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diff --git a/sample_notebooks/ShantanuBhosale/chapter40.ipynb b/sample_notebooks/ShantanuBhosale/chapter40.ipynb new file mode 100644 index 00000000..e297a2c2 --- /dev/null +++ b/sample_notebooks/ShantanuBhosale/chapter40.ipynb @@ -0,0 +1,1989 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 40: D.C TRANSMISSION AND DISTRIBUTION\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.1 ,Page No :- 1574" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "percentage saving in copper is = 50.0 %.\n" + ] + } + ], + "source": [ + "#A DC 2-wire feeder supplies a constant load with a sending-end voltage of 220V.Calculate the saving in copper\n", + "#if this voltage is doubled with power transmitted remaining the same.\n", + "##################################################################################################################\n", + "\n", + "\n", + "\n", + "#Given\n", + "V1 = 220.0\n", + "V2 = 440.0\n", + "##Let us assume the wire has##\n", + "#length -> length \n", + "#area -> area\n", + "#current density -> cd\n", + "#power -> P\n", + "P = 10000.0 #assumption\n", + "length = 1000.0 #assumption \n", + "cd = 5.0 #assumption\n", + "#The values are assumed as these terms cancel out while calculating percentage.\n", + "I1 = P/V1\n", + "area = I1/cd\n", + "#Vol of Cu required for 220V ->vol1\n", + "vol1 = 2*area*length\n", + "\n", + "\n", + "I2 = P/V2\n", + "area = I2/cd\n", + "#Vol of Cu required for 440V ->vol2\n", + "vol2 = 2*area*length\n", + "\n", + "#percentage saving of copper is\n", + "per_cent = ((vol1-vol2)/vol1)*100\n", + "print 'percentage saving in copper is ',per_cent,'%.'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.2 ,Page No :- 1577" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum voltage drop from one end is = 12.0 V.\n", + "Maximum voltage drop from both end is = 3.0 V.\n" + ] + } + ], + "source": [ + "#A uniform 2-wire d.c distributor 200 metres long is loaded with 2 amperes/metre.Resistance of\n", + "#single wire is 0.3 ohm/kilometre.Calculate the maximum voltage drop if the distributor is fed\n", + "#(a)from one end (b)from both ends with equal voltages.\n", + "#################################################################################################\n", + "\n", + "#Given\n", + "length = 200.0 #metres\n", + "#current per unit length is\n", + "cur = 2.0 #amp/metre\n", + "#resistance per unit length is\n", + "res = 0.3/1000 #ohm/metre\n", + "\n", + "#total resistance is\n", + "R = res*length #ohm\n", + "#total current is\n", + "I = cur*length #amp\n", + "#Total drop over a distributor fed from one end is given by\n", + "drop1 = (1/2.0)*I*R #volts\n", + "#Total drop over a distributor fed from both ends is given by\n", + "drop2 = (1/8.0)*I*R\n", + "print 'Maximum voltage drop from one end is = ',drop1,'V.'\n", + "print 'Maximum voltage drop from both end is = ',drop2,'V.'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.3 ,Page No :- 1577" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cross sectional area of distributor = 116.412 cm^2\n" + ] + } + ], + "source": [ + "#A 2-wire d.c distributor AB is 300 metres long.It is fed at point A.The various loads and\n", + "#their positions are given below.\n", + "# At point distance from A in metres concentrated load in A\n", + "# C 40 30\n", + "# D 100 40 \n", + "# E 150 100\n", + "# F 250 50\n", + "#If the maximum permissible voltage drop is not to exceed 10V,find the cross-sectional\n", + "#area of the distributor.Take resistivity = 1.78*10^(-8) ohm-m.\n", + "###########################################################################################\n", + "\n", + "\n", + "#Given\n", + "resistivity = 1.78e-8 #ohm-m\n", + "drop_max = 10.0 #V\n", + "#loads and their positions\n", + "I1 = 30.0 #A\n", + "l1 = 40.0 #m\n", + "I2 = 40.0 #A\n", + "l2 = 100.0 #m\n", + "I3 = 100.0 #A\n", + "l3 = 150.0 #m\n", + "I4 = 50 #A\n", + "l4 = 250 #m\n", + "#We know that R = resistivity*length/Area\n", + "#Also max drop = I1*R1 + I2*R2 + I3*R3 + I4*R4 , using this\n", + "area = 2*(I1*l1 + I2*l2 + I3*l3 + I4*l4)*resistivity/drop_max #m^2\n", + "area = area*1000000 #cm^2 \n", + "print 'Cross sectional area of distributor =',area,'cm^2'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.4 ,Page No :- 1578" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hence drop at minimum potential where load is 70 A is = 48.4 V.\n" + ] + } + ], + "source": [ + "#A 2-wire d.c distributor F1F2 1000 metres long is loaded as under:\n", + "#Distance from F1(in metres): 100 250 500 600 700 800 850 920\n", + "#Load in amperes: 20 80 50 70 40 30 10 15\n", + "#The feeding points F1 and F2 are maintained at the same potential.Find which point will have the\n", + "#minimum potential and what will be the drop at this point?Take the cross-section of the conductors\n", + "#as 0.35 cm^2 and specific resistance of copper as 1.764*10^(-6) ohm-cm\n", + "#####################################################################################################\n", + "\n", + "#Given\n", + "import numpy as np\n", + "resistivity = 1.764e-8 #ohm-m\n", + "area = 0.35e-4 #m^2 \n", + "#loads and their positions taking from F1\n", + "I1 = 20 #A\n", + "l1 = 100 #m\n", + "I2 = 80 #A\n", + "l2 = 150 #m\n", + "I3 = 50 #A\n", + "l3 = 250 #m\n", + "I4 = 70 #A\n", + "l4 = 100 #m\n", + "I5 = 40 #A\n", + "l5 = 100 #m\n", + "I6 = 30 #A\n", + "l6 = 50 #m\n", + "I7 = 10 #A\n", + "l7 = 70 #m\n", + "I8 = 15 #A\n", + "l8 = 80 #m \n", + "\n", + "#sum of loads from F1\n", + "load1 = I1*l1 + I2*(l1+l2) + I3*(l1+l2+l3) #A-m\n", + "load2 = I8*l8 + I7*(l7+l8) + I6*(l6+l7+l8) + I5*(l5+l6+l7+l8) #A-m\n", + "\n", + "#guessing the point of minimum potential\n", + "if load1>load2:\n", + " load2_new = load2 + I4*(l4+l5+l6+l7+l8)\n", + " if load2_new>load1:\n", + " pivot = I4\n", + "\n", + "#solving 2 equations simultaneously\n", + "# x + y = 70(pivot) & 47000(load1) + 600(l1+l2+l3)x = 20,700(load2) + 400(l5+l6+l7+l8)y)\n", + "line1 = l1+l2+l3+l4 #m\n", + "line2 = l4+l5+l6+l7+l8 #m \n", + "\n", + "a = [[1,1],[line1,-line2]]\n", + "b = [pivot,load2-load1]\n", + "soln = np.linalg.solve(a,b) #soln is array with its elements[x,y]\n", + "#drop at minimum potential per conductor (in A-m)\n", + "drop_m = load1 + soln[0]*line1 #A-m\n", + "\n", + "#resistance per metre = resistivity/Area\n", + "res = resistivity/area #ohm/m\n", + "\n", + "#Hence, drop in voltage per conductor is\n", + "drop = drop_m*res #V \n", + "#drop due to both is\n", + "drop = drop*2 #V\n", + "\n", + "print 'Hence drop at minimum potential where load is',pivot,'A is =',round(drop,2),'V.'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.5 ,Page No :- 1579" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current entering at A is = 88.6 A \n", + "The current entering at B is = 211.4 A.\n" + ] + } + ], + "source": [ + "#The resistance of a cable is 0.1ohm per 1000 metre for both conductors.It is loaded as shown in Fig.40.14(a).