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| author | Trupti Kini | 2017-03-10 23:30:19 +0600 |
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| committer | Trupti Kini | 2017-03-10 23:30:19 +0600 |
| commit | e217ca98f7ad102376abf21791c15b24439ac04f (patch) | |
| tree | aece270607bc625853e8fe66ae42b49bcbd81837 /A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter41.ipynb | |
| parent | 4ff7f8fe804a5ed609522a76661dfa513d927fb8 (diff) | |
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Added(A)/Deleted(D) following books
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter40.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter40.ipynb
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter41.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter41.ipynb
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter43.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter43.ipynb
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter44.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter44.ipynb
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter45.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter45.ipynb
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter46.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter46.ipynb
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter47.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter47.ipynb
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter49.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter49.ipynb
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter50.ipynb -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter50.ipynb
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/screenshots/figure_1.png -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/screenshots/figure_1.png
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/screenshots/figure_2.png -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/screenshots/figure_2.png
R A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/screenshots/figure_3.png -> A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/screenshots/figure_3.png
A BSc_1st_Year_Physics_by_P._Balabhaskar,_N._Srinivasa_Rao,_B._Sanjeeva_Rao/README.txt
A Electric_Drives_Concepts_And_Applications_by_Vedam_Subrahmanyam/README.txt
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diff --git a/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter41.ipynb b/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter41.ipynb new file mode 100644 index 00000000..0a494d2d --- /dev/null +++ b/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter41.ipynb @@ -0,0 +1,3477 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 41 : A.C. Transmission and Distribution" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.1 , PAGE NO :- 1613" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cu for 3-phase system = 0.29 * Cu for dc system.\n" + ] + } + ], + "source": [ + "'''A 3-phase, 4-wire system is used for lighting. Compare the amount of copper required with that needed for a\n", + "2-wire D.C. system with same line voltage. Assume the same losses and balanced load. The neutral is one half\n", + "the cross-section of one of the respective outers.'''\n", + "\n", + "from sympy import Symbol\n", + "\n", + "#(a) Two-wire DC\n", + "#We know that, I = P/V .Therefore\n", + "I = Symbol('I')\n", + "#Also let the resistance be R1\n", + "R1 = Symbol('R1')\n", + "#power loss is\n", + "loss1 = 2*(I*I)*R1 \n", + "\n", + "#(b) Three-phase,4-wire\n", + "\n", + "#We know that, I2 = P/3*V .Therefore I2 = I/3\n", + "#Also let the resistance be R2\n", + "R2 = Symbol('R2')\n", + "#power loss is\n", + "loss2 = 3*(I/3*I/3)*R2\n", + "\n", + "#loss1/loss2 = 2*I^2*R1/(I^2*R2*1/3) .Let ratio of resistances is R1/R2 = r1_r2\n", + "#As loss1 = loss2\n", + "r1_r2 = 1.0/6\n", + "\n", + "#Let the ratio of areas of conductors be a1_a2 As R o< 1/A\n", + "a2_a1 = 1/r1_r2\n", + "\n", + "#Cu loss of 3-phase/Cu loss of dc system = 3.5*A2*l/2*A1*l\n", + "ratio = (3.5/2)/a2_a1\n", + "\n", + "print \"Cu for 3-phase system = \",round(ratio,2),\" * Cu for dc system.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.2 , PAGE NO :- 1613" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wt of copper for 3-conductors = 9349.58 kg.\n" + ] + } + ], + "source": [ + "'''Estimate the weight of copper required to supply a load of 100 MW at upf by a 3-phase, 380-kV system over a distance\n", + "of 100 km. The neutral point is earthed. The resistance of the conductor is 0.045 ohm/cm^2/km. The weight of copper\n", + "is 0.01 kg/cm^3. The efficiency of transmission can be assumed to be 90 percent.'''\n", + "\n", + "#Power loss in the line\n", + "loss1 = (1 - 0.9) * 100.0 #MW\n", + "#Line current Il = P/vl*cosQ\n", + "Il = 100 * 1.0e+6/(1.732*380*(1.0e+3)*1) #A\n", + "#Since I^2R loss in 3-conductors is loss1, loss per conductor is\n", + "loss_c = loss1*1.0e+6/3 #W\n", + "\n", + "#Resistance per conductor Using loss = I^2*R\n", + "R_c = loss_c/(Il*Il) #ohm\n", + "#Resistance per conductor per km\n", + "R_km = R_c/100 #ohm\n", + "#Conductor cross-section\n", + "Vol = 0.045/R_km #m^3\n", + "#Volume of copper per meter run\n", + "Vol = Vol*100 #cm^3\n", + "#Weight of copper for 3-conductor for 100 km length\n", + "wt = 3 * (Vol * 0.01) * 100 * 1000 #kg\n", + "\n", + "print \"Wt of copper for 3-conductors = \",round(wt,2),\"kg.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.3 , PAGE NO :- 1614" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Additional power transmitted = 80.5 %.\n" + ] + } + ], + "source": [ + "'''A d.c. 2-wire distribution system is converted into a.c. 3-phase, 3-wire system by adding a third conductor of\n", + "the same size as the two existing conductors. If voltage between conductors and percentage power loss remain the same, calculate\n", + "the percentage additional balanced load which can now be carried by the conductors at 0.95 p.f.'''\n", + "\n", + "from sympy import Symbol\n", + "\n", + "#(a)DC 2-wire system\n", + "\n", + "#Let us assume\n", + "V = 1.0\n", + "R = 1.0\n", + "I1 = Symbol('I1')\n", + "#Power transmitted\n", + "P1 = V*I1\n", + "#Power loss\n", + "loss1 = 2*(I1**2)*R \n", + "#%power loss = power loss/Power transmitted\n", + "ploss1 = loss1/P1\n", + "\n", + "#(b)3-phase, 3-wire system\n", + "\n", + "I2 = Symbol('I2')\n", + "#Power transmitted P = 1.732*VI*cosQ\n", + "P2 = 1.732*V*I2*(0.95)\n", + "#Power loss\n", + "loss2 = 3*(I2**2)*R\n", + "#%power loss = power loss/Power transmitted\n", + "ploss2 = loss2/P2\n", + "\n", + "#As ploss1 == ploss2\n", + "I2 = 2*(0.95)*I1/1.732\n", + "P2 = 1.732*V*I2*(0.95)\n", + "#Add. power transmitted\n", + "ptrans = (P2 - P1)/P1 *100 #%\n", + "\n", + "print \"Additional power transmitted = \",round(ptrans,2),\"%.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.4 , PAGE NO :- 1614" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Additional power transmitted = 30.0 MW.\n" + ] + } + ], + "source": [ + "'''A 2-phase, 3-wire a.c. system has a middle conductor of same cross-sectional area as the outer and supplies a load\n", + "of 20 MW. The system is converted into 3-phase, 4-wire system by running a neutral wire. Calculate the new power which\n", + "can be supplied if voltage across consumer terminal and percentage line losses remain the same. Assume balanced load.'''\n", + "\n", + "from sympy import Symbol\n", + "\n", + "#(a)2-phase 3-wire system\n", + "\n", + "#Let us assume\n", + "V = 1.0\n", + "R = 1.0\n", + "cosQ = 1.0\n", + "I1 = Symbol('I1')\n", + "#Power transmitted\n", + "P1 = 2*V*I1*cosQ\n", + "#Power loss\n", + "loss1 = 2*(I1**2)*R \n", + "#%power loss = power loss/Power transmitted\n", + "ploss1 = loss1/P1\n", + "\n", + "#(b)3-phase, 4-wire system\n", + "\n", + "I2 = Symbol('I2')\n", + "#Power transmitted P = 1.732*VI*cosQ\n", + "P2 = 3*V*I2*cosQ\n", + "#Power loss\n", + "loss2 = 3*(I2**2)*R\n", + "#%power loss = power loss/Power transmitted\n", + "ploss2 = loss2/P2\n", + "\n", + "#As ploss1 == ploss2\n", + "I2 = I1\n", + "P2 = 3*V*I2*cosQ\n", + "#New Power that can be supplied P1/P2 = 20/x\n", + "pnew = 20*P2/P1 #MW\n", + "\n", + "print \"Additional power transmitted = \",round(pnew,2),\"MW.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.5 , PAGE NO :- 1617" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L = 1.02 uH/m.\n", + "L = 10.92 uH/m.\n" + ] + } + ], + "source": [ + "'''What is the inductance per loop metre of two parallel conductors of a single phase system if each has a diameter of 1\n", + "cm and their axes are 5 cm apart when conductors have a relative permeability of (a) unity and (b) 100. The relative\n", + "permeability of the surrounding medium is unity in both cases. End effects may be neglected and the current may be assumed\n", + "uniformly distributed over cross-section of the wires.'''\n", + "\n", + "import math as m\n", + "# (a)\n", + "# u = u0\n", + "u0 = 4*3.14*1.0e-7 #H/m\n", + "ui = 1.0\n", + "L = u0/3.14*(m.log(5.0/0.5) + ui/4)*1.0e+6 #uH/m\n", + "print \"L = \",round(L,2),\"uH/m.\"\n", + "# (b)\n", + "# u = u0\n", + "u0 = 4*3.14*1.0e-7 #H/m\n", + "ui = 100.0\n", + "L = u0/3.14*(m.log(5.0/0.5) + ui/4)*1.0e+6 #uH/m\n", + "print \"L = \",round(L,2),\"uH/m.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.6 , PAGE NO :- 1617" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "loop impedance is = 19.83 ohm.\n" + ] + } + ], + "source": [ + "'''A 20-km single-phase transmission line having 0.823 cm diameter has two line conductors separated by 1.5 metre.\n", + "The conductor has a resistance of 0.311 ohm per kilometre. Find the loop impedance of this line at 50 Hz.'''\n", + "\n", + "import math as m\n", + "\n", + "#Given\n", + "length = 20.0*1000 #m\n", + "u = 4*3.14*1.0e-7 #H\n", + "ui = 1.0\n", + "D = 1.5 #m\n", + "r = 0.823/2*1.0e-2 #m\n", + "#inductance is\n", + "L = length*(u/3.14*(m.log(D/r) + ui/4 )) #H\n", + "#reactance is\n", + "X = 2*3.14*50.0*L #ohm\n", + "#loop resistance\n", + "R = 2*length*0.311/1000 #ohm\n", + "#impedance is\n", + "Z = m.sqrt(X*X + R*R) #ohm\n", + "\n", + "print \"loop impedance is = \",round(Z,2),\"ohm.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.7 , PAGE NO :- 1618" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Capacitance (excluding ground effect) = 0.00928 uF/km\n", + "Capacitance (including ground effect) = 0.00932 uF/km\n" + ] + } + ], + "source": [ + "'''The conductors in a single-phase transmission line are 6 m above ground. Each conductor has a diameter of 1.5 cm\n", + "and the two conductors are spaced 3 m apart. Calculate the capacitance per km of the line\n", + "(i) excluding ground effect and (ii) including the ground effect.'''