVA with 0.8 pf lagging is 97.32 percent.\n",
+ "(c)The efficiency at half rated kVA with unity pf is 97.47 percent.\n",
+ "(d)The kVA load for maximum efficieny is 137 kVA.\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.26,Page number: 409 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the impedance on the high voltage side.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=50e03 #Power rating of the transformer(in VA)\n",
+ "V1=2400.0 #Voltage of primary winding(in Volts)\n",
+ "V2=240.0 #Voltage of secondary winding(in Volts)\n",
+ "f=50.0 #Frequency rating of the transformer(in Hertz)\n",
+ "LV=240.0 #Low tension voltage(in Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "I2=VA/V2\n",
+ "Z_L=V2/I2\n",
+ "K=V2/V1\n",
+ "Zeq=Z_L/(K*K)\n",
+ "I_high=K*I2\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The load impedance connected to the LV side is %.3f Ohms.\" %(Z_L)\n",
+ "print \"(b)The load impedance referred to to the high voltage side is %.2f Ohms.\" %(Zeq)\n",
+ "print \"(c)The current referred to the high voltage side is %.3f A.\" %(I_high) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The load impedance connected to the LV side is 1.152 Ohms.\n",
+ "(b)The load impedance referred to to the high voltage side is 115.20 Ohms.\n",
+ "(c)The current referred to the high voltage side is 20.833 A.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.27,Page number: 409"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the kVA output of the transformer.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "VA=10e03 #Power rating of the transformer(in VA)\n",
+ "V1=2300.0 #Voltage of HT winding(in Volts)\n",
+ "V2=230.0 #Voltage of LT winding(in Volts)\n",
+ "f=50.0 #Frequency rating of the transformer(in Hertz)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "I_HT=VA/V1\n",
+ "I_LT=VA/V2\n",
+ "I2=I_HT+I_LT\n",
+ "I1=I_LT\n",
+ "kVA_out=(V1*I2)/1000.0\n",
+ "VA_c=(V1*I1)\n",
+ "VA_i=V1*(I2-I1)\n",
+ "K=V1/(V1+V2)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The current distribution in the windings: load current=%.2f A and the input current is %.2f A.\" %(I2,I1)\n",
+ "print \"(b)The kVA output is %.2f kVA.\" %(kVA_out)\n",
+ "print \"(c)The volt-amperes transferred conductively is %.2f kVA\" %(VA_c/1000.0) \n",
+ "print \" The volt-amperes transferred inductively is %.2f kVA\" %(VA_i/1000.0)\n",
+ "print \"(d)The saving in copper as compared to the two-winding transformer is %.2f per cent.\" %(K*100)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The current distribution in the windings: load current=47.83 A and the input current is 43.48 A.\n",
+ "(b)The kVA output is 110.00 kVA.\n",
+ "(c)The volt-amperes transferred conductively is 100.00 kVA\n",
+ " The volt-amperes transferred inductively is 10.00 kVA\n",
+ "(d)The saving in copper as compared to the two-winding transformer is 90.91 per cent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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