\n", + "#Find the current supplied at A and at B.If both the ends are supplied at 200 V\n", + "##############################################################################################################\n", + "\n", + "#Given\n", + "#resistance per metre\n", + "res = 0.1/1000 #ohm/m\n", + "#loads and their positions taking from A\n", + "I1 = 50.0 #A\n", + "l1 = 500.0 #m\n", + "I2 = 100.0 #A\n", + "l2 = 700.0 #m\n", + "I3 = 150.0 #A\n", + "l3 = 300.0 #m\n", + "l4 = 250.0 #m \n", + "\n", + "#Assuming I flows from A to B\n", + "# equation is res*[500*i + 700(i-50) + 300(i-150) + 250(i-300)] = 0\n", + "current_i = (I1*l2+(I1+I2)*l3 + (I1+I2+I3)*l4)/(l1+l2+l3+l4)\n", + "current_AC = current_i\n", + "current_CD = current_i-I1\n", + "current_DE = current_CD-I2\n", + "current_EB = current_DE-I3\n", + "if current_EB<0:\n", + " current_EB = -current_EB;\n", + "print 'The current entering at A is = ',round(current_i,1),'A '\n", + "print 'The current entering at B is = ',round(current_EB,1),'A.' " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.6 ,Page No :- 1580" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current supplied at A is = 88.6 A.\n", + "Current supplied at B is = -211.4 A.\n", + "Current in AC is = 88.6 A.\n", + "Current in CD is = 38.6 A.\n", + "Current in DE is = -61.4 A.\n", + "Current in EB is = -211.4 A.\n", + "Drop over AC is = 4.4 V.\n", + "Drop over CD is = 2.7 V.\n", + "Drop over DE is = -1.8 V.\n", + "Voltage at C is = 195.6 V.\n", + "Voltage at D is = 192.9 V.\n", + "Voltage at E is = 194.7 V.\n" + ] + } + ], + "source": [ + "#The resistance of two conductors of a 2-conductor distributor shown in Fig.39.15 is 0.1ohm per 1000m\n", + "#for both conductors.Find (a)the current supplied at A(b)the current supplied at B\n", + "#(c)the current in each section (d)the voltages at C,D and E.Both A and B are maintained at 200V.\n", + "######################################################################################################\n", + "\n", + "#Given\n", + "#resistance per metre\n", + "res = 0.1/1000 #ohm/m\n", + "#loads and their positions taking from A\n", + "I1 = 50.0 #A\n", + "l1 = 500.0 #m\n", + "I2 = 100.0 #A\n", + "l2 = 700.0 #m\n", + "I3 = 150.0 #A\n", + "l3 = 300.0 #m\n", + "l4 = 250.0 #m \n", + "\n", + "#Assuming I flows from A to B\n", + "# equation is res*[500*i + 700(i-50) + 300(i-150) + 250(i-300)] = 0\n", + "current_i = (I1*l2+(I1+I2)*l3 + (I1+I2+I3)*l4)/(l1+l2+l3+l4)\n", + "current_AC = current_i\n", + "current_CD = current_i-I1\n", + "current_DE = current_CD-I2\n", + "current_EB = current_DE-I3\n", + "print \"Current supplied at A is = \",round(current_i,1),\"A.\"\n", + "print \"Current supplied at B is = \",round(current_EB,1),\"A.\"\n", + "print \"Current in AC is = \",round(current_i,1),\"A.\"\n", + "print \"Current in CD is = \",round(current_CD,1),\"A.\"\n", + "print \"Current in DE is = \",round(current_DE,1),\"A.\"\n", + "print \"Current in EB is = \",round(current_EB,1),\"A.\"\n", + "#Drop in volts is (resistance/metre)*length*current\n", + "drop_AC = res*l1*current_AC #V\n", + "drop_CD = res*l2*current_CD #V \n", + "drop_DE = res*l3*current_DE #V\n", + "print \"Drop over AC is = \",round(drop_AC,1),\"V.\"\n", + "print \"Drop over CD is = \",round(drop_CD,1),\"V.\"\n", + "print \"Drop over DE is = \",round(drop_DE,1),\"V.\"\n", + "\n", + "#Voltages at C,D,E are\n", + "volt_C = 200-drop_AC #V\n", + "volt_D = volt_C-drop_CD #V\n", + "volt_E = volt_D-drop_DE #V\n", + "print 'Voltage at C is = ',round(volt_C,1),'V.'\n", + "print 'Voltage at D is =',round(volt_D,1),'V.'\n", + "print 'Voltage at E is = ',round(volt_E,1),'V.'\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.7 ,Page No :- 1581" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Therefore point of minimum potential is D and minimum potential is = 246.0 V.\n" + ] + } + ], + "source": [ + "#A 200 m long distributor is fed from both ends A and B at the same voltage of 250V.The\n", + "#concentrated loads of 50,40,30 and 25 A are coming on the distributor at distances of 50,75,\n", + "#100 and 150 m respectively from end A.Determine the minimum potential and locate its positions.\n", + "#Also,determine the current in each section of the distributor.It is given that the resistance\n", + "#of the distributor is 0.08ohm per 100 metres for go and return.\n", + "##################################################################################################\n", + "\n", + "\n", + "#Given\n", + "#resistance per metre\n", + "res = 0.08/100 #ohm/m\n", + "V_A = 250.0 #V\n", + "V_B = 250.0 #V\n", + "#load currents and their positions\n", + "I_C = 50.0 #A\n", + "I_D = 40.0 #A\n", + "I_E = 30.0 #A\n", + "I_F = 25.0 #A\n", + "l_AC = 50.0 #m\n", + "l_CD = 75.0 - l_AC #m\n", + "l_DE = 100.0 - l_CD - l_AC #m\n", + "l_EF = 150.0 - l_DE - l_CD - l_AC #m\n", + "l_FB = 200.0-150.0\n", + "#Assuming I flows from A to B\n", + "# equation is res*[50*i + 25(i-50) + 25(i-90) + 50(i-120)+50(i-145)] = 0\n", + "current_i = (l_CD*I_C + l_DE*(I_C+I_D)+l_EF*(I_C+I_D+I_E) + l_FB*(I_C+I_D+I_E+I_F))/200.0\n", + "current_AC = current_i\n", + "current_CD = current_i-I_C\n", + "current_DE = current_CD-I_D\n", + "current_EF = current_DE-I_E\n", + "current_FB = current_EF-I_F\n", + "#As from figure in the book , point D is at minimum potential\n", + "if (current_CD>0) & (current_DE<0):\n", + " point = \"D\"\n", + " \n", + "#drop in volts = resistance/metre*sum(length*current) \n", + "drop_d = res*(l_AC*current_AC + l_CD*current_CD)\n", + "min_pot = V_A-drop_d\n", + "print \"Therefore point of minimum potential is\",point,\"and minimum potential is = \",round(min_pot,1),\"V.\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.8 ,Page No :- 1582" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage at point C is = 250.13 V.\n", + "Voltage at point D is = 247.73 V.\n" + ] + } + ], + "source": [ + "#Each conductor of a 2-core distributor,500 metres long has a cross-sectional area\n", + "#of 2 cm^2.The feeding point A is supplied at 255V and the feeding point B at\n", + "#250V and load currents of 120A and 160A are taken at points C and D which are\n", + "#150 metres and 350 metres respectively from the feeding point A.Calculate the\n", + "#voltage at each load.Specific Resistance of copper is 1.7*10^(-6) ohm-cm\n", + "##################################################################################\n", + "\n", + "#Given\n", + "area = 2e-4 #m^2\n", + "resistivity = 1.7e-8 #ohm-m\n", + "#load currents and their positions\n", + "i_c = 120.0 #A\n", + "i_d = 160.0 #A\n", + "l_ac = 150.0 #m\n", + "l_cd = 200.0 #m\n", + "l_db = 150.0 #m\n", + "V_a = 255.0 #V\n", + "V_b = 250.0 #V\n", + "#Resistance = resistivity*length/Area\n", + "#It is a 2 core distributor.Therefore,resistance per metre is\n", + "res = 2*resistivity/area #ohm/m\n", + "#drop over whole distributor is equal to\n", + "drop = V_a - V_b #V\n", + "#Therefore equation of total drop can be written as\n", + "# resistivity*(150i+200(i-120)+150(i-280))=5\n", + "current_i = (drop/res + l_cd*i_c + l_db*(i_c+i_d))/(l_ac+l_cd+l_db) #A\n", + "current_ac = current_i #A\n", + "current_cd = current_ac-i_c #A\n", + "current_db = current_cd-i_d #A\n", + "\n", + "#Voltage at C = 255-drop over AC\n", + "volt_c = V_a-res*l_ac*current_ac #V\n", + "#Voltage at D = 250-drop over DB \n", + "volt_d = V_b -res*l_db*abs(current_db) #V\n", + "print \"Voltage at point C is = \",round(volt_c,2),\"V.\"\n", + "print \"Voltage at point D is = \",round(volt_d,2),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.9 ,Page No :- 1583" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volatge at point Q is = 225.25 V.\n", + "Voltage at point B is = 236.56 V.