\n", + "\n", + "import math as m\n", + "#Given\n", + "D = 3.0 #m (distance between conductors)\n", + "r = 1.5/2*1e-2 #m (radius)\n", + "h = 6.0 #m (height from ground)\n", + "eps = 8.85e-12 #epsilon (constant)\n", + "\n", + "#(i)Capacitance per km excluding ground effect\n", + "Cn = 2*3.14*eps/(m.log(D/r))*1.0e+9 #uF/km\n", + "print \"Capacitance (excluding ground effect) = \",round(Cn,5),\"uF/km\"\n", + "\n", + "#(ii)Capacitance including ground effect\n", + "Cn = 2*3.14*eps/(m.log(D/(r*m.sqrt(1 + D*D/(4*h*h)))))*1.0e+9 #uF/km\n", + "print \"Capacitance (including ground effect) = \",round(Cn,5),\"uF/km\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.8 , PAGE NO :- 1620" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sending-end voltage = 3859.08 V.\n" + ] + } + ], + "source": [ + "'''A single-phase line has an impedance of 5ang(60) and supplies a load of 120 A,3,300 V at 0.8 p.f. lagging.\n", + "Calculate the sending-end voltage and draw a vector diagram.'''\n", + "\n", + "import cmath as cm\n", + "import math as m\n", + "#Given\n", + "Er = cm.rect(3300.0,0) #V (Voltage) \n", + "Z = cm.rect(5.0,3.14/3) #ohm (Impedance)\n", + "pf = 0.8 #power factor\n", + "theta = m.acos(pf) #Q (power factor angle)\n", + "\n", + "I = cm.rect(120.0,-theta) #A (current)\n", + "\n", + "#Voltage drop\n", + "V = (I)*(Z) #V\n", + "\n", + "#Sending-end voltage is\n", + "Es = Er + V #V\n", + "\n", + "Es = m.sqrt(Es.real**2 + Es.imag**2)#V\n", + "\n", + "print \"Sending-end voltage = \",round(Es,2),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPE 41.9 , PAGE NO :- 1620" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sending end voltage is = 33.71 V.\n", + "Power factor = 0.796 lag\n" + ] + } + ], + "source": [ + "'''An overhead, single-phase transmission line delivers 1100 kW at 33 kV at 0.8 p.f. lagging. The total resistance of the line is\n", + "10ohm and total inductive reactance is 15ohm . Determine\n", + "(i) sending-end voltage (ii) sending-end p.f. and (iii) transmission efficiency.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "#Given\n", + "P = 1100.0 #kW (Power delivered)\n", + "V = 33.0 #kV (Voltage)\n", + "pf = 0.8 # (Power factor)\n", + "R = 10.0 #ohm (Resistance)\n", + "X = 15.0 #ohm (Reactance) \n", + "#Full-load line current is\n", + "I = P/(V*pf) #A\n", + "theta = m.acos(pf) #Q (power factor angle)\n", + "I = cm.rect(I,-theta) #ohm (impedance)\n", + "#Line-loss\n", + "loss = I*I*R/1000 #kW\n", + "\n", + "\n", + "#(iii)Transmission efficiency\n", + "eff = (P/(P+loss))*100 #%\n", + "#Line voltage drop is IZ\n", + "Z = R + 1j * X\n", + "\n", + "\n", + "#Sending end voltage is\n", + "Es = V + I*Z/1000 #V\n", + "Es1 = m.sqrt(Es.real**2 + Es.imag**2) #V\n", + "print \"Sending end voltage is = \",round(Es1,2),\"V.\"\n", + "\n", + "\n", + "#Sending end pf angle is\n", + "theta2 = theta + cm.phase(Es)\n", + "pf2 = m.cos(theta2) #power factor\n", + "\n", + "print \"Power factor = \",round(pf2,3),\"lag\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.10 , PAGE NO :- 1621" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max. length in km is = 13.59 km.\n" + ] + } + ], + "source": [ + "'''What is the maximum length in km for a 1-phase transmission line having copper conductors of 0.775 cm^2 cross-section\n", + "over which 200 kW at unity power factor and at 3300 V can be delivered ? The efficiency of transmission is 90 per cent.\n", + "Take specific resistance as (1.725 * 10–8) ohm-m.'''\n", + "\n", + "#Given\n", + "A = 0.775e-4 #m^2 (Area of copper conductor)\n", + "P = 200.0 #kW (Power)\n", + "V = 3300.0 #V (Voltage)\n", + "pf = 1.0 # (Power factor)\n", + "rho = 1.725e-8 #ohm-m (Specific Resistance)\n", + "\n", + "#Sending-end power is\n", + "Es = P/0.9 #kW\n", + "#Line losses\n", + "loss = Es - P #kW\n", + "#Line current\n", + "I = P/(V*pf)*1000 #A\n", + "\n", + "\n", + "#If R is resistance of consuctor then 2*I^2*R = loss\n", + "R = loss/(2*I*I)*1000 #ohm\n", + "\n", + "#Now, using R = rho*l/A . The length is\n", + "l = R*A/rho #m \n", + "l = l/1000 #km\n", + "print \"Max. length in km is = \",round(l,2),\"km.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.11 , PAGE NO :- 1621" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load current = 209.19 A\n", + "Voltage at sending end is = 2419.76 V.\n", + "Sending power = 526.03 kW.\n", + "Sending end reactive power = 701.38 kVAR.\n", + "Sending end volt ampere kVA = 876.72 kVA.\n" + ] + } + ], + "source": [ + "'''An industrial load consisting of a group of induction motors which aggregate 500 kW at 0.6 power factor lagging is\n", + "supplied by a distribution feeder having an equivalent impedance of (0.15 + j0.6) ohm. The voltage at the load end of\n", + "the feeder is 2300 volts.\n", + "(a) Determine the load current.\n", + "(b) Find the power, reactive power and voltampere supplied to the sending end of the feeder.\n", + "(c) Find the voltage at the sending end of the feeder.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "#Given\n", + "P = 500.0 #kW (Power)\n", + "V = 2300.0 #V (Voltage)\n", + "pf = 0.6 # (Power factor)\n", + "#(a)line current\n", + "I = P*1000/(V*pf*1.732) #A (Current)\n", + "theta = m.acos(pf) # (Power factor angle) \n", + "I1 = cm.rect(I,-theta) #A\n", + "Z = 0.15 + 1j*0.6 #ohm (Impedance)\n", + "\n", + "#Voltage drop is\n", + "drop = I1*Z #V\n", + "#Sending end voltage is\n", + "Es = V + drop #V\n", + "Es = abs(Es)\n", + "\n", + "#Sending end pf angle is\n", + "theta2 = theta + cm.phase(Es)\n", + "pf2 = m.cos(theta2) #power factor\n", + "pf21 = m.sin(theta2) #sinQ component\n", + "\n", + "#Sending power = root(3)*Vl*Il*cosQ\n", + "Ps = 1.732*Es*I*pf2/1000 #kW\n", + "\n", + "#Sending end reactive power = root(3)*Vl*Il*sinQ\n", + "Prs = 1.732*Es*I*pf21/1000 #kVAR\n", + "\n", + "#Sending end volt ampere kVA = root(3)*Vl*Il\n", + "Pvs = 1.732*Es*I/1000 #kVA\n", + "\n", + "print \"load current = \",round(I,2),\"A\"\n", + "print \"Voltage at sending end is = \",round(Es,2),\"V.\"\n", + "print \"Sending power = \",round(Ps,2),\"kW.\"\n", + "print \"Sending end reactive power = \",round(Prs,2),\"kVAR.\"\n", + "print \"Sending end volt ampere kVA = \",round(Pvs,2),\"kVA.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.12 , PAGE NO :- 1622" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R = 3.24 ohm.\n", + "Reactance X = 6.11 ohm.\n" + ] + } + ], + "source": [ + "'''A 33-kV, 3-phase generating station is to supply 10 MW load at 31 kV and 0.9 power factor lagging over a 3-phase\n", + "transmission line 3 km long. For the efficiency of the line to be 96% , what must be the resistance and reactance of the line?'''\n", + "\n", + "import math as m\n", + "#Given\n", + "#Power Output\n", + "pout = 10.0 #MW (Power output)\n", + "eff = 0.96 # (Efficiency)\n", + "pin = pout/eff #MW (Power input)\n", + "\n", + "#Total loss\n", + "loss = pin - pout #MW\n", + "\n", + "#Full-load current I = P/V*pf*root(3)\n", + "I = pout*1e+6/(31.0e+3*0.9*1.732) #A\n", + "\n", + "#If R is resistance per phase,then 3*I*I*R = loss\n", + "R = loss*1e+6/(3*I*I) #ohm\n", + "\n", + "#Now, Vs per phase is\n", + "Vs = 33/1.732 #kV\n", + "#Vr per phase is\n", + "Vr = 31/1.732 #kV\n", + "#Using Vs = Vr + I(RcosQ + XsinQ )\n", + "X = ((Vs - Vr)/I*1000 - R*0.9)/m.sqrt(1 - 0.9*0.9)\n", + "\n", + "print \"Resistance R = \",round(R,2),\"ohm.\"\n", + "print \"Reactance X = \",round(X,2),\"ohm.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## EXAMPLE 41.13 , PAGE NO :- 1622" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage at recieving end = 59.91 V.\n", + "Angle between voltages = 3.12 degrees.\n", + "Transmission efficiency is = 92.59 %.\n" + ] + } + ], + "source": [ + "'''A balanced Y-connected load of (300 + j100) ohm is supplied by a 3-phase line 40 km long with an impedance of\n", + "(0.6 + j0.7) ohm per km (line-to-neutral). Find the voltage at the receiving end when the voltage at the sending\n", + "end is 66 kV. What is the phase angle between these voltages? Also, find the transmission efficiency of the line.'''\n", + "\n", + "import math as m\n", + "from sympy import Symbol,solve,Eq\n", + "#Resistance for 40 km conductor length\n", + "R = 40 * 0.6 #ohm\n", + "#Reactance for 40 km conductor length\n", + "X = 40 * 0.7 #ohm \n", + "#Total resistance/phase\n", + "R1 = R + 300 #ohm\n", + "#Total reactance/phase\n", + "X1 = X + 100 #ohm\n", + "#Total impedance/phase\n", + "Z = m.sqrt(R1**2 + X1**2) #ohm\n", + "#Line current\n", + "Il = 66000.0/1.732/Z #A \n", + "\n", + "#Now,\n", + "theta = m.atan(100.0/300.0)\n", + "cosQ = m.cos(theta)\n", + "sinQ = m.sin(theta)\n", + "#Voltage drop in conductor resistance\n", + "dropR = Il*R #V\n", + "#Voltage drop in conductor reactance\n", + "dropX = Il*X #V\n", + "\n", + "#Let us assume recieving end voltage as Vr\n", + "Vr = Symbol('Vr')\n", + "#Sending-end voltage is\n", + "Vs1 = 66000.0/1.732 #V\n", + "Vs2 = (Vr + dropR*cosQ + dropX*sinQ)**2 + (dropX*cosQ - dropR*sinQ)**2 #V\n", + "eq = Eq(Vs1*Vs1,Vs2)\n", + "Vr = solve(eq)\n", + "Vr1 = Vr[1] #V\n", + "\n", + "#Line-voltage across load\n", + "Vrl = Vr1*1.732/1000 #kV\n", + "print \"Voltage at recieving end = \",round(Vrl,2),\"V.\"\n", + "\n", + "#Angle between voltages\n", + "a_b = (dropX*cosQ - dropR*sinQ)/(Vr1 + dropR*cosQ + dropX*sinQ)\n", + "theta2 = m.atan(a_b)*180/3.14 #angle\n", + "print \"Angle between voltages = \",round(theta2,2),\"degrees.\"\n", + "#Transmission Efficiency\n", + "eff = (300.0/R1)*100\n", + "print \"Transmission efficiency is = \",round(eff,2),\"%.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.14 , PAGE NO :- 1623" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sending end voltage is = 12.12 V.\n", + "Power factor = 0.76\n", + "Transmission Efficiency = 93.93 %\n", + "Voltage Regulation = 10.2 %\n" + ] + } + ], + "source": [ + "'''Define ‘regulation’ and ‘efficiency’ of a short transmission line.A 3-phase, 50-Hz, transmission line having resistance of \n", + "5ohm per phase and inductance of 30 mH per phase supplies a load of 1000 kW at 0.8 lagging and 11 kV at the receiving end. Find.\n", + "(a) sending end voltage and power factor (b) transmission efficiency (c) regulation.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#Recieving-end Voltage\\phase\n", + "Vr = 11000.0/1.732 #V\n", + "#Line current\n", + "Il = 1000.0e+3/(1.732*11000.0*0.8) #A\n", + "#Inductive reactance\n", + "Xl = 2*3.14*50*30.0e-3 #ohm\n", + "R = 5.0 #ohm\n", + "#Impedance\n", + "Z = R + 1j*Xl #ohm\n", + "#drop per conductor\n", + "theta = m.atan(0.8)\n", + "Il1 = cm.rect(Il,-theta)\n", + "drop = Il1*(Z) #ohm\n", + "\n", + "#(a)Sending end voltage\n", + "Vs = Vr + drop #V\n", + "Vs1 = abs(Vs)*1.732/1000 #kV\n", + "#For power factor\n", + "theta2 = theta + cm.phase(Vs)\n", + "#Power factor\n", + "pf = m.cos(theta2)\n", + "\n", + "#(b)\n", + "#Power loss\n", + "loss = 3*Il*Il*R/1000 #kW\n", + "#Input Power\n", + "pin = 1000.0 + loss #kW\n", + "#Transmission efficiency\n", + "eff = 1000.0/pin *100 #% \n", + "# (c)% Voltage regulation\n", + "reg = (Vs1 - 11.0)/11.0*100 #%\n", + "\n", + "print \"Sending end voltage is = \",round(Vs1,2),\"V.