diff options
Diffstat (limited to 'Electrical_Machines_-_1_by_Tarlok_Singh/Chap3.ipynb')
| -rwxr-xr-x | Electrical_Machines_-_1_by_Tarlok_Singh/Chap3.ipynb | 1687 |
1 files changed, 1687 insertions, 0 deletions
diff --git a/Electrical_Machines_-_1_by_Tarlok_Singh/Chap3.ipynb b/Electrical_Machines_-_1_by_Tarlok_Singh/Chap3.ipynb new file mode 100755 index 00000000..cd64dbc4 --- /dev/null +++ b/Electrical_Machines_-_1_by_Tarlok_Singh/Chap3.ipynb @@ -0,0 +1,1687 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter3 - Transformer(Single Phase)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.1 page 167" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import ceil\n", + "#Caption:Calculate the number of turns on both primary and secondary winding in a single phase transformer\n", + "\n", + "E_1=500 #primary induced emf(in Volts)\n", + "E_2=250 #secondary induced emf(in Volts)\n", + "F=50 #supply frequency(in Hz)\n", + "B_max=1.2 #maximum flux density(in T)\n", + "a=0.009 #cross sectional area(in square meter)\n", + "F_x=B_max*a #maximum value of flux(in Wb)\n", + "N_1=ceil(E_1/(4.44*F*F_x)) #number of primary turns\n", + "print 'number of primary turns in transformer =',N_1 \n", + "N_2=ceil(E_2/(4.44*F*F_x)) #number of secondary turns\n", + "print 'number of secondary turns in transformer =',N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "number of primary turns in transformer = 209.0\n", + "number of secondary turns in transformer = 105.0\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.2 page 167" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a loss less transformer calculate \n", + "#(a)Number of turns on high voltage side (b)The primary current (c)The secondary current\n", + "\n", + "V_1=110 #primary voltage(in Volts)\n", + "V_2=220 #secondary voltage(in Volts)\n", + "N_1=130 #low voltage side turns\n", + "N_2=N_1*V_2/V_1 #high voltage side turns\n", + "print 'Number of turns on high voltage side =',N_2 \n", + "O_p=1 #Output of the transformer in KVA\n", + "I_2=O_p*1000/V_2 #Secondary current(in Amp)\n", + "I_1=O_p*1000/V_1 #Primary current(in Amp)\n", + "print 'Primary current = %0.2f Amp'%I_1\n", + "print 'Secondary current = %0.2f Amp'%I_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of turns on high voltage side = 260.0\n", + "Primary current = 9.09 Amp\n", + "Secondary current = 4.55 Amp\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.3 Page 168" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the secondary voltage and the volts per turn\n", + "\n", + "N_1=800 #Primary turns in a transformer\n", + "N_2=200 #Secondary turns in a transformer\n", + "V_1=100 #Primary voltage\n", + "V_2=V_1*(N_2/N_1) #Secondary voltage\n", + "print 'Secondary voltage = %0.2f Volts '%V_2 \n", + "V_n=V_1/N_1 #Volts per turn\n", + "print 'Volts per turn =',V_n " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Secondary voltage = 25.00 Volts \n", + "Volts per turn = 0.125\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.4 Page 169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the primary current in a single phase transformer\n", + "\n", + "V_1=230 #Primary voltage(in Volts)\n", + "V_2=460 #Secondary voltage(in Volts)\n", + "I_2=10 #secondary current(in Amp)\n", + "I_1=I_2*(V_2/V_1) #Primary current(in Amp)\n", + "print 'Primary current = %0.2f Amp'%I_1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Primary current = 20.00 Amp\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.5 Page 169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate (a)the approximate values of the primary and secondary currents (b)the approimate number of primary turns\n", + "\n", + "V_1=6600 #Primary voltage(in Volts)\n", + "V_2=400 #Secondary voltage(in Volts)\n", + "N_2=100 #secondary turns\n", + "O_p=200 #output of a transformer in KVA\n", + "I_1=O_p*1000/V_1 #Full load primary current(in Amp)\n", + "print 'Full load primary current = %0.1f Amp'%I_1 \n", + "I_2=O_p*1000/V_2 #Full load secondary current(in Amp)\n", + "print 'Full load secondary current = %0.f Amp '%I_2 \n", + "N_1=N_2*(V_1/V_2) #Number of primary turns\n", + "print 'Number of primary turns =',N_1 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Full load primary current = 30.3 Amp\n", + "Full load secondary current = 500 Amp \n", + "Number of primary turns = 1650.0\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.6 page 169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Find voltage in secondary terminal\n", + "V_1=100 #Primary voltage(in Volts)\n", + "N_1=20 #Number of primary turns\n", + "N_2=40 #Number of secondary turns\n", + "V_2=(N_2/N_1)*V_1 #voltage of the secondary terminal(in Volts)\n", + "print 'voltage of the secondary terminal = %0.2f V'%V_2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage of the secondary terminal = 200.00 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.7 page 170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the r.m.s value of the induced emf in the secondary coil.