\n" + ] + } + ], + "source": [ + "#A 2-wire distributor 500 metres long is fed at P at 250V and loads of 40A,20A,60A,30A are tapped off\n", + "#off from points A,B,C and D which are at distances of 100 metres,150 metres,300 metres and 400 metres\n", + "#from P respectively.The distributor is also uniformly loaded at the rate of 0.1A/m.If the resistance of\n", + "#the distributor per metre(go and return) is 0.0005 ohm,calculate the voltage at(i)pointQ and(ii)point B.\n", + "###########################################################################################################\n", + "\n", + "#Given\n", + "V_P = 250.0 #V\n", + "resistance = 0.0005 #ohm/m\n", + "\n", + "#loads and their positions\n", + "i_a = 40.0 #A\n", + "i_b = 20.0 #A\n", + "i_c = 60.0 #A\n", + "i_d = 30.0 #A\n", + "l_pa = 100.0 #m\n", + "l_ab = 150.0-100.0 #m\n", + "l_bc = 300.0-150.0 #m\n", + "l_cd = 400.0-300.0 #m\n", + "#uniform dstributed load\n", + "cur_uni = 0.1 #A/m\n", + "\n", + "\n", + "#considering drop due to concentrated loading\n", + "drop_pa = (i_a+i_b+i_c+i_d)*l_pa*resistance #V\n", + "drop_ab = (i_b+i_c+i_d)*l_ab*resistance #V \n", + "drop_bc = (i_c+i_d)*l_bc*resistance #V\n", + "drop_cd = i_d*l_cd*resistance #V\n", + "tot_drop = drop_pa + drop_ab + drop_bc + drop_cd #V\n", + "\n", + "#considering drop due to uniform loading(drop = irl^2/2) l = 500m\n", + "drop_uni = cur_uni*resistance*(500.0*500.0)/2 #V\n", + "\n", + "V_Q = V_P - (tot_drop + drop_uni) #V\n", + "#for point B\n", + "#drop due to concentrated loading\n", + "drop_b = drop_pa + drop_ab #V\n", + "#drop due to uniform loading (drop = ir(lx-x^2/2)) l=500m x=150m\n", + "drop_ub = cur_uni*resistance*(500*(l_pa+l_ab)-(l_pa+l_ab)*(l_pa+l_ab)/2) #V\n", + "\n", + "V_B = V_P - (drop_b + drop_ub) #V\n", + "\n", + "print \"Volatge at point Q is = \",round(V_Q,2),\"V.\"\n", + "print \"Voltage at point B is = \",round(V_B,2),\"V.\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.10 ,Page No :- 1583" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in section AC is = 53.75 A.\n", + "Current in section CD is = 33.75 A.\n", + "Current in section DE is = -6.25 A.\n", + "Current in section EF is = -31.25 A.\n", + "Current in section FB is = -61.25 A.\n", + "Minimum voltage is at point D and minimum voltage is = 233.18 V.\n" + ] + } + ], + "source": [ + "#A distributor AB is fed from both ends.At feeding point A,the voltage is maintained at 236V and at B at 237V.\n", + "#The total length of the distributor is 200 metres and loads are tapped off as under:\n", + "#(i) 20A at 50 metres from A (ii) 40A at 75 metres from A. (iii)25A at 100 metres from A (iv)30A at 150 metres from A\n", + "#The reistance per kilometre of one conductor is 0.4ohm.Calculate the currents in the various sections of the distributor,\n", + "#the minimum voltage and the point at which it occurs.\n", + "###########################################################################################################################\n", + "\n", + "\n", + "#Given\n", + "#resistance per metre\n", + "res = 2*0.4/1000 #ohm/m\n", + "V_a = 236.0 #V\n", + "V_b = 237.0 #V\n", + "#loads and their positions\n", + "i_c = 20.0 #A\n", + "i_d = 40.0 #A\n", + "i_e = 25.0 #A\n", + "i_f = 30.0 #A\n", + "l_ac = 50.0 #m\n", + "l_cd = 25.0 #m\n", + "l_de = 25.0 #m\n", + "l_ef = 50.0 #m\n", + "l_fb = 50.0 #m\n", + "#Voltage drop equation res*(50i + 25(i-20)+25(i-60) + 50(i-85) + 50(i-115)=-1)\n", + "current_i = ((V_a-V_b)/res + l_cd*(i_c)+l_de*(i_c+i_d)+l_ef*(i_c+i_d+i_e)+l_fb*(i_c+i_d+i_e+i_f))/200.0\n", + "current_ac = current_i\n", + "current_cd = current_ac-i_c\n", + "current_de = current_cd-i_d\n", + "current_ef = current_de-i_e\n", + "current_fb= current_ef-i_f\n", + "if current_cd>0:\n", + " if current_de<0:\n", + " point = \"D\"\n", + "#Minimum potential is at D as shown in figure\n", + "drop = res*(current_ac*l_ac + current_cd*l_cd)\n", + "V_d = V_a-drop\n", + "print \"Current in section AC is = \",round(current_ac,2),\"A.\"\n", + "print \"Current in section CD is = \",round(current_cd,2),\"A.\"\n", + "print \"Current in section DE is = \",round(current_de,2),\"A.\"\n", + "print \"Current in section EF is = \",round(current_ef,2),\"A.\"\n", + "print \"Current in section FB is = \",round(current_fb,2),\"A.\"\n", + "print \"Minimum voltage is at point\",point,\"and minimum voltage is = \",round(V_d,2),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.11 ,Page No :- 1584" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current supplied by feeder at point A is 46.29 A and that by point B is 109.71 A.\n", + "Voltage at point B is = 240.55 V.\n", + "Voltage at point C is = 239.63 V.\n", + "Voltage at point D is = 239.42 V.\n", + "Voltage at point E is = 239.38 V.\n" + ] + } + ], + "source": [ + "#A distributor cable AB is fed at its ends A and B.Loads of 12,24,72 and 48 A are taken from the cable at\n", + "#points C,D,E and F.The resistances of sections AC,CD,DE,EF and FB of the cable are 8,6,4,10 and 5 milliohm\n", + "#respecively(for the go and return conductors together). The potential difference at point A is 240V,the p.d\n", + "#at the load F is also to be 240V.Calculate the voltages at the feeding point B,the current supplied by each\n", + "#feeder and the p.d.s at the loads C,D and E.\n", + "##############################################################################################################\n", + "\n", + "#Given\n", + "V_a = 240.0 #V \n", + "V_f = 240.0 #V\n", + "#loads and their resistances.\n", + "i_c = 12.0 #A\n", + "i_d = 24.0 #A\n", + "i_e = 72.0 #A\n", + "i_f = 48.0 #A\n", + "\n", + "r_ac = 8e-3 #ohm\n", + "r_cd = 6e-3 #ohm\n", + "r_de = 4e-3 #ohm\n", + "r_ef = 10e-3 #ohm\n", + "r_fb = 5e-3 #ohm\n", + "\n", + "#Voltage drop accross AF is zero.Therefore equation 8i +6(i-12) + 4(i-36)+10(i-108)*10^(-3)\n", + "current_i = (r_cd*i_c + r_de*(i_c+i_d) + r_ef*(i_c+i_d+i_e))/(28.0e-3) #A\n", + "#currents in different sections\n", + "current_ac = current_i #A\n", + "current_cd= current_ac-i_c #A\n", + "current_de = current_cd-i_d #A\n", + "current_ef = current_de-i_e #A \n", + "current_fb = current_ef-i_f #A\n", + "#voltage at different points are\n", + "V_b = V_f - current_fb*r_fb #V\n", + "V_c = V_a - current_ac*r_ac #V\n", + "V_d = V_c - current_cd*r_cd #V\n", + "V_e = V_d - current_de*r_de #V \n", + "\n", + "print \"Current supplied by feeder at point A is\",round(current_ac,2),\"A and that by point B is\",round(abs(current_fb),2),\"A.\"\n", + "print \"Voltage at point B is = \",round(V_b,2),\"V.\"\n", + "print \"Voltage at point C is = \",round(V_c,2),\"V.\"\n", + "print \"Voltage at point D is = \",round(V_d,2),\"V.\"\n", + "print \"Voltage at point E is = \",round(V_e,2),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.12 ,Page No :- 1585" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current supplied at P is = 143.75 A.\n", + "The current supplied at Q is = 116.25 A.\n", + "Power dissipated in distributor is = 847.34 W.\n" + ] + } + ], + "source": [ + "#A two-wire d.c sdistributor PQ,800 metre long is loaded as under:\n", + "#Distance from P(metres): 100 250 500 600 700\n", + "#Loads in amperes: 20 80 50 70 40\n", + "#The feeding point at P is maintained at 248V and that at Q at 245V.The total resistance of\n", + "#the distributor(lead and return) is 0.1 ohm.Find (a)the current supplied at P and Q and\n", + "#(b)the power dissipated in the distributor.