\"\n", + "print \"Power factor = \",round(pf,2)\n", + "print \"Transmission Efficiency = \",round(eff,2),\"%\"\n", + "print \"Voltage Regulation = \",round(reg,2),\"%\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.15 , PAGE NO :- 1624" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Active power = 111397.0 kW.\n", + "Reactive power = 53952.0 kVAR.\n" + ] + } + ], + "source": [ + "'''A short 3-φ line with an impedance of (6 + j8) ohm per line has sending and receiving end line voltages of 120 and\n", + "110 kV respectively for some receiving-end load at a p.f. of 0.9. Find the active power and the reactive power at the\n", + "receiving end.'''\n", + "\n", + "import math as m\n", + "\n", + "#Given\n", + "R = 6.0 #ohm\n", + "X = 8.0 #ohm\n", + "cosQ = 0.9\n", + "sinQ = m.sqrt(1 - 0.9*0.9)\n", + "#Sending end-voltage\n", + "Vs = 120.0/1.732*1000 #V\n", + "#Recieving end-voltage\n", + "Vr = 110.0/1.732*1000 #V\n", + "#Now Vs = Vr + IRcosQ + IXsinQ.Therefore,line current is\n", + "I = (Vs - Vr)/(R*cosQ + X*sinQ) #A\n", + "\n", + "#Active Power at recieving end\n", + "act_pwr = 1.732*110.0*I*cosQ #kW\n", + "#Reactive Power at recieving end\n", + "rct_pwr = 1.732*110.0*I*sinQ #kVAR\n", + "\n", + "print \"Active power = \",round(act_pwr),\"kW.\"\n", + "print \"Reactive power = \",round(rct_pwr),\"kVAR.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.16 , PAGE NO :- 1624" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "-------pf = 0.707------- \n", + "% regulation = 15.18\n", + "Efficiency = 93.8\n", + "-------pf = 0.9--------- \n", + "% regulation = 9.36\n", + "Efficiency = 96.08\n" + ] + } + ], + "source": [ + "'''A 3-phase, 20 km line delivers a load of 10 MW at 11 kV having a lagging p.f. of 0.707 at the receiving end.\n", + "The line has a resistance of 0.02 ohm/km phase and an inductive reactance of 0.07 ohm/km/phase. Calculate the regulation\n", + "and efficiency of the line. If, now, the receiving end p.f. is raised to 0.9 by using static capacitors, calculate the\n", + "new value of regulation and efficiency.'''\n", + "\n", + "import cmath as cm\n", + "import math as m\n", + "\n", + "print \"-------pf = 0.707------- \"\n", + "#(i) When pf = 0.707 (lag)\n", + "pf = 0.707\n", + "#line current \n", + "Il = 10.0e+6/(1.732*11000*pf) #A\n", + "#Vr per phase\n", + "Vr = 11000.0/1.732 #V\n", + "#Total resistance/phase for 20km\n", + "R = 20*0.02 #W\n", + "#Total reactance/phase for 20km\n", + "X = 20*0.07 #W\n", + "\n", + "#Total impedance/phase\n", + "Z = R + 1j*X #ohm\n", + "#If Vr is taken as reference vector,then drop per phase is\n", + "theta = m.acos(pf)\n", + "Il1 = cm.rect(Il,-theta) #A\n", + "#drop/phase\n", + "drop = Il1*Z #V\n", + "#Sending end voltage\n", + "Vs = Vr + drop #V\n", + "Vs1 = abs(Vs) #V\n", + "#% regulation\n", + "reg = (Vs1 - Vr)/Vr*100\n", + "print \"% regulation = \",round(reg,2)\n", + "\n", + "#Total line loss\n", + "loss = 3*Il*Il*R/1.0e+6 #MW\n", + "#Total output\n", + "tot_op = 10.0 + loss #MW\n", + "#Efficiency\n", + "eff = 10.0/tot_op*100\n", + "print \"Efficiency = \",round(eff,2)\n", + "#-----------------------------------------------------------------------------------------------------------------------#\n", + "print \"-------pf = 0.9--------- \"\n", + "#(ii) When pf = 0.9 (lag)\n", + "pf = 0.9\n", + "#line current \n", + "Il = 10.0e+6/(1.732*11000*pf) #A\n", + "\n", + "#If Vr is taken as reference vector,then drop per phase is\n", + "theta = m.acos(pf)\n", + "Il1 = cm.rect(Il,-theta) #A\n", + "#drop/phase\n", + "drop = Il1*Z #V\n", + "#Sending end voltage\n", + "Vs = Vr + drop #V\n", + "Vs1 = abs(Vs) #V\n", + "#% regulation\n", + "reg = (Vs1 - Vr)/Vr*100\n", + "print \"% regulation = \",round(reg,2)\n", + "\n", + "#Total line loss\n", + "loss = 3*Il*Il*R/1.0e+6 #MW\n", + "#Total output\n", + "tot_op = 10.0 + loss #MW\n", + "#Efficiency\n", + "eff = 10.0/tot_op*100\n", + "print \"Efficiency = \",round(eff,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.17 , PAGE NO :- 1625" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sending end-voltage is = 11.86 kV.\n", + "Total loss = 68.57 kW.\n", + "Reduction in loss = 24.69 kW.\n" + ] + } + ], + "source": [ + "'''A load of 1,000 kW at 0.8 p.f. lagging is received at the end of a 3-phase line 10 km long. The resistance and inductance\n", + "of each conductor per km are 0.531 W and 1.76 mH respectively. The voltage at the receiving end is 11 kV at 50 Hz.\n", + "Find the sending-end voltage and the power loss in the line. What would be the reduction in the line loss if the p.f.\n", + "of the load were improved to unity?'''\n", + "\n", + "import cmath as cm\n", + "import math as m\n", + "\n", + "#Line current\n", + "Il = 1000.0 * 1000/(1.732 * 11 * 1000 * 0.8) #A\n", + "#Voltage/phase\n", + "V = 11000/1.732 #V\n", + "X = 2*3.14*50* 1.76e-3 *10 #ohm\n", + "R= 0.531 * 10 #ohm\n", + "Z = R + 1j*X\n", + "#Voltage drop/phase\n", + "theta = m.acos(0.8)\n", + "Il1 = cm.rect(Il,-theta)\n", + "drop = Il1*Z #V\n", + "#Sending end voltage is\n", + "Vs = V + drop #V\n", + "#line-to-line sending-end voltage\n", + "Vs1= abs(Vs)*1.732/1000 #kV\n", + "#Total loss \n", + "loss = 3*Il*Il*R/1000 #kW\n", + "\n", + "#Line current for unity p.f.\n", + "Il2 = 1000/(11*1.732) #A\n", + "#New losses\n", + "new_loss = 3*Il2*Il2*R/1000 #kW\n", + "\n", + "#Reduction in loss\n", + "red_loss = loss - new_loss #kW\n", + "\n", + "print \"Sending end-voltage is = \",round(Vs1,2),\"kV.\"\n", + "print \"Total loss = \",round(loss,2),\"kW.\"\n", + "print \"Reduction in loss = \",round(red_loss,2),\"kW.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.18 , PAGE NO :- 1625" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of line = 69.55 km.\n" + ] + } + ], + "source": [ + "'''Estimate the distance over which a load of 15,000 kW at 0.85 p.f. can be delivered by a 3-phase transmission line\n", + "having conductors of steel-cored aluminium each of resistance 0.905 W per kilometre. The voltage at the receiving end\n", + "is to be 132 kV and the loss in transmission is to be 7.5% of the load.'''\n", + "\n", + "#Line current\n", + "Il = 15000/(132 * 1.732 * 0.85) #A\n", + "#Total loss\n", + "loss = (7.5/100)*15000 #kW\n", + "\n", + "#If R is the resistance of one conductor, then 3*I^2*R = loss\n", + "R = loss*1000/(3*Il*Il) #ohm \n", + "\n", + "#Length of the line\n", + "length = R/0.905 #km.\n", + "print \"Length of line = \",round(length,2),\"km.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## EXAMPLE 41.19 , PAGE NO :- 1625" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power factor = 0.797\n", + "Efficiency of transmission line is = 97.15\n" + ] + } + ], + "source": [ + "'''A 3-phase line has a resistance of 5.31 ohm and inductance of 0.0176 H. Power is transmitted at 33 kV, 50-Hz from one end\n", + "and the load at the receiving end is 3,600 kW at 0.8 p.f. lagging. Find the line current, receiving-end voltage,\n", + "sending-end p.f. and efficiency of transmisson.'''\n", + "\n", + "from sympy import Symbol,solve,Eq\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#Let us assume that Vr is receiving end voltage\n", + "Vr = Symbol('Vr')\n", + "#Power delivered/phase = Vr*I*cosQ.Therefore I is\n", + "I = (3600.0/3)*1000/(0.8*Vr)\n", + "#Sending end voltage/phase =\n", + "Vs = 33000.0/1.732 #V\n", + "R = 5.31 #ohm\n", + "X = 2*3.14*0.0176*50 #ohm\n", + "\n", + "#Now,\n", + "cosQ = 0.8\n", + "sinQ = m.sqrt(1 - 0.8*0.8)\n", + "\n", + "#As we know, Vs = Vr + IRcosQ + IXsinQ\n", + "Vs2 = Vr + I*R*cosQ + I*X*sinQ #V\n", + "eq = Eq(Vs,Vs2)\n", + "Vr = solve(eq)\n", + "Vr1 = Vr[1] #V\n", + "#Line voltage at receiving end\n", + "Vrl = Vr1*1.732/1000 #kV\n", + "I = (3600.0/3)*1000/(0.8*Vr1) #A\n", + "\n", + "Vs = Vr1 + I*(cosQ - 1j*sinQ)*(R + 1j*X) #V\n", + "#Power factor\n", + "theta = m.acos(0.8) + cm.phase(Vs)\n", + "pf2 = m.cos(theta)\n", + "print \"Power factor = \",round(pf2,3)\n", + "#Power lost in line is\n", + "loss = 3*I*I*R/1000 #kW\n", + "#Power at sending end is\n", + "tot_pwr = 3600.0 + loss #kW\n", + "#Eficiency of transmission is\n", + "eff = 3600.0/tot_pwr*100\n", + "print \"Efficiency of transmission line is = \",round(eff,2) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.20 , PAGE NO :- 1626" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total Maximum power = 18.0 MW.\n", + "Total kVAR supplied = 28801.69 kW.\n" + ] + } + ], + "source": [ + "'''A 3-phase short transmission line has resistance and reactance per phase of 15 ohm and 20 ohm respectively. If the sending-end\n", + "voltage is 33 kV and the regulation of the line is not to exceed 10%, find the maximum power in kW which can be transmitted\n", + "over the line. Find also the kVAR supplied by the line when delivering the maximum power.'''\n", + "\n", + "import math as m\n", + "\n", + "#Given\n", + "Vs = 33000.0/1.732 #V (Sending end voltage)\n", + "Vr = Vs/(1 + 10.0/100) #V (Receiving end voltage)\n", + "Z = m.sqrt(15**2 + 20**2) #ohm (impedance)\n", + "R = 15.0 #ohm (resistance)\n", + "X = 20.0 #ohm (reactance) \n", + "#Maximum Power transmitted is given by\n", + "Pmax = (Vr/Z)**2*(Z*(Vs/Vr) - R) #watts/phase\n", + "\n", + "#Total max. power\n", + "ptot = Pmax*3/1e+6 #MW\n", + "\n", + "#kVAR supplied per phase is given by\n", + "pkvar = (Vr/Z)**2*X/1e+3 #kW\n", + "#Total kVAR supplied\n", + "kvartot = 3*pkvar #kW\n", + "\n", + "print \"Total Maximum power = \",round(ptot,2),\"MW.\"\n", + "print \"Total kVAR supplied = \",round(kvartot,2),\"kW.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.21 , PAGE NO :- 1627" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sending end voltage = 67.69 kV.\n", + "Efficiency = 98.8 %\n", + "Max. value of Q for 3-phases are = 346840.0 kVA.\n" + ] + } + ], + "source": [ + "'''A 3-phase, 50-Hz generating station supplies a load of 9,900 kW at 0.866 p.f. (lag) through a short overhead transmission line.\n", + "Determine the sending-end voltage if the receiving-end voltage is 66 kV and also the efficiency of transmission.\n", + "The resistance per km is 4ohm and inductance 40 mH. What is the maximum power in kVA that can be transmitted through\n", + "the line if both the sending and receiving-end voltages are kept at 66 kV and resistance of the line is negligible.?'''\n", + "\n", + "import math as m\n", + "#Resistance\n", + "R = 4.0 #ohm\n", + "#Reactance\n", + "X = 40.0e-3*(2*3.14*50)#ohm\n", + "#Impedance\n", + "Z = m.sqrt(R*R + X*X) #ohm\n", + "#Line current\n", + "I = 9900.0/(1.732*66*0.866) #A\n", + "#Receiving end voltage\n", + "Vr = 66000.0/1.732 #V\n", + "#Now,\n", + "cosQ = 0.866\n", + "sinQ = m.sqrt(1 - 0.866**2)\n", + "#Sending end voltage is\n", + "Vs = Vr + I*R*cosQ + I*X*sinQ #V\n", + "#Line value of sending end voltage\n", + "Vs1 = Vs*1.732/1000 #kV\n", + "print \"Sending end voltage = \",round(Vs1,2),\"kV.\"\n", + "#Total line loss\n", + "loss = 3*I*I*R/1000 #kW\n", + "#Efficiency is\n", + "eff = 9900.0/(9900.0 + loss)*100#%\n", + "print \"Efficiency = \",round(eff,2),\"%\"\n", + "\n", + "#Max. value of Q for 3-phases are (As Vs = Vr R is negligible)\n", + "Z = X #ohm \n", + "max_value = (3*Vr*Vr)/(Z*Z)*X*1e-3 #kVA\n", + "print \"Max. value of Q for 3-phases are = \",round(max_value,-1),\"kVA.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.22 , PAGE NO :- 1627" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power factor at sending end = 0.79\n", + "Transmission efficiency = 94.73 %\n" + ] + } + ], + "source": [ + "'''3-phase load of 2000 kVA,0.8 p.f. is supplied at 6.6 kV, 50-Hz by means of a 33 kV transmission line 20 km long and a 5 : 1\n", + "transformer. The resistance per km of each conductor is 0.4 ohm and reactance 0.5 ohm. The resistance and reactance of the\n", + "transformer primary are 7.5 ohm and 13.2 ohm, whilst the resistance of the secondary is 0.35 ohm and reactance 0.65 ohm.\n", + "Find the voltage necessary at the sending end of transformission line when 6.6 kV is maintained at the load-end and find\n", + "the sending-end power factor. Determine also the efficiency of transmission.'''\n", + "\n", + "import math as m\n", + "#Impedance of high voltage line\n", + "Zh = 8.0 + 1j*10.0 #ohm\n", + "#Impedance of transformer(primary side)\n", + "Zt = 7.5 + 1j*13.2 #ohm\n", + "#Total impedance on high tension side\n", + "Ztot_h = Zh + Zt #ohm\n", + "#Impedance as referred to secondary side\n", + "Zsec = Ztot_h/(5**2) #ohm\n", + "\n", + "#Total impedance on high tension side\n", + "Ztot_l = Zsec + (0.35 + 1j*0.65) #ohm\n", + "\n", + "#Now,kVA load per phase\n", + "load = 2000.0/3 #kVA \n", + "#Receiving-end voltage per phase\n", + "Vr = 6.6/1.732 #kV\n", + "#current in line is\n", + "I = load/Vr #A\n", + "#Now,\n", + "cosQ = 0.8\n", + "sinQ = m.sqrt(1 - 0.8*0.8)\n", + "#Drop per conductor\n", + "drop = I*(Zsec.real*cosQ + Zsec.imag*sinQ) #V\n", + "#Sending end voltage is\n", + "Vs = Vr + drop/1000 #kV\n", + "#Sending end voltage referred from high voltage side\n", + "Vs = Vs*5 #kV\n", + "#Line sending end voltage\n", + "Vsl = Vs*1.732 #kV\n", + "\n", + "#If theta is phase angle at sending end then\n", + "tantheta = (sinQ + I*Zsec.imag/(Vr*1000))/(cosQ + I*Zsec.real/(Vr*1000))\n", + "theta = m.atan(tantheta)\n", + "pf = m.cos(theta)\n", + "print \"Power factor at sending end = \",round(pf,2)\n", + "#power loss/phase\n", + "loss = (I*I)*0.97/1000 #kW\n", + "#power at the receiving end/phase\n", + "power = 2000.0*cosQ/3 #kW\n", + "#Transmission efficiency\n", + "eff = power/(power + loss)*100\n", + "print \"Transmission efficiency = \",round(eff,2),\"%\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.23 , PAGE NO :- 1629" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sending end current is 240.3 A.\n", + "Sending end voltage is 79598.0 V.\n", + "Voltage regulation = 20.6 %.\n" + ] + } + ], + "source": [ + "'''A (medium) single-phase transmission line 50 km long has the following constants :\n", + "\n", + "resistance/km = 0.5 ohm ; reactance/km = 1.6 ohm\n", + "susceptance/km = 28 * 10−6 S ; receiving-end line voltage = 66,000 V\n", + "Assuming that total capacitance of the line is located at receiving end alone, determine the\n", + "sending-end voltage, the sending-end current and regulation. The line is delivering 15,000 kW at\n", + "0.8 p.f. lagging. Draw a vector diagram to illustrate your answer.'''\n", + "\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#Load current at receiving end Using I = P/VcosQ\n", + "Ir = 15.0e+6/(66e+3*0.8) #A\n", + "\n", + "#Total resistance is\n", + "R = 0.5*50.0 #ohm\n", + "#Total reactance is\n", + "X = 1.6*50.0 #ohm\n", + "#Susceptance\n", + "B = 28e-6*50.0 #Siemens\n", + "#Capacitive admittance\n", + "Y = B #Siemens\n", + "#Sending end current Is is vector sum of load current Ir and capacitive current Ic\n", + "Er = 66000.0\n", + "Ic = 1j*Er*Y #A\n", + "theta = m.acos(0.8)\n", + "Irl = cm.rect(Ir,-theta) #A\n", + "\n", + "#Sending end current is\n", + "Is = Irl + Ic #A\n", + "Z = R + 1j*X #ohm\n", + "print \"Sending end current is\",round(abs(Is),2),\"A.\"\n", + "#line drop\n", + "drop = Is*Z #V\n", + "#Sending end voltage is\n", + "Es = Er + drop #V\n", + "print \"Sending end voltage is\",round(abs(Es)),\"V.\"\n", + "#Voltage regulation is\n", + "reg = (abs(Es)-abs(Er))/abs(Er)*100 #%\n", + "print \"Voltage regulation = \",round(reg,2),\"%.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.24 , PAGE NO :- 1631" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Line value of sending end voltage = 155.43 kV.\n", + "Power factor = 0.774 lagging.\n" + ] + } + ], + "source": [ + "'''A 3-phase, 50-Hz overhead transmission line 100 km long with 132 kV between lines at the receiving end has the\n", + "following constants :\n", + "\n", + "resistance/km/phase = 0.15 ohm inductance/km/phase = 1.20 mH\n", + "capacitance/km/phase = 0.01 mF\n", + "\n", + "Determine, using an approximate method of allowing for capacitance, the voltage, current and\n", + "p.f. at the sending end when the load at the receiving end is 72 MW at 0.8 p.f. lagging. Draw vector\n", + "diagram for the circuit assumed.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "#For a 100-km length of line\n", + "#Resistance of line is\n", + "R = 0.15*100 #ohm\n", + "Xl = (2*3.14*50)*(1.2e-3)*100 #ohm\n", + "Xc = 1/(2*3.14*50*0.1e-3)*100 #ohm\n", + "\n", + "#Using nominal T-method\n", + "Vr = 132/1.732 #kV\n", + "#Load current\n", + "Ir = (72e+6)/(1.732*132e+3*0.8) #A\n", + "#Load current is\n", + "theta = m.acos(0.8)\n", + "Irl = cm.rect(Ir,-theta) #A\n", + "#Impedance Zbc is\n", + "Zbc = R/2 + 1j*Xl/2 #ohm\n", + "#Drop/phase over BC is\n", + "drop = Irl*Zbc #V\n", + "#Now, voltage V1 is\n", + "V1 = Vr*1000 + drop #V\n", + "\n", + "#From fig. Ic is\n", + "Ic = V1/(-1j*Xc) #A\n", + "\n", + "#Sending end current is\n", + "Is = Ic + Irl #A\n", + "\n", + "#Impedance Zab is\n", + "Zab = R/2 + 1j*Xl/2 #ohm\n", + "#Drop/phase over AB is\n", + "drop2 = Is*Zab #V\n", + "#Sending end voltage is\n", + "Vs = V1 + drop2 #V\n", + "#Line value of sending end voltage is\n", + "Vsl = 1.732*abs(Vs)/1000 #kV\n", + "print \"Line value of sending end voltage =\",round(Vsl,2),\"kV.\"\n", + "#Phase angle between Vs and Is is\n", + "angle = abs(cm.phase(Vs)) + abs(cm.phase(Is))\n", + "pf = m.cos(angle) #(lag)\n", + "print \"Power factor = \",round(pf,3),\"lagging.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.25 , PAGE NO :- 1632" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Line value of sending end voltage = 117.15 kV.\n", + " sending end current = 110.11 A.\n", + "Efficiency = 95.95 %.\n" + ] + } + ], + "source": [ + "'''A 3-phase, 50-Hz transmission line, 100 km long delivers 20 MW at 0.9 p.f. lagging and at 110 kV. The resistance and\n", + "reactance of the line per phase per km are 0.2 ohm and 0.4 ohm respectively while the capacitive admittance is 2.5 * 10e−6 S per\n", + "km. Calculate (a) the voltage and current at the sending end and (b) the efficiency of transmission. Use the nominal T-method.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#Resistance for 100 km is\n", + "R = 0.2*100 #ohm\n", + "#Reactance for 100 km is\n", + "X = 0.4*100 #ohm\n", + "#Capacitive admittance for 100 km is\n", + "Y = 2.5e-6*100 #Siemens\n", + "#Receiving end voltage Er is\n", + "Er = 110.0/1.732 #kV\n", + "Ir = 20.0e+6/(1.732*110e+3*0.9) #A\n", + "#Now\n", + "cosQ = 0.9\n", + "sinQ = m.sqrt(1 - 0.9*0.9)\n", + "theta = m.acos(0.9) \n", + "Irl = cm.rect(Ir,-theta) #A\n", + "\n", + "#Impedance is\n", + "Zbc = R/2 + 1j*X/2 #ohm\n", + "#Voltage drop between point B and C is\n", + "dropbc = Irl*Zbc #V\n", + "V1 = Er*1000 + dropbc #V\n", + "\n", + "#Current through capacitor \n", + "Ic = V1*1j*Y #A\n", + "\n", + "#Sending end current is\n", + "Is = Ic + Irl #A\n", + "Zab = Zbc\n", + "#Voltage drop between point A and B is\n", + "dropab = Irl*Zab #V\n", + "Vs = V1 + dropab #V\n", + " \n", + "#Line value of sending end voltage is\n", + "Vsl = 1.73*abs(Vs)/1000 #kV\n", + "print \"Line value of sending end voltage =\",round(Vsl,2),\"kV.\"\n", + "print \" sending end current =\",round(abs(Is),2),\"A.\"\n", + " \n", + "#Phase angle between Vs and Is is\n", + "angle = abs(cm.phase(Vs)) + abs(cm.phase(Is))\n", + "pf = m.cos(angle) #(lag)\n", + "\n", + "#Input power\n", + "pin = 1.73*abs(Vsl)*abs(Is)*pf/1e+3 #MW\n", + "\n", + "#Efficiency\n", + "eff = 20.0/pin*100\n", + "print \"Efficiency = \",round(eff,2),\"%.\" \n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.26 , PAGE NO :- 1635" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Line value of sending-end voltage 174.83 kV.\n", + "power factor = 0.75\n" + ] + } + ], + "source": [ + "'''A 3-phase transmission line,100 km long has following constants:\n", + "resistance per km per phase = 0.28 ohm ; inductive reactance per km per phase = 0.63 ohm .\n", + "Capacitive susceptance per km per phase = 4 * 10e-6 siemens.\n", + "If the load at the receiving end is 75 MVA at 0.8 p.f. lagging with 132 kV between lines calculate sending-end voltage,\n", + "current and p.f. Use nominal-pi-method.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#For 100 km length line\n", + "#Resistance/phase\n", + "R = 0.28*100 #ohm\n", + "#Inductive Reactance/phase\n", + "Xl = 0.63*100 #ohm\n", + "#Capacitive Susceptance/phase\n", + "Y = 4.0e-6*100 #S\n", + "#Capacitive Susceptance at each end\n", + "Y = 1j*Y/2 #S\n", + "#Receiving end voltage Vr is\n", + "Vr = 132e+3/1.732 #V\n", + "#Receiving end current Ir is\n", + "Ir = 75.0e+6/(1.732*132e+3*0.8) #A\n", + "theta = m.acos(0.8)\n", + "Irl = cm.rect(Ir,-theta) #A\n", + "#Current through capacitance is\n", + "Ic = Vr*Y #A\n", + "#Now\n", + "Il = Ic + Irl #A\n", + "\n", + "#Drop per conductor is\n", + "Zl = R +1j*Xl #ohm\n", + "drop = Il*Zl #V\n", + "\n", + "#sending end voltage\n", + "Vs = Vr + drop\n", + "#Line value of sending-end voltage\n", + "Vsl = abs(Vs)*1.732/1000 #kV\n", + "print \"Line value of sending-end voltage\",round(Vsl,2),\"kV.\"\n", + "#\n", + "Ic2 = Vs*Y #A\n", + "Is = Ic2 + Il\n", + "\n", + "#Angle between VS and IS\n", + "angle = abs(cm.phase(Vs)) + abs(cm.phase(Is)) \n", + "pf = m.cos(angle)\n", + "#cos41.4 = 0.75 \n", + "print \"power factor = \",round(pf,2) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.27 , PAGE NO :- 1637" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Regulation = 15.18 %\n", + "Transmission efficiency = 95.02 %.\n" + ] + } + ], + "source": [ + "'''A 100-km long, three-phase, 50-Hz transmission line has resistance/phase/km = 0.1 ohm ; reactance/phase/km = 0.5 ohm ; \n", + "susceptance/phase/km = 10 * 10−6 siemens.If the line supplies a load of 20 MW at 0.9 p.f. lagging at 66 kV at the receiving end,\n", + "calculate by nominal ‘p’ method, the regulation and efficiency of the line.Neglect leakage.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "#For a 100 km line\n", + "#resistance/phase\n", + "R = 0.1 * 100 #ohm\n", + "#inductive reactance/phase\n", + "Xl = 0.5 * 100 #ohm\n", + "#Capacitive susceptance/phase\n", + "Yc = 10.0 * 1.0e-6 * 100 #siemens\n", + "#Susceptance at each end\n", + "Yc = 1j*Yc/2 #siemens\n", + "#Voltage at receiving end is\n", + "Vr = 66.0e+3/1.732 #V\n", + "Ir = 20.0e+6/(1.732*66e+3*0.9) #A\n", + "theta = m.acos(0.9)\n", + "Irl = cm.rect(Ir,-theta) #A\n", + "Ic1 = Vr*Yc #A\n", + "Il = Irl + Ic1 #A\n", + "\n", + "#drop/conductor\n", + "Zl = R + 1j*Xl #ohm\n", + "drop = Il*Zl #V\n", + "#Sending end voltage is\n", + "Vs = Vr + drop #V\n", + "#Line value of sending end voltage\n", + "Vsl = abs(Vs)*1.732/1000 #kV\n", + "Ic2 = Vs*Yc #A\n", + "#Sending end current is\n", + "Is = Ic2 + Il #A \n", + "\n", + "#Phase angle between Vs and Is is\n", + "angle = abs(cm.phase(Vs)) + abs(cm.phase(Is))\n", + "pf = m.cos(angle) #(lag)\n", + "\n", + "\n", + "#(i) Regulation is\n", + "reg = (abs(Vsl) - 66.0)/66.0 * 100 \n", + "print \"Regulation = \",round(reg,2),\"%\"\n", + "#(ii)Efficiency\n", + "#Input power is\n", + "pin = 1.732*Vsl*abs(Is)*pf*1000.0/1.0e+6 #MW\n", + "\n", + "eff = 20.0/pin*100 #%\n", + "\n", + "print \"Transmission efficiency = \",round(eff,2),\"%.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.28 , PAGE NO :- 1638" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Line value of sending end voltage = 122.24 kV.\n", + "Sending end current = 195.65 A.\n", + "Transmission efficiency = 95.36 %.\n", + "Receiving end voltage = 63148.47 V.\n", + "Receiving end current = 19.0 A.\n" + ] + } + ], + "source": [ + "'''(a) A 50-Hz, 3-phase, 100-km long line delivers a load of 40 MVA at 110 kV and 0.7 p.f. lag. The line constants\n", + "(line to neutral) are :\n", + "resistance of 11 ohms, inductive reactance of 38 ohms and capacitive susceptance of 3 * 10−4 siemens. Find the sending-end\n", + "voltage, current,power factor and efficiency of power transmission.\n", + "(b) draw the vector diagram.\n", + "(c) If the sending-end voltage is held constant and load is removed, calculate the receiving-end voltage and current.'''\n", + "\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#resistance/phase\n", + "R = 11.0 #ohm\n", + "#inductive reactance/phase\n", + "Xl = 38.0 #ohm\n", + "#Capacitive susceptance/phase\n", + "Yc = 3.0e-4 #siemens\n", + "#Susceptance at each end\n", + "Yc = 1j*Yc/2 #siemens\n", + "#Voltage at receiving end is\n", + "Vr = 110.0e+3/1.732 #V\n", + "Ir = 40.0e+6/(1.732*110e+3) #A\n", + "theta = m.acos(0.7)\n", + "Irl = cm.rect(Ir,-theta) #A\n", + "Ic1 = Vr*Yc #A\n", + "Il = Irl + Ic1 #A\n", + "\n", + "#drop/conductor\n", + "Zl = R + 1j*Xl #ohm\n", + "drop = Il*Zl #V\n", + "#Sending end voltage is\n", + "Vs = Vr + drop #V\n", + "#Line value of sending end voltage\n", + "Vsl = abs(Vs)*1.732/1000 #kV\n", + "Ic2 = Vs*Yc #A\n", + "#Sending end current is\n", + "Is = Ic2 + Il #A \n", + "print \"Line value of sending end voltage = \",round(Vsl,2),\"kV.\"\n", + "print \"Sending end current = \",round(abs(Is),2),\"A.\"\n", + "#Phase angle between Vs and Is is\n", + "angle = abs(cm.phase(Vs)) + abs(cm.phase(Is))\n", + "pf = m.cos(angle) #(lag)\n", + "\n", + "#Efficiency\n", + "#Input power is\n", + "pin = 1.732*Vsl*abs(Is)*pf*1000.0/1.0e+6 #MW\n", + "\n", + "eff = (40.0*0.7)/pin*100 #%\n", + "\n", + "print \"Transmission efficiency = \",round(eff,2),\"%.\"\n", + "\n", + "\n", + "#(c)Under no-load condition, current in the conductor is Ic1\n", + "#Drop/phase =\n", + "drop = Ic1*Zl #V\n", + "#Sending end voltage is\n", + "Vs = Vr + drop #V\n", + "Ic2 = Vs*Yc #A\n", + "Is = Ic1 + Ic2 #A\n", + "print \"Receiving end voltage = \",round(abs(Vs),2),\"V.\"\n", + "print \"Receiving end current = \",round(abs(Is),2),\"A.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.29 , PAGE NO :- 1640" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sending end voltage is = 59558.07 V.\n", + "Sending end current is = 282.23 A.\n", + "Sending end power is = 48.99 MW.\n", + "Transmission efficiency = 81.65 %\n" + ] + } + ], + "source": [ + "'''Find the following for a single-circuit transmission line delivering a load of 50 MVA at 110 kV and p.f. 0.8 lagging :\n", + "(i) sending-end voltage, (ii) sending-end current, (iii) sending-end power, (iv) efficiency of\n", + "transmission. (Given A = D = 0.98 ∠3º , B = 110 ∠75º ohm, C = 0.0005 ∠80º ohm).'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#Receiving end voltage Vr is\n", + "Vr = 110.0/1.732 #kV\n", + "Ir = 50.0e+6/(1.732*110e+3) #A\n", + "theta = m.acos(0.8)\n", + "Irl = cm.rect(Ir,theta) #A\n", + "\n", + "#Now,\n", + "A = D = cm.rect(0.98,3*3.14/180)\n", + "B = cm.rect(110.0,75*3.14/180)\n", + "C = cm.rect(0.0005,80*3.14/180)\n", + "\n", + "#(i)Sending end voltage is\n", + "Vs = A*Vr*1000 + B*Irl #V\n", + "print \"Sending end voltage is = \",round(abs(Vs),2),\"V.\"\n", + "#(ii)Sending end current is\n", + "Is = C*Vr*1000 + D*Irl #A\n", + "print \"Sending end current is = \",round(abs(Is),2),\"A.\"\n", + "#Angle between Vs and Is is\n", + "angle = cm.phase(Is) - cm.phase(Vs)\n", + "pf = m.cos(angle)\n", + "\n", + "#(iii)Sending end power is\n", + "s_pwr = 3*abs(Vs)*abs(Is)*pf/1.0e+6 #MW\n", + "print \"Sending end power is = \",round(s_pwr,2),\"MW.\"\n", + "#Receiving end power is\n", + "r_pwr = 50.0*0.8 #MW\n", + "#(iv)Transmission efficiency\n", + "eff = r_pwr/s_pwr*100 #%\n", + "print \"Transmission efficiency = \",round(eff,2),\"%\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.30 , PAGE NO :- 1640" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A = (0.9325+0.016875j)\n", + "B = (22.5+90j)\n", + "C = (-1.265625e-05+0.001449375j)\n", + "D = (0.9325+0.016875j)\n", + "Voltage regulation = 34.77 %.\n" + ] + } + ], + "source": [ + "'''A 150 km, 3-φ, 110-V, 50-Hz transmission line transmits a load of 40,000 kW at 0.8 p.f. lag at receiving end.\n", + "resistance/km/phase = 0.15 ohm, reactance/km/phase = 0.6 ohm ; susceptance/km/phase = 10e−5 S\n", + "(a) determine the A, B, C and D constants of the line . (b) find regulation of the line.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#For a 150km length line\n", + "R = 0.15*150.0 #ohm\n", + "X = 0.6*150.0 #ohm\n", + "Y = 1j*1.0e-5*150.0 #S\n", + "Z = R + 1j*X #ohm\n", + "\n", + "#(a)\n", + "A = D = 1 + Y*Z/2\n", + "B = Z\n", + "C = Y*(1 + Y*Z/4)\n", + "print \"A = \" ,A\n", + "print \"B = \" ,B\n", + "print \"C = \" ,C\n", + "print \"D = \" ,D\n", + "#(b)\n", + "Vr = 110/1.732*1000 #V\n", + "Ir = 40.0e+6/(1.732*110*0.8*1000)#A\n", + "theta = m.acos(0.8)\n", + "Irl = cm.rect(Ir,-theta) #A\n", + "#Sending end voltage is\n", + "Vs = A*Vr + B*Irl #V\n", + "\n", + "#Now, Vs = A*Vro => Vro = Vs/A\n", + "Vro = Vs/A #V\n", + "\n", + "#Voltage regulation\n", + "reg = (abs(Vro) - abs(Vr))/abs(Vr)*100\n", + "print \"Voltage regulation = \",round(reg,2),\"%.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.31 , PAGE NO :- 1641" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Regulation = 29.97 %\n", + "Charging current Ic = (-7.1+127.64j) A.\n" + ] + } + ], + "source": [ + "'''A 132-kV, 50-Hz, 3-phase transmission line delivers a load of 50 MW at 0.8 p.f. lagging at receiving-end.\n", + "The generalised constants of the transmission line are A = D = 0.95 ∠1.4º ; B = 96 ∠ 7.8º ; C = 0.0015 ∠90º\n", + "Find the regulation of the line and the charging current. Use nominal T-method.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#Given\n", + "A = D = cm.rect(0.95,1.4*3.14/180)\n", + "B = cm.rect(96.0,78*3.14/180)\n", + "C = cm.rect(0.0015,90*3.14/180)\n", + "\n", + "#Receiving end voltage\n", + "Ir = 50e+6/(1.732*132e+3*0.8) #A\n", + "theta = m.acos(0.8)\n", + "Irl = cm.rect(Ir,-theta) #A\n", + "Vr = 132e+3/1.732 #V\n", + "#Sending end voltage\n", + "Vs = A*Vr + B*Irl #V\n", + "#Sending end current\n", + "Is = C*Vr + D*Irl #A\n", + "Vro = abs(Vs)/abs(A)\n", + "\n", + "#% Regulation\n", + "reg = (abs(Vro) - abs(Vr))/abs(Vr)*100 #%\n", + "\n", + "print \"Regulation = \",round(reg,2),\"%\"\n", + "#Charging current\n", + "Ic = Is - Irl #A\n", + "print \"Charging current Ic = \",complex(round(Ic.real,2),round(Ic.imag,2)),\"A.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## EXAMPLE 41.32 , PAGE NO :- 1641" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A = A1*A2 + B1*C2\n", + "C = A2*C1 + C2*D1\n", + "B = A1*B2 + B1*D2\n", + "D = B2*C1 + D1*D2\n" + ] + } + ], + "source": [ + "'''A 3-phase transmission line consists of two lines 1 and 2 connected in series,line 1 being at the sending end and\n", + "2 at the receiving end. The respective auxiliary constants of the two lines are :\n", + "A1, B1, C1, D1 and A2, B2, C2, D2. Find the A, B, C, D constants of the whole line which is equivalent to two\n", + "series-connected lines.'''\n", + "\n", + "from sympy import Symbol\n", + "A1 = Symbol('A1')\n", + "B1 = Symbol('B1')\n", + "C1 = Symbol('C1')\n", + "D1 = Symbol('D1')\n", + "\n", + "A2 = Symbol('A2')\n", + "B2 = Symbol('B2')\n", + "C2 = Symbol('C2')\n", + "D2 = Symbol('D2')\n", + "\n", + "Vr = Symbol('Vr')\n", + "Ir = Symbol('Ir')\n", + "\n", + "#--------For A &C---------------#\n", + "Ir = 0\n", + "Vr = 1\n", + "#For line no. 2\n", + "V = A2*Vr + B2*Ir\n", + "I = C2*Vr + D2*Ir \n", + "\n", + "#For line no. 1\n", + "Vs = A1*V + B1*I\n", + "Is = C1*V + D1*I\n", + "\n", + "#For A\n", + "print \"A = \",Vs\n", + "print \"C = \",Is\n", + "\n", + "#--------For B & D---------------#\n", + "Ir = 1\n", + "Vr = 0\n", + "#For line no. 2\n", + "V = A2*Vr + B2*Ir\n", + "I = C2*Vr + D2*Ir \n", + "\n", + "#For line no. 1\n", + "Vs = A1*V + B1*I\n", + "Is = C1*V + D1*I\n", + "\n", + "#For A\n", + "print \"B = \",Vs\n", + "print \"D = \",Is\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.33 , PAGE NO :- 1644" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Critical voltage = 136.01 kV.\n" + ] + } + ], + "source": [ + "'''Find the disruptive critical voltage for a transmission line having :\n", + "conductor spacing = 1 m ; conductor (stranded) radius = 1 cm\n", + "barometric pressure = 76 cm of Hg ; temperature = 40ºC\n", + "Air break-down potential gradient (at 76 cm of Hg and at 25ºC) = 21.1 kV (r.m.s.)/cm.'''\n", + "\n", + "import math as m\n", + "\n", + "#Given\n", + "g0 = 21.1 #kV/cm (air breakdown potential gradient)\n", + "m0 = 0.85 #(assumed)\n", + "d = 3.92*76/(273 + 40)\n", + "r = 1.0 #cm\n", + "D = 100.0 #cm\n", + "\n", + "#Vc = 2.3*m0*g0*d*r*log(D/r)\n", + "Vc = 2.3*m0*g0*d*r*m.log10(D/r) #kV\n", + "#Line voltage\n", + "Vc = 1.732*Vc\n", + "print \"Critical voltage = \",round(Vc,2),\"kV.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.34 , PAGE NO :- 1644" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Critical voltage = 152.4 kV.\n", + "visual corona voltage = 157.67 kV.\n" + ] + } + ], + "source": [ + "'''Find the disruptive critical and visual corona voltage of a grid-line operating at 132 kV.\n", + "conductor dia = 1.9 cm ; conductor spacing = 3.81 m\n", + "temperature = 44ºC ; barometric pressure = 73.7 cm\n", + "conductor surface factor : fine weather = 0.8 ; rough weather = 0.66.'''\n", + "\n", + "\n", + "import math as m\n", + "\n", + "#Given\n", + "\n", + "m0 = 0.8 #\n", + "d = 3.92*73.7/(273 + 44)\n", + "r = 1.9 #cm\n", + "D = 381.0 #cm\n", + "\n", + "#Vc = 2.3*m0*g0*d*r*log(D/r)\n", + "Vc = 48.8*m0*d*r/2*m.log10(2*D/r) #kV\n", + "#Line voltage\n", + "Vc = 1.732*Vc\n", + "print \"Critical voltage = \",round(Vc,2),\"kV.\"\n", + "\n", + "#(b)\n", + "#Vv = 2.3*g0*mv*d*r*(1 + 0.3/root(d*r))*log10(D/r)\n", + "mv = 0.66\n", + "Vv = 48.8*mv*d*r*(1 + 0.3/m.sqrt(d*r))*m.log10(D/r)\n", + "print \"visual corona voltage = \",round(Vv,2),\"kV.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## EXAMPLE 41.35 , PAGE NO :- 1644" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Critical disruptive voltage = 53.34 kV/km.\n", + "Corona loss = 121.49 kW.\n" + ] + } + ], + "source": [ + "'''A certain 3-phase equilateral transmission line has a total corona loss of 53 kW at 106 kV and a loss of 98 kW at 110.9 kV.\n", + "What is the disruptive critical voltage between lines? What is the corona loss at 113 kV?'''\n", + "\n", + "from sympy import Symbol,solve,Eq\n", + "\n", + "#total corona loss P o< (V- Vc)^2\n", + "# P1/P2 = (V1 - Vc)^2/(V2 - Vc)^2\n", + "\n", + "V1 = 106/1.732 #kV\n", + "V2 = 110.9/1.732 #kV\n", + "#Let Vc be distruptive critical voltage\n", + "Vc = Symbol('Vc')\n", + "#ratio P1/P2 =\n", + "P1_P2a = 53.0/98.0\n", + "#Ratio (V1 - Vc)^2/(V2 - Vc)^2\n", + "P1_P2b = (V1 - Vc)**2/(V2 - Vc)**2\n", + "#Equating the ratios\n", + "eq = Eq(P1_P2a,P1_P2b)\n", + "Vc = solve(eq)\n", + "Vc1 = Vc[0] #kV/km\n", + "\n", + "#Now, W/98 = (Vb - Vc)^2/(V2 - Vc)^2\n", + "W = 98*(113/1.732 - Vc1)**2/(V2 - Vc1)**2 #kW\n", + "print \"Critical disruptive voltage = \",round(Vc1,2),\"kV/km.\"\n", + "print \"Corona loss = \",round(W,2),\"kW.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.36 , PAGE NO :- 1645" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "corona power loss per km = 1.76 kW/km/phase.\n" + ] + } + ], + "source": [ + "'''A 3-phase, 50-Hz, 220-kV transmission line consists of conductors of 1.2 cm radius spaced 2 metres at the corners of an\n", + "equilateral triangle. Calculate the corona power loss per km of the line at a temperature of 20ºC and barometric pressure of \n", + "72.2 cm.Take the surface factors of the conductor as 0.96.'''\n", + "\n", + "import math as m\n", + "\n", + "#As we know P = 241*(f+25)/d*root(r/D)*(V-Vc)^2\n", + "#Here,\n", + "d = 3.92*(72.2)/(273 + 20)\n", + "#Given\n", + "m0 = 0.96 # (surface factors) \n", + "D = 200.0 #cm (distance btwn condr.)\n", + "r = 1.2 #cm (radius of condr.)\n", + "#Critical Voltage\n", + "Vc = 48.8*m0*d*r*m.log10(D/r) #kV/phase\n", + "V = 220.0/1.732 #kV/phase\n", + "P = 241*(50+25)/d*m.sqrt(r/D)*(V-Vc)**2*1.0e-5 #kW/km/phase\n", + "#Total loss for 3-phase\n", + "loss = 3*P #kW/km/phase\n", + "print \"corona power loss per km = \",round(loss,2),\"kW/km/phase.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.37 , PAGE NO :- 1648" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sheath diameter = 8.29 cm.\n" + ] + } + ], + "source": [ + "'''A single-core lead-covered cable is to be designed for 66-kV to earth. Its conductor radius is 1.0 cm and its three insulating\n", + "materials A, B, C have permittivities of 5.4 and 3 respectively with corresponding maximum safe working stress of 38 kV per cm\n", + "(r.m.s. value), 26-kV per cm and 20-kV per cm respectively. Find the minimum diameter of the lead sheath.'''\n", + "\n", + "import math as m\n", + "from sympy import solve,Symbol,Eq\n", + "\n", + "#Given\n", + "gA = 38.0 #kV/cm (working stress on A)\n", + "gB = 26.0 #kV/cm (working stress on B)\n", + "gC = 20.0 #kV/cm (working stress on C)\n", + "ea = 5.0 # (rel. permittivity of A)\n", + "eb = 4.0 # (rel. permittivity of B)\n", + "ec = 3.0 # (rel. permittivity of C)\n", + "ra = 1.0 #cm (radius A)\n", + "#Working stress O< 1/(rel. permittivty)*(condr. radius)\n", + "#gA/gB = rb*eb/ra*ea\n", + "#Radius of B is\n", + "rb = (gA/gB)*(ra*ea/eb) #cm \n", + "#gA/gC = rc*ec/ra*ea\n", + "#Radius of C is\n", + "rc = (gA/gC)*(ra*ea/ec) #cm\n", + "\n", + "#Now , V = g*r*2.3*log10(r1/r)\n", + "\n", + "V1 = gA*ra*2.3*m.log10(rb/ra) #kV\n", + "V2 = gB*rb*2.3*m.log10(rc/rb) #kV\n", + "V = 66.0 #kV\n", + "V3 = V - V1 - V2 #V\n", + "\n", + "#Let the radius of sheath be rs\n", + "#V3 = gC*rc*2.3*m.log10(rs/rc) #kV . Therefore, rs is\n", + "\n", + "rs = rc * 10**(V3/(gC*rc*2.3)) #cm\n", + "print \"Sheath diameter = \",round(2*rs,2),\"cm.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.38 , PAGE NO :- 1649" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Charging current = 4.31 A.\n" + ] + } + ], + "source": [ + "'''The capacitances per kilometer of a 3-phase cable are 0.63 μF between the three cores bunched and the sheath and 0.37 μF\n", + "between one core and the other two connected to sheath. Calculate the charging current taken by eight kilometres of this cable\n", + "when connected to a 3-phase, 50-Hz, 6,600-V supply.'''\n", + "\n", + "#According to question 0.63 = 3*Cs\n", + "Cs = 0.63/3 #uF/km\n", + "#Also 0.37 = 2*C1 + Cs\n", + "C1 = (0.37 - Cs)/2 #uF/km\n", + "#For 8 km\n", + "Cs = Cs*8 #uF\n", + "C1 = C1*8 #uF\n", + "\n", + "Cn = Cs + 3*C1 #uF\n", + "Vp = 6600.0/1.732 #V\n", + "w = 2*3.14*50.0 #rad/s\n", + "#Charging current is\n", + "Ic = Vp*(w*Cn*1e-6) #A\n", + "print \"Charging current = \",round(Ic,2),\"A.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.39 , PAGE NO :- 1649" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Charging for 15 km = 48.0 A.\n" + ] + } + ], + "source": [ + "'''A 3-core, 3-phase belted cable tested for capacitance between a pair of cores on single phase with the third core earthed,\n", + "gave a capacitance of 0.4 mF per km. Calculate the charging current for 1.5 km length of this cable when connected to 22 kV, \n", + "3-phase,50-Hz supply.'''\n", + "\n", + "\n", + "Cl = 0.4 #uF (Capacitance)\n", + "Vl = 22000.0 #V (line voltage)\n", + "w = 2*3.14*50 #rad/s\n", + "#Charging current\n", + "Ic = (2.0/1.732)*Vl*Cl*w*1.0e-6 #A/km\n", + "#Charging current for 15 km\n", + "Ic = 15*Ic #A\n", + "print \"Charging for 15 km = \",round(Ic),\"A.\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.40 , PAGE NO :- 1649" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Capacitance (i) = 0.33 uF.\n", + "Capacitance (ii) = 0.13 uF.\n", + "capacitance between 2 shorted conductors and the other = 0.49 uF.\n" + ] + } + ], + "source": [ + "'''A 3-core, 3-phase metal-sheathed cable has (i) capacitance of 1 μF between shorted conductors and sheath and (ii) capacitance\n", + "between two conductors shorted with the sheath and the third conductor 0.6 μF. Find the capacitance\n", + "(a) between any two conductors (b) between any two shorted conductors and the third conductor.'''\n", + "\n", + "#(a)\n", + "#(i)As we know 3*Cs = 1.0\n", + "Cs = 1.0/3 #uF\n", + "#(ii) 2*C1 + Cs = 0.6 #uF\n", + "C1 = (0.6 - Cs)/2 #uF\n", + "print \"Capacitance (i) = \",round(Cs,2),\"uF.\"\n", + "print \"Capacitance (ii) = \",round(C1,2),\"uF.\"\n", + "#(b)\n", + "#capacitance between 2 shorted conductors and the other is given by\n", + "Cn = 2*C1 + (2.0/3)*Cs #uF\n", + "print \"capacitance between 2 shorted conductors and the other = \",round(Cn,2),\"uF.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.41 , PAGE NO :- 1650" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage drop from A to B = (7.47+2.78j) V.\n" + ] + } + ], + "source": [ + "'''A 2-wire a.c. feeder 1 km long supplies a load of 100 A at 0.8 p.f. lag 200 volts\n", + "at its far end and a load of 60 A at 0.9 p.f. lag at its mid-point. The resistance and reactance per km\n", + "(lead and return) are 0.06 ohm and 0.08 ohm respectively. Calculate the voltage drop along the\n", + "distributor from sending end to mid-point and from mid-point to far end.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#Voltage at C\n", + "Vc = 200.0 #V\n", + "theta = m.acos(0.8)\n", + "Ic = 100.0 #A\n", + "Icl = cm.rect(Ic,-theta) #A\n", + "#Loop impedance of feeder BC is\n", + "Z = (0.06 + 1j*0.08)/2 #ohm\n", + "#Voltage drop in BC \n", + "drop = Icl*Z #V\n", + "#Voltage at B is\n", + "Vb = Vc + drop #V\n", + "#Current at B is\n", + "Ib = 60.0 #A\n", + "theta = m.acos(0.9)\n", + "Ibl = cm.rect(Ib,-theta) #A\n", + "#Current in feeder AB is\n", + "Iab = Ibl + Icl #A\n", + "#Drop in AB is\n", + "dropab = Iab*Z #V\n", + "print \"Voltage drop from A to B = \",complex(round(dropab.real,2),round(dropab.imag,2)),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.42 , PAGE NO :- 1651" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "drop = 5.0 V.\n" + ] + } + ], + "source": [ + "'''A single-phase a.c. distributor 500 m long has a total impedance of (0.02 + j 0.04) ohm and is fed from one end at\n", + "250V. It is loaded as under :\n", + "(i) 50 A at unity power factor 200 m from feeding point.\n", + "(ii) 100 A at 0.8 p.f. lagging 300 m from feeding point.\n", + "(iii) 50 A at 0.6 p.f. lagging at the far end.\n", + "Calculate the total voltage drop and voltage at the far end.'''\n", + "\n", + "#Using 3rd method\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#The center of gravity(C.G) of load is at the following distance from feeding end\n", + "\n", + "cg = (50.0*200 + 100.0*300 + 50.0*500)/(50 + 100 + 50) #m\n", + "\n", + "#Value of resistance upto CG\n", + "R = cg*0.02/500 #ohm\n", + "#Value of reactance upto CG\n", + "X = cg*0.04/500 #ohm\n", + "#Average pf\n", + "pf = (50*1 + 100.0*0.8 + 50.0*0.6)/200\n", + "cosQ = pf\n", + "sinQ = m.sqrt(1 - pf*pf)\n", + "#Drop is\n", + "Itot = 50.0 + 100.0 + 50.0 #A\n", + "drop = 200.0*(R*cosQ + X*sinQ) #V\n", + "print \"drop = \",round(drop),\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.43 , PAGE NO :- 1652" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Supply voltage = 345.0 V.\n", + "Supply Voltage lead B by = 4.59 degrees.\n" + ] + } + ], + "source": [ + "'''A single-phase distributor, one km long has resistance and reactance per conductor of 0.2 ohm and 0.3 ohm respectively. At the\n", + "far end, the voltage VB = 240 V and the current is 100 A at a power factor of 0.8 lag. At the mid-point A of the distributor \n", + "current of 100 A is tapped at a power factor of 0.6 lag with reference to the voltage VA at the mid-point. Calculate the supply\n", + "voltage VS for the distributor and the phase angle between VS and VB.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#Given\n", + "Vb = 240.0 #V (Voltage at B)\n", + "Ib = 100.0 #A (current)\n", + "theta = m.acos(0.8)\n", + "Ibl = cm.rect(Ib,-theta) #A\n", + "X = 0.2 + 1j*0.3 #ohm\n", + "#Drop over AB\n", + "dropab = Ibl*(X) #V\n", + "\n", + "#Voltage at A\n", + "Va = Vb + dropab #V\n", + "\n", + "#Phase angle between A and B\n", + "ang = cm.phase(Va)\n", + "theta2 = m.acos(0.6)\n", + "theta2 = theta2 - ang\n", + "Ia = 100.0 #A\n", + "Ial = cm.rect(Ia,-theta2) #A\n", + "\n", + "#Total I\n", + "I = Ial + Ibl #A\n", + "#Drop in section\n", + "dropsa = I*X #V\n", + "\n", + "#Voltage at S\n", + "Vs = Va + dropsa #V\n", + "print \"Supply voltage = \",round(abs(Vs)),\"V.\"\n", + "print \"Supply Voltage lead B by = \",round(cm.phase(Vs)*180/3.14,2),\"degrees.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.44 , PAGE NO :- 1653" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ia = (25-24j) A.\n" + ] + } + ], + "source": [ + "'''A 1-phase ring distributor ABC is fed at A. The loads at B and C are 20 A at 0.8 p.f. lagging and 15 A at 0.6 p.f.\n", + "lagging respectively, both expressed with reference to voltage at A. The total impedances of the sections AB, BC and\n", + "CA are (1 + j1), (1 + j2) and (1 + j3) ohm respectively. Find the total current fed at A and the current in each section.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "\n", + "#Current in AB is\n", + "theta = m.acos(0.8) \n", + "Iab = cm.rect(20,-theta)\n", + "\n", + "#Current in AC is\n", + "theta = m.acos(0.6) \n", + "Iac = cm.rect(15,-theta)\n", + "\n", + "#Impedances of dif. section\n", + "Zab = 1 + 1j #ohm\n", + "Zbc = 1 + 1j*2 #ohm\n", + "Zca = 1 + 1j*3 #ohm\n", + "\n", + "#Drop over AB is\n", + "dropab = Iab*Zab #V\n", + "\n", + "#Drop over AC is\n", + "dropac = Iac*Zca #V\n", + "\n", + "#Potential diff. B and C is\n", + "pd_bc = dropac - dropab #V\n", + "\n", + "#Equivalent Thevenin's thm impedance across bc\n", + "Ztot = Zab + Zca + Zbc #ohm\n", + "\n", + "#Current in BC\n", + "Ibc = pd_bc/Ztot #A\n", + "\n", + "#Total current in AB is\n", + "Iab2 = Iab + Ibc #A\n", + "\n", + "#Total current in BC is\n", + "Ibc2 = Iac - Ibc #A\n", + "\n", + "#Total current fed at point A\n", + "Ia = Iab2 + Ibc2 #A\n", + "\n", + "print \"Ia = \",Ia,\"A.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.45 , PAGE NO :- 1654" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I1 = (19.5-16.5j) A.\n", + "I2 = (14.5-19.5j) A.\n", + "I3 = (3.5-4.5j) A.\n", + "Vb = (347.5-22.5j) V.\n", + "Vc = (327-24j) V.\n" + ] + } + ], + "source": [ + "'''A 2-wire ring distributor ABC is supplied at A at 400 V. Point loads of 20 A at a p.f. of 0.8 lagging and 30 A at a p.f. 0.6\n", + "lagging are tapped off at B and C respectively. Both the power factors refer to the voltage at A. The respective go-and-return \n", + "impedances of sections AB,BC and CA are (1 + j2) ohm, (2 + j3) ohm and (1 + j3) ohm. Calculate the current flowing through each\n", + "section and the potentials at B and C. Use Superposition theorem.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "\n", + "#Current in AB is\n", + "theta = m.acos(0.8) \n", + "Iab = cm.rect(20,-theta)\n", + "\n", + "#Current in AC is\n", + "theta = m.acos(0.6) \n", + "Iac = cm.rect(30,-theta)\n", + "\n", + "#Impedances of dif. section\n", + "Zab = 1 + 1j*2 #ohm\n", + "Zbc = 2 + 1j*3 #ohm\n", + "Zca = 1 + 1j*3 #ohm\n", + "\n", + "#Let the currents in AB and AC is Ia1 and Ia2\n", + "Ia1 = Iab*(Zbc + Zca)/(Zbc + Zca + Zab) #A\n", + "\n", + "\n", + "Ia2 = Iab - Ia1 #A\n", + "\n", + "#Let the currents in AB and AC is Ia11 and Ia22\n", + "Ia11 = Iac*(Zca)/(Zbc + Zca + Zab) #A\n", + "Ia22 = Iac - Ia11 #A\n", + "\n", + "#As per superposition thm,\n", + "I1 = Ia1 + Ia11 #A\n", + "I2 = Ia2 + Ia22 #A\n", + "I3 = Ia11 - Ia2 #A\n", + "\n", + "#Potential Of B is\n", + "Vb = 400.0 - I1*Zab #V\n", + "#Potential of C is \n", + "Vc = 400.0 - I2*Zca #V\n", + "\n", + "print \"I1 = \",I1,\"A.\"\n", + "print \"I2 = \",I2,\"A.\"\n", + "print \"I3 = \",I3,\"A.\"\n", + "print \"Vb = \",Vb,\"V.\"\n", + "print \"Vc = \",Vc,\"V.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.46 , PAGE NO :- 1655" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Iab = (139.76-42.81j) A\n", + "Ibc = (99.76-12.81j) A\n", + "Icd = (-20.24-12.81j) A\n", + "Ida = (-80.86+22.2j) A\n" + ] + } + ], + "source": [ + "'''A 3-phase ring main ABCD, fed from end A, supplies balanced loads of 50 A at 0.8 p.f. lagging at B, 120 A at u.p.f. at C and\n", + "70 A at 0.866 p.f. lagging at D, the load currents being referred to the voltage at point A.The impedance per phase of the various\n", + "line sections are :\n", + "section AB = (1 + j0.6)ohm ; section BC = (1.2 + j0.9) ohm\n", + "section CD = (0.8 + j0.5)ohm ; section DA = (3 + j2) ohm\n", + "Determine the currents in the various sections.'''\n", + "\n", + "from sympy import Symbol\n", + "import math as m\n", + "import cmath as cm\n", + "import numpy as np\n", + "\n", + "#Let the current in section AB = (x + jy)A\n", + "x = Symbol('x')\n", + "y = Symbol('y')\n", + "Iab = x + 1j*y #A\n", + "#Current in BC\n", + "theta = m.acos(0.8)\n", + "Icl = cm.rect(50.0,-theta) #A\n", + "Ibc = Iab - Icl #A\n", + "#Current in CD\n", + "Icd = Ibc - 120 #A\n", + "#Current in DA\n", + "theta = m.acos(0.866)\n", + "Idl = cm.rect(70.0,-theta) #A\n", + "Ida = Icd - Idl #A\n", + "#Applying kirchoff's law\n", + "V = (1 + 1j*0.6)*Iab + (1.2 + 1j*0.9)*Ibc + (0.8 + 1j*0.5)*Icd + (3 + 1j*2)*Ida #V\n", + "#Equating V to 0 we get following two equations\n", + "# 6*x - 4*y + 1009.8 = 0 - 1 \n", + "# 4*x + 6*y - 302.2 = 0 - 2\n", + "\n", + "\n", + "a = np.array([[6,-4], [4,6]])\n", + "b = np.array([1009.8,302.2])\n", + "vec = np.linalg.solve(a, b)\n", + "x = vec[0]\n", + "y = vec[1]\n", + "\n", + "\n", + "Iab = x + 1j*y #A\n", + "#Current in BC\n", + "Ibc = Iab - Icl #A\n", + "#Current in CD\n", + "Icd = Ibc - 120 #A\n", + "#Current in DA\n", + "Ida = Icd - Idl #A\n", + "\n", + "print \"Iab = \",complex(round(Iab.real,2),round(Iab.imag,2)),\"A\"\n", + "print \"Ibc = \",complex(round(Ibc.real,2),round(Ibc.imag,2)),\"A\"\n", + "print \"Icd = \",complex(round(Icd.real,2),round(Icd.imag,2)),\"A\"\n", + "print \"Ida = \",complex(round(Ida.real,2),round(Ida.imag,2)),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.47 , PAGE NO :- 1656" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power transmitted by 1st line = 5283.01 kW.\n", + "Power transmitted by 2nd line = 6716.99 kW.\n", + "Total power(check) = 12000.0 kW.\n" + ] + } + ], + "source": [ + "'''A total load of 12,000 kW at a power factor of 0.8 lagging is transmitted to a substation by two overhead\n", + "three-phase lines connected in parallel. One line has a conductor resistance of 2 ohm per conductor and reactance\n", + "(line to neutral) of 1.5 ohm, the corresponding values for the other line being 1.5 and 1.2 ohm respectively. Calculate\n", + "the power transmitted by each overhead line.'''\n", + "\n", + "import math as m\n", + "import cmath as cm\n", + "\n", + "#Let us assume a line voltage of 1000 kV for convinience.\n", + "Z1 = 2 + 1j*1.5 #ohm \n", + "Z2 = 1.5 + 1j*1.2 #ohm\n", + "#Total load current\n", + "Il = 12000/(1.732*1000*0.8) #A\n", + "theta = m.acos(0.8)\n", + "Il1 = cm.rect(Il,-theta) #A\n", + "\n", + "#Now\n", + "I1 = Il1*Z2/(Z1 + Z2) #A\n", + "\n", + "#Power transmitted by 1st line\n", + "P1 = 1.732*I1.real*1000 #kW\n", + "\n", + "#Now\n", + "I2 = Il1*Z1/(Z1 + Z2) #A\n", + "\n", + "#Power transmitted by 1st line\n", + "P2 = 1.732*I2.real*1000 #kW\n", + "\n", + "#Total power\n", + "power = P1 + P2 #kW\n", + "print \"Power transmitted by 1st line = \",round(P1,2),\"kW.\"\n", + "print \"Power transmitted by 2nd line = \",round(P2,2),\"kW.\"\n", + "print \"Total power(check) = \",round(power,2),\"kW.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## EXAMPLE 41.48 , PAGE NO :- 1658" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V1 = 7.54 kV\n", + "V2 = 8.29 kV\n", + "V3 = 9.87 kV\n", + "V4 = 12.45 kV\n", + "string efficiency = 76.55 %\n" + ] + } + ], + "source": [ + "'''For a string insulator with four discs, the capacitance of the disc is ten times the capacitance between the pin and earth.\n", + "Calculate the voltage across each disc when used on a 66-kV line. Also, calculate the string efficiency.'''\n", + "\n", + "import math as m\n", + "from sympy import Symbol,solve,Eq\n", + "#Let C be self capacitance and kC be capacitance b/w each link and earth,\n", + "#Given\n", + "k = 10.0\n", + "#Let voltage across 1st disc be\n", + "V1 = Symbol('V1')\n", + "V2 = (1+k)/k*V1 #V\n", + "V3 = (1 + 3/k + 1/k**2)*V1 #V\n", + "V4 = (1 + 6/k + 5/k**2 + 1/k**3)*V1 #V\n", + "#Voltage V is\n", + "Va = V1 + V2 + V3 + V4 #kV\n", + "Vb = 66/1.73 #kV\n", + "eq = Eq(Va,Vb)\n", + "V1 = solve(eq)\n", + "V1l = V1[0] #kV\n", + "V2 = (1+k)/k*V1l #V\n", + "V3 = (1 + 3/k + 1/k**2)*V1l #V\n", + "V4 = (1 + 6/k + 5/k**2 + 1/k**3)*V1l #V\n", + "print \"V1 = \",round(V1l,2),\"kV\"\n", + "print \"V2 = \",round(V2,2),\"kV\"\n", + "print \"V3 = \",round(V3,2),\"kV\"\n", + "print \"V4 = \",round(V4,2),\"kV\"\n", + "#string efficiency\n", + "eff = 66.0/(4*1.732*V4)*100 #%\n", + "print \"string efficiency = \",round(eff,2),\"%\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.49 , PAGE NO :- 1659" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V1 = 5.51 kV\n", + "V2 = 6.12 kV\n", + "V3 = 7.42 kV\n", + "string efficiency = 85.63 %\n" + ] + } + ], + "source": [ + "'''Explain what is meant by string efficiency and how it can be improved. Each line of a 3-phase, 33-kV system is suspended by a\n", + "string of 3 identical insulator discs. The capacitance of each disc is 9 times the capacitance to ground. Find voltage distribution\n", + "across each insulator and the string efficiency. Suggest a method for improving the string efficiency.'''\n", + "\n", + "import math as m\n", + "from sympy import Symbol,solve,Eq\n", + "#Let C be self capacitance and kC be mutual capacitance b/w each\n", + "#Given\n", + "k = 9.0\n", + "#Let voltage across 1st disc be\n", + "V1 = Symbol('V1')\n", + "V2 = (1+k)/k*V1 #V\n", + "V3 = (1 + 3/k + 1/k**2)*V1 #V\n", + "\n", + "#Voltage V is\n", + "Va = V1 + V2 + V3 #kV\n", + "Vb = 33/1.732 #kV \n", + "eq = Eq(Va,Vb)\n", + "V1 = solve(eq)\n", + "V1l = V1[0] #kV\n", + "V2 = (1+k)/k*V1l #V\n", + "V3 = (1 + 3/k + 1/k**2)*V1l #V\n", + "\n", + "print \"V1 = \",round(V1l,2),\"kV\"\n", + "print \"V2 = \",round(V2,2),\"kV\"\n", + "print \"V3 = \",round(V3,2),\"kV\"\n", + "\n", + "#string efficiency\n", + "eff = 33.0/(3*1.732*V3)*100 #%\n", + "print \"string efficiency = \",round(eff,2),\"%\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.50 , PAGE NO :- 1660" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Bus-bar voltage of station A = 33.79 kV\n", + "phase angle = 1.75 degrees\n" + ] + } + ], + "source": [ + "'''The bus-bar voltages of two stations A and B are 33 kV and are in phase. If station A sends 8.5 MW power at u.p.f.\n", + "to station B through an interconnector having an impedance of (3 + j4) ohm , determine the bus-bar voltage of station A\n", + "and the phase angle shift between the busbar voltages.'''\n", + "\n", + "import cmath as cm\n", + "\n", + "#Voltage of station B is\n", + "Vb = 33000.0/1.732 #V/phase\n", + "\n", + "#Power transferred\n", + "pin = 8.5 #MW\n", + "\n", + "#Current in interconnector\n", + "I = pin*1e+6/(1.732*33000)\n", + "\n", + "#Voltage Va is\n", + "Va = Vb + I*(3+1j*4)\n", + "#Bus-bar voltage of station A is\n", + "Va1 = 1.732*abs(Va)/1000 #kV\n", + "\n", + "print \"Bus-bar voltage of station A = \",round(Va1,2),\"kV\"\n", + "print \"phase angle = \",round(cm.phase(Va)*180/3.14,2),\"degrees\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.51 , PAGE NO :- 1666" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Clearance between conductor and water at mid-way = 62.0 m.\n" + ] + } + ], + "source": [ + "'''A transmission line conductor at a river crossing is supported from two towers at heights of 70 m above water level.\n", + "The horizontal distance between towers is 300 m. If the tension in conductor is 1,500 kg, find the clearance at a point\n", + "midway between the towers. The size of conductor is 0.9 cm2. Density of conductor material is 8.9 g/cm3 and suspension\n", + "length of the string is 2 metres.'''\n", + "\n", + "#Given\n", + "l = 300.0/2 #m (distance of mid pt from one tower)\n", + "T = 1500.0 #kg-wt (tension)\n", + "\n", + "#density = mass/volume. Therefore Weight (W) = rho*l*A\n", + "W = (0.9e-4)*(8.9e+3) #kg-wt\n", + "\n", + "#Sag in the line\n", + "sag = (W*l**2)/(2*T) #m\n", + "\n", + "\n", + "#Clearance between conductor and water at mid-way\n", + "cler = 70.0 - sag - 2.0 #m\n", + "print \"Clearance between conductor and water at mid-way = \",round(cler),\"m.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.52 , PAGE NO :- 1666" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total weight per metre run of line = 2.46 kg-wt/m.\n" + ] + } + ], + "source": [ + "'''The effective diameter of a line is 1.96 cm and it weighs 90 kg per 100 metre length. What would be the additional loading due\n", + "to ice of radial thickness 1.25 cm and a horizontal wind pressure of 30 kg/m^2 of projected area? Also, find the total weight \n", + "per metre run of the line.Density of ice is 920 kg/m^3.'''\n", + "\n", + "import math as m\n", + "\n", + "#Given\n", + "W = 90.0/100 #kg/m (weight of condr)\n", + "#Weight due to ice is Wi = 3.14*rho*R*(D+R)\n", + "Wi = 3.14*920.0*0.0125*(0.0196 + 0.0125) #kg/m (Weight of ice)\n", + "#Weight due to wind is Ww = P*(D + 2*R)\n", + "Ww = 30.0*(0.0196 + 2*0.0125) #kg/m\n", + "\n", + "#Total weight is given by\n", + "\n", + "Wt = m.sqrt((W+Wi)**2 + Ww**2) #kg-wt/m\n", + "\n", + "print \"Total weight per metre run of line = \",round(Wt,2),\"kg-wt/m.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.53 , PAGE NO :- 1666" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "sag = 3.27 m.\n" + ] + } + ], + "source": [ + "'''A transmission line has a span of 150 metres between supports, the supports being at the same level. The conductor has a\n", + "cross-sectional area of 2 cm^2. The ultimate strength is 5,000 kg/cm^2. The specific gravity of the material is 8.9. If the wind\n", + "pressure is 1.5 kg/m length of the conductor, calculate the sag at the centre of the conductor if factor of safety is 5.'''\n", + "\n", + "\n", + "import math as m\n", + "\n", + "#Safety factor = Ultimate stress/Working stress\n", + "sf = 5.0 #Safety factor\n", + "wstress = 5000.0/sf #kg/cm^2\n", + "#Tension\n", + "A = 2.0 #cm^2 (cross-sectional area)\n", + "T = wstress*A #kg-wt\n", + "#Volume of 1m of conductor\n", + "V = A*100.0 #cm^3\n", + "#Wt of 1m of material Using density = mass/volume\n", + "W = 8.9*200.0/1000.0 #kg-wt\n", + "#W is 1.78 and not 1.98\n", + "\n", + "#Total weight per metre\n", + "Wt = m.sqrt(W**2 + 1.5**2)\n", + "\n", + "#Sag = W*l^2/2T\n", + "l = 150.0/2 #m (length from pole to centre)\n", + "sag = Wt*l**2/(2*T) #m\n", + "\n", + "\n", + "print \"sag = \",round(sag,2),\"m.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.54 , PAGE NO :- 1667" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Verticle Sag = 5.38 m.\n" + ] + } + ], + "source": [ + "'''A transmission line has a span of 200 metres between level supports. The conductor has a cross-sectional area of 1.29 cm^2,\n", + "weighs 1,170 kg/km and has a breaking stress of 4,218 kg/cm2. Calculate the sag for a factor of safety of 5 allowing a wind pressure\n", + "of 122 kg per m^2 of projected area. What is the vertical sag?'''\n", + "\n", + "import math as m\n", + "\n", + "#Safety factor = ultimate stress/working stress \n", + "wstress = 4218.0/5 #kg/cm2 (working stress)\n", + "#Working tension\n", + "A = 1.29 #cm^2 (cross-sectional area) \n", + "T = wstress*A #kg-wt\n", + "W = 1170.0/1000 #kg-wt/m\n", + "#Let us now find diameter of the conductor from the equation\n", + "# A = 3.14 * d^2/4\n", + "d = m.sqrt(4*A/3.14)/100 #m \n", + "#Projected area of the conductor per metre length\n", + "proArea = d*1 #m^2\n", + "#Weight of wind exerted\n", + "Ww = 122*proArea #kg-wt/m\n", + "\n", + "#Total weight\n", + "Wt = m.sqrt(W**2 + Ww**2) #kg-wt/m\n", + "\n", + "#Slant sag\n", + "l = 200.0/2 #m\n", + "sag = Wt*l**2/(2*T) #m\n", + "\n", + "#Now, tan θ = Ww/W\n", + "tanQ = Ww/W\n", + "#vertical sag\n", + "vsag = sag*m.cos(m.atan(tanQ)) #m\n", + "print \"Verticle Sag = \",round(vsag,2),\"m.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.55 , PAGE NO :- 1667" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max. sag produced = 3.93 m.\n", + "Verticle sag = 2.36 m.\n" + ] + } + ], + "source": [ + "'''A transmission line has a span of 214 metres. The line conductor has a cross-section of 3.225 cm^2 and has an ultimate \n", + "breaking strengh of 2,540 kg/cm^2. Assuming that the line is covered with ice and provides a combined copper and ice load of\n", + "1.125 kg/m while the wind pressure is 1.5 kg/m run (i) calculate the maximum sag produced. Take a factor of safety of 3\n", + "(ii) also determine the vertical sag.'''\n", + "\n", + "import math as m\n", + "\n", + "#(i) Maximum sag\n", + "#Here,\n", + "W = 1.125 #kg-wt/m\n", + "Ww = 1.5 #kg-wt/m\n", + "Wt = m.sqrt(1.125**2 + 1.5**2) #kg-wt/m\n", + "#Safety factor = ultimate stress/working stress\n", + "wstress = 2540.0/3 #kg-wt/cm2\n", + "#Permissible tension\n", + "A = 3.225 #cm^2 (cross-sectional area) \n", + "T = wstress*(3.225) #kg-wt\n", + "#Sag is\n", + "l = 214.0/2 #m \n", + "sag = Wt*l**2/(2*T) #m\n", + "\n", + "#(ii)Verticle Sag\n", + "#Now,\n", + "tanQ = Ww/W\n", + "cosQ = m.cos(m.atan(tanQ))\n", + "#vertical sag\n", + "vsag = sag*cosQ #m\n", + "print \"Max. sag produced = \",round(sag,2),\"m.\"\n", + "print \"Verticle sag = \",round(vsag,2),\"m.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.56 , PAGE NO :- 1658" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum clearance = 23.02 m.\n", + "Hence, clearance point = 30.7 m.\n" + ] + } + ], + "source": [ + "'''Two towers of height 30 and 90 m respectively support a transmission line conductor at water crossing. The horizontal\n", + "distance between the towers is 500 m. If the tension in the conductor is 1,600 kg, find the minimum clearance of the\n", + "conductor and the clearance of the conductor mid-way between the supports. Weight of the conductor is 1.5 kg/m. Bases of\n", + "the towers can be considered to be at the water level.'''\n", + "\n", + "\n", + "#Given\n", + "l = 500.0/2 #m (mid-way of towers)\n", + "h = 90.0 - 30.0 #m (height)\n", + "T = 1600.0 #kg-wt (tension)\n", + "W = 1.5 #kg-wt/m (weight)\n", + "\n", + "#Point of maximum sag is\n", + "x1 = l - h*T/(2*W*l)\n", + "\n", + "#Sag is\n", + "sag = W*x1**2/(2*T) #m\n", + "\n", + "#Minimum clearance is\n", + "cler = 30.0 - sag #m\n", + "print \"Minimum clearance = \",round(cler,2),\"m.\"\n", + "#Taking this point as reference , sag of mid-pt is\n", + "sag2 = W*(l-x1)**2/(2*T) #m\n", + "\n", + "#Mid-point Clearance\n", + "cler2 = cler + sag2\n", + "print \"Hence, clearance point = \",round(cler2,2),\"m.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## EXAMPLE 41.57 , PAGE NO :- 1658" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Clearance of conductor from water surface = 63.89 m.\n" + ] + } + ], + "source": [ + "'''An overhead transmission line at a river crossing is supported from two towers at heights of 50 m and 100 m above the\n", + "water level, the horizontal distance between the towers being 400 m. If the maximum allowable tension is 1,800 kg and the\n", + "conductor weighs 1 kg/m, find the clearance between the conductor and water at a point mid-way between the towers.'''\n", + "\n", + "#Given\n", + "h = 100.0 - 50.0 #m (height)\n", + "l = 400.0/2 #m (mid-way length)\n", + "T = 1800.0 #kg-wt (tension)\n", + "W = 1.0 #kg-wt/m (Weigth)\n", + "#Point of max sag\n", + "x1 = l - h*T/(2*W*l) #m\n", + "x2 = l + h*T/(2*W*l) #m\n", + "\n", + "#Distance of mid-pt from the above point is\n", + "d1 = (x2 - x1)/2 #m\n", + "\n", + "#Therefore height of P above max sag point is\n", + "h1 = W*d1**2/(2*T) #m\n", + "h2 = W*x2**2/(2*T) #m\n", + "\n", + "#Therefore point P from top is\n", + "Psag = h2 - h1 #m\n", + "\n", + "#Clearance above ground is\n", + "cler = 100.0 - Psag #m\n", + "\n", + "print \"Clearance of conductor from water surface = \",round(cler,2),\"m.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## EXAMPLE 41.58 , PAGE NO :- 1659" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Horizontal Tension = 2373.42 kg-wt/m.\n" + ] + } + ], + "source": [ + "'''A conductor is strung across a river, being supported at the two ends at heights of 20 m and 16 m respectively, from the bed\n", + "of the river. The distance between the supports is 375 m and the weight of the conductor = 1.2 kg/m.If the clearance of the\n", + "conductor from the river bed be 9 m, find the horizontal tension in the conductor. Assume a parabolic configuration and that\n", + "there is no wind or ice loading.'''\n", + "\n", + "from sympy import Symbol,solve,Eq\n", + "\n", + "#Given\n", + "l = 375.0/2 #m (mid-pt of towers)\n", + "h = 20.0 - 16.0 #m (height)\n", + "W = 1.2 #kg-wt/m\n", + "\n", + "#Let T be the tension in conductor\n", + "T = Symbol('T')\n", + "x1 = l - h*T/(2*W*l) #m\n", + "\n", + "#Minimum clearance\n", + "cler = 9.0 #m \n", + "sag = 16.0 - cler #m\n", + "\n", + "#Now ,sag is also\n", + "sag2 = (W*x1*x1)/(2*T) #m\n", + "\n", + "#Equating the two equations\n", + "eq = Eq(sag,sag2)\n", + "T = solve(eq)\n", + "T1 = T[0] #kg-wt/m\n", + "print \"Horizontal Tension = \",round(T1,2),\"kg-wt/m.\"" + ] + } + ], + "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 +} |