\n", + "\n", + "F_x=0.02 #Maximum value of flux(in Wb)\n", + "N_2=55 #Number of secondary turns\n", + "F=50 #Supply frequency(in Hz)\n", + "E_2=4.44*F*F_x*N_2 #r.m.s value of induced emf in the secondary (in Volts)\n", + "print 'r.m.s value of induced emf in the secondary = %0.1f Volts '%E_2 \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "r.m.s value of induced emf in the secondary = 244.2 Volts \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.8 page 170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In single phase transformer Calculate (a)The maximum flux density in the core and (b)induced emf in the secondary\n", + "\n", + "N_1=80 #Primary turns\n", + "N_2=240 #secondary turns\n", + "f=50 #Supply frequency(in Hz)\n", + "E_1=240 #Supply voltage(in Volts)\n", + "F_max=E_1/(4.44*f*N_1) #Maximum value of the flux in the core\n", + "a=200 #Cross sectional area of core(in cm**2)\n", + "A=a*10**-4 #Cross sectional area of core(in m**2)\n", + "B_max=F_max/A #Peak value of flux density in the core(in T)\n", + "print 'Peak value of flux density in the core = %0.4f T '%B_max \n", + "E_2=E_1*(N_2/N_1) #Induced emf in the secondary winding(in Volts)\n", + "print 'Induced emf in the secondary winding = %0.f Volts '%E_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak value of flux density in the core = 0.6757 T \n", + "Induced emf in the secondary winding = 720 Volts \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.9 page 171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate (a)Primary and secondary currents on full load (b)the maximum value of flux (c)the number of primary turns.\n", + "\n", + "O_p=200 #Rated output (in KVA)\n", + "V_1=3300 #Primary voltage (in Volts)\n", + "V_2=240 #Secondary voltage (in Volts)\n", + "N_2=100 #Secondary turns\n", + "f=50 #supply frequency(in Hz)\n", + "I_1=O_p*1000/V_1 #Primary current(in Amp)\n", + "print 'Primary current on full load = %0.2f Amp'%I_1 \n", + "I_2=O_p*1000/V_2 #secondary current(in Amp)\n", + "print 'secondary current on full load = %0.2f Amp '%I_2 \n", + "F_x=V_2/(4.44*f*N_2) #Maximum value of flux(in Wb)\n", + "print 'Maximum value of flux = %0.4f Wb '%F_x \n", + "N_1=N_2*(V_1/V_2) #Primary turns\n", + "print 'Primary turns =',N_1 \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Primary current on full load = 60.61 Amp\n", + "secondary current on full load = 833.33 Amp \n", + "Maximum value of flux = 0.0108 Wb \n", + "Primary turns = 1375.0\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.10 page 175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer primary side is open , Find (a)core loss, (b)loss component of current ,(c)Magnetising current.\n", + "\n", + "\n", + "O_p=50 #output (in KVA)\n", + "V_2=230 #Secondary voltage(in Volts)\n", + "P=187 #meter's power reading (in watt)\n", + "I_w=P/V_2 #Loss component of current(in Amp)\n", + "I_o=6.5 #meter's current reading (in Amp)\n", + "I_m=((I_o)**2-(I_w)**2) #Magnetising current(in Amp)\n", + "P=V_2*I_w #Core loss(in watt)\n", + "print 'Core loss = %0.2f Watt '%P \n", + "print 'Loss component of current = %0.3f Amp '%I_w \n", + "print 'Magnetising current = %0.2f Amp'%I_m " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Core loss = 187.00 Watt \n", + "Loss component of current = 0.813 Amp \n", + "Magnetising current = 41.59 Amp\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.11 page 175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, degrees, acos, sin, cos\n", + "#Caption:In a transformer Calculate \n", + "#(a)Magnetising component of no load current (b)the iron loss (c)the maximum value of flux in the core.\n", + "\n", + "V_1=460 #supply voltage(in Volts)\n", + "I_o=15 #No load current (in Amp)\n", + "p=degrees(acos(0.2)) #power angle(when power factor is 0.2)\n", + "n=sin(p*pi/180) \n", + "I_m=I_o*n #Magnetising component of no load current \n", + "print 'Magnetising component of no load current = %0.2f Amp '%I_m \n", + "L_ir=V_1*I_o*cos(p*pi/180)/1000 #the iron loss in transformer(in kWatt)\n", + "print 'the iron loss in transformer = %0.2f kW '%L_ir \n", + "E_1=V_1 #at no load condition\n", + "N_1=550 #primary winding\n", + "f=50 #supply frequency (in Hz)\n", + "F_m=E_1/(4.44*N_1*f) #the maximum value of flux in the core(in Wb)\n", + "print 'the maximum value of flux in the core = %0.4f Wb'%F_m " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetising component of no load current = 14.70 Amp \n", + "the iron loss in transformer = 1.38 kW \n", + "the maximum value of flux in the core = 0.0038 Wb\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.12 page 176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer, Find (a)The magnetising current (b)The iron loss current \n", + "\n", + "p=degrees(acos(0.22)) #power angle(when power factor is 0.22)\n", + "I_o=0.3 #no load current(in Amp)\n", + "I_m=I_o*sin(p*pi/180) #Magnetising current(in Amp)\n", + "print 'Magnetising current = %0.4f Amp'%I_m \n", + "I_w=I_o*cos(p*pi/180) #Iron loss current (in Amp)\n", + "print 'Iron loss current = %0.3f Amp '%I_w " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetising current = 0.2926 Amp\n", + "Iron loss current = 0.066 Amp \n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.13 page 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Find the active and reactive components of a single phase transformer.\n", + "\n", + "V_1=440 #supply Voltage(in Volts)\n", + "p=degrees(acos(0.3) )#power angle when power factor is 0.3\n", + "P_o=80 #power input to the hv winding(in Watt)\n", + "I_o=P_o/(V_1*cos(p*pi/180)) #No load current (in Amp)\n", + "I_w=I_o*cos(p*pi/180) #Active component of no load current (in Amp)\n", + "print 'Active component of no load current = %0.3f Amp '%I_w \n", + "I_m=I_o*sin(p*pi/180) #Reactive component of no load current (in Amp)\n", + "print 'Reactive component of no load current = %0.3f Amp '%I_m " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Active component of no load current = 0.182 Amp \n", + "Reactive component of no load current = 0.578 Amp \n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.14 page 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the current taken by the primary winding.\n", + "\n", + "V_1=6600 #Primary voltage(in Volts)\n", + "V_2=400 #Secondary voltage(in Volts)\n", + "I_o=0.7 #NO load current(in Amp)\n", + "p_o=0.24 #No load power factor\n", + "q_o=degrees(acos(p_o)) #power angle when no load power factor is 0.24\n", + "I_2=100 #Secondary current(in Amp)\n", + "p_2=0.8 #Secondary power factor\n", + "q_2=degrees(acos(p_2)) #power angle when secondary power factor is 0.8\n", + "K=V_2/V_1 #ratio of primary to secondary voltage\n", + "I_1=K*I_2 #primary current(in Amp)\n", + "Q=q_o-q_2 #resultant power angle\n", + "I=((I_o)**2 + (I_1)**2+2*I_o*I_1*cos(Q*pi/180))**(1/2) #Resultant current taken by the primary(in Amp)\n", + "print 'Resultant current taken by the primary = %0.2f Amp '%I " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resultant current taken by the primary = 6.62 Amp \n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.15 page 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the current taken by the primary winding.\n", + "\n", + "V_1=440 #Primary voltage(in Volts)\n", + "V_2=110 #Secondary voltage(in Volts)\n", + "I_o=5 #NO load current(in Amp)\n", + "p_o=0.2 #No load power factor\n", + "q_o=degrees(acos(p_o)) #power angle when no load power factor is 0.24\n", + "I_2=120 #Secondary current(in Amp)\n", + "p_2=0.8 #Secondary power factor\n", + "q_2=degrees(acos(p_2)) #power angle when secondary power factor is 0.8\n", + "K=V_2/V_1 #ratio of primary to secondary voltage\n", + "I_1=K*I_2 #primary current(in Amp)\n", + "Q=q_o-q_2 #resultant power angle\n", + "I=((I_o)**2 + (I_1)**2+2*I_o*I_1*cos(Q*pi/180))**(1/2) #Resultant current taken by the primary(in Amp)\n", + "print 'Resultant current taken by the primary = %0.2f Amp '%I " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resultant current taken by the primary = 33.90 Amp \n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.16 page 187" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate Equivalent resistance \n", + "\n", + "V_1=4400 #Primary voltage (in Volts)\n", + "V_2=220 #Secondary voltage (in Volts)\n", + "R_1=3.45 #Primary resistance (in Ohm)\n", + "X_1=5.2 #Primary reactances (in Ohm)\n", + "R_2=0.009 #secondary resistance (in Ohm)\n", + "X_2=0.015 #secondary reactance (in Ohm)\n", + "K=V_2/V_1 #voltage ratio\n", + "R_o1=R_1+(R_2/K**2) #Equivalent resistance as referred to primary(in Ohm)\n", + "print 'Equivalent resistance as referred to primary = %0.2f Ohm '%R_o1 \n", + "R_o2=R_2+(R_1*K**2) #Equivalent resistance as referred to secondary(in Ohm)\n", + "print 'Equivalent resistance as referred to secondary = %0.4f Ohm '%R_o2 \n", + "X_o1=X_1+(X_2/K**2) #Equivalent reactance as referred to primary(in Ohm)\n", + "print 'Equivalent reactance as referred to primary = %0.2f Ohm'%X_o1 \n", + "X_o2=X_2+X_1*(K**2) #Equivalent reactance as referred to secondary(in Ohm)\n", + "print 'Equivalent reactance as referred to secondary = %0.3f Ohm '%X_o2 \n", + "Z_o1=((R_o1)**2 + (X_o1)**2)**(1/2) #Equivalent impedence as referred to primary(in Ohm)\n", + "print 'Equivalent impedence as referred to primary = %0.2f Ohm'%Z_o1 \n", + "Z_o2=((R_o2)**2 + (X_o2)**2)**(1/2) #Equivalent impedence as referred to secondary(in Ohm)\n", + "print 'Equivalent impedence as referred to secondary = %0.3f Ohm'%Z_o2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent resistance as referred to primary = 7.05 Ohm \n", + "Equivalent resistance as referred to secondary = 0.0176 Ohm \n", + "Equivalent reactance as referred to primary = 11.20 Ohm\n", + "Equivalent reactance as referred to secondary = 0.028 Ohm \n", + "Equivalent impedence as referred to primary = 13.23 Ohm\n", + "Equivalent impedence as referred to secondary = 0.033 Ohm\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.17 page 190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the secondary terminal voltage at full load.\n", + "\n", + "V_1=2000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=400 #Secondary voltage at no load (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=5 #Primary resistance(in Ohm)\n", + "R_2=0.2 #Secondary resistance(in Ohm)\n", + "X_1=12 #Primary reactance(in Ohm)\n", + "X_2=0.48 #Secondary reactance(in Ohm)\n", + "R_o2=R_2 + (K**2)*R_1 #Equivalent resistance as referred to secondary(in Ohm)\n", + "X_o2=X_2 + (K**2)*X_1 #Equivalent reactance as referred to secondary(in Ohm)\n", + "O_p=10 #Rated output(i KVA)\n", + "I_2=O_p*1000/V_2 #Full load secondary current(in Amp)\n", + "p=0.8 #power factor\n", + "a=degrees(acos(p)) #power angle\n", + "V_d=I_2*R_o2*cos(a*pi/180) + I_2*X_o2*sin(a*pi/180) #Voltage drop (in Volts)\n", + "V=V_2-V_d #Secondary terminal voltage at full load (in Volts)\n", + "print 'Secondary terminal voltage at full load = %0.2f Volts '%V " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Secondary terminal voltage at full load = 377.60 Volts \n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.18 page 191" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the secondary terminal voltage at full load.\n", + "\n", + "V_1=11000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=400 #Secondary voltage at no load (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=4 #Primary resistance(in Ohm)\n", + "R_2=0.2 #Secondary resistance(in Ohm)\n", + "X_1=10 #Primary reactance(in Ohm)\n", + "X_2=0.4 #Secondary reactance(in Ohm)\n", + "R_o2=R_2 + (K**2)*R_1 #Equivalent resistance as referred to secondary(in Ohm)\n", + "X_o2=X_2 + (K**2)*X_1 #Equivalent reactance as referred to secondary(in Ohm)\n", + "O_p=10 #Rated output(in KVA)\n", + "I_2=O_p*1000/V_2 #Full load secondary current(in Amp)\n", + "Z_o2=((R_o2)**2 +(X_o2)**2)**(1/2) #Equivalent impedance as referred to secondary(in Ohm)\n", + "V=V_2-I_2*Z_o2 #Secondary terminal voltage at full load (in Volts)\n", + "print 'Secondary terminal voltage at full load = %0.3f Volts '%V " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Secondary terminal voltage at full load = 388.465 Volts \n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.19 page 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Determine equivalent resistance \n", + "\n", + "V_1=2000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=400 #Secondary voltage at no load (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=5.2 #Primary resistance(in Ohm)\n", + "R_2=0.2 #Secondary resistance(in Ohm)\n", + "X_1=12.5 #Primary reactance(in Ohm)\n", + "X_2=0.5 #Secondary reactance(in Ohm)\n", + "R_o2=R_2 + (K**2)*R_1 #Equivalent resistance as referred to secondary(in Ohm)\n", + "print 'Equivalent resistance as referred to secondary = %0.2f Ohm '%R_o2 \n", + "X_o2=X_2 + (K**2)*X_1 #Equivalent reactance as referred to secondary(in Ohm)\n", + "print 'Equivalent reactance as referred to secondary = %0.2f Ohm '%X_o2 \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent resistance as referred to secondary = 0.41 Ohm \n", + "Equivalent reactance as referred to secondary = 1.00 Ohm \n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.20 page 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine equivalent resistance and reactance \n", + "\n", + "V_1=2000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=200 #Secondary voltage at no load (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=2 #Primary resistance(in Ohm)\n", + "R_2=0.025 #Secondary resistance(in Ohm)\n", + "X_1=4 #Primary reactance(in Ohm)\n", + "X_2=0.04 #Secondary reactance(in Ohm)\n", + "R_eq12=(K**2)*R_1 #equivalent resistance of primary referred to secondary(in Ohm)\n", + "print 'equivalent resistance of primary referred to secondary = %0.2f Ohm '%R_eq12 \n", + "X_eq12=(K**2)*X_1 #equivalent reactance of primary referred to secondary(in Ohm)\n", + "print 'equivalent reactance of primary referred to secondary = %0.2f Ohm '%X_eq12 \n", + "R_e2=(K**2)*R_1 + R_2 #total resistance of primary referred to secondary(in Ohm)\n", + "print 'total resistance of primary referred to secondary = %0.3f Ohm '%R_e2 \n", + "X_e2=(K**2)*X_1 + X_2 #total reactance of primary referred to secondary(in Ohm)\n", + "print 'total reactance of primary referred to secondary = %0.3f Ohm '%X_e2 \n", + "R_eq21=R_2/(K**2) #equivalent resistance of secondary referred to primar(in Ohm)\n", + "print 'equivalent resistance of secondary referred to primary = %0.2f Ohm '%R_eq21 \n", + "X_eq21=X_2/(K**2) #equivalent reactance of secondary referred to primar(in Ohm)\n", + "print 'equivalent reactance of secondary referred to primary = %0.2f Ohm '%X_eq21 \n", + "R_e1=R_1 + R_2/(K**2) #total resistance of secondary referred to primary(in Ohm)\n", + "print 'total resistance of secondary referred to primary = %0.2f Ohm '%R_e1 \n", + "X_e1=X_1 + X_2/(K**2) #total reactance of secondary referred to primary(in Ohm)\n", + "print 'total reactance of secondary referred to primary = %0.2f Ohm '%X_e1 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "equivalent resistance of primary referred to secondary = 0.02 Ohm \n", + "equivalent reactance of primary referred to secondary = 0.04 Ohm \n", + "total resistance of primary referred to secondary = 0.045 Ohm \n", + "total reactance of primary referred to secondary = 0.080 Ohm \n", + "equivalent resistance of secondary referred to primary = 2.50 Ohm \n", + "equivalent reactance of secondary referred to primary = 4.00 Ohm \n", + "total resistance of secondary referred to primary = 4.50 Ohm \n", + "total reactance of secondary referred to primary = 8.00 Ohm \n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.21 Page 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine Primary resistance and total resistance of transformer referred to primary.\n", + "\n", + "V_1=2000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=220 #Secondary voltage at no load (in Volts)\n", + "R_1=1.06 #Primary resistance(in Ohm)\n", + "R_2=0.013 #Secondary resistance(in Ohm)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_eq1=(K**2)*R_1 #Primary resistance referred to secondary(in Ohm)\n", + "print 'Primary resistance referred to secondary = %0.4f Ohm '%R_eq1 \n", + "R_eq2=R_2/(K**2) #secondary resistance referred to primary(in Ohm)\n", + "print 'secondary resistance referred to primary = %0.4f Ohm '%R_eq2 \n", + "R_e1=R_1 + R_eq2 #total resistance of transformer referred to primary(in Ohm)\n", + "print 'total resistance of transformer referred to primary = %0.4f Ohm '%R_e1 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Primary resistance referred to secondary = 0.0128 Ohm \n", + "secondary resistance referred to primary = 1.0744 Ohm \n", + "total resistance of transformer referred to primary = 2.1344 Ohm \n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.22 page 200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Find Full load regulation at a power factor (a) 0.8 lagging (b) unity (c) 0.8 leading.\n", + "\n", + "V_1=6600 #Primary voltage (in Volts)\n", + "V_2=250 #Secondary voltage (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=10 #Primary resistance (in Ohm)\n", + "R_2=0.02 #Secondary resistance (in Ohm)\n", + "X_o1=35 #Total leakage reactance referred to the primary winding(in Ohm)\n", + "R_o2=R_2 + R_1*(K**2) #Equivalent resistance reffered to the secondary(in Ohm)\n", + "X_o2=X_o1*(K**2) #Equivalent reactance reffered to the secondary(in Ohm)\n", + "O_p=40 #Rated output (in KVA)\n", + "I_2=O_p*1000/250 #Secondary current(in Amp)\n", + "P_1=0.8 #Power factor\n", + "q=(acos(P_1)) \n", + "V_r1=((I_2*R_o2*cos(q) + I_2*X_o2*sin(q))/V_2)*100 #Voltage regulation at 0.8 lagging power factor(in %)\n", + "print 'Voltage regulation at 0.8 lagging power factor = %0.3f %%'%V_r1 \n", + "V_r0=(I_2*R_o2/V_2)*100 #Voltage regulation at unity power factor(in %)\n", + "print 'Voltage regulation at unity power factor = %0.2f %%'%V_r0 \n", + "V_r2=((I_2*R_o2*cos(q) - I_2*X_o2*sin(q))/V_2)*100 #Voltage regulation at 0.8 leading power factor(in %)\n", + "print 'Voltage regulation at 0.8 leading power factor = %0.2f %%'%V_r2 \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage regulation at 0.8 lagging power factor = 3.687 %\n", + "Voltage regulation at unity power factor = 2.20 %\n", + "Voltage regulation at 0.8 leading power factor = -0.17 %\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.23 page 201" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the regulation of transformer\n", + "\n", + "#(I_2*R_a2/E_2)*100=1\n", + "Rs_d=1 #Percentage resistive drop\n", + "#(I_2*X_a2/E_2)*100=5\n", + "Re_d=5 #Percentage reactive drop\n", + "#power factor=cosd(q1)=0.8 (lagging)\n", + "q1=degrees(acos(0.8) )\n", + "#Voltage regulation= ((I_2*R_a2*cosd(q1)+I_2*X_a2*sind(q1))/100)*100\n", + "#V_r=(I_2*R_a2/E_2)*100*cosd(q1)+(I_2*X_a2/E_2)*100*sind(q1)\n", + "V_r1=Rs_d*cos(q1*pi/180)+Re_d*sin(q1*pi/180) #Voltage regulation when power factor is 0.8 lagging\n", + "print 'Voltage regulation when power factor is 0.8 lagging = %0.2f %%'%V_r1 \n", + "q2=-degrees(acos(0.8) )\n", + "#V_r2=(I_2*R_a2/E_2)*100*cosd(q2)+(I_2*X_a2/E_2)*100*sind(q2) #Voltage regulation when power factor is 0.8 leading\n", + "V_r2=Rs_d*cos(q2*pi/180)+Re_d*sin(q2*pi/180) #Voltage regulation when power factor is 0.8 leading\n", + "print 'Voltage regulation when power factor is 0.8 leading = %0.2f %%'%V_r2," + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage regulation when power factor is 0.8 lagging = 3.80 %\n", + "Voltage regulation when power factor is 0.8 leading = -2.20 %\n" + ] + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.25 page 208" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the efficiency at unity power at Full load and Half load.\n", + "\n", + "P_c=400 #Full load copper loss(in Watts)\n", + "P_C=P_c/1000 #Full load copper loss(in KW)\n", + "P_i=350 #Full load iron loss(in Watts)\n", + "P_I=P_i/1000 #Full load iron loss(in KW)\n", + "P_f=1 #Power factor\n", + "KVA=25 #Rating of the transformer\n", + "O_p=KVA*P_f #Output at full load condition(in KW)\n", + "L_1=P_C+P_I #Losses at full load condition(in KW)\n", + "I_p=O_p+L_1 #Input at full load condition(in KW)\n", + "E_fL=(O_p/I_p)*100 #Efficiency in full load condition\n", + "print 'Efficiency of the tranformer at full load condition = %0.2f %%'%E_fL \n", + "#At half load condition\n", + "O_P=(1/2)*KVA*P_f #Output of the transformer at half load condition\n", + "L_2=((1/2)**2)*P_C+P_I #Losses at half load condition(in KW)\n", + "I_P=O_P+L_2 #Input at half load condition\n", + "E_hL=(O_P/I_P)*100 #Efficiency of the transformer at half load condition(in %)\n", + "print 'Efficiency of the transformer at half load condition = %0.2f %%'%E_hL " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the tranformer at full load condition = 97.09 %\n", + "Efficiency of the transformer at half load condition = 96.53 %\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.26 page 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the single phase transformer efficiency for 75% of the full load output at power factor unity and 0.8 lagging.\n", + "\n", + "P_i=1.5 #Core loss(in KW)\n", + "P_c=4.5 #Full load copper loss(in KW)\n", + "P_C=((3/4)**2)*P_c #Copper loss at 75% of full load(in KW)\n", + "P_t=P_i+P_C #Total loss at 75% of full load output(in KW)\n", + "KVA=300 #Rating of the transformer(in KVA)\n", + "P_f1=1 #power factor value when it is unity\n", + "P_f2=0.8 #power factor value when it is 0.8 lagging\n", + "O_p1=0.75*KVA*P_f1 #Output at 75% of full load and at unity power factor(in KW)\n", + "E_f1=(O_p1/(O_p1+P_t))*100 #Efficiency of the transformer for 75% of full load output at power factor unity(in %)\n", + "print 'Efficiency of the transformer for 75%% of full load output at power factor unity = %0.2f %%'%E_f1 \n", + "O_p2=0.75*KVA*P_f2 #Output at 75% of full load and at 0.8 lagging power factor(in KW)\n", + "E_f2=(O_p2/(O_p2+P_t))*100 #Efficiency of the transformer for 75% of full load output at power factor 0.8 lagging(in %)\n", + "print 'Efficiency of the transformer for 75%% of full load output at power factor 0.8 lagging = %0.2f %%' %E_f2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer for 75% of full load output at power factor unity = 98.24 %\n", + "Efficiency of the transformer for 75% of full load output at power factor 0.8 lagging = 97.81 %\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.27 page 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate efficiency at full load, unity power factor. \n", + "\n", + "P_f1=1 #power factor unity\n", + "P_f2=0.8 #power factor 0.8 lagging or leading\n", + "KVA=25 #Rating of the transformer(in KVA)\n", + "O_p1=KVA*P_f1 #Output at unity power factor(in KW)\n", + "P_c=400 #copper losses(in Watt)\n", + "P_C=P_c/1000 #copper losses(in KW)\n", + "P_i=320 #iron losses(in Watt)\n", + "P_I=P_i/1000 #iron losses(in KW)\n", + "P_T=P_I+P_C #total losses(in KW)\n", + "I_p1=O_p1+P_T #Input (in KW)\n", + "E_f1=(O_p1/I_p1)*100 #Efficiency of the transformer at full load and unity power factor(in %)\n", + "print 'Efficiency of the transformer at full load and unity power factor = %0.2f %%' %E_f1\n", + "O_p2=KVA*P_f2 #output at 0.8 lagging power factor(in KW)\n", + "I_p2=O_p2+P_T #input incase of 0.8 power factor(in KW)\n", + "E_f2=(O_p2/I_p2)*100 #Efficiency of the transformer at full load and 0.8 lagging power factor(in %)\n", + "print 'Efficiency of the transformer at full load and 0.8 lagging power factor = %0.2f %%' %E_f2\n", + "E_f3=E_f2 #At 0.8 leading power factor. since there is no change in input and output, so efficiency is unchanged\n", + "print 'At 0.8 leading power factor. since there is no change in input and output, so efficiency is unchanged = %0.2f %%' %E_f3" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer at full load and unity power factor = 97.20 %\n", + "Efficiency of the transformer at full load and 0.8 lagging power factor = 96.53 %\n", + "At 0.8 leading power factor. since there is no change in input and output, so efficiency is unchanged = 96.53 %\n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.28 page 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the efficiency of the transformer on Half load and Full load.\n", + "KVA=100 #rating of the(in KVA)\n", + "P_f=0.8 #power factor\n", + "O_p=(1/2)*KVA*P_f #Output at half load(in KW)\n", + "P_i=700 #iron loss (in Watt)\n", + "P_i1=P_i/1000 #iron loss at half and full load(in KW)\n", + "P_c=400 #copper losses (in Watt)\n", + "P_c1=((1/2)**2)*P_c/1000 #copper losses at half load condition (in KW)\n", + "P_t=P_c1+P_i1 #Total losses in half load condition(in KW)\n", + "E_f=(O_p/(O_p+P_t))*100 #Efficiency of the transformer on half load in percentage\n", + "print 'Efficiency of the transformer on half load = %0.2f %%' %E_f\n", + "O_P=KVA*P_f #Output in case of full load(in KW)\n", + "P_c2=P_c/1000 #Copper losss at full load condition(in KW)\n", + "P_T=P_c2+P_i1 #Total losses in full load condition(in KW)\n", + "E_F=(O_P/(O_P+P_T))*100 #Efficiency of the transformer on full load condition in percentage\n", + "print 'Efficiency of the transformer on full load condition = %0.2f %%' %E_F" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer on half load = 98.04 %\n", + "Efficiency of the transformer on full load condition = 98.64 %\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.29 page 213" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the efficiency at full load unity power factor\n", + "\n", + "KVA=25 #Rating of the transformer(in KVA)\n", + "P_c=400 #Full load copper loss(in Watt)\n", + "P_c1=P_c/1000 #Full load copper loss(in KW)\n", + "P_i=350 #Iron loss(in Watt)\n", + "P_i1=P_i/1000 #Iron loss (in KW)\n", + "P_f=1 #Power factor unity\n", + "P_f1=0.8 #Power factor 0.8 lagging\n", + "O_p1=KVA*P_f #Output at full load and unity power factor(in KW)\n", + "P_t1=P_c1+P_i1 #Total losses at full load and unity power factor(in KW)\n", + "I_p1=O_p1+P_t1 #Input at full load and unity power factor(in KW)\n", + "E_f1=(O_p1/I_p1)*100 #Efficiency of the transformer at full load and unity power factor(in KW)\n", + "print 'Efficiency of the transformer at full load and unity power factor = %0.2f kW'%E_f1 \n", + "O_p2=(1/2)*KVA*P_f1 #Output At half full load, 0.8 power factor lag.(in KW)\n", + "P_c2=((1/2)**2)*P_c1 #Copper loss At half full load, 0.8 power factor lag.(in KW)\n", + "P_t2=P_c2+P_i1 #Total loss At half full load, 0.8 power factor lag.(in KW)\n", + "I_p2=O_p2+P_t2 #Input At half full load, 0.8 power factor lag.(in KW)\n", + "E_f2=(O_p2/I_p2)*100 #Efficiency of the transformer at half full load and 0.8 lagging power factor(in KW)\n", + "print 'Efficiency of the transformer at half full load and 0.8 lagging power factor = %0.2f kW'%E_f2 \n", + "#Maximum efficiency occurs when Copper loss = Iron loss,let the maximum efficiency occcur at x times full load\n", + "x=(P_i/P_c)**(1/2) \n", + "L=x*KVA #load in KVA corresponding to maximum efficiency\n", + "print 'the load for maximum efficiency = %0.2f kVA '%L " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer at full load and unity power factor = 97.09 kW\n", + "Efficiency of the transformer at half full load and 0.8 lagging power factor = 95.69 kW\n", + "the load for maximum efficiency = 23.39 kVA \n" + ] + } + ], + "prompt_number": 61 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.30 page 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the efficiency of full load current\n", + "\n", + "KVA=50 #Rating of the transformer(in KVA)\n", + "V_1=6600 #Primary voltage(in Volt)\n", + "V_2=200 #Secondary voltage(in VOlt)\n", + "I_1=KVA*1000/6600 #Full load primary current(in Amp)\n", + "P_f1=1 #power factor at unity\n", + "P_f2=0.8 #Power factor at 0.8\n", + "O_p1=KVA*P_f1 #Output at unity power factor(in KW)\n", + "P_i=650 #Iron loss(in Watt)\n", + "P_i1=P_i/1000 #Iron loss (in KW)\n", + "P_c=885 #Copper loss(in Watt)\n", + "P_c1=P_c/1000 #Copper loss(in KW)\n", + "I_p1=O_p1+P_c1+P_i1 #Input at unity power factor(in KW)\n", + "E_f1=(O_p1/I_p1)*100 #Efficiency of the transformer at unity power factor\n", + "print 'Efficiency of the transformer at unity power factor = %0.f %% '%E_f1 \n", + "O_p2=KVA*P_f2 #Output at 0.8 power factor(in KW) \n", + "#Maximum efficiency occurs when Copper loss = Iron loss,let the maximum efficiency occcur at x times full load\n", + "x=(P_i1/P_c1)**(1/2) \n", + "print 'the maximum efficiency occurs at the full load of = %0.3f'%x \n", + "O_P=x*KVA*P_f2 #Output at maximum efficiency(in KW)\n", + "E_F=(O_P/(O_P+P_i1+P_c1))*100 #Maximum Efficiency of the transformer at 0.8 power factor\n", + "print 'Maximum Efficiency of the transformer = %0.2f %%' %E_F\n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer at unity power factor = 97 % \n", + "the maximum efficiency occurs at the full load of = 0.857\n", + "Maximum Efficiency of the transformer = 95.71 %\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.31 page 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In the single phase transformer,find the ratio of iron and copper loss such that maximum efficiency occurs at 75% of full load.\n", + "\n", + "x=75/100 #the value of load which is 75% of full load\n", + "P_r=x**2 #Ratio of the iron loss and copper loss for maximum efficiency\n", + "print 'Ratio of the iron and copper loss for the maximum efficiency =',P_r " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of the iron and copper loss for the maximum efficiency = 0.5625\n" + ] + } + ], + "prompt_number": 67 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.32 page 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the iron losses and the full load copper losses.\n", + "\n", + "KVA=400 #Rating of the transformer(in KVA)\n", + "E_f1=0.9877 #Transformer efficiency when delivering full load at 0.8 power factor\n", + "P_f1=0.8 #power factor at full load\n", + "P_f2=1 #power factor at half load\n", + "O_p1=KVA*P_f1 #Output on full load when power factor is 0.8(in KW)\n", + "I_p1=(O_p1/E_f1) #Input on full load when power factor is 0.8(in KW)\n", + "P_t1=I_p1-O_p1 #Total losses on full load when power factor is 0.8(in KW)\n", + "O_p2=(1/2)*KVA*P_f2 #Output at half load when power factor is 1\n", + "E_f2=0.9913 #Transformer efficiency when delivering half load at 1 power factor\n", + "I_p2=O_p2/E_f2 #Input at half load when power factor is 1\n", + "P_t2=I_p2-O_p2#Total losses at half load when power factor is 1\n", + "#P_t1=P_c+P_i\n", + "#P_t2=(1/4)P_c+P_i\n", + "P_c=(4/3)*(P_t1-P_t2) #Full load and copper losses\n", + "P_i=(1/3)*(4*P_t2-P_t1) #iron losses\n", + "print 'full load and copper losses = %0.3f kW'%P_c \n", + "print 'iron loss = %0.3f kW '%P_i " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full load and copper losses = 2.973 kW\n", + "iron loss = 1.012 kW \n" + ] + } + ], + "prompt_number": 70 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.33 page 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import ceil\n", + "#Caption:In a transformer find all day efficiency\n", + "\n", + "KVA=15 #Rating of the transformer(in KVA)\n", + "E_f=0.98 #Efficiency of the transformer\n", + "P_F=1 #for unity power factor\n", + "O_P=KVA*P_F #Output of the transformer at unity power factor(in KW)\n", + "I_P=O_P/E_f #Input to the transformer(in KW)\n", + "P_T=I_P-O_P #Total losses(in KW)\n", + "#At Maximum efficiency\n", + "P_C=P_T/2 #copper loss for maximum efficiency(in KW)\n", + "P_I=P_C #iron losss at maximum efficiency copper loss=iron loss\n", + "L_1=2 #load for 12 hours (in KW)\n", + "L_2=12 #load for 6 hours (in KW)\n", + "L_3=18 #load for next 6 hours (in KW)\n", + "P_f1=0.5 #Power factor at L_1 load\n", + "P_f2=0.8 #Power factor at L_2 load\n", + "P_f3=0.9 #Power factor at L_3 load\n", + "T_1=12 #Time when L_1 working(in hours)\n", + "T_2=6 #Time when L_2 working(in hours)\n", + "T_3=6 #Time when L_3 working(in hours)\n", + "O_p1=L_1*T_1+L_2*T_2+L_3*T_3 #All day output(in KWh)\n", + "P_i1=P_I*24 #Iron losses for 24 hours(in KWh)\n", + "P_c1=T_1*P_C*((L_1/P_f1)/KVA)**2+T_2*P_C*((L_2/P_f2)/KVA)**2+T_3*P_C*((L_3/P_f3)/KVA)**2 #Copper loss for 24 hours(in KWh)\n", + "P_t=P_c1+P_i1 #Total losses of transformer for 24 hours(in KWh)\n", + "I_p1=O_p1+P_t #All day input(in KWh)\n", + "E_f1=(O_p1/I_p1)*100 #All day efficiency of transformer\n", + "print 'All day efficiency of transformer = %0.2f %% '%ceil(E_f1) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All day efficiency of transformer = 97.00 % \n" + ] + } + ], + "prompt_number": 73 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.34 page 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a transformer find all day efficiency\n", + "\n", + "KVA=1500 #Rating of the transformer(in KVA)\n", + "P_C=4.5 #copper loss for maximum efficiency(in KW)\n", + "P_I=3.2 #iron losss at maximum efficiency copper loss=iron loss\n", + "L_1=1200 #load for 6 hours (in KW)\n", + "L_2=900 #load for next 10 hours (in KW)\n", + "L_3=300 #load for next 4 hours (in KW)\n", + "L_4=0 #load for next 4 hours (in KW)\n", + "P_f1=0.8 #Power factor at L_1 load\n", + "P_f2=0.75 #Power factor at L_2 load\n", + "P_f3=0.8 #Power factor at L_3 load\n", + "P_f4=0 #Power factor at L_4 load\n", + "T_1=6 #Number of hours when L_1 working(in hours)\n", + "T_2=10 #Number of hours when L_2 working(in hours)\n", + "T_3=4 #Number of hours when L_3 working(in hours)\n", + "T_4=4 #Number of hours when L_4 working(in hours)\n", + "O_p1=L_1*T_1+L_2*T_2+L_3*T_3 #All day output(in KWh)\n", + "P_i1=P_I*24 #Iron losses for 24 hours(in KWh)\n", + "P_c1=T_1*P_C*((L_1/P_f1)/KVA)**2+T_2*P_C*((L_2/P_f2)/KVA)**2+T_3*P_C*((L_3/P_f3)/KVA)**2 #Copper loss for 24 hours(in KWh)\n", + "P_t=P_c1+P_i1 #Total losses of transformer for 24 hours(in KWh)\n", + "I_p1=O_p1+P_t #All day input(in KWh)\n", + "E_f1=(O_p1/I_p1)*100 #All day efficiency of transformer\n", + "print 'All day efficiency of transformer = %0.2f %%' %(E_f1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All day efficiency of transformer = 99.24 %\n" + ] + } + ], + "prompt_number": 74 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.35 page 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a transformer find all day efficiency\n", + "\n", + "KVA=15 #Rating of the transformer(in KVA)\n", + "P_c=0.35 #Full load copper loss (in KW)\n", + "P_i=0.25 #iron losss \n", + "P_I=P_i*24 #Iron loss per day\n", + "L_1=1/4 #load for 9 hours of full load\n", + "L_2=1 #load for 7 hours of full load\n", + "L_3=3/4 #load for next 6 hours of full load\n", + "P_f1=0.6 #Power factor at L_1 load\n", + "P_f2=0.8 #Power factor at L_2 load\n", + "P_f3=1 #Power factor at L_3 load\n", + "T_1=9 #Time when L_1 working(in hours)\n", + "T_2=7 #Time when L_2 working(in hours)\n", + "T_3=6 #Time when L_3 working(in hours)\n", + "P_c1=((1/4)**2)*P_c #Copper loss at 1/4 load\n", + "P_C1=9*P_c1 #Copper loss for 9 hours at 1/4 load\n", + "P_c2=P_c #Copper loss at full load\n", + "P_C2=7*P_c2 #Copper loss for 7 hours at full load\n", + "P_c3=((3/4)**2)*P_c #Copper loss at 3/4 load\n", + "P_C3=6*P_c3 #Copper loss for 6 hours at 3/4 load\n", + "P_C=P_C1+P_C2+P_C3 #Copper loss per day(in KW)\n", + "P_T=P_C+P_I #Iron loss per day(in KW)\n", + "O_P=L_1*KVA*P_f1*T_1+L_2*KVA*P_f2*T_2+L_3*KVA*P_f3*T_3 #Total output per day(in KWh)\n", + "I_P=O_P+P_T #Total input(in KWh)\n", + "E_F=(O_P/I_P)*100 #All day efficiency(in %)\n", + "print 'All day efficiency = %0.2f %%' %E_F" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All day efficiency = 94.59 %\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.36 page 225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the economy of copper in auto transformer and current distribution in primary\n", + "\n", + "V_1=500 #Primary voltage\n", + "V_2=400 #Secondary voltage\n", + "I_2=100 #Secondary voltage\n", + "I_1=V_2*I_2/V_1 #Primary current\n", + "print 'Current in primary winding = %0.2f Amp '%I_1\n", + "K=V_2/V_1 #Transformer ratio\n", + "S=K*100 #Saving (in %)\n", + "print 'Economy of copper = %0.2f %%'%S" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current in primary winding = 80.00 Amp \n", + "Economy of copper = 80.00 %\n" + ] + } + ], + "prompt_number": 76 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.37 page 225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a auto transformer determine Transformation ratio secondary current primary current\n", + "\n", + "V_1=250 #Primary voltage(in Voltage)\n", + "V_2=125 #Secondary voltage(in Voltage)\n", + "K=V_2/V_1 #Transformation ratio\n", + "N_1=250 #Primary turns\n", + "print 'Transformation ratio =',K\n", + "P_f=1 #Unity power factor\n", + "L=5 #Value of load(in KW)\n", + "I_2=L*1000/(V_2*P_f) #Secondary current(in Amp)\n", + "print 'Secondary current = %0.2f Amp '%I_2 \n", + "I_1=K*I_2 #Primary current(in Amp)\n", + "print 'Primary current = %0.2f Amp '%I_1 \n", + "N_2=K*N_1 #Secondary turns\n", + "print 'number of turns across secondary winding =',N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transformation ratio = 0.5\n", + "Secondary current = 40.00 Amp \n", + "Primary current = 20.00 Amp \n", + "number of turns across secondary winding = 125.0\n" + ] + } + ], + "prompt_number": 77 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.38 page 29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In an auto transformer determine current in different section and KVA output and power transferred\n", + "\n", + "KVA=10 \n", + "V_1=2400 #Voltage in first winding\n", + "V_2=240 #Voltage in second winding\n", + "I_1=KVA*1000/V_1 #Current rating of 2400 Volts winding\n", + "I_2=KVA*1000/V_2 #Current rating of 240 Volts winding\n", + "I_l=I_1+I_2 #Total load current\n", + "print 'Current rating of 2400 Volts winding = %0.2f Amp '%I_1 \n", + "print 'Current rating of 240 Volts winding = %0.2f Amp '%I_2 \n", + "print 'Total load current = %0.2f Amp '%I_l \n", + "KVA_r=V_1*I_l/1000 #KVA output rating\n", + "print 'KVA output rating = %0.2f KVA '%KVA_r \n", + "P_i=V_1*I_1 #power transferred inductively\n", + "P_c=V_1*I_2 #power transferred conductively\n", + "print 'power transferred inductively = %0.2f VA '%P_i \n", + "print 'power transferred conductively = %0.2f VA '%P_c \n", + "N_1=2640 #Number Primary winding in case of two winding transformer\n", + "N_2=2400 #Number Secondary winding in case of two winding transformer\n", + "K=N_1/N_2 #Transformer ratio\n", + "Saving=(1/K)*100 #Saving in copper\n", + "print 'Saving in copper = %0.2f %%'%Saving \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current rating of 2400 Volts winding = 4.17 Amp \n", + "Current rating of 240 Volts winding = 41.67 Amp \n", + "Total load current = 45.83 Amp \n", + "KVA output rating = 110.00 KVA \n", + "power transferred inductively = 10000.00 VA \n", + "power transferred conductively = 100000.00 VA \n", + "Saving in copper = 90.91 %\n" + ] + } + ], + "prompt_number": 79 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.39 Page 230" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a transformer calculate Power output and power transformed and power conducted\n", + "KVA=25 \n", + "V_s=2200 #Source voltage\n", + "V_1=2200 #Voltage in first winding\n", + "V_2=220 #Voltage in second winding\n", + "I_1=KVA*1000/V_1 #Current rating of 2200 V winding\n", + "I_2=KVA*1000/V_2 #Current rating of 220 V winding\n", + "V_o=V_1+V_2 #Output voltage\n", + "I_l=I_1+I_2 #Input line current\n", + "I_o=I_2 #Output current of auto transformer\n", + "KVA_r=V_o*I_2/1000 #KVA rating\n", + "P_f1=0.8 #\n", + "P_o=KVA*P_f1 #Power output at full load and 0.8 power factor\n", + "KVA_t=V_1*I_1/1000 #KVA transformed \n", + "print 'KVA transformed = %0.2f kVA'%KVA_t\n", + "P_t=KVA_t*P_f1 #Power transformed(in KW)\n", + "print 'Power transformed = %0.2f kW'%P_t\n", + "KVA_c=V_s*I_o/1000 #KVA conducted(in KVA)\n", + "P_c=KVA_c*P_f1 #Power conducted(in KW)\n", + "print 'Power conducted = %0.2f kW'%P_c\n", + "#E_f=Output/(Output+Losses)\n", + "#Losses=((1/E_f)-1)*Output\n", + "E_f=0.9 #Efficiency of the two winding transformer\n", + "P_f2=0.85 #New power factor of the two einding transformer\n", + "O_p1=KVA_t*1000*P_f2 #Output of the two winding transformer\n", + "L=((1/E_f)-1)*O_p1 #losses in a 2-winding transformer\n", + "#Losses in auto transformer=losses in 2-winding transformer\n", + "O_P=KVA_r*1000*P_f2 #output of the auto transformer\n", + "E_F=(O_P/(O_P+L))*100 #Efficiency of the auto transformer(in %)\n", + "print 'Efficiency of the auto transformer = %0.2f %%'%E_F\n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "KVA transformed = 25.00 kVA\n", + "Power transformed = 20.00 kW\n", + "Power conducted = 200.00 kW\n", + "Efficiency of the auto transformer = 99.00 %\n" + ] + } + ], + "prompt_number": 81 + } + ], + "metadata": {} + } + ] +} |