path: root/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_VOL-III_by_B.L.Thareja/chapter41.ipynb

{
 "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.\""
   ]
  }
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