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diff --git a/Principles_of_Power_System/chapter6_1.ipynb b/Principles_of_Power_System/chapter6_1.ipynb deleted file mode 100644 index a1cd7a3c..00000000 --- a/Principles_of_Power_System/chapter6_1.ipynb +++ /dev/null @@ -1,1112 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:357707f074867e12cc217076fc35399c9b1e6483f5b92a23e0c3be75457e81b5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 6: Power Factor Improvement"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.1, Page Number: 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "kW1 = 300\n",
- "pf1 = 0.6 #power factor\n",
- "pf2 = 1\n",
- "\n",
- "#Calculation:\n",
- "kVA = kW1/pf1\n",
- "kW2 = kVA*pf2\n",
- "kW = kW2-kW1\n",
- "\n",
- "#Result:\n",
- "print \"Increased power supplied by the alternator is\",kW,\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Increased power supplied by the alternator is 200.0 kW\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.2, Page Number: 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "V = 400 #rated voltage(V)\n",
- "Im = 31.7 #motor curtrent(A)\n",
- "pf1 = 0.7 #initial power factor\n",
- "pf2 = 0.9 #raised power factor\n",
- "f = 50 #Hz\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Referring to the phasor diagram,\n",
- "Ima = Im*pf1 #Active component of Im(A)\n",
- "I = Ima/pf2 #total current(A)\n",
- "Imr = Im*math.sqrt(1-pf1**2) #Reactive component of Im(A)\n",
- "Ir = I*math.sqrt(1-pf2**2) #Reactive component of I(A)\n",
- "Ic = Imr-Ir #A\n",
- "\n",
- "C = Ic/(V*2*math.pi*f)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Required value of capacitor is\",round(C,7),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required value of capacitor is 9.46e-05 uF\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.3, Page Number: 110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "#(i):\n",
- "kW1 = 20\n",
- "pf1 = 1\n",
- "\n",
- "#(ii):\n",
- "kW2 = 100\n",
- "pf2 = 0.707\n",
- "\n",
- "#(iii):\n",
- "kW3 = 50\n",
- "pf3 = 0.9\n",
- "\n",
- "#Calculation:\n",
- "kVA1 = kW1/pf1\n",
- "kVA2 = kW2/pf2\n",
- "kVA3 = kW3/pf3\n",
- "\n",
- "kVAR1 = kVA1*0\n",
- "kVAR2 = -kVA2*math.sin(math.acos(0.707))\n",
- "kVAR3 = kVA3*math.sin(math.acos(0.9))\n",
- "\n",
- "kW = kW1+kW2+kW3\n",
- "kVAR = kVAR1+kVAR2+kVAR3\n",
- "kVA = math.sqrt(kW**2+kVAR**2)\n",
- "pf = kW/kVA\n",
- "\n",
- "#Result:\n",
- "print \"Total kW is\",kW\n",
- "print \"Total kVA is\",round(kVA)\n",
- "print \"Power factor is\",round(pf,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total kW is 170\n",
- "Total kVA is 186.0\n",
- "Power factor is 0.913\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.4, Page Number: 111"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "pf1 = 0.75 #Original p.f lag\n",
- "P = 5 #motor input(kW)\n",
- "pf2 = 0.9 #final p.f lag\n",
- "\n",
- "#Calculation:\n",
- "kVAR = P*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))\n",
- "R = kVAR/3\n",
- "\n",
- "#Result:\n",
- "print \"Rating of capacitors is\",round(R,3),\"kVAR\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rating of capacitors is 0.663 kVAR\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.5, Page Number: 111"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "Vph = 400 #rated voltage(V)\n",
- "n = .93 #efficiency\n",
- "P = 74.6 #power output(kW)\n",
- "pf1 = 0.75 #power faactor\n",
- "pf2 = 0.95 #raised power factor\n",
- "n1 = 4 #no. of capacitors in each branch\n",
- "V1 = 100 #rating of capacitors(V)\n",
- "f = 50 #frequency(Hz)\n",
- "\n",
- "#Calculation:\n",
- "Pi = round(P/n) #motor input(kW)\n",
- "phy1 = math.acos(pf1) \n",
- "phy2 = math.acos(pf2)\n",
- "#Leading kVAR taken by the condenser bank\n",
- "kVAR = round(Pi*(math.tan(phy1)-math.tan(phy2)),2) \n",
- "kVAR1 = round(kVAR/3,2) #Leading kVAR taken by each of three sets\n",
- "\n",
- "#Fig. above shows the delta* connected condenser bank.\n",
- "C = symbols('C')\n",
- "Icp = 2*math.pi*f*C*Vph #Phase current of capacitor(A)\n",
- "\n",
- "c = solve(Vph*Icp/1000-kVAR1,C)[0] #capacitance(F)\n",
- "\n",
- "#Result:\n",
- "print \"Capacitance of each capacitor is\",round(c*4*10**6,1),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of each capacitor is 1173.8 uF\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.6, Page Number: 112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "P = 800 #load(kW)\n",
- "pf1 = 0.8 #initial power factor\n",
- "pf2 = 0.9 #improved power factor\n",
- "t = 3000 #working hours\n",
- "c1 = 100 #1st part tariff(Rs/kVA)\n",
- "c2 = 20 #2nd part tariff(paise/kWh)\n",
- "c3 = 60 #cost of capacitors(Rs/kVAR)\n",
- "n = 10 #interest & depreciation on capacitors(%)\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "#Leading kVAR taken by the capacitors:\n",
- "kVAR = P*(math.tan(phy1)-math.tan(phy2))\n",
- "\n",
- "#Annual cost before p.f. correction:\n",
- "kVAm1 = P/pf1 #max demand\n",
- "T1 = kVAm1*c1+P*t*c2/100 #total annual cost(Rs)\n",
- "\n",
- "#Annual cost after p.f. correction:\n",
- "kVAm2 = P/pf2 #max demand\n",
- "T2 = c1*kVAm2+P*t*c2/100+n*c3*kVAR/100\n",
- "\n",
- "T = T1-T2 #annual saving(Rs)\n",
- "\n",
- "#Result:\n",
- "print \"Annual saving is Rs\",round(T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Annual saving is Rs 9836.0\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.7, Page Number: 112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "P1 = 200 #factory load(kW)\n",
- "pf1 = 0.85 #initial power factor\n",
- "pf2 = 0.9 #improved power factor\n",
- "t = 2500 #working hours\n",
- "c1 = 150 #1st part tariff(Rs/kVA)\n",
- "c2 = 5 #2nd part tariff(paise/kWh)\n",
- "c3 = 420 #cost of capacitors(Rs/kVAR)\n",
- "Pcl = 100 #capacitor loss(W/kVAR)\n",
- "n = 10 #interest & depreciation on capacitors(%)\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "#suppose the leading kVAR taken by the capacitors is x:\n",
- "x = symbols('x')\n",
- "Pcl1 = Pcl*x/1000 #capacitor loss(kW)\n",
- "P2 = P1+Pcl1 #total power(kW)\n",
- "x1 = solve((P1*math.tan(phy1)-P2*math.tan(phy2))-x,x)[0]\n",
- "\n",
- "\n",
- "#Annual cost before p.f. correction:\n",
- "kVAm1 = P1/pf1 #max demand\n",
- "T1 = kVAm1*c1+P1*t*c2/100 #total annual cost(Rs)\n",
- "\n",
- "\n",
- "#Annual cost after p.f. correction:\n",
- "kVAm2 = P1/pf2 #max demand\n",
- "T2 = c1*kVAm2+P1*t*c2/100+n*c3*x1/100+round(Pcl*x1*t*c2/100000)\n",
- "\n",
- "T = T1-T2 #annual saving(Rs)\n",
- "\n",
- "#Result:\n",
- "print \"Annual saving is Rs\",round(T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Annual saving is Rs 553.0\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.8, Page Number: 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "pf = 0.8 #power factor\n",
- "kVA = 750 # monthly demand\n",
- "c1 = 8.50 #monthly power rate(Rs/kVA per month)\n",
- "R = 250 #rating of capacitors(kVA)\n",
- "C2 = 20000 #installed cost of equipment(Rs)\n",
- "r = 10 #fixed charge rate(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "phy = math.acos(pf)\n",
- "kW = kVA*math.cos(phy) #kW component of demand\n",
- "kVAR = kVA*math.sin(phy) #kVAR component of demand\n",
- "kVAR1 = kVAR-R #Leading kVAR taken by the capacitors\n",
- "kVA1 = math.sqrt(kVAR1**2+kW**2) #kVA after p.f. improvement\n",
- "kVA11 = round(kVA-kVA1,1) #Reduction in kVA\n",
- "T1 = c1*kVA11*12 #Yearly saving on kVA charges\n",
- "T2 = r*C2/100 #Fixed charges/year(Rs)\n",
- "T = T1-T2 #net saaving(Rs)\n",
- "\n",
- "#Result:\n",
- "print \"The annual saving is Rs\",T"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual saving is Rs 9985.0\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.9, Page Number: 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "pf1 = 0.8 #initial power factor\n",
- "pf2 = 0.9 #improved power factor\n",
- "L1 = 200 #load(kW)\n",
- "L2 = 80 #motor load(kW)\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "L = L1+L2 #combined load(kW)\n",
- "#In Fig. above, \u0394 OAB is the power triangle for load,\n",
- "#\u0394 ODC for combined load and \u0394 BEC for the motor.\n",
- "#(i):\n",
- "kVAR = L1*math.tan(phy1)-L*math.tan(phy2) #Leading kVAR taken by the motor\n",
- "#(ii):\n",
- "kVArat = math.sqrt(L2**2+kVAR**2) #kVA rating of motor\n",
- "#(iii):\n",
- "pf = L2/kVArat #p.f. of motor\n",
- "\n",
- "#Result:\n",
- "print \"The leading kVAR taken by the motor is\",round(kVAR,2)\n",
- "print \"kVA rating of the motor is\",round(kVArat,2)\n",
- "print \"Power factor at which the motor operates is\",round(pf,3),\"leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The leading kVAR taken by the motor is 14.39\n",
- "kVA rating of the motor is 81.28\n",
- "Power factor at which the motor operates is 0.984 leading\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.10, Page Number: 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "#for induction motor:\n",
- "kW1 = 37.3 \n",
- "pf1 = 0.8 #lagging\n",
- "n1 = 0.85\n",
- "\n",
- "#for synchronous motor:\n",
- "kW2 = 18.65\n",
- "pf = 0.9 #leading\n",
- "n2 = 0.9\n",
- "\n",
- "#for lighting load:\n",
- "kW3 = 10\n",
- "pf3 = 1\n",
- "c1 = 60 #Rs/kVA of max demand\n",
- "c2 = 5 #paie per kWh\n",
- "t = 2000 #working hours\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "P1 = kW1/n1 #power input to induction motor(kW)\n",
- "Q1 = P1*math.tan(math.acos(pf1)) #Lagging kVAR taken by induction motor\n",
- "\n",
- "P2 = kW2/n2 #Input power to synchronous motor(kW)\n",
- "Q2 = P2**math.tan(math.acos(pf1)) #Leading kVAR taken by synchronous motor\n",
- "\n",
- "#Since lighting load works at unity p.f., its lagging kVAR = 0.\n",
- "Q3 = 0 \n",
- "\n",
- "kVAR = Q1-Q2 #net kVAR\n",
- "kW = round(P1+P2+kW3,1) #total active power\n",
- "kVA = round(math.sqrt(kW**2+kVAR**2))\n",
- "\n",
- "T = c1*kVA+c2*kW*t/100\n",
- "\n",
- "#Result:\n",
- "print \"The annual electical charges is Rs\",round(T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual electical charges is Rs 12140.0\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.11, Page Number: 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "#(i)for a lighting load:\n",
- "P1 = 500 #kW\n",
- "\n",
- "\n",
- "#(ii) for a load:\n",
- "P2 = 400 #kW\n",
- "pf2 = 0.707 #power factor lagging\n",
- "\n",
- "\n",
- "#(iii)for a load:\n",
- "P3 = 800 #kW\n",
- "pf3 = 0.8 #power factor leading\n",
- "\n",
- "#(iv)for a load:\n",
- "P4 = 500 #kW\n",
- "pf4 = 0.6 #power factor lagging\n",
- "\n",
- "\n",
- "#(v)a synchronous motor driving a d.c. generator:\n",
- "P5 = 540 #power rated at generator(kW)\n",
- "n = 0.9 #overall efficiency\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Total lagging kVAR taken by loads (ii) and (iv)\n",
- "kVAR24 = P2*math.tan(math.acos(pf2))+P4*math.tan(math.acos(pf4))\n",
- "\n",
- "#Leading kVAR taken by the load (iii)\n",
- "kVAR3 = P3*math.tan(math.acos(pf3))\n",
- "\n",
- "#Leading kVAR to be taken by synchronous motor\n",
- "kVAR5 = kVAR24-kVAR3\n",
- "Pi = P5/n #motor input(kW)\n",
- "phy = math.atan(kVAR5/Pi) #phase angle of synchronous motor\n",
- "pf = math.cos(phy)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The power factor of synchronous motor is\",round(pf,2,),\"leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power factor of synchronous motor is 0.79 leading\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.12, Page Number: 115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "P1 = 100 #power of synchronous motor(hp)\n",
- "P2 = 200 #power aggregated by induction motor(hp)\n",
- "pf2 = 0.707 #lagging power factor for induction motor\n",
- "n2 = 0.82 #efficiency of induction motor\n",
- "P3 = 30 #lighting load(kW)\n",
- "c1 = 100 #1st part tariff(Rs per kVA per annum)\n",
- "c2 = 0.06 #2nd part tariff Rs per kWh\n",
- "pfa = 0.8 #power factor factor initially(lagging)\n",
- "na = 0.93 #efficiency of synchronous motor initially\n",
- "pfb = 0.8 #power factor factor changed later(leading)\n",
- "nb = 0.93 #efficiency of synchronous motor changed later\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#(a) When synchronous motor runs at p.f. 0\u00b78 lagging.\n",
- "kW1 = round(P1*735.5/(na*1000)) #input to synchronous motor(kW)\n",
- "kVAR1 = round(kW1*math.tan(math.acos(pfa)),2) #Lagging kVAR taken by the synchronous motor(kVAR)\n",
- "kW2 = round(P2*735.5/(n2*1000),1) #input to induction motor(kW)\n",
- "kVAR2 = kW2*math.tan(math.acos(pf2)) #Lagging kVAR taken by the induction motor(kVAR) \n",
- "#Since lighting load works at unity p.f., its lagging kVAR is zero.\n",
- "kVARt = kVAR1+kVAR2 #Total lagging kVAR\n",
- "kWt = kW1+kW2+P3 #Total active power(kW)\n",
- "kVAt = round(math.sqrt(kWt**2+kVARt**2),1) #total kVA\n",
- "C1 = kVAt*c1 #annual kVA demand charges\n",
- "E1 = kWt*8760 #Energy consumed/year(kWh)\n",
- "C2 = round(0.06*E1) #annual energy charges(Rs)\n",
- "T1 = C1+C2 #total Annual bill(Rs)\n",
- "\n",
- "\n",
- "#(b) When synchronous motor runs at p.f. 0\u00b78 leading.\n",
- "kVARn = kVAR2-kVAR1 #Net lagging kVAR\n",
- "kWtt = kWt #total active power(kW)\n",
- "kVAtt = math.sqrt(kVARn**2+kWtt**2) #total kVA\n",
- "C11 = c1*kVAtt #annual kVA charges(Rs)\n",
- "C22 = C2 #Annual energy charges(Rs)\n",
- "T2 = C11+C22 #total annual bill(Rs)\n",
- "Ts = T1-T2 #annual saving(Rs)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The annual saving is Rs\",round(Ts/100)*100"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual saving is Rs 6200.0\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.13, Page Number: 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "pf1 = 0.75 #Power factor of the factory(lagging)\n",
- "x = 72 #Max. demand charges(Rs per kVA per annum)\n",
- "d = 10 #interest & depreciation of capital investment(%)\n",
- "M = 175 #max. demand(kW)\n",
- "c2 = 120 #cost of phase advancing equipment(Rs/kVAR)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "y = c2*d/100 #Expenditure on phase advancing equipment(Rs/kVAR/annum)\n",
- "pf2 = math.sqrt(1-(y/x)**2) #power factor\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Most economical p.f. at which factory should operate is\",round(pf2,3),\"lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Most economical p.f. at which factory should operate is 0.986 lagging\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.14, Page Number: 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "L = 400 #avg. demand(kW)\n",
- "pf1 = 0.8 #original power factor(lagging)\n",
- "LF = 0.5 #annual load factor\n",
- "c1 = 50 #first part tariff(Rs/kVA per annum)\n",
- "c2 = 0.05 #second part tariff(Rs/kWh)\n",
- "pf2 = 0.95 #improved power factor\n",
- "c3 = 100 #phase advancement equipment cost(Rs/kVAR)\n",
- "d = 10 #annual interest and depreciation(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "M = L/LF #maximum demand(kW)\n",
- "#Leading kVAR taken by phase advancement equipment:\n",
- "kVAR1 = M*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))\n",
- "y = d*c3/100 #Expenditure on phase advancing equipment(Rs/kVAR/annum)\n",
- "kVA1 = M/pf1 #Max. kVA demand at pf1 power fctor\n",
- "kVA2 = round(M/pf2) #Max. kVA demand at pf2 power fctor\n",
- "Cs = c1*(kVA1-kVA2) #Annual saving in maximum demand charges\n",
- "Ce = y*kVAR1 #Annual expenditure on phase advancing equipment\n",
- "NS = Cs-Ce #net annual saving\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The capacity of the phase advancing equipment is\",round(kVAR1),\"kVAR\"\n",
- "print \"(ii)The annual saving is Rs\",round(NS)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The capacity of the phase advancing equipment is 337.0 kVAR\n",
- "(ii)The annual saving is Rs 4529.0\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.15, Page Number: 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "L = 50 #avg. demand(kW)\n",
- "pf1 = 0.75 #original power factor(lagging)\n",
- "LF = 0.5 #annual load factor\n",
- "c1 = 100 #first part tariff(Rs/kVA of max demand per annum)\n",
- "c2 = 0.05 #second part tariff(Rs/kWh)\n",
- "c3 = 600 #cost of loss free capacitors(Rs/kVAR)\n",
- "d = 10 #depreciation & interest(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "y = c3*d/100 #Expenditure on capacitors(Rs/kVAR/annum)\n",
- "pfe = math.sqrt(1-(y/c1)**2) #Most economical power factor\n",
- "M = L/LF #Max kW demand\n",
- "kVAm1 = M/pf1 #The maximum kVA demand at pf1 power factor\n",
- "kVAm2 = M/pfe #The maximum kVA demand at pfe power factor\n",
- "Cs = c1*(kVAm1-kVAm2) #annual saving(Rs)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Most economical power factor is\",round(pfe,3),\"lagging\"\n",
- "print \"Annual saving is Rs\",round(Cs)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Most economical power factor is 0.8 lagging\n",
- "Annual saving is Rs 833.0\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.16, Page Number: 120"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "L = 200 #avg. demand(kW)\n",
- "pf1 = 0.8 #original power factor(lagging)\n",
- "LF = 0.5 #annual load factor\n",
- "c1 = 100 #first part tariff(Rs/kVA of max demand per annum)\n",
- "c2 = 0.05 #second part tariff(Rs/kWh)\n",
- "c3 = 500 #cost of phase advancing plant(Rs/kVAR)\n",
- "d = 10 #depreciation & interest(%)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "y = c3*d/100 #Expenditure on capacitors(Rs/kVAR/annum)\n",
- "pfe = math.sqrt(1-(y/c1)**2) #Most economical power factor\n",
- "Pc = L*(math.tan(math.acos(pf1))-math.tan(math.acos(pfe))) #Capacity of phase advancing plant\n",
- "E = L*5000 #units consumed per year(kWh)\n",
- "C2 = c2*E #annual energy charges(Rs)\n",
- "C1 = y*Pc #Annual cost of phase advancing plant(Rs)\n",
- "Cm = c1*L/pfe #Max. demand charge(Rs)\n",
- "C = C2+C1+Cm #annual charge(Rs)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The value to which the power factor be improved is\",round(pfe,3),\"lagging\"\n",
- "print \"(ii) The capacity of the phase advancing plant is\",round(Pc,2),\"kVAR\"\n",
- "print \"(iii)The new bill for energy is Rs\",round(C,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The value to which the power factor be improved is 0.866 lagging\n",
- "(ii) The capacity of the phase advancing plant is 34.53 kVAR\n",
- "(iii)The new bill for energy is Rs 74820.5\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.17, Page Number: 120"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "E = 80000 #units consumed per year\n",
- "kVAm1 = 500 #maximum kVA demand\n",
- "pf1 = 0.707 #initial power factor(lagging)\n",
- "c1 = 120 #first part tariff(Rs/kVA of max demand per annum)\n",
- "c2 = 0.025 #second part tariff(Rs/kWh)\n",
- "c3 = 50 #cost of phase advancing plant(Rs/kVAR)\n",
- "pf2 = 0.9 #increased powe factor(Rs)\n",
- "d = 10 #depreciation on advancing plant(%)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "C1 = c1*kVAm1+c2*E #annual cost of supply(Rs)\n",
- "P = kVAm1*pf1 #max. kW demand\n",
- "#Leading kVAR taken by phase advancing equipment:\n",
- "kVARe = P*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))\n",
- "C2 = round(kVARe*c3*d/100) #Annual cost of phase advancing equipment(Rs)\n",
- "#When p.f. is raised from 0\u00b7707 lag to 0\u00b79 lag\n",
- "kVAm2 = P/pf2 #new maximum kVA demand\n",
- "kVAr = kVAm1-kVAm2 #Reduction in kVA demand\n",
- "Cs = round(c1*kVAr) #annual saving(Rs)\n",
- "#As the units consumed remain the same, therefore, saving will\n",
- "#be equal to saving in M.D. charges minus annual cost of \n",
- "#phase advancing plant.\n",
- "Cs2 = Cs-C2 #annual charges(Rs)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The annual cost of supply is Rs\",C1\n",
- "print \"Annual saving is Rs\",Cs2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual cost of supply is Rs 62000.0\n",
- "Annual saving is Rs 11955.0\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.18, Page Number: 123"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "pf = 0.7 #power factor when plant working at its max. capacity(lagging)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#For calculating the cost of increasing plant capacity. \n",
- "#We have, referring to Fig. 6.15, the increase in kVA capacity is BD.\n",
- "\n",
- "#Now OE*cos phy2 = OD*cos phy1\n",
- "#or OB*cos phy2 = OD*cos phy1 (Since OE = OB)\n",
- "#or OD = OB * cos phy2/cos phy1 \n",
- "# = OB * 0\u00b785/0\u00b77\n",
- "# = 1\u00b72143 OB\n",
- "\n",
- "#Increase in the kVA capacity of the plant is\n",
- "#BD = OD \u2212 OB = 1\u00b72143 \u00d7 OB \u2212 OB = 0\u00b72143 OB\n",
- "#Total cost of increasing the plant capacity\n",
- "# = Rs 800 \u00d7 0\u00b72143 \u00d7 OB\n",
- "# = Rs 171\u00b744 \u00d7 OB ...(i)\n",
- "\n",
- "cos_phy1 = 0.7; sin_phy1 = 0.714\n",
- "cos_phy1 = 0.85; sin_phy1 = 0.527\n",
- "\n",
- "#Leading kVAR taken by p.f. correction equipment is\n",
- "# ED = CD \u2212 CE = OD*sin phy1 \u2212 OE*sin phy2\n",
- "# = 1\u00b72143 * OB*sin phy1 \u2212 OB*sin phy2\n",
- "# = OB*(1\u00b72143 * 0\u00b7714 \u2212 0\u00b7527) = 0\u00b734 * OB\n",
- "#Suppose the cost per kVAR of the equipment be Rs y.\n",
- "#So, total cost of p.f. correction equipment\n",
- "# = Rs 0\u00b734 \u00d7 OB \u00d7 y ...(ii)\n",
- "#The cost per kVAR of the equipment that would justify its \n",
- "#installation is when exp. (i) = exp. (ii)\n",
- "#i.e.,\n",
- "# 171\u00b744 \u00d7 OB = 0\u00b734 \u00d7 OB \u00d7 y\n",
- "y = 171.44/0.34 #Rs/kVAR\n",
- "\n",
- "#If the losses in p.f. correction equipment are neglected, \n",
- "#then its kVAR = kVA.\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The maximum cost per kVA of p.f. correction equipment\"\n",
- "print \"that can be paid is Rs\",round(y,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum cost per kVA of p.f. correction equipment\n",
- "that can be paid is Rs 504.2\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.19, Page Number: 124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "pf = 0.7 #power factor when plant working at its max. capacity(lagging)\n",
- "\n",
- "#Calculation:\n",
- "#Cost of increasing plant capacity\n",
- "# BD = OD \u2212 OB\n",
- "# = OB * 0.866/0.70-OB\n",
- "# = OB*(1\u00b7237 \u2212 1)\n",
- "# = 0\u00b7237 * OB\n",
- "#\u2234 Annual cost of increasing the plant capacity\n",
- "# = Rs 10 * 0\u00b7237 * OB\n",
- "# = Rs. 2.37 * OB ...(i)\n",
- "\n",
- "#Cost of phase advancing equipment. Leading kVAR taken by\n",
- "#phase advancing equipment,\n",
- "# ED = CD \u2212 CE\n",
- "# = OD*sin phy1 \u2212 OE*sin phy2\n",
- "# = 1.237 * OB * sin phy1 \u2212 OB*sin phy2\n",
- "# = OB*(1\u00b7237 * 0\u00b7174 \u2212 0\u00b75) = 0\u00b7383 \u00d7 OB\n",
- "#Let the cost per kVAR of the equipment be Rs y.\n",
- "#Annual cost of phase advancing equipment\n",
- "# = Rs 0.1 * y * 0\u00b7383 * OB ...(ii)\n",
- "#For economical use, the two costs should be equal i.e.,\n",
- "# exp. (i) = exp. (ii).\n",
- "#or 0\u00b71 * y * 0\u00b7383 * OB = 2\u00b737 * OB\n",
- "y = 2.37/(0.1*0.383) #Rs\n",
- "\n",
- "\n",
- "#If the losses in the phase advancing equipment are neglected,\n",
- "#then its kVAR = kVA.\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The maximum cost per kVA of phase advancing equipment\"\n",
- "print \"that can be paid is Rs\",round(y,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum cost per kVA of phase advancing equipment\n",
- "that can be paid is Rs 61.88\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |