{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 03 : Transformers" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.1, Page No 148" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# To determine no load power factor,core loss current and magnetising current\n", "# and no load ckt parameters of transformer\n", "\n", "Pi=50.0\n", "V1=230.0\n", "Io=2\n", "\n", "#Calculations\n", "pf=Pi/(V1*Io)\n", "print(\"No load power factor %.2f v \" %pf)\n", "Im=Io*math.sin(math.radians(math.degrees(math.acos(pf))))\n", "print(\"Magnetising current(A) %.2f v \" %Im)\n", "Ii=Io*pf\n", "print(\"core loss current(A) %.2f v \" %Ii)\n", "Gi=Pi/V1**2\n", "print(\"Gi(mho) %.2f v \" %Gi)\n", "Bm=Im/V1\n", "\n", "#Results\n", "print(\"Bm(mho) %.2f v \" %Bm)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "No load power factor 0.00 v \n", "Magnetising current(A) 2.00 v \n", "core loss current(A) 0.00 v \n", "Gi(mho) 0.00 v \n", "Bm(mho) 0.01 v \n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.2, Page No 149" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# To calculate no load current and its pf and no load power drawn from mains\n", "\n", "\n", "E=200 \n", "f=50 \n", "N1=150 # no of turns\n", "b1=.1 \n", "b2=.05 \n", "phi_max=E/(4.44*f*N1) \n", "print(\"flux(Wb)%.2f v \" %phi_max)\n", "B_max=phi_max/(b1*b2) \n", "print(\"B_max(T) = %.2f v \" %B_max)\n", "\n", "H_max=250 #According to B_max, H_max is 250AT/m\n", "l_c=.2*(3.0+3.5) #length of core\n", "AT_max=H_max*l_c \n", "print(\"AT_max = %.2f v \" %AT_max)\n", "T_max=N1 \n", "I_mmax=AT_max/T_max \n", "I_mrms=I_mmax/2**.5 \n", "print(\"I_mrms(A) = %.2f v \" %I_mrms)\n", "v=2*(20*10*5)+2*(45*10*5) \n", "d=.0079 #density of core material\n", "w=v*d \n", "cl=3 #core loss/kg\n", "closs=w*cl \n", "print(\"core loss(W) = %.2f v \" %closs)\n", "I_i=closs/E \n", "print(\"I_i(A) = %.2f v \" %I_i)\n", "def rect2polar(x,y):\n", " r=math.sqrt(x**2+y**2)\n", " return r\n", "\n", "def rect2polar2(x,y):\n", " pff=math.cos(math.radians(math.degrees(math.acos(y/x))))\n", " return pff\t\n", "\n", "I_o=rect2polar(I_i,-I_mmax) \n", "pf=rect2polar(I_i,-I_mmax) \n", "print(\"no load current(A) = %.2f v \" %I_o)\n", "print(\"no load power factor = %.2f v \" %pf)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "flux(Wb)0.01 v \n", "B_max(T) = 1.20 v \n", "AT_max = 325.00 v \n", "I_mrms(A) = 1.53 v \n", "core loss(W) = 154.05 v \n", "I_i(A) = 0.77 v \n", "no load current(A) = 2.30 v \n", "no load power factor = 2.30 v \n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.3, Page No 149" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# To calculate primary and scondary side impedences,current and their pf and real power\n", "# and calculate terminal voltage\n", "N_1=150.0 \n", "N_2=75.0 \n", "\n", "a=N_1/N_2 \n", "\n", "Z_2=[5,30] #polar(magnitude,phase diff)\n", "I_2=[0,0]\n", "I_1=[0,0]\n", "print(Z_2,'secondary impedence(ohm)') \n", "Z_1=[a**2*Z_2[0],Z_2[1]] \n", "print(Z_1,'primary impedence(ohm)') \n", "\n", "V_1=[200,0] #polar(magnitde,phase diff)\n", "V_2=[V_1[0]/a,V_1[1]] \n", "print(V_2,'secondary terminal voltage(V)') \n", "\n", "I_2[0]=V_2[0]/Z_2[0] \n", "I_2[1]=V_2[1]-Z_2[1] \n", "print(I_2,'I_2=') \n", "pf=math.cos(math.radians(I_2[1]))\n", "print(pf,'pf lagging=') \n", "\n", "I_1[0]=I_2[0]/a \n", "I_1[1]=I_2[1] \n", "print(I_1,'I_1(A)') \n", "pf=math.cos(math.radians(I_1[1]))\n", "print(pf,'pf lagging=') \n", "\n", "P_2=V_2[0]*I_2[0]*math.cos(math.radians(I_2[1])) \n", "print(P_2,'secondary power output(W)=') \n", "#P_1=primary power output\n", "P_1=P_2 #as the transormer is lossless\n", "print(P_1,'primary power output(W)=') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "([5, 30], 'secondary impedence(ohm)')\n", "([20, 30], 'primary impedence(ohm)')\n", "([100, 0], 'secondary terminal voltage(V)')\n", "([20, -30], 'I_2=')\n", "(0.8660254037844387, 'pf lagging=')\n", "([10, -30], 'I_1(A)')\n", "(0.8660254037844387, 'pf lagging=')\n", "(1732.0508075688774, 'secondary power output(W)=')\n", "(1732.0508075688774, 'primary power output(W)=')\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.4 Page No 150" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# To calculate primary current and its pf\n", "\n", "def polar2rect(r,theta):\n", "\tx=r*math.cos(math.radians(theta))\n", "\treturn x\n", "\n", "def polar2rect2(r,theta):\n", "\ty=r*math.sin(math.radians(theta))\n", "\treturn y\n", "\t\n", "def rect2polar(x,y):\n", "\tr=math.sqrt(x**2+y**2) \n", "\treturn r\n", "\n", "def rect2polar2(x,y):\n", "\ttheta=math.degrees(math.atan(y/x))\n", "\treturn theta\n", "\n", "#Calculations\n", "I_2=[10,-30] \n", "I_2r=[0,0] \n", "I_2r[0]=polar2rect(I_2[0],I_2[1]) \n", "I_2r[1]=polar2rect2(I_2[0],I_2[1]) \n", "\n", "I_0=[1.62,-71.5] \n", "I_0r=[0,0] \n", "I_0r[0]=polar2rect(I_0[0],I_0[1]) \n", "I_0r[1]=polar2rect2(I_0[0],I_0[1]) \n", "\n", "I_1r=I_0r+I_2r \n", "I_1[0]=rect2polar(I_1r[0],I_1r[1]) \n", "I_1[1]=rect2polar2(I_1r[0],I_1r[1]) \n", "print(I_1[0],'primary current(A)=') \n", "pf=math.cos(math.radians(I_1[1]))\n", "print(pf,'power factor=') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(1.62, 'primary current(A)=')\n", "(0.3173046564050921, 'power factor=')\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.5, Page No 152" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "# Equivalent circuit referred to(i)HV side (ii)LV side\n", "\n", "N_1=2000 \n", "N_2=200 \n", "\n", "a=N_1/N_2 \n", "\n", "#Calculations\n", "Z_2=p=complex(0.004,0.005) #low voltage impedence\n", "Z_2hv=a**2*Z_2 \n", "print(Z_2hv,'Z_2 referred to hv side(ohm)') #when referred to hv side\n", "\n", "Y_0=complex(0.002,-0.015) #shunt branch admittance\n", "Y_0hv=Y_0/a**2 \n", "print(Y_0hv,'Y_0 referred to hv side(mho)') \n", "\n", "Z_1=complex(0.42,-0.52) #low voltage impedence\n", "Z_1lv=Z_1/a**2 \n", "print(Z_1lv,'Z_1 referred to lv side(ohm)') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "((0.4+0.5j), 'Z_2 referred to hv side(ohm)')\n", "((2e-05-0.00015j), 'Y_0 referred to hv side(mho)')\n", "((0.0042-0.0052j), 'Z_1 referred to lv side(ohm)')\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.6 Page No 163" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "# To find the voltage at the load end of the transformer when load is drawing transformer current\n", "\n", "I=20/2 #rated load current(hv side)\n", "\n", "Z1=[.25,1.4] #impedence of feeder (REAL,IMAGINERY)\n", "Z2=[.82,1.02] #impedence of transformer (REAL,IMAGINERY)\n", "\n", "Z=Z1+Z2 \n", "print(Z,'Z(ohm)') \n", "\n", "pf=.8 \n", "phi=math.degrees(math.acos(pf))\n", "#from phasor diagram\n", " \n", "#Calculations\n", "R=Z[0]\n", "X=Z[0]\n", "AF=I*X*math.cos(math.radians(phi))\n", "FE=I*R*math.sin(math.radians(phi))\n", "AE=AF-FE \n", "OA=2000 \n", "OE=math.sqrt(OA**2-AE**2) \n", "\n", "BD=I*R*math.cos(math.radians(phi))\n", "DE=I*X*math.sin(math.radians(phi))\n", "\n", "BE=BD+DE \n", "V1=OE \n", "print(V1,'V1(V)') \n", "V2=V1-BE \n", "print(V2,'V2(V)') \n", "\n", "loadvol=V2/10 #referred to LV side\n", "print(loadvol,'load voltage(V)') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "([0.25, 1.4, 0.82, 1.02], 'Z(ohm)')\n", "(1999.999937499999, 'V1(V)')\n", "(1996.499937499999, 'V2(V)')\n", "(199.6499937499999, 'load voltage(V)')\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.7, Page No 164" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# Approx equivalent ckt referred to hv and lv sides resp,\n", "\n", "#open ckt test data with HV side open\n", "ocv=200.0 \n", "oci=4.0 \n", "ocp=120.0 \n", "#short ckt test data with LV side open\n", "scv=60.0 \n", "sci=10.0 \n", "scp=300.0 \n", "#OC test(LV side)\n", "\n", "#Calculations\n", "Y_o=oci/ocv \n", "print(\"Y_o %.4f v \" %Y_o)\n", "G_i=ocp/ocv**2 \n", "print(\"G_i %.4f v \" %G_i)\n", "B_m=math.sqrt(Y_o**2-G_i**2) \n", "print(\"B_m %.4f v \" %B_m) \n", "#SC test(HV side)\n", "Z=scv/sci \n", "print(\"Z(ohm) %.4f v \" %Z) \n", "R=scp/sci**2 \n", "print(\"R(ohm) %.4f v \" %R) \n", "X=math.sqrt(Z**2-R**2) \n", "print(\"X(ohm) %.4f v \" %X)\n", "\n", "N_H=2000 \n", "N_L=200 \n", "a=N_H/N_L #transformation ratio\n", "\n", "#Equivalent ckt referred to HV side\n", "G_iHV=G_i/a**2 \n", "print(\"G_i(HV)mho %.4f v \" %G_iHV) \n", "B_mHV=B_m/a**2 \n", "print(\"B_m(HV)mho %.4f v \" %B_mHV) \n", "\n", "#Equivalent ckt referred to LV side\n", "RLV=R/a**2 \n", "print(\"R(LV)ohm %.4f v \" %RLV) \n", "XLV=X/a**2 \n", "print(\"X(LV)ohm %.4f v \" %XLV) " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Y_o 0.0200 v \n", "G_i 0.0030 v \n", "B_m 0.0198 v \n", "Z(ohm) 6.0000 v \n", "R(ohm) 3.0000 v \n", "X(ohm) 5.1962 v \n", "G_i(HV)mho 0.0000 v \n", "B_m(HV)mho 0.0002 v \n", "R(LV)ohm 0.0300 v \n", "X(LV)ohm 0.0520 v \n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.8 Page No 165" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# to calculate (a)open ckt current,power and pf when LV excited at rated voltage\n", "# (b) voltage at which HV side is excited, ip power and its pf\n", "\n", "r=150000 #rating(VA)\n", "V1=2400.0 \n", "V2=240.0 \n", "a=V1/V2 \n", "\n", "R_1=.2 \n", "X_1=.45 \n", "R_i=10000 \n", "R_2=2*10**-3 \n", "X_2=4.5*10**-3 \n", "X_m=1600 \n", "#Referring the shunt parameters to LV side\n", "\n", "#Calculations\n", "R_iLV=R_i/a**2 \n", "X_mLV=X_m/a**2 \n", "I_oLV=[V2/100.0,V2/16.0] \n", "I_o=math.sqrt(I_oLV[0]**2+I_oLV[1]**2) \n", "print(I_o,'I_o(A)') \n", "pf=math.cos(math.radians(math.degrees(math.atan(I_oLV[1]/I_oLV[0]))))\n", "print(\"pf = %.2f v \" %pf) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(15.190786681406594, 'I_o(A)')\n", "pf = 0.16 v \n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.10 Page No 165" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "#To find exciting current and expess impedence in pu in both HV and LV sides\n", "\n", "V_BHV=2000.0 \n", "I_BHV=10.0 \n", "Z_BHV=V_BHV/I_BHV \n", "\n", "V_BLV=200.0 \n", "I_BLV=100.0 \n", "Z_BLV=V_BLV/I_BLV \n", "\n", "I_o=3.0 \n", "a=V_BHV/V_BLV \n", "\n", "\n", "#Calculations\n", "I_oLV=I_o/100 \n", "print(I_oLV,'I_o(LV)pu=') \n", "I_oHV=I_o/(a*10) \n", "print(I_oHV,'I_o(HV)pu=') \n", "\n", "Z=complex(8.2,10.2) \n", "ZHV=Z/Z_BHV \n", "print(ZHV,'Z(HV)pu=') \n", "z=Z/a**2 \n", "ZLV=z/Z_BLV \n", "print(ZLV,'Z(LV)pu=') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(0.03, 'I_o(LV)pu=')\n", "(0.03, 'I_o(HV)pu=')\n", "((0.040999999999999995+0.051j), 'Z(HV)pu=')\n", "((0.040999999999999995+0.051j), 'Z(LV)pu=')\n" ] } ], "prompt_number": 29 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.11 Page No 172" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "\n", "V_2=200 \n", "I_2=100 \n", "pf=.8 \n", "P_o=V_2*I_2*pf #power output\n", "\n", "P_i=120.0 \n", "P_c=300.0 \n", "k=1 \n", "\n", "#Calculations\n", "P_L=P_i+k**2*P_c #total losses\n", "\n", "n=1-(P_L/(P_o+P_L)) \n", "print(\"n percent = %.2f v \" %(n*100))\n", "\n", "K=math.sqrt(P_i/P_c) #max efficiency\n", "\n", "n_max=1-(2*P_i/(P_o*K+2*P_i)) #pf=.8\n", "\n", "#Results\n", "print(\"n_max percent = %.2f v \" %(n_max*100))" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "n percent = 97.44 v \n", "n_max percent = 97.68 v \n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.13, Page No 176" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# Comparing all-day efficiencies for diff given load cycles\n", "\n", "r=15.0 # kva rating\n", "n_max=.98 \n", "pf=1.0 \n", "P_o=20.0 \n", "P_i=r*(1-n_max)/2 \n", "k=r*pf/P_o \n", "P_c=P_i/(k**2) \n", "\n", "def power(P_o,h):\n", " k=P_o/20\n", " P_c=P_i*P_o/r\n", " W_o=P_o*h\n", " W_in=(P_o+P_i+(k**2)*P_c)*h\n", " return W_o,W_in\n", "\n", "#(a)full load of 20kva 12hrs/day and no load rest of the day\n", "\n", "#Calculations\n", "a=[20,12] \n", "W_oa=[0,0]\n", "W_ina=[0,0]\n", "[W_oa[0],W_ina[0]]=power(a[0],a[1]) \n", "aa=[0,12] \n", "[W_oa[1],W_ina[1]]=power(aa[0],aa[1]) \n", "print(W_oa,'W_o(kWh) for a') \n", "print(W_ina,'W_in(kWh) for a') \n", "n_ada=sum(W_oa)/sum(W_ina) \n", "print(n_ada*100,'n_allday(a) in %age') \n", "\n", "#(b)full load of 20kva 4hrs/day and .4 of full load rest of the day\n", "b=[20,4] \n", "W_ob=[0,0]\n", "W_inb=[0,0]\n", "[W_ob[0],W_inb[0]]=power(b[0],b[1]) \n", "bb=[8,20] \n", "[W_ob[1],W_inb[1]]=power(bb[0],bb[1]) \n", "print(W_ob,'W_o(kWh) for b') \n", "print(W_inb,'W_in(kWh) for b') \n", "n_adb=sum(W_ob)/sum(W_inb) \n", "\n", "#Results\n", "print(n_adb*100,'n_allday(b) in %age')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "([240, 0], 'W_o(kWh) for a')\n", "([244.2, 1.8000000000000016], 'W_in(kWh) for a')\n", "(97.5609756097561, 'n_allday(a) in %age')\n", "([80, 160], 'W_o(kWh) for b')\n", "([81.39999999999999, 163.0], 'W_in(kWh) for b')\n", "(98.19967266775778, 'n_allday(b) in %age')\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.14, Page No 186" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "\n", "# To calculate volatage regulation, volatage at load terminals and operating efficiency\n", "\n", "S=20*1000 \n", "V1=200 \n", "V2=2000 \n", "I1=S/V1 \n", "I2=S/V2 \n", "Rh=3 \n", "Xh=5.2 \n", "pf=0.8 \n", "\n", "#Calculations\n", "phi=math.degrees(math.acos(pf))\n", "Vha=V2+I2*(Rh*math.cos(math.radians(phi))+Xh*math.sin(math.radians(phi))) #lagging\n", "Vrega=(Vha-V2)*100/V2 \n", "print(Vrega,'vol-reg lagging(%)') \n", "Vhb=V2+I2*(Rh*math.cos(math.radians(phi))-Xh*math.sin(math.radians(phi))) #leading\n", "Vregb=(Vhb-V2)*100/V2 \n", "print(Vregb,'vol-reg leading(%)') \n", "V11=V2-I2*(Rh*math.cos(math.radians(phi))+Xh*math.sin(math.radians(phi))) \n", "v1=V11/I2 \n", "print(v1,'V_L(V)') \n", "ploss=120+10*10*3 \n", "pop=v1*I1*math.cos(math.radians(phi)) \n", "eff=(1-(ploss/(ploss+pop)))*100.0\n", "\n", "#Results\n", "print(eff,'eff(%)') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(2.759999999999991, 'vol-reg lagging(%)')\n", "(-0.36000000000000226, 'vol-reg leading(%)')\n", "(194.48, 'V_L(V)')\n", "(97.37145145947028, 'eff(%)')\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.15, Page No 187" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# To determine voltage regulation and efficiency\n", "r=150*1000.0 #rating in va\n", "v1=2400.0 \n", "v2=240.0 \n", "a=v2/v1 \n", "R_hv=.2+.002/a**2 \n", "X_hv=.45+.0045/a**2 \n", "I_2fl=r/v2 \n", "pf=0.8 #lagging\n", "\n", "#Calculations\n", "phi=math.degrees(math.acos(pf))\n", "I_2=I_2fl*a \n", "vd=I_2*(R_hv*math.cos(math.radians(phi))+X_hv*math.sin(math.radians(phi))) \n", "V2=v1 \n", "vr=(vd/V2)*100 \n", "print(vr,'vol reg(%)') \n", "V1=v1+vd \n", "P_out=r*pf \n", "P_c=(I_2**2)*R_hv #copper loss\n", "P_i=(V1**2)/10000 \n", "P_L=P_c+P_i \n", "n=P_out/(P_out+P_L) \n", "print(n*100,'eff(%)') \n", "I_o=[0,0]\n", "I2=[0,0]\n", "I_o[0]=V1/(10*1000) \n", "I_o[1]=-V1/(1.6*1000) #inductive effect\n", "I2[0]=I_2*(math.cos(math.radians(phi))) \n", "I2[1]=I_2*(-math.sin(math.radians(phi))) \n", "I_1=I_o+I2 \n", "b=math.sqrt(I_1[0]**2+I_1[1]**2) \n", "print(b,'I_1(A)') \n", "pff=math.cos(math.radians(math.degrees(math.atan(I_1[1]/I_1[0]))))\n", "\n", "#Results\n", "print(pff,'pf') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(2.239583333333333, 'vol reg(%)')\n", "(98.22813719947018, 'eff(%)')\n", "(1.5530997008125598, 'I_1(A)')\n", "(0.15799050110667276, 'pf')\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.16, Page No 197" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# to calculate voltage ratings,kva ratings and efficieny of autotransformer\n", "\n", "AB=200.0\n", "BC=2000.0\n", "V_1=BC \n", "print(V_1,'V_1(V)') \n", "V_2=AB+BC \n", "print(V_2,'V_2(V)') \n", "r=20000 #rating of transformer\n", "I_2=r/AB \n", "I_1=I_2+10 \n", "\n", "#Calculations\n", "rr=V_2*I_2/1000 #kva rating of autotransformer\n", "print(rr,'kva rating') \n", "ri=V_1*(I_1-I_2)/1000 #kva inductive\n", "rc=rr-ri \n", "print(ri,'kva transferred inductively') \n", "print(rc,'kva transferred conductively') \n", "W_c=120 #core loss\n", "W_cu=300 #cu loss\n", "W_t=W_c+W_cu #total loss\n", "pf=0.8 \n", "W=V_2*I_2*pf #full load output\n", "n=1-(W_t/W) \n", "\n", "#Results\n", "print(n*100,'eff(%)') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(2000.0, 'V_1(V)')\n", "(2200.0, 'V_2(V)')\n", "(220.0, 'kva rating')\n", "(20.0, 'kva transferred inductively')\n", "(200.0, 'kva transferred conductively')\n", "(99.76136363636363, 'eff(%)')\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.17, Page No 198" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# To determine the rating and full load efficiency of autotransformer\n", "\n", "#when used as transformer\n", "v1=240.0\n", "v2=120.0\n", "r=12000.0\n", "I1=r/v1 \n", "I2=r/v2 \n", "\n", "#Calculations\n", "#when connected as autotransformer\n", "V1=240.0\n", "V2=v1+v2 \n", "rr=I2*V2 \n", "print(rr,'rating of autotransformer(va)') \n", "\n", "pf=1 \n", "P_o=r*pf #output power\n", "n=.962 #efficiency at upf\n", "P_L=P_o*(1-n)/n \n", "\n", "pff=.85 #if pf=.85\n", "Po=rr*pff \n", "nn=1/(1+P_L/Po) \n", "\n", "#Results\n", "print(nn*100,'efficiency(%) at .85 pf is') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(36000.0, 'rating of autotransformer(va)')\n", "(98.4745694673036, 'efficiency(%) at .85 pf is')\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.18, Page No 198" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# To calculate sec. line voltage, line current and output va\n", "\n", "print('(a)Y/D conn') \n", "V_LY=6600 \n", "V_PY=V_LY/math.sqrt(3) \n", "a=12 \n", "V_PD=V_PY/a \n", "V_LD=V_PD \n", "print(V_LD,'sec line voltage(V)') \n", "\n", "I_PY=10 \n", "I_PD=I_PY*a \n", "I_LD=I_PD*math.sqrt(3) \n", "print(I_LD,'sec. line current(A)') \n", "r=math.sqrt(3)*V_LD*I_LD \n", "print(r,'output rating(va)') \n", "\n", "#Calculations\n", "print('(b)D/Y conn') \n", "I_LD=10 \n", "I_PD=I_LD/math.sqrt(3) \n", "I_LY=I_PD*a \n", "print(I_LY,'sec. line current(A)') \n", "V_PD=6600 \n", "V_PY=V_PD/a \n", "V_LY=V_PY*math.sqrt(3.0)\n", "\n", "#Results\n", "print(V_LY,'sec line voltage(V)') \n", "r=math.sqrt(3)*V_LY*I_LY \n", "print(r,'output rating(va)') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)Y/D conn\n", "(317.5426480542942, 'sec line voltage(V)')\n", "(207.84609690826525, 'sec. line current(A)')\n", "(114315.3532995459, 'output rating(va)')\n", "(b)D/Y conn\n", "(69.2820323027551, 'sec. line current(A)')\n", "(952.6279441628825, 'sec line voltage(V)')\n", "(114315.3532995459, 'output rating(va)')\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.19, Page No 223" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "\n", "# To compute all the currents and voltages in all windings of Y/D transformer\n", "\n", "S=complex(500,100) #load is 500MW and 100MVar\n", "s=abs(S) \n", "r=s/3 #MVA rating of each single ph transformer\n", "\n", "V1=22 #D side\n", "V2=345 #Y side\n", "a=V2/(math.sqrt(3)*V1) #voltage rating of each single phase\n", "print('Y side') \n", "V_A=(V2/math.sqrt(3))*complex(math.cos(math.radians(0)),math.sin(math.radians(0))) \n", "V_B=(V2/math.sqrt(3))*complex(math.cos(math.radians(-120)),math.sin(math.radians(-120)))\n", "V_C=(V2/math.sqrt(3))*complex(math.cos(math.radians(-240)),math.sin(math.radians(-240)))\n", "\n", "\n", "#Calculations\n", "V_AB=V_A-V_B \n", "print(V_AB,'V_AB(V)') \n", "V_BC=V_B-V_C \n", "print(V_BC,'V_BC(V)') \n", "V_CA=V_C-V_A \n", "print(V_CA,'V_CA(V)') \n", "\n", "IA=S/(3*V_A) \n", "print(IA,'IA(A)') \n", "IB=S/(3*V_B) \n", "print(IB,'IB(A)') \n", "IC=S/(3*V_C) \n", "print(IC,'IC(A)') \n", "print('D side') \n", "V_ab=V_A/a \n", "print(V_ab,'V_ab(V)') \n", "V_bc=V_B/a \n", "print(V_bc,'V_bc(V)') \n", "V_ca=V_C/a \n", "print(V_ca,'V_ca(V)') \n", "\n", "I_ab=a*IA \n", "I_bc=a*IB \n", "I_ca=a*IC \n", "Ia=I_ab-I_bc \n", "print(Ia,'Ia(A)') \n", "Ib=I_bc-I_ca \n", "\n", "#Results\n", "print(Ib,'Ib(A)') \n", "Ic=I_ca-I_ab \n", "print(Ic,'Ic(A)')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Y side\n", "((298.77876430563134+172.50000000000003j), 'V_AB(V)')\n", "((1.2789769243681803e-13-345j), 'V_BC(V)')\n", "((-298.77876430563146+172.49999999999997j), 'V_CA(V)')\n", "((0.8367395205646748+0.16734790411293496j), 'IA(A)')\n", "((-0.5632972965142212+0.6409637291029529j), 'IB(A)')\n", "((-0.2734422240504539-0.8083116332158876j), 'IC(A)')\n", "D side\n", "((22.000000000000004+0j), 'V_ab(V)')\n", "((-10.999999999999996-19.052558883257653j), 'V_bc(V)')\n", "((-11.00000000000001+19.052558883257646j), 'V_ca(V)')\n", "((12.675796066340055-4.288071240791203j), 'Ia(A)')\n", "((-2.624319405407383+13.121597027036948j), 'Ib(A)')\n", "((-10.051476660932671-8.833525786245746j), 'Ic(A)')\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.20, Page No 223" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# to find the load voltage when it draws rated current from transformer\n", "\n", "# here pu method is used\n", "r=20 #kva rating of three 1-ph transformer\n", "MVA_B=r*3/1000.0\n", "v2=2*math.sqrt(3.0) #in kv voltage base on hv side\n", "v1=.2 #in kv voltage base on lv side\n", "\n", "#Calculations\n", "z1=complex(.0004,.0015) #feeder impedence\n", "Z1=z1*MVA_B/v1**2 # lv line(pu)\n", "z2=complex(.13,.95) #load impedence\n", "Z2=z2*MVA_B/v2**2 # hv line(pu)\n", "z_T=complex(.82,1.02) \n", "ZTY=z_T*MVA_B/v2**2 # star side(pu)\n", "\n", "Ztot=Z1+Z2+ZTY \n", "V1=1 #sending end voltage [pu]\n", "I1=1 #rated current(pu)\n", "pf=.8 \n", "V2=V1-I1*((Ztot.real)*pf+(Ztot.imag)*.6) #load voltage(pu)\n", "V2v=V2*v1\n", "\n", "#Results\n", "print(V2v,'load voltage(kv)') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(0.197692, 'load voltage(kv)')\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.21, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# to calculate fault currentin feeder lines,primary and secondary lines of receiving end transformers\n", "\n", "r=60.0 #kva rating of 3-ph common base\n", "s=200.0 #kva rating of 3ph transformer\n", "#sending end\n", "X_Tse=.06*r/s #.06= reactance of transformer based on its own rating\n", "#in 2 kv feeder\n", "V_B=2000.0/math.sqrt(3) #line to neutral\n", "I_B=r*1000.0/(math.sqrt(3)*2000) \n", "Z_B=V_B/I_B \n", "X_feeder=0.7/Z_B #feeder reactance=0.7\n", "\n", "#Calculations\n", "#receiving end\n", "X_Tre=0.0051 \n", "X_tot=X_Tse+X_feeder+X_Tre \n", "V_se=20/20 \n", "I_fc=V_se/X_tot #feeder current\n", "\n", "I_f=I_fc*I_B \n", "\n", "#Results\n", "print(I_f,'current in 2kv feeder(A)') \n", "I_t1=I_f/math.sqrt(3) \n", "print(I_t1,'current in 2kv winding of transformer(A)') \n", "I_t2=I_t1*10 \n", "print(I_t2,'current in 200kv winding of transformer(A)') \n", "I_l=I_t2*math.sqrt(3) \n", "print(I_l,'current at load terminals(A)') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(515.4913117764517, 'current in 2kv feeder(A)')\n", "(297.6190476190477, 'current in 2kv winding of transformer(A)')\n", "(2976.190476190477, 'current in 200kv winding of transformer(A)')\n", "(5154.913117764517, 'current at load terminals(A)')\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.22, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# To calculate voltage and kva rating of 1-ph transformer\n", "\n", "V_p=33.0 #primary side voltage(V)\n", "V_s=11.0 #secondary side voltage(V)\n", "\n", "#Calculations\n", "V_p1=V_p/math.sqrt(3) #per ph primary side voltage(V)\n", "V_p2=V_s/math.sqrt(3) #per ph secondary side voltage(V)\n", "\n", "r=6000.0 #kva rating 3-ph\n", "s=r/3.0 #per phase\n", "\n", "#Results\n", "print('Y/Y conn') \n", "print(V_p1,'primary side ph voltage(V)') \n", "print(V_p2,'secondary side ph voltage(V)') \n", "print(s,'kva rating of transformer') \n", "\n", "print('Y/D conn') \n", "print(V_p1,'primary side ph voltage(V)') \n", "print(V_s,'secondary side ph voltage(V)') \n", "print(s,'kva rating of transformer') \n", "\n", "print('D/Y conn') \n", "print(V_p,'primary side ph voltage(V)') \n", "print(V_p2,'secondary side ph voltage(V)') \n", "print(s,'kva rating of transformer') \n", "\n", "print('D/D conn') \n", "print(V_p,'primary side ph voltage(V)') \n", "print(V_s,'secondary side ph voltage(V)') \n", "print(s,'kva rating of transformer') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Y/Y conn\n", "(19.05255888325765, 'primary side ph voltage(V)')\n", "(6.3508529610858835, 'secondary side ph voltage(V)')\n", "(2000.0, 'kva rating of transformer')\n", "Y/D conn\n", "(19.05255888325765, 'primary side ph voltage(V)')\n", "(11.0, 'secondary side ph voltage(V)')\n", "(2000.0, 'kva rating of transformer')\n", "D/Y conn\n", "(33.0, 'primary side ph voltage(V)')\n", "(6.3508529610858835, 'secondary side ph voltage(V)')\n", "(2000.0, 'kva rating of transformer')\n", "D/D conn\n", "(33.0, 'primary side ph voltage(V)')\n", "(11.0, 'secondary side ph voltage(V)')\n", "(2000.0, 'kva rating of transformer')\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.23, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# to calculate (a)reactance in ohms(b)line voltage,kva rating,series reactance for Y/Y and Y/D conn\n", "\n", "Xpu=0.12 # of 1-ph transformer\n", "\n", "def Xohm(kv,MVA):\n", "\tX=(Xpu*kv**2)/MVA\n", "\treturn X\n", "\n", "print('(a)') \n", "MVAa=75*10**-3 \n", "Vhv=6.6 \n", "Vlv=.4 \n", "\n", "#Calculations\n", "Xhv=Xohm(Vhv,MVAa) \n", "print(Xhv,'X(ohm)of hv side') \n", "Xlv=Xohm(Vlv,MVAa) \n", "print(Xlv,'X(ohm)of lv side') \n", "\n", "print('(b)') \n", "print('Y/Y') \n", "MVAb=MVAa*3 \n", "Vhv=6.6*math.sqrt(3) \n", "print(Vhv,'V_hv(kV)') \n", "Vlv=.4*math.sqrt(3) \n", "print(Vlv,'V_lv(kV)') \n", "Xhv=Xohm(Vhv,MVAb) \n", "print(Xhv,'X(ohm)of hv side') \n", "Xlv=Xohm(Vlv,MVAb) \n", "print(Xlv,'X(ohm)of lv side') \n", "\n", "print('Y/D') \n", "MVAb=MVAa*3 \n", "Vhv=6.6*math.sqrt(3) \n", "print(Vhv,'V_hv(kV)') \n", "Vlv=.4 \n", "print(Vlv,'V_lv(kV)') \n", "Xhv=Xohm(Vhv,MVAb) \n", "\n", "#Results\n", "print(Xhv,'X(ohm)of hv side') \n", "Xlv=Xohm(Vlv,MVAb) \n", "print(Xlv,'X(ohm)of lv side') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)\n", "(69.69599999999998, 'X(ohm)of hv side')\n", "(0.25600000000000006, 'X(ohm)of lv side')\n", "(b)\n", "Y/Y\n", "(11.431535329954588, 'V_hv(kV)')\n", "(0.6928203230275509, 'V_lv(kV)')\n", "(69.69599999999998, 'X(ohm)of hv side')\n", "(0.256, 'X(ohm)of lv side')\n", "Y/D\n", "(11.431535329954588, 'V_hv(kV)')\n", "(0.4, 'V_lv(kV)')\n", "(69.69599999999998, 'X(ohm)of hv side')\n", "(0.08533333333333334, 'X(ohm)of lv side')\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.24, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "#find how 2 transformers connected in parallel share the load\n", "\n", "Z1=complex(.012,.06) \n", "Z2=2*complex(.014,.045) \n", "Z=Z1+Z2 \n", "r=800 #kva rating\n", "pf=.8 \n", "\n", "#Calculations\n", "S_L=r*(complex(pf,-1*math.cos(math.radians(math.degrees(math.acos(pf)))))) \n", "S_1=S_L*Z2/Z \n", "print(S_1,'load by first transformer(kVA)') \n", "S_2=S_L*Z1/Z \n", "print(S_2,'load by second transformer(kVA)') \n", "\n", "S_2rated=300 \n", "S_Lmax=S_2rated*abs(Z)/abs(Z1) \n", "print(S_Lmax,'max load by both transformer(kVA)') \n", "\n", "r1=600 #kva\n", "V=440 \n", "Z1actual=Z1*V/(r1*1000/V) \n", "Z2actual=Z2*V/(r1*1000/V) \n", "Zactual=Z1actual+Z2actual \n", "Z_Lact=V**2/(S_L*1000) \n", "\n", "V1=445 \n", "I1=(V1*Z2actual-10*Z_Lact)/(Z1actual*Z2actual+Z_Lact*Zactual) \n", "I2=(V1*-1*Z1actual-10*Z_Lact)/(Z1actual*Z2actual+Z_Lact*Zactual) \n", "S1=V*I1/1000 \n", "print(S1,'kVA of first transformer') \n", "S2=V*I2/1000.0\n", "\n", "#Results\n", "print(S2,'kVA of second transformer') \n", "Pout=abs(S1)*math.cos(math.radians(math.degrees(math.atan((S1.imag)/(S1.real)))))+abs(S2)*math.cos(math.radians(math.degrees(math.atan((S2.imag)/(S2.real))))) \n", "print(Pout,'total output power(kW)') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "((372.3153526970954-404.18257261410787j), 'load by first transformer(kVA)')\n", "((267.6846473029046-235.81742738589213j), 'load by second transformer(kVA)')\n", "(761.1352856601269, 'max load by both transformer(kVA)')\n", "((328.8923360471516-318.290570499325j), 'kVA of first transformer')\n", "((-271.1465723386868+316.1011277711098j), 'kVA of second transformer')\n", "(600.0389083858385, 'total output power(kW)')\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.25, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "#find pu value of the equivalent ckt,steady state short ckt current and voltages\n", "\n", "r=5.0 #MVA rating\n", "V_Bp=6.35 #for primary\n", "I_Bp=r*1000/V_Bp \n", "V_Bs=1.91 #for secondary\n", "I_Bs=r*1000/V_Bs \n", "#from resp tests\n", "V1=.0787 \n", "I1=.5 \n", "V2=.1417 \n", "I2=.5 \n", "V3=.1212 \n", "I3=.5 \n", "\n", "#Calculations\n", "X12=V1/I1 \n", "X13=V2/I2 \n", "X23=V3/I3 \n", "X1=I1*(X12+X13-X23) \n", "X2=I2*(X23+X12-X13) \n", "X3=I3*(X13+X23-X12) \n", "print(X1,'X1(pu)') \n", "print(X2,'X2(pu)') \n", "print(X3,'X3(pu)') \n", "\n", "V1=1 \n", "I_sc=V1/X13 \n", "I_scp=I_sc*I_Bp \n", "print(I_scp,'sc current primary side(A)') \n", "I_sct=I_sc*r*1000.0*1000/(400/math.sqrt(3.0)) \n", "\n", "#Results\n", "print(I_sct,'sc current tertiary side(A)') \n", "V_A=I_sc*X3 \n", "V_Aact=V_A*1.91*math.sqrt(3) \n", "print(V_Aact,'V_A(actual) line to line(kV)') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(0.09919999999999998, 'X1(pu)')\n", "(0.05820000000000003, 'X2(pu)')\n", "(0.18420000000000003, 'X3(pu)')\n", "(2778.410637978651, 'sc current primary side(A)')\n", "(76396.03067964349, 'sc current tertiary side(A)')\n", "(2.1502243444618827, 'V_A(actual) line to line(kV)')\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.26, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# to calculate line currents of 3 ph side\n", "\n", "N1=6600.0 \n", "N2=100.0 \n", "a=N1/N2 \n", "b=(math.sqrt(3)/2)*a \n", "P=400.0 #kW\n", "pfa=.707 \n", "pfb=1 \n", "V=100 \n", "\n", "#Calculations\n", "Ia=P*1000/(V*pfa) \n", "Ib=P*2*1000/(V*pfb) \n", "I_A=Ia/b \n", "print(I_A,'I_A(A)') \n", "I_BC=Ib/a \n", "I_B=I_BC-49.5*complex(pfa,pfa) \n", "\n", "#Results\n", "print(abs(I_B),'I_B(A)') \n", "I_C=I_BC+49.5*complex(pfa,-1*pfa) \n", "print(abs(I_C),'I_C(A)') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(98.98423028410711, 'I_A(A)')\n", "(93.04777457436566, 'I_B(A)')\n", "(160.08088066112694, 'I_C(A)')\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.27, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "#to calculate magnitude and phase of secondary current\n", "\n", "X1=505.0 #uohm\n", "X2=551.0 #uohm\n", "R1=109.0 #uohm\n", "R2=102.0 #uohm\n", "Xm=256.0 #mohm\n", "I1=250.0 #A\n", "\n", "#Calculations\n", "I22=complex(0,Xm*1000)*I1/(complex(R1,X2+Xm*1000)) \n", "N1=250.0\n", "N2=5.0 \n", "I2=I22*(N2/N1) \n", "print(abs(I2),'current magnitude(A)') \n", "print(math.degrees(math.atan((I2.imag)/(I2.real))),'phase(degree)') \n", "print('now Rb is introduced in series') \n", "Rbb=200 #uohm\n", "Rb=(N2/N1)**2*Rbb \n", "I22=complex(0,Xm*1000)*I1/(complex((R1+Rb),X2+Xm*1000)) \n", "I2=I22*(N2/N1) \n", "\n", "#Results\n", "print(abs(I2),'current magnitude(A)') \n", "print(math.degrees(math.atan((I2.imag)/(I2.real))),'phase(degree)') \n", "print('no chnage as Rb is negligible') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(4.989260944110411, 'current magnitude(A)')\n", "(0.02434307249297878, 'phase(degree)')\n", "now Rb is introduced in series\n", "(4.989260943449164, 'current magnitude(A)')\n", "(0.024360938966050547, 'phase(degree)')\n", "no chnage as Rb is negligible\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.28, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "#to calculate sec voltage magnitude and ph\n", " \n", "a=6000.0/100 #turn ratio\n", "R1=780.0 \n", "R2=907.0 \n", "X1=975.0 \n", "X2=1075.0 \n", "Xm=443.0*1000 \n", "print('sec open') \n", "#Zb=inf \n", "V1=6500.0\n", "\n", "#Calculations\n", "V22=complex(0,Xm)*V1/complex(R1,Xm) \n", "V2=V22/a \n", "print(abs(V2),'voltage magnitude(V)') \n", "print(math.degrees(math.atan((V2.imag)/(V2.real))),'phase(deg)') \n", "\n", "print('when Zb=Rb') \n", "Rb=1 \n", "Rbb=Rb*a**2 \n", "Zm=complex(0,Xm/1000)*Rbb/complex(0,Xm/1000)+Rbb \n", "R=complex(R1/1000,X1/1000)+Zm \n", "Vm=Zm*V1/R \n", "V2=Vm/a \n", "print(abs(V2),'voltage magnitude(V)') \n", "print(math.degrees(math.atan((V2.imag)/(V2.real))),'phase(deg)') \n", "\n", "print('when Zb=jXb') \n", "Rb=complex(0,1) \n", "Rbb=Rb*a**2 \n", "Zm=complex(0,Xm/1000)*Rbb/complex(0,Xm/1000)+Rbb \n", "R=complex(R1/1000,X1/1000)+Zm \n", "Vm=Zm*V1/R \n", "V2=Vm/a \n", "\n", "#Results\n", "print(abs(V2),'voltage magnitude(V)') \n", "print(math.degrees(math.atan((V2.imag)/(V2.real))),'phase(deg)') \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "sec open\n", "(108.33316540930123, 'voltage magnitude(V)')\n", "(0.10088185516424111, 'phase(deg)')\n", "when Zb=Rb\n", "(108.32159750052864, 'voltage magnitude(V)')\n", "(-0.007757962982324285, 'phase(deg)')\n", "when Zb=jXb\n", "(108.31866454530893, 'voltage magnitude(V)')\n", "(0.006206202333075708, 'phase(deg)')\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.29, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "#to calculate L1 and L2 and coupling cofficient\n", "\n", "a=10.0 \n", "V_p=200.0 \n", "I_p=4.0 \n", "Xm=V_p/I_p \n", "f=50 \n", "\n", "#Calculations\n", "L1=Xm/(2*math.pi*f) \n", "print(L1,'L1(H)') \n", "V_s=1950.0 \n", "w_max=V_s/(math.sqrt(2)*math.pi*f) \n", "M=w_max/(math.sqrt(2)*I_p) \n", "\n", "v_s=2000 \n", "i_s=.41 \n", "w_max=math.sqrt(2)*i_s*M \n", "E1=math.sqrt(2)*math.pi*f*w_max \n", "L2=v_s/(math.sqrt(2)*math.pi*f*math.sqrt(2)*i_s) \n", "\n", "#Results\n", "print(L2,'L2(H)') \n", "k=M/(math.sqrt(L1)*math.sqrt(L2)) \n", "print(k,'coupling coeff') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(0.15915494309189535, 'L1(H)')\n", "(15.527311521160522, 'L2(H)')\n", "(0.9871122656516838, 'coupling coeff')\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.30, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "# to calculate leakage inductance, magnetisisng inductance,mutual inductance and self-inductance\n", " \n", "V1=2400.0\n", "V2=240.0 \n", "a=V1/V2 \n", "R1=.2 \n", "X1=.45 \n", "Rl=10000.0 \n", "R2=2*10**-3 \n", "X2=4.5*10**-3 \n", "Xm=1600.0 \n", "f=50.0 \n", "\n", "#Calculations\n", "l1=X1/(2*math.pi*f) \n", "print(l1,'leakage inductance ie l1(H)') \n", "l2=X2/(2*math.pi*f) \n", "print(l2,'l2(H)') \n", "Lm1=Xm/(2*math.pi*f) \n", "print(Lm1,'magnetising inductance(H)') \n", "L1=Lm1+l1 \n", "print(L1,'self-inductance ie L1(H)') \n", "M=Lm1/a \n", "L2=l2+M/a \n", "print(L2,'L2(H)') \n", "k=M/math.sqrt(L1*L2) \n", "\n", "#Results\n", "print(k,'coupling factor') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(0.001432394487827058, 'leakage inductance ie l1(H)')\n", "(1.4323944878270581e-05, 'l2(H)')\n", "(5.092958178940651, 'magnetising inductance(H)')\n", "(5.094390573428478, 'self-inductance ie L1(H)')\n", "(0.050943905734284776, 'L2(H)')\n", "(0.9997188290793214, 'coupling factor')\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.31, Page No 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "#to calculate %voltage reg and efficiency\n", "\n", "P=500000.0 \n", "V1=2200.0 \n", "V2=1100.0 \n", "V0=110.0 \n", "I0=10.0 \n", "P0=400.0 \n", "\n", "#Calculations\n", "Y0=I0/V0 \n", "Gi=P0/(V0**2) \n", "Bm=math.sqrt(Y0**2-Gi**2) \n", "Vsc=90 \n", "Isc=20.5 \n", "Psc=808 \n", "Z=Vsc/Isc \n", "R=Psc/Isc**2 \n", "X=math.sqrt(Z**2-R**2) \n", "TR=V1/V2 \n", "Gi_HV=Gi/TR**2 \n", "Bm_HV=Bm/TR**2 \n", "R_LV=R/TR**2 \n", "X_LV=X/TR**2 \n", "I2=P/V2 \n", "pf=.8 \n", "Th=math.acos(pf) \n", "dV=I2*(R_LV*math.cos(Th)+X_LV*math.sin(Th)) \n", "VR=(dV/V2)*100 \n", "print(VR,'voltage regulation(%)') \n", "Pi=P0 \n", "Pc=Psc \n", "n=P*100/(P+Pi+Pc) \n", "\n", "#Results\n", "print(n,'eff(%)') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(40.35372116523329, 'voltage regulation(%)')\n", "(99.75898229876618, 'eff(%)')\n" ] } ], "prompt_number": 22 } ], "metadata": {} } ] }