\n", + "##################################################################################################\n", + "\n", + "#Given\n", + "V_p = 248.0 #V\n", + "V_q = 245.0 #V\n", + "res = 0.1/800 #ohm/m \n", + "#loads and their positions\n", + "i1 = 20.0 #A\n", + "i2 = 80.0 #A\n", + "i3 = 50.0 #A\n", + "i4 = 70.0 #A\n", + "i5 = 40.0 #A\n", + "l1 = 100.0 #m\n", + "l2 = 250.0-100.0 #m\n", + "l3 = 500.0 -250.0 #m\n", + "l4 = 600.0-500.0 #m\n", + "l5 = 700.0-600.0 #m\n", + "l6 = 800.0-700.0 #m\n", + "#drop accross the distributor :- 1/8000(100i + 150(i-20) + 250(i-100)+ 100(i-150)+100(i-220)+100(i-260) )=3\n", + "current_i = ((V_p-V_q)/res + l2*i1+l3*(i1+i2)+l4*(i1+i2+i3)+l5*(i1+i2+i3+i4)+l6*(i1+i2+i3+i4+i5))/800.0\n", + "current_p = current_i #A\n", + "current_2 = current_p-i1 #A\n", + "current_3 = current_2-i2 #A\n", + "current_4 = current_3-i3 #A\n", + "current_5 = current_4-i4 #A\n", + "current_q = current_5-i5 #A\n", + "#Power loss = sum(I^2R)\n", + "loss = res*(current_p*current_p*l1 + current_2*current_2*l2 + current_3*current_3*l3 + current_4*current_4*l4 + current_5*current_5*l5 + current_q*current_q*l6)\n", + "print \"The current supplied at P is = \",round(current_p,2),\"A.\"\n", + "print \"The current supplied at Q is = \",round(abs(current_q),2),\"A.\"\n", + "print \"Power dissipated in distributor is =\",round(loss,2),\"W.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.13 ,Page No :- 1586" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The point of minimum potential is D and minimum potential is = 231.76 V.\n", + "Current fed at the end A is = 366.0 A.\n", + "Current fed at the end B is = 454.0 A.\n" + ] + } + ], + "source": [ + "#The two conductors of a d.c distributor cable 1000m long have a total resistance of 0.1 ohm.\n", + "#The ends A and B are fed at 240V.The cable is uniformly loaded at 0.5 A per metre length\n", + "#and has concentrated loads of 120A,60A,100A and 40A at points distant 200,400,700 and 900m.\n", + "#respectively from the end A.Calculate (i)the point of minimum potential on the distributor\n", + "#(ii)the value of minimum potential and (iii) currents fed at the ends A and B.\n", + "###############################################################################################\n", + "\n", + "#Given\n", + "V_a = 240.0 #V\n", + "V_b = 240.0 #V\n", + "res = 0.1/1000 #ohm/m\n", + "#concentrated loads and their positions\n", + "i_c = 120.0 #A\n", + "i_d = 60.0 #A\n", + "i_e = 100.0 #A\n", + "i_f = 40.0 #A\n", + "l_ac = 200.0 #m\n", + "l_cd = 400.0-200.0 #m\n", + "l_de = 700.0-400.0 #m\n", + "l_ef = 900.0-700.0 #m\n", + "l_fb = 1000.0-900.0 #m\n", + "#Uniform loading\n", + "cur_uni = 0.5 #A/m\n", + "#Equation for drop from A to B -> (1/10000)*(200i + 200(i-120)+ 300(i-180)+200(i-280)+100(i-320))=0\n", + "current_i = (l_cd*i_c + l_de*(i_c+i_d)+l_ef*(i_c+i_d+i_e)+l_fb*(i_c+i_d+i_e+i_f))/1000\n", + "\n", + "#point of minimum potential\n", + "current_ac = current_i #A\n", + "current_cd = current_ac-i_c #A\n", + "current_de = current_cd-i_d #A\n", + "current_ef = current_de-i_e #A\n", + "current_fb = current_ef-i_f #A\n", + "\n", + "if current_cd>0:\n", + " if current_de<0:\n", + " point = \"D\"\n", + "#As from figure it is inferred that point of minimum potential is D.\n", + "#Therefore,uniform load from point A to D(supplied from A)\n", + "cur_uni_A = cur_uni*(l_ac + l_cd) #A\n", + "cur_A = cur_uni_A + current_ac #A\n", + "#Therefore,uniform load from point B to D(supplied from B)\n", + "cur_uni_B = cur_uni*(l_de + l_ef + l_fb) #A\n", + "cur_B = cur_uni_B + abs(current_fb) #A\n", + "\n", + "#drop at D due to concentrated load(from A)\n", + "drop_con = res*(current_ac*l_ac + current_cd*l_cd)\n", + "#drop at D due to uniform load(from A)[formula-> irl^2/2]\n", + "drop_uni = cur_uni*res*(l_ac+l_cd)*(l_ac+l_cd)/2\n", + "#total drop is\n", + "drop_tot = drop_con + drop_uni\n", + "\n", + "#potential at D is\n", + "V_d = V_a - drop_tot\n", + "print \"The point of minimum potential is\",point,\"and minimum potential is = \",round(V_d,2),\"V.\"\n", + "print \"Current fed at the end A is = \",round(cur_A,2),\"A.\"\n", + "print \"Current fed at the end B is = \",round(cur_B,2),\"A.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.14 ,Page No :- 1587" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage V is = 8.62 V.\n", + "Cross-sectional Area A is = 2.78 cm^2.\n", + "Cross-sectional Area A1 is = 0.26 cm^2.\n", + "Cross-sectional Area A2 is = 2.24 cm^2.\n" + ] + } + ], + "source": [ + "#It is proposed to lay out a d.c distribution system comprising three sections-the first section consists\n", + "#of a cable from the sub-station to a point distant 800 metres from which two cables are taken,one 350 metres\n", + "#long supplying a load of 22kW and the other 1.5 kilometre long and supplying a load of 44kW.Calculate the\n", + "#cross-sectional area of each cable so that the total weight of copper required shall be minimum if the maximum\n", + "#drop of voltage along the cable is not to exceed 5% of the normal voltage of 440V at the consumer's premises.\n", + "#Take specific resistance of copper at working temperature equal to 2*10e-7 ohm-cm.\n", + "###################################################################################################################\n", + "\n", + "#Given\n", + "resistivity = 2*10e-7 #ohm-cm\n", + "dist = 800.0*100 #cm\n", + "#Current taken from 350m section\n", + "cur_1 = 22000.0/440\n", + "#Current taken from 1.5km section\n", + "cur_2 = 44000.0/440\n", + "#Therefore,current in first section\n", + "cur = cur_1 + cur_2\n", + "#Let us assume\n", + "#V->voltage drop accross first section\n", + "#R->resistance of the first section\n", + "#A->cross-sectional area of te first section\n", + "\n", + "from sympy import Eq, var, solve\n", + "var('V') \n", + "#Now , R = V/I\n", + "R = V/cur\n", + "# A = resistivity*l/R -> A = resistivity*l*I/V \n", + "A = resistivity*dist/R\n", + "#Max allowable drop\n", + "max_drop = (5.0/100)*440.0\n", + "#Voltage drop along other sections\n", + "vol_drop = max_drop - V\n", + "#Cross-sectional area of 350 m A = resistivity*l/R \n", + "A1 = resistivity*350.0*100*cur_1/(vol_drop)\n", + "#Cross-sectional area of 1.5km A = resistivity*l/R \n", + "A2 = resistivity*1500.0*100*cur_2/(vol_drop)\n", + "\n", + "\n", + "#Now,Total weight is propotional to total volume \n", + "W = 800.0*A + 350.0*A1+1500.0*A2\n", + "Diff = W.diff(V)\n", + "eq = Eq(Diff,0)\n", + "\n", + "V = solve(eq)\n", + "#We get 2 values of V of which Negative is not possible.Therefore,\n", + "V = float(V[1])\n", + "A = resistivity*dist*cur/V\n", + "vol_drop = max_drop - V\n", + "A1 = resistivity*350.0*100*cur_1/vol_drop\n", + "A2 = resistivity*1500.0*100*cur_2/vol_drop\n", + "print \"Voltage V is = \",round(V,2),\"V.\"\n", + "print \"Cross-sectional Area A is = \",round(A,2),\"cm^2.\"\n", + "print \"Cross-sectional Area A1 is = \",round(A1,2),\"cm^2.\"\n", + "print \"Cross-sectional Area A2 is = \",round(A2,2),\"cm^2.\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.15 ,,Page No :- 1588" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The point of minimum potential is at 261.74 m from A.\n", + "The minimum potential is = 247.34 V.\n" + ] + } + ], + "source": [ + "#A d.c two-wire distributor AB is 450m long and is fed at both ends at 250 volts.It is loaded as follows:20A at 60m from A,\n", + "#40A at 100m from A and a uniform loading of 1.5A/m from 200 to 450m from A.The resistance of each conductor is\n", + "#0.05ohm/km.Find the point of minimum potential and its potential.\n", + "####################################################################################################################\n", + "\n", + "#Given\n", + "V_a = 250.0 #V\n", + "V_b = 250.0 #V\n", + "res = 0.05/1000 #ohm/m\n", + "cur_uni = 1.5 #A/m (uniform loading)\n", + "#loads and positions\n", + "i_c = 20.0 #A\n", + "i_d = 40.0 #A\n", + "l_ac = 60.0 #m\n", + "l_cd = 40.0 #m\n", + "l_de = 100.0 #m\n", + "l_eb = 250.0 #m\n", + "\n", + "#Let us assume that point of minimum potential is D and let i be current in section CD.\n", + "#Therefore,current from B is (40-i).If r is resistance then\n", + "#(20+i)*60r + i*40r = (40-i)*350r + 1.5*r*250^2/2 [drop over AD = drop over BD as V_a = V_b]\n", + "\n", + "cur_i = (i_d*(l_de+l_eb)*res + cur_uni*res*l_eb*l_eb/2 - i_c*l_ac*res)/((l_ac+l_cd+l_de+l_eb)*res) #A\n", + "\n", + "#cur_i > 40 i.e 40-i is negative,it means D is not point of minimum potential.Let F be point of minimum potential(between DB)\n", + "#current in section DF is\n", + "cur_df = cur_i-i_d #A\n", + "\n", + "#distance EF\n", + "dist_ef = cur_df/cur_uni #m\n", + "\n", + "#distance of F from A is\n", + "dist = l_ac + l_cd + l_de + dist_ef #m\n", + "\n", + "#total drop over AF is [(20+i)*60r + i*40r+ (i-40)*161.7r - 1.5*r*61.7^2/2\n", + "drop_af = 2*res*((i_c+cur_i)*l_ac + cur_i*l_cd + cur_df*(l_de+dist_ef)-cur_uni*dist_ef*dist_ef/2) #V\n", + "#potential at F\n", + "V_f = V_a - drop_af #V\n", + "print \"The point of minimum potential is at\",round(dist,2),\"m from A.\"\n", + "print \"The minimum potential is = \",round(V_f,2),\"V.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.16 ,Page No :- 1588" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current fed at A is = 225.0 A.\n", + "Current fed at B is = 475.0 A.\n", + "Point of minimum potential from B is = 475.0 metres.\n", + "Voltage at minimum potential is = 230.72 V.\n" + ] + } + ], + "source": [ + "#A two-wire d.c distributor AB,1000 metres long,is supplied from both ends,240V at A and 242V at B.There is a\n", + "#concentrated load of 200A at a distance of 400 metre from A and a uniformly distrubuted load of 1.0A/m between\n", + "#the mid-point and end B.Determine (i)the currents fed at A and B(ii)the point of minimum potential and\n", + "#(iii)voltage at this point.Take cable resistance as 0.005 ohm per 100 metre each core.\n", + "#####################################################################################################################\n", + "\n", + "#Given\n", + "#resistance per 100 metres\n", + "res = 2*0.005/100 #ohm/m\n", + "cur_uni = 1.0 #A/m\n", + "cur_con = 200.0 #A\n", + "len_uni = 500.0\n", + "#Let us assume that Ib current flows from point B.\n", + "#Considering a element dx in BD(500 metres) at a distance of X units(100 m each)\n", + "#voltage drop over dx = (1-100*x)*res*dx\n", + "#voltage drop over BD by integrating is = 0.05*Ib - 12.5\n", + "#voltage drop over DC = (Ib-500)*0.01\n", + "#voltage drop over CA = (Ib-700)*0.01*4\n", + "#total drop over AB = \n", + "tot_drop = 242.0-240.0\n", + "#summation of drops from AC + CD + DB\n", + "from sympy import Eq, var, solve\n", + "var('Ib') \n", + "sum = (Ib-500)*0.01 +(Ib-700)*0.01*4 + 0.05*Ib - 12.5\n", + "\n", + "eq = Eq(sum,tot_drop)\n", + "\n", + "Ib = solve(eq)\n", + "Ib = float(Ib[0])\n", + "#Total current\n", + "cur_tot = len_uni*cur_uni + cur_con\n", + "Ia = cur_tot - Ib #A\n", + "#Current in distributed load\n", + "cur_dis = Ia-cur_con #A\n", + "#point of minimum potential from D is\n", + "distD = cur_dis/cur_uni\n", + "#Therefore distance from B is\n", + "distB = len_uni-distD\n", + "#Therefore voltage drop is\n", + "from scipy.integrate import quad\n", + "\n", + "def integrand(x):\n", + " return (Ib-100*x)*res*100\n", + "\n", + "ans, err = quad(integrand, 0, (distB/100))\n", + "#Therefore potential of M is\n", + "pot_M = 242.0-ans #V\n", + "print \"Current fed at A is = \",Ia,\"A.\"\n", + "print \"Current fed at B is = \",Ib,\"A.\"\n", + "print \"Point of minimum potential from B is = \",distB,\"metres.\"\n", + "print \"Voltage at minimum potential is = \",round(pot_M,2),\"V.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.17 ,Page No :- 1590" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage at B is = 236.9 V.\n", + "Voltage at C is = 235.98 V.\n", + "Voltage at D is = 237.45 V.\n" + ] + } + ], + "source": [ + "#A 400-metre ring distributor has loads as shown in Fig. 40.29(a) where distances are in metres.The resistance\n", + "#of each conductor is 0.2 ohm per 1000 metres and the loads tapped off at points B,C,D are as shown.If the\n", + "#distributor is fed at A,find voltages at B,C and D.\n", + "#################################################################################################################\n", + "\n", + "#Given\n", + "\n", + "res = 0.2/1000 #ohm/m\n", + "V_a = 240.0 #V\n", + "#loads and positions\n", + "i_b = 100.0 #A\n", + "i_c = 70.0 #A\n", + "i_d = 50.0 #A\n", + "l_ab = 60.0 #m\n", + "l_bc = 80.0 #m\n", + "l_cd = 90.0 #m\n", + "l_da = 70.0 #m\n", + "\n", + "#total drop ->70i + 90(i-50)+80(i-120)+60(i-220)=0\n", + "cur_i = (l_cd*i_d + l_bc*(i_d+i_c) + l_ab*(i_d+i_c+i_b))/(l_ab+l_bc+l_cd+l_da)\n", + "#drops in different sections\n", + "drop_da = 2*cur_i*l_da*res\n", + "drop_cd = 2*(cur_i-i_d)*l_cd*res\n", + "drop_bc = 2*abs(cur_i-i_d-i_c)*l_bc*res\n", + "drop_ab = 2*abs(cur_i-i_d-i_c-i_b)*l_ab*res\n", + "\n", + "#voltages at different points\n", + "V_d = V_a - drop_da\n", + "V_c = V_d - drop_cd\n", + "V_b = V_a - drop_ab\n", + "print \"Voltage at B is = \",round(V_b,2),\"V.\"\n", + "print \"Voltage at C is = \",round(V_c,2),\"V.\"\n", + "print \"Voltage at D is = \",round(V_d,2),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.18 ,Page No :- 1591" + ] + }, + { + "cell_type": "code", + "execution_count": 53, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage at B is = 394.2 V.\n", + "Voltage at C is = 393.42 V.\n", + "Current in section BC is = 43.33 A.\n" + ] + } + ], + "source": [ + "#In a direct current ring main,a voltage of 400V is maintained at A.At B,500 metres away from A,a load of 150A is taken\n", + "#and at C,300 metres from B,a load of 200A is taken.The distance between A and C is 700 metres.The resistance of each\n", + "#conductor of the mains is 0.03ohm per 1000 metres.Find the voltage at B and C and also find the current in the section BC.\n", + "##############################################################################################################################\n", + "\n", + "#Given\n", + "V_a = 400.0 #V\n", + "res = 0.03/1000 #ohm/m\n", + "#loads and positions\n", + "i_b = 150.0 #A\n", + "i_c = 200.0 #A\n", + "l_ab = 500.0 #m\n", + "l_bc = 300.0 #m\n", + "l_ca = 700.0 #m\n", + "\n", + "#total drop-> 500i + 300(i-150) + 700(i-350) = 0\n", + "cur_i = (l_bc*i_b + l_ca*(i_b+i_c))/(l_ab+l_bc+l_ca)\n", + "#current in different sections\n", + "cur_ab = cur_i\n", + "cur_bc = cur_i-i_b\n", + "cur_ca = abs(cur_bc-i_c)\n", + "#drops in different sections\n", + "drop_ab = 2*cur_ab*l_ab*res\n", + "drop_bc = 2*cur_bc*l_bc*res\n", + "#voltages in different sections\n", + "V_b = V_a-drop_ab\n", + "V_c = V_b-drop_bc\n", + "print \"Voltage at B is = \",round(V_b,2),\"V.\"\n", + "print \"Voltage at C is = \",round(V_c,2),\"V.\"\n", + "print \"Current in section BC is = \",round(cur_bc,2),\"A.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.19 ,Page No :- 1591" + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in AB,BC,CD,DE,EA is 29.04 A, 19.04 A, 0.96 A, 30.96 A, 40.96 A respectively.\n", + "\n", + "Voltage at B,C,D,E is 217.1 V, 216.14 V, 216.15 V, 216.93 V respectively\n", + "\n", + "Current in AB,BC,DE,CE,EA is 27.72 A, 17.72 A, 32.28 A, 9.76 A, 42.28 A respectively.\n", + "\n", + "Voltage at B,C,D,E is 217.23 V, 216.34 V, 216.02 V, 216.83 V respectively\n" + ] + } + ], + "source": [ + "#A d.c ring main ABCDE is fed at point A from a 220-V supply and the resistances(including both lead and return)\n", + "#of the various sections are as follows(in ohms):AB=0.1;BC=0.05;CD=0.01;DE=0.025 and EA=0.075.The main supplies\n", + "#loads of 10A at B; 20A at C; 30A at D and 10A at E.Find the magnitude and direction of the current flowing in each\n", + "#section and the voltage at each load point.\n", + "#If the points C and E are further linked together by a conductor of 0.05 ohm resistance and the output currents\n", + "#from the mains remain unchanged,find the new distribution of the current and voltage in the network.\n", + "#####################################################################################################################\n", + "\n", + "#Given\n", + "\n", + "V_a = 220.0 #V\n", + "#resistances of different sections\n", + "r_ab = 0.1 #ohm\n", + "r_bc = 0.05 #ohm\n", + "r_cd = 0.01 #ohm\n", + "r_de = 0.025 #ohm\n", + "r_ea = 0.075 #ohm\n", + "#loads\n", + "i_b = 10.0 #A\n", + "i_c = 20.0 #A\n", + "i_d = 30.0 #A\n", + "i_e = 10.0 #A\n", + "#total drop -> 0.1i + 0.05(i-10) + 0.01(i-30) + 0.025(i-60) + 0.075(i-70)=0\n", + "cur_i = (r_bc*i_b + r_cd*(i_b+i_c) + r_de*(i_b+i_c+i_d) + r_ea*(i_b+i_c+i_d+i_e))/(r_ab+r_bc+r_cd+r_de+r_ea)\n", + "#current in different sections\n", + "cur_ab = cur_i\n", + "cur_bc = cur_ab-i_b\n", + "cur_cd = cur_bc-i_c\n", + "cur_de = cur_cd-i_d\n", + "cur_ea = cur_de-i_e\n", + "\n", + "#drops in different sections\n", + "drop_ab = cur_ab*r_ab\n", + "drop_bc = cur_bc*r_bc\n", + "drop_de = abs(cur_de)*r_de\n", + "drop_ea = abs(cur_ea)*r_ea\n", + "#voltages at different points\n", + "V_b = V_a - drop_ab\n", + "V_c = V_b - drop_bc\n", + "V_e = V_a - drop_ea\n", + "V_d = V_e - drop_de\n", + "print \"Current in AB,BC,CD,DE,EA is\",round(cur_ab,2),\"A,\",round(cur_bc,2),\"A,\",round(abs(cur_cd),2),\"A,\",round(abs(cur_de),2),\"A,\",round(abs(cur_ea),2),\"A respectively.\" \n", + "print \"\"\n", + "print \"Voltage at B,C,D,E is\",round(V_b,2),\"V,\",round(V_c,2),\"V,\",round(V_d,2),\"V,\",round(V_e,2),\"V respectively\"\n", + "print \"\"\n", + "#part-2\n", + "#Potential difference between end points of interconnector(CE)\n", + "V_ce = V_e-V_c\n", + "#Resistance between CE ,as shown in figure\n", + "r1 = r_ab+r_bc+r_ea\n", + "r2 = r_de + r_cd\n", + "res_ce = r1*r2/(r1+r2)+ 0.05\n", + "\n", + "#Current in interconnector [I = V/R Ohm's Law]\n", + "cur_ce = V_ce/res_ce\n", + "#Current goes from E to C as E is at higher potential.\n", + "\n", + "#The current in other sections will also change.\n", + "#let us assume i1 along ED, voltage round the closed mesh EDC is zero.\n", + "#total drop -> -0.025*i1-0.01*(i1-30)+0.05*9.75 = 0\n", + "\n", + "cur_i1 = (0.05*cur_ce + r_cd*i_d)/(r_cd+r_de)\n", + "\n", + "current_ea = i_e+cur_i1+cur_ce\n", + "current_ab = (i_b+i_c+i_d+i_e)-current_ea\n", + "current_bc = current_ab-i_b\n", + "current_de = current_ea-i_e\n", + "#new drops\n", + "drop_ab = current_ab*r_ab\n", + "drop_bc = current_bc*r_bc\n", + "drop_ea = current_ea*r_ea\n", + "drop_de = current_de*r_de\n", + "\n", + "#new potentials\n", + "V_b = V_a - drop_ab\n", + "V_c = V_b - drop_bc\n", + "V_e = V_a - drop_ea\n", + "V_d = V_e - drop_de\n", + "\n", + "print \"Current in AB,BC,DE,CE,EA is\",round(current_ab,2),\"A,\",round(current_bc,2),\"A,\",round(current_de,2),\"A,\",round(cur_ce,2),\"A,\",round(current_ea,2),\"A respectively.\"\n", + "print \"\"\n", + "print \"Voltage at B,C,D,E is\",round(V_b,2),\"V,\",round(V_c,2),\"V,\",round(V_d,2),\"V,\",round(V_e,2),\"V respectively\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.20 ,Page No :- 1594" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage across 3 ohm load is = 244.9 V.\n", + "Voltage across 4 ohm load is = 247.9 V.\n" + ] + } + ], + "source": [ + "#In a 3-wire distribution system,the supply voltage is 250V on each side.The load on one side is a 3 ohm\n", + "#resistance and on the other, a 4 ohm resistance.The resistance of each of the 3 conductors is 0.05 ohm.\n", + "#Find the load voltages.\n", + "#########################################################################################################\n", + "\n", + "import numpy as np\n", + "#Given\n", + "#Resistances\n", + "res_1 = 3.0 #ohm\n", + "res_2 = 4.0 #ohm\n", + "res_con = 0.05 #ohm\n", + "V_sup = 250.0 #V\n", + "\n", + "#Let the assumed directions of unknown currents be as shown in figure.\n", + "#KVL for ABCD\n", + "# (3+0.05)x + 0.05(x-y) = 250 -------------- eqn 1\n", + "a = res_1 + 2*res_con\n", + "b = -(res_con)\n", + "#KVL for DCEFD\n", + "# 0.05(y-x) + (4+0.05)y = 250 -------------- eqn 2\n", + "c = res_2+ 2*res_con \n", + "#Solving eqn 1 and eqn2\n", + "m = [[a,b],[b,c]]\n", + "n = [V_sup,V_sup]\n", + "soln = np.linalg.solve(m,n) #soln is array with its elements[x,y]\n", + "#Calculating the load voltages\n", + "#V1 = 250-0.05*x-0.05(x-y)\n", + "vol1 = V_sup - res_con*soln[0]-res_con*(soln[0]-soln[1]) #V\n", + "#V2 = 250 + 0.05(x-y)- 0.05y\n", + "vol2 = V_sup + res_con*(soln[0]-soln[1]) - res_con*soln[1] #V\n", + "print \"Voltage across 3 ohm load is = \",round(vol1,1),\"V.\"\n", + "print \"Voltage across 4 ohm load is = \",round(vol2,1),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.21 ,Page No :- 1594" + ] + }, + { + "cell_type": "code", + "execution_count": 57, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Potential Difference across AB is = 248.62 V.\n", + "Potential Difference across QK is = 247.83 V.\n", + "Potential Difference across CD is = 248.4 V.\n", + "Potential Difference across FE is = 247.65 V.\n" + ] + } + ], + "source": [ + "#A 3-wire d.c distributor PQ,250 metres long,is supplied at end P at 500/250V and is loaded as under:\n", + "#Positive side: 20A 150 metres from P ; 30A 250 metres from P.\n", + "#Negative side: 24A 100 metres from P ; 36A 220 metres from P.\n", + "#The resistance of each outer wire is 0.02 ohm per 100 metres and the cross-section of the middle wire\n", + "#is one-half of the outer.Find the voltage across each load point.\n", + "##########################################################################################################\n", + "\n", + "#Given\n", + "V_PN = 250.0 #V\n", + "V_NR = 250.0 #V\n", + "res_out = 0.02/100 #ohm/m\n", + "res_mid = 2*res_out #ohm/m (Area of middle wire is half.As, R = rho*l/A .Therefore,Resistance doubles)\n", + "\n", + "#Given Currents\n", + "i_ab = 20.0 #A\n", + "i_qk = 30.0 #A\n", + "i_cd = 24.0 #A\n", + "i_fe = 36.0 #A\n", + "\n", + "#Currents in different sections\n", + "i_pa = i_ab+i_qk #A\n", + "i_aq = i_qk #A\n", + "i_fk = i_qk #A\n", + "i_bf = i_fe-i_qk #A\n", + "i_bc = i_ab-i_bf #A\n", + "i_cn = i_cd-i_bc #A\n", + "i_de = i_fe #A\n", + "i_dr = i_cd+i_fe #A\n", + "\n", + "\n", + "#lengths of different sections\n", + "l_pa = 150.0 #m\n", + "l_aq = 100.0 #m\n", + "l_kf = 250.0-220.0 #m\n", + "l_bc = 150.0-100.0 #m\n", + "l_bf = 220.0-150.0 #m\n", + "l_cn = 100.0 #m\n", + "l_de = 220.0-100.0 #m\n", + "l_dr = 100.0 #m\n", + "\n", + "#Resistances of different sections\n", + "r_pa = l_pa*res_out #ohm\n", + "r_aq = l_aq*res_out #ohm\n", + "r_kf = l_kf*res_mid #ohm\n", + "r_bc = l_bc*res_mid #ohm\n", + "r_bf = l_bf*res_mid #ohm\n", + "r_cn = l_cn*res_mid #ohm\n", + "r_de = l_de*res_out #ohm\n", + "r_dr = l_dr*res_out #ohm\n", + "\n", + "#Drop across different sections\n", + "drop_pa = r_pa*i_pa #V\n", + "drop_aq = r_aq*i_aq #V\n", + "drop_kf = r_kf*i_fk #V\n", + "drop_bc = r_bc*i_bc #V\n", + "drop_bf = r_bf*i_bf #V\n", + "drop_cn = r_cn*i_cn #V\n", + "drop_de = r_de*i_de #V\n", + "drop_dr = r_dr*i_dr #V\n", + "\n", + "#Voltages across different sections\n", + "vol_ab = V_PN - drop_pa - drop_bc + drop_cn #V\n", + "vol_qk = vol_ab - drop_aq - drop_kf + drop_bf #V\n", + "vol_cd = V_NR - drop_cn - drop_dr #V \n", + "vol_fe = vol_cd + drop_bc - drop_bf - drop_de #V\n", + "\n", + "print \"Potential Difference across AB is = \",round(vol_ab,2),\"V.\"\n", + "print \"Potential Difference across QK is = \",round(vol_qk,2),\"V.\"\n", + "print \"Potential Difference across CD is = \",round(vol_cd,2),\"V.\"\n", + "print \"Potential Difference across FE is = \",round(vol_fe,2),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.22 ,Page No :- 1597" + ] + }, + { + "cell_type": "code", + "execution_count": 58, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total load on main generator is = 155.0 kW.\n", + "Load on Balancer 1 is = 22.5 kW.\n", + "Load on Balancer 2 is = 27.5 kW.\n" + ] + } + ], + "source": [ + "#A d.c 3-wire system with 500-V between outers has lighting load of 100kW on the positive and 50kW on the\n", + "#negative side.If,at this loading,the balancer machines have each a loss of 2.5kW,Calculate the kW loading\n", + "#of each balancer machine and the total load on the system.\n", + "###########################################################################################################\n", + "\n", + "#Given\n", + "V_out = 500.0 #V\n", + "load_p = 100.0 #kW (positive side)\n", + "load_n = 50.0 #KW (negative side)\n", + "load_b = 2.5 #kW (balancer machine)\n", + "#total load on main generator\n", + "load_tot = load_p + load_n + 2*load_b #kW\n", + "#Output current of main generator\n", + "cur_out = load_tot*1000/V_out #W/V->A\n", + "#load current on positive side\n", + "cur_p = load_p*1000/(V_out/2) #A\n", + "#load current on negative side\n", + "cur_n = load_n*1000/(V_out/2) #A\n", + "#Current through neutral(Out of balance)\n", + "cur_o = cur_p-cur_n #A\n", + "\n", + "#Currents of balancer\n", + "cur_b1 = cur_p-cur_out #A\n", + "cur_b2 = cur_o - cur_b1 #A\n", + "\n", + "#Load on balancer\n", + "load_b1 = (V_out/2)*cur_b1/1000 #kW\n", + "load_b2 = (V_out/2)*cur_b2/1000 #kW\n", + "\n", + "print \"Total load on main generator is = \",round(load_tot,2),\"kW.\"\n", + "print \"Load on Balancer 1 is = \",round(load_b1,2),\"kW.\"\n", + "print \"Load on Balancer 2 is = \",round(load_b2,2),\"kW.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.23 ,Page No :- 1598" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total load on main generator is = 1216.0 kW.\n", + "Current through Balancer 1 is = 168.0 A.\n", + "Current through Balancer 2 is = 232.0 A.\n" + ] + } + ], + "source": [ + "#In a 500/250-V d.c 3-wire system,there is a current of 2000A on the +ve side, 1600A on the negative side\n", + "#and a load of 300 kW across the outers.The loss in each balancer set is 8 kW.Calculate the current in each\n", + "#armature of the balancer set and total load on the main generator.\n", + "#############################################################################################################\n", + "\n", + "#Given\n", + "V_out = 500.0 #V\n", + "cur_p = 2000.0 #A (current on positive side)\n", + "cur_n = 1600.0 #A (current on negative side)\n", + "load_ext = 300.0 #kW (across outers)\n", + "load_b = 8.0 #kW (loss in balancer set)\n", + "#loading on positive side\n", + "load_p = (cur_p*(V_out/2))/1000 #kW\n", + "#loading on negative side\n", + "load_n = (cur_n*(V_out/2))/1000 #kW\n", + "#Total loading on main generator\n", + "load_tot = load_p + load_n + 2*load_b + load_ext #kW\n", + "\n", + "#current on main generator -> I = W/V\n", + "cur_tot = load_tot*1000/V_out #A\n", + "\n", + "#current through neutral(out of balance)\n", + "cur_o = cur_p-cur_n #A\n", + "\n", + "#current through external resistance\n", + "cur_ext = load_ext*1000/V_out #A\n", + "\n", + "#current through balancer sets\n", + "cur_b1 = (cur_p+cur_ext)-cur_tot #A\n", + "cur_b2 = cur_o - cur_b1 #A\n", + "\n", + "print \"Total load on main generator is = \",round(load_tot,2),\"kW.\"\n", + "print \"Current through Balancer 1 is = \",round(cur_b1,2),\"A.\"\n", + "print \"Current through Balancer 2 is = \",round(cur_b2,2),\"A.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.24 ,Page No :- 1598" + ] + }, + { + "cell_type": "code", + "execution_count": 60, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current supplied by generator is = 7000.0 A.\n", + "Current in positive side is = 6000.0 A.\n", + "Current in negative side is = 8000.0 A.\n", + "Current in neutral is = 2000.0 A.\n", + "Current through armature 1 is = 1000.0 A.\n", + "Current through armature 2 is = 1000.0 A.\n" + ] + } + ], + "source": [ + "#On a 3-wire d.c distribution system with 500V between outers,there is a load of 1500kW on the positive\n", + "#side and 2000 kW on the negative side.Calculate the current in the neutral and in each of the balancer\n", + "#armatures and the total current supplied by the generator.Neglect losses.\n", + "##########################################################################################################\n", + "\n", + "#Given\n", + "V_out = 500.0 #V\n", + "load_p = 1500.0 #kW (load on positive side)\n", + "load_n = 2000.0 #kW (load on negative side)\n", + "#total loading on main generator\n", + "load_tot = load_p + load_n #kW\n", + "#current supplied by generator\n", + "cur_tot = load_tot*1000/V_out #A\n", + "#current on positive side\n", + "cur_p = load_p*1000/(V_out/2) #A\n", + "#current on negative side\n", + "cur_n = load_n*1000/(V_out/2) #A\n", + "#current in neutral(out of balance)\n", + "cur_o = abs(cur_p-cur_n) #A\n", + "#current through armatures\n", + "cur_b1 = cur_tot-cur_p #A\n", + "cur_b2 = cur_o-cur_b1 #A\n", + "\n", + "print \"Current supplied by generator is = \",cur_tot,\"A.\"\n", + "print \"Current in positive side is = \",cur_p,\"A.\"\n", + "print \"Current in negative side is = \",cur_n,\"A.\"\n", + "print \"Current in neutral is = \",cur_o,\"A.\"\n", + "print \"Current through armature 1 is = \",cur_b1,\"A.\"\n", + "print \"Current through armature 2 is = \",cur_b2,\"A.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.25 ,Page No :- 1599" + ] + }, + { + "cell_type": "code", + "execution_count": 61, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in balancer set 1 is = 22.0 A.\n", + "Current in balancer set 2 is = 28.0 A.\n", + "Output of main generator is = 119.5 kW.\n" + ] + } + ], + "source": [ + "#A 125/250 V,3-wire distributor has an out-of-balance current of 50 A and larger load of 500 A.The balancer\n", + "#set has a loss of 375 W in each machine.Calculate the current in each of the balancer machines and output\n", + "#of main generator.\n", + "############################################################################################################\n", + "\n", + "#Given\n", + "V_out = 250.0 #V\n", + "#Currents\n", + "cur_p = 500.0 #A\n", + "cur_o = 50.0 #A\n", + "cur_n = cur_p - cur_o #A\n", + "#larger Load\n", + "load_p = cur_p*(V_out/2)/1000 #kW\n", + "#smaller Load\n", + "load_n = cur_n*(V_out/2)/1000 #kW\n", + "#Balancer loss\n", + "loss_b = 2*375.0/1000 #kW\n", + "#total load on generator\n", + "load_tot = load_p + load_n + loss_b\n", + "#current from main generator -> VI = W\n", + "cur_tot = load_tot*1000/V_out #A\n", + "\n", + "#Current in balancer sets\n", + "cur_b1 = cur_p - cur_tot #A\n", + "cur_b2 = cur_o - cur_b1 #A\n", + "print \"Current in balancer set 1 is = \",cur_b1,\"A.\"\n", + "print \"Current in balancer set 2 is = \",cur_b2,\"A.\"\n", + "print \"Output of main generator is = \",load_tot,\"kW.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.26 ,Page No :- 1599" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total load on main generator is = 1210.0 kW.\n", + "Load on Balancer set 1 is = 20.0 kW.\n", + "Load on balancer set 2 is = 30.0 kW.\n" + ] + } + ], + "source": [ + "#The load on d.c 3-wire system with 500 V between outers consists of lighting current of 1500 A on the\n", + "#positive side and 1300 A on the negative side while motors connected across the outers absorb 500kW.\n", + "#Assuming that at this loading,the balancer machines have each a loss of 5kW,calculate the load on the\n", + "#main generator and on each of the balancer machines.\n", + "##########################################################################################################\n", + "\n", + "#Given\n", + "cur_p = 1500.0 #A\n", + "cur_n = 1300.0 #A\n", + "V_out = 500.0 #V\n", + "load_ext = 500.0 #kW\n", + "loss_b = 2*5.0 #kW\n", + "\n", + "#current through external load\n", + "cur_ext = load_ext*1000/V_out #A\n", + "#larger load\n", + "load_p = cur_p*(V_out/2)/1000 #kW\n", + "#smaller load\n", + "load_n = cur_n*(V_out/2)/1000 #kW\n", + "#total load on generator\n", + "load_tot = load_p + load_n + loss_b + load_ext #kW\n", + "#current from generator -> VI = W\n", + "cur_tot = load_tot*1000/V_out #A\n", + "#current through neutral(out of balance)\n", + "cur_o = cur_p-cur_n #A\n", + "#current through balancer sets\n", + "cur_b1 = (cur_p+cur_ext)-cur_tot #A\n", + "cur_b2 = cur_o-cur_b1 #A\n", + "#load of balancer sets\n", + "load_b1 = cur_b1*(V_out/2)/1000 #kW\n", + "load_b2 = cur_b2*(V_out/2)/1000 #kW\n", + "\n", + "print \"Total load on main generator is = \",load_tot,\"kW.\"\n", + "print \"Load on Balancer set 1 is = \",load_b1,\"kW.\"\n", + "print \"Load on balancer set 2 is = \",load_b2,\"kW.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.27 ,Page No :- 1599" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage across Balancer 1 is = 230.0 A.\n", + "Voltage across Balancer 2 is = 250.0 A.\n", + "Load current on main generator is = 1110.0 A.\n" + ] + } + ], + "source": [ + "#A d.c 3-wire system with 480 V across outers supplies 1200 A on the positive and 1000 A on the negative side.\n", + "#The balancer machines have each an armature resistances of 0.1W and take 10 A on no-load.Find\n", + "#(a)the voltage across each balancer and\n", + "#(b)the total load on the main generator and the current loading of each balancer machine.\n", + "#The balancer field windings are in series across the outers\n", + "################################################################################################################\n", + "\n", + "#Given\n", + "V_out = 480.0 #V\n", + "#currents\n", + "cur_p = 1200.0 #A\n", + "cur_n = 1000.0 #A\n", + "cur_o = cur_p - cur_n #A (out of balance)\n", + "#armature resistance \n", + "res_arm = 0.1 #ohm\n", + "#no-load current\n", + "cur_nold = 10.0 #A\n", + "\n", + "#Let us assume current Im flows through mtoring machine,then (200-Im) flows through generating machine.\n", + "#Let Vg and Vm be potential difference of 2 machines.\n", + "\n", + "#Total losses in sets = no-load losses + Cu losses in two machines\n", + "#loss_set = V_out*cur_nold + 0.1*Im^2+ 0.1*(200-Im)^2\n", + "#Vm*Im = Vg*Ig + loss_set\n", + "#Now, Vm = Eb+Im*Ra Vg = Eb-Ig*Ra\n", + "Eb = V_out/2-res_arm*cur_nold\n", + "\n", + "#Therefore, Vm = 239 + Im*0.1 and Vg = 239 - (200-Im)*0.1\n", + "#Hence,Equation is \n", + "#(239+0.1*Im)*Im = [239 - (200-Im)*0.1]*(200-Im) + loss_set\n", + "#Simplified -> 239Im = 239*(200-Im)+4800\n", + "\n", + "#Solving this equation\n", + "from sympy import Eq, var, solve\n", + "var('Im') \n", + "eq = Eq(Eb*(2*Im-cur_o),V_out*cur_nold)\n", + "Im = solve(eq)\n", + "Im = int(Im[0])\n", + "Ig = cur_o-Im\n", + "#Voltage across balancers\n", + "\n", + "Vm = Eb + Im*res_arm #V\n", + "Vg = Eb - Ig*res_arm #V \n", + "\n", + "#Load on main generator\n", + "cur_load = cur_p - Ig #A\n", + "print \"Voltage across Balancer 1 is = \",round(Vg,2),\"A.\"\n", + "print \"Voltage across Balancer 2 is = \",round(Vm,2),\"A.\"\n", + "print \"Load current on main generator is = \",round(cur_load,2),\"A.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.28 ,Page No :- 1600" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage on positive side is = 283.0 V.\n", + "Voltage on negative side is = 177.0 V.\n" + ] + } + ], + "source": [ + "#A d.c 3-wire system with 460V between outers supplies 250kW on the positive and 400kW on the negative side,\n", + "#the voltages being balanced.Calculate the voltage on the positive and negative side,the voltages being balanced.\n", + "#Calculate the voltage on the positive and negative sides repectively,if the neutral wire becomes disconnected\n", + "#from balancer set.\n", + "#################################################################################################################\n", + "\n", + "#Given\n", + "V_mid = 230.0 #V\n", + "V_out = 460.0 #V\n", + "#loads\n", + "load_p = 250.0 #kW\n", + "load_n = 400.0 #kW\n", + "#resistance on positive side -> (V^2/R) = W\n", + "res_p = (V_mid*V_mid)/(load_p*1000) #ohm\n", + "\n", + "#resistance on negative side -> (V^2/R) = W\n", + "res_n = (V_mid*V_mid)/(load_n*1000) #ohm\n", + "\n", + "#Voltages after disconnecting balancer set\n", + "vol_p = (res_p/(res_p+res_n))*V_out #V\n", + "vol_n = V_out - vol_p #V\n", + "\n", + "print \"Voltage on positive side is = \",round(vol_p),\"V.\"\n", + "print \"Voltage on negative side is = \",round(vol_n),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "EXAMPLE 40.29 ,Page No :- 1601" + ] + }, + { + "cell_type": "code", + "execution_count": 66, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Terminal potential difference of the booster is = 180.0 V.\n", + "Output of booster is = 21.6 kW.\n" + ] + } + ], + "source": [ + "#A 2-wire system has the voltage at the supply end maintained at 500.The line is 3 km long.If the full-load\n", + "#current is 120 A,what must be the booster voltage and output in order that the far end voltage may also be 500 V.\n", + "#Take the resistance of the cable at the working temperature as 0.5ohm/kilometre.\n", + "####################################################################################################################\n", + "\n", + "#Total resistance of line\n", + "res_tot = 0.5*3 #ohm\n", + "#Full load current\n", + "cur_full = 120.0 #A\n", + "\n", + "#drop in the line-> V=IR\n", + "drop = res_tot*cur_full #V\n", + "\n", + "#Output of booster ->VI = W\n", + "output = drop*cur_full/1000 #kW\n", + "\n", + "print \"Terminal potential difference of the booster is = \",drop,\"V.\"\n", + "print \"Output of booster is = \",round(output,2),\"kW.\"" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |