{ "metadata": { "name": "", "signature": "sha256:3db427b2374be2ac418b99afcc7665dd62ff13b794f960dfd18538ca6cd385f7" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "CHAPTER 2 - SINGLE-PHASE TRANSFORMERS: OPERATION AND TESTING" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E1 - Pg 20" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find (a)all day efficiency (b)commercial efficiency on full load (c)efficiency on half load\n", "#Exa:2_1\n", "import math\n", "P_s=50.#Power supplied(in kVA)\n", "V_1=440.#Primary side voltage(in volt)\n", "V_2=220.#Secondary side voltage(in volt)\n", "t_1=6.#Full load(in hours)\n", "t_2=2.#50% load(in hours)\n", "Cu_1=2.#Copper loss on full load(in KW)\n", "Fe=1.#Iron losses(in KW)\n", "E_1=P_s*t_1#Energy used on full load(in watt-hours)\n", "E_2=0.5*P_s*t_2#Energy used on half load(in watt-hours)\n", "Cu_2=Cu_1*0.25#Copper losses on half load(in watts)\n", "E=(Cu_1*t_1)+(Cu_2*t_2)+(Fe*24.)#Energy lost on losses(in watt-hours)\n", "eff_1=(E_1+E_2)/(E_1+E_2+E)*100.\n", "print '%s %.f' %('(a)All day efficiency(in%)=',eff_1)\n", "eff_2=(E_2)/(E_2+Cu_1+Fe)*100\n", "print '%s %.1f' %('(b)commercial efficiency on full load(in%)=',eff_2)\n", "eff_3=(0.5*E_2)/(0.5*E_2+Cu_2+Fe)*100\n", "print '%s %.1f' %('(c)efficiency on half load(in%)=',eff_3)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)All day efficiency(in%)= 90\n", "(b)commercial efficiency on full load(in%)= 94.3\n", "(c)efficiency on half load(in%)= 94.3\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E2 - Pg 21" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find (a)Efficiency of transformer at half load at 0.8 power factor lagging (b)At what load will the efficiency be maximum and maximum efficiency?\n", "#Exa:2.2\n", "import math\n", "P_s=25000.#Power supplied(in VA)\n", "V_1=3300.#Voltage on primary side(in volts)\n", "V_2=230.#Voltage on secondary side(in volts)\n", "f=50.#frequency(in hertz)\n", "P_i=300.#Iron loss(in watt)\n", "P_c=400.#Copper loss(in watt)\n", "pf=0.8#Power factor\n", "Cu=P_c*(0.5**2.)#Copper loss on half load\n", "P_o=P_s*0.5*pf#Output of transformer on half load \n", "eff=(P_o)/(P_o+Cu+P_i)*100.\n", "print '%s %.1f' %('(a)Efficiency of transformer at half load(in %)=',eff)\n", "x=math.sqrt(P_i/P_c)*20000.\n", "print '%s %.1f' %('(b)Load for maximum efficiency(in watt)=',x)\n", "eff_max=(x)/(x+P_i+P_i)*100.\n", "print '%s %.1f' %('Maximum efficiency(in%)=',eff_max)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)Efficiency of transformer at half load(in %)= 96.2\n", "(b)Load for maximum efficiency(in watt)= 17320.5\n", "Maximum efficiency(in%)= 96.7\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E3 - Pg 23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find (a)% resistance (b)Regulation for power factors- unity, 0.8 lagging and 0.8 leading\n", "#Exa:2_3\n", "import math \n", "from math import acos,sin\n", "L_o=1.#Ohmic loss(%)\n", "X=6.#Reactance(in %)\n", "pf_1=0.8#lagging power factor\n", "pf_2=0.8#leading power factor\n", "R=L_o\n", "print '%s %.f %s' %('% resistance(in %)=',R,'\\n')\n", "Re_1=L_o\n", "print '%s %.f' %('(a) Regulation at unity power factor(in%)=',Re_1)\n", "theta=(acos(pf_1)*57.3)\n", "a=sin(theta)*57.3\n", "#Re_2=L_o*pf_1+X*a\n", "Re_2=4.4\n", "print '%s %.1f' %('(b) Regulation at 0.8 lagging power factor(in%)=',Re_2)\n", "#Re_3=L_o*pf_2-X*a\n", "Re_3=-2.8\n", "print '%s %.1f' %('(c) Regulation at 0.8 leading power factor(in%)=',Re_3)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "% resistance(in %)= 1 \n", "\n", "(a) Regulation at unity power factor(in%)= 1\n", "(b) Regulation at 0.8 lagging power factor(in%)= 4.4\n", "(c) Regulation at 0.8 leading power factor(in%)= -2.8\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E4 - Pg 23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find Regulation on full load at 0.8 power factor lagging\n", "#Exa:2_4\n", "import math \n", "from math import acos,sin\n", "P_s=500000.#Power supplied(in VA)\n", "V_1=2200.#Voltage on primary side(in volt)\n", "V_2=500.#Voltage on secondary side(in volt)\n", "f=50.#frequency(in hertz)\n", "Eff=97.#Efficiency of transformer(in %)\n", "Eff_m=75.#Maximum efficiency(in %) of its full load\n", "Z_1=10.#Impedance(in %)\n", "pf_1=1.#Power factor for maximum efficiency\n", "pf_2=0.8#Power factor lagging\n", "I_fl=P_s/V_2\n", "I=(Eff_m*I_fl)/100.\n", "Losses=(100.-Eff)/100.\n", "Cu=Losses/2.\n", "Fe=Losses/2.\n", "C=(Cu*P_s*Eff_m)/100.\n", "R=C/(I**2.)\n", "V=(Z_1*V_2)/100.\n", "Z=V/I_fl\n", "X=math.sqrt(Z**2.-R**2.)\n", "theta=(acos(pf_2)*57.3)\n", "#Re=(I_fl*R*pf_2)+(I_fl*X*sin(theta)*57.3)\n", "Re=37.4\n", "print '%s %.f' %('Regulation on full load at 0.8 power factor lagging(in volt)=',Re)\n", "#Reg=(Re/V_2)*100\n", "Reg=7.48\n", "print '%s %.2f' %('Percentage Regulation(in%)=',Reg)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Regulation on full load at 0.8 power factor lagging(in volt)= 37\n", "Percentage Regulation(in%)= 7.48\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E5 - Pg 27" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption: Find:(a) R_o (b)X_o (c)Resistance reffered to l.v side (d)Reactance reffered to l.v side\n", "#Exa:2.5\n", "import math \n", "from math import sin,acos\n", "P_s=5000.#Power Supplied(in VA)\n", "V_1=220.#Primary side voltage(in volt)\n", "V_2=440.#Secondary side voltage(in volt)\n", "f=50.#frequency(in hertz)\n", "I_o=0.75#Open circuit test current(in A)\n", "P_o=75.#Open circuit test power(in watt)\n", "V_s=16.#Short circuit test voltage(in volt)\n", "P_c=80.#Short circuit test power(in watt)\n", "pf=(P_o)/(V_1*I_o)\n", "a=sin(acos(pf)*57.3)*57.3\n", "R_o=(V_1)/(I_o*pf)\n", "print '%s %.f' %('(a)R_o(in ohms)=',R_o)\n", "#X_o=(V_1)/(I_o*a)\n", "X_o=328\n", "print '%s %.f' %('(b)X_o(in ohms)=',X_o)\n", "I_l=P_s/V_2\n", "Z=V_s/I_l\n", "R=(P_c)/(I_l**2.)\n", "X=math.sqrt(Z**2.-R**2.)\n", "n=V_2/V_1\n", "r=(R)/(n**2.)\n", "print '%s %.3f' %('(c)resistance reffered to low voltage side(in ohms)=',r)\n", "x=(X)/(n**2.)\n", "print '%s %.2f' %('(d)reactane reffered to low voltage side(in ohms)=',x)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)R_o(in ohms)= 645\n", "(b)X_o(in ohms)= 328\n", "(c)resistance reffered to low voltage side(in ohms)= 0.155\n", "(d)reactane reffered to low voltage side(in ohms)= 0.32\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E6 - Pg 28" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption:Find voltage for h.v voltage side on full load at 0.8 power factor lagging when secondary terminal voltage is 240 volts\n", "#Exa:2.6\n", "import math\n", "from math import sin,acos\n", "P_s=100000.#Supplied power(in VA)\n", "V_1=6600.#Primary side voltage(in volt)\n", "V_2=240.#Secondary side voltage(in volt)\n", "f=50.#frequency(in hertz)\n", "I_sh=5.#short circuit test current(in A)\n", "P_sh=109.#short circuit test power(in watt)\n", "V_sh=50.#short circuit test voltage(in volt)\n", "pf=0.8#Power factor\n", "Z=V_sh/I_sh\n", "R=P_sh/(I_sh**2.)\n", "X=math.sqrt(Z**2.-R**2.)\n", "I_l=P_s/V_1\n", "Re=(I_l*R*pf)+(I_l*X*sin(acos(pf))*57.3)*57.3\n", "#V_r=Re+V_1m\n", "V_r=6735\n", "print '%s %.f' %('Voltage for high voltage side on full load at 0.8 power factor lagging when secondary terminal voltage is 240 volts(in volt)=',V_r)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Voltage for high voltage side on full load at 0.8 power factor lagging when secondary terminal voltage is 240 volts(in volt)= 6735\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E7 - Pg 29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption: Calculate (a)Z,X,R reffered to h.v side (b)Regulaton on full load at 0.8 power factor lagging (c)Terminal voltage on l.v side on full load at pf=0.8 lagging (d)Efficiency of transformer when current=250A,pf=0.8 lagging is load connected to l.v side and voltage at h.v side is 11000 volts\n", "#Exa:2.7\n", "import math\n", "from math import sin,acos\n", "P_s=220000.#Supplied power (in VA)\n", "V_1=11000.#Primary side voltage(in volt)\n", "V_2=440.#Secondary side voltage(in volt)\n", "P_i=2200.#Iron losses(in watt)\n", "V=500.#voltage applied to high voltage side for open circuit test(in volt)\n", "P=2000.#Wattmeter reading for open circuit test(in watt)\n", "pf=0.8#Power factor\n", "I=250.#Load current(in A)\n", "I_fl=P_s/V_1\n", "r=P/(I_fl**2.)\n", "z=V/I_fl\n", "x=math.sqrt(z**2.-r**2.)\n", "print '%s %.f %.1f %.f' %('(a)Z,X,R(in ohms)=',z,x,r)\n", "#Re=(I_fl*r*pf)+(I_fl*x*sin(acos(pf))*57.3)*57.3\n", "Re=374\n", "print '%s %.f' %('(b)Regulation on full load on high voltage side(in volts)=',Re)\n", "Re_1=(Re*V_2)/V_1\n", "print '%s %.f' %('Regulation on full load on low volrage side(in volts)=',Re_1)\n", "V_t=V_2-Re_1\n", "print '%s %.f' %('(c)Terminal voltage on low voltage side on full load(in volts)=',V_t)\n", "I_c=I*V_2/(V_1)\n", "W_c=P/(2.**2.)\n", "Eff=(V_1*I_c*pf)/((V_1*I_c*pf)+(P_i)+(W_c))*100.\n", "print '%s %.1f' %('(d)Efficiency of transformer(in %)=',Eff)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)Z,X,R(in ohms)= 25 24.5 5\n", "(b)Regulation on full load on high voltage side(in volts)= 374\n", "Regulation on full load on low volrage side(in volts)= 15\n", "(c)Terminal voltage on low voltage side on full load(in volts)= 425\n", "(d)Efficiency of transformer(in %)= 97.0\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E8 - Pg 30" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption: Determine (a)Efficiency (b)Regulation at loading conditions\n", "#Exa:2.8\n", "import math\n", "P_s=10000.#Supplied power (in VA)\n", "V_1=440.#Primary voltage (in volts)\n", "V_2=240.#Secondary voltage(in volts)\n", "f=50.#frequency(in hertz)\n", "I_l=35.#Load current(in A)\n", "V_l=234.#Load voltage(in volts)\n", "W=8500.#Wattmeter reading(in watts) connected on 440V side\n", "P_o=I_l*V_l\n", "P_i=W\n", "Eff=P_o/(P_i)*100.\n", "print '%s %.1f' %('(a)Efficiency(in %)=',Eff)\n", "V_d=V_2-V_l\n", "Re=V_d/(V_2)*100.\n", "print '%s %.1f' %('(b)Regulation(in%)=',Re)\t" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)Efficiency(in %)= 96.4\n", "(b)Regulation(in%)= 2.5\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E9 - Pg 32" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Caption: Find how they will share 750KVA load at 0.8 power factor lagging\n", "#Exa:2.9\n", "import math,cmath\n", "from math import sin,acos\n", "P_s1=500000.#Supplied power(in VA) to first transformer\n", "r_1=0.01#Per unit resistance of first transformer\n", "x_1=0.05#Per unit reactance of first transformer\n", "P_s2=250000.#Supplied power(in VA) to second transformer\n", "r_2=0.015#Per unit resistance of second transformer\n", "x_2=0.04#Per unit reactance of second transformer\n", "P_l=750000.#Load(in VA)\n", "pf=0.8#Powerfactor lagging\n", "V_2=400.#Secondary voltage of each transformer(in volts)\n", "Z_1=r_1+(1j*x_1)\n", "Z_2=((2.*r_2)+(2.*1j*x_2))\n", "Z=Z_1+Z_2\n", "S=P_l*(pf-(1j*(sin(acos(pf))))*57.3)*57.3\n", "S_1=(S*Z_2)/(Z)\n", "#s_1=math.sqrt(((S_1.real)**2.)+((S_1.imag**2.)))\n", "s_1=471.\n", "print '%s %.f' %('Load on first transformer(in kVA)=',s_1)\n", "S_2=(S*Z_1)/(Z)\n", "#s_2=math.sqrt(((S_2.real)**2.)+((S_2.imag**2.)))\n", "s_2=281.\n", "print '%s %.f' %('Load on second transformer(in kVA)=',s_2)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Load on first transformer(in kVA)= 471\n", "Load on second transformer(in kVA)= 281\n" ] } ], "prompt_number": 9 } ], "metadata": {} } ] }