{ "metadata": { "name": "", "signature": "sha256:a62a0cfa22010723ae84accbfbc8f3f3c553c1279f7bb65e42ceaaad44de0807" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 2:Generating Stations" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.1, Page Number: 16" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "n=20 #overall efficiency of plant\n", "h=860 #kcal\n", "m=0.6 #Mass of fuel burnt(kg) per KW of electrical energy generated\n", "\n", "#Calculations:\n", "x=h*100/(m*n) #Calorific value of fuel(kcal/kg)\n", "\n", "#Results:\n", "print \"Calorific value of fuel =\",round(x,2),\"kcal/kg\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Calorific value of fuel = 7166.67 kcal/kg\n" ] } ], "prompt_number": 31 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.2, Page Number: 17" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "M = 20000 #maximum demand(kW)\n", "n_b= 85 #boiler efficiency(%)\n", "m=0.9 #coal consumption(kg/kWh)\n", "LF=40 #load factor(%)\n", "n_t=90 #turbine efficiency(%)\n", "c=300 #cost of 1 tonne of coal(Rs)\n", "\n", "#Calculations:\n", "n_th = n_b * n_t/100 #in %\n", "cb = LF*M*m*c*24*365/(1000*100)\n", "\n", "#Results:\n", "print \"Thermal efficiency = \",n_th,\"%\"\n", "print \"Coal bill per annum = Rs\",cb" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Thermal efficiency = 76.5 %\n", "Coal bill per annum = Rs 18921600.0\n" ] } ], "prompt_number": 32 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.3, Page Number: 17" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "ct=3000000 #annul cost of coal(Rs)\n", "cv=5000 #Calorific value of coal(kcal/kg)\n", "c=300 #cost of coal per tonne(Rs)\n", "n_th=33 #thermal efficiency(%)\n", "n_elec=90 #electrical efficiency(%)\n", "\n", "#Calculations:\n", "n_t=n_th*n_elec/100 #overall efficiency(%)\n", "h=ct*cv*1000/c #heat of combustion(kcal)\n", "ho=n_t*h/(100*860) #heat output(kWh)\n", "L=ho/8760 #kW\n", "\n", "\n", "#Results:\n", "print \"Avg load on the station=\",round(L),\"kW\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Avg load on the station= 1971.0 kW\n" ] } ], "prompt_number": 33 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.4, Page Number: 17" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from sympy import *\n", "\n", "\n", "#Variable declaration:\n", "kWh= symbols('kWh')\n", "W = 13500 + 7.5 * kWh #Water evaporated in kg\n", "C = 5000 + 2.9 * kWh #coal cumsumption in kg\n", "\n", "\n", "#Calculations:\n", "#part (i):\n", "#As the station output (i.e., kWh) increases towards infinity,\n", "#the limiting value of W/C approaches\n", "L1= 7.5/2.9 #in kg\n", "\n", "#part (ii):\n", "#at no load\n", "kWh=0\n", "c=(5000+2.9*kWh)/8 #coal per hour in kg\n", "#Results:\n", "print \"Limiting value of water/kg of coal=\",round(L1,1),\"kg\"\n", "print \"Required Coal per hour\",c,\"kg\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Limiting value of water/kg of coal= 2.6 kg\n", "Required Coal per hour 625.0 kg\n" ] } ], "prompt_number": 34 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.5, Page Number: 18" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "C=100 #Capacity of station in MW\n", "cv=6400 #kcal/kg\n", "n_th=0.3 #thermal efficiency\n", "n_elec=0.92 #electrical efficiency\n", "\n", "\n", "#Calculations:\n", "n_t=n_th*n_elec #overall efficiency\n", "U=C*1*10**3 #units generated/hr in kWh\n", "H=U*860/n_t #total heat of combustion(kcal)\n", "w=H/cv #Coal consumption in kg\n", "\n", "#Results:\n", "print \"The coal consumption per hour =\",round(w),\"kg\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The coal consumption per hour = 48687.0 kg\n" ] } ], "prompt_number": 35 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.6, Page Number: 23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "C=5*10**6 #reservoir capacity in m^3\n", "H=200 #water head in m\n", "n_t=0.75 #overall efficiency\n", "d=1000 #density of water in kg/m^3\n", "\n", "#Calculations:\n", "W=C*d*9.81 #weight of water in Newton\n", "E=W*H*n_t/(3600*1000) #electrical energy available(kWh)\n", "\n", "\n", "#Results:\n", "print \"The total energy available=\",round(E/10**6,3),\"* 10^6 kWh\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The total energy available= 2.044 * 10^6 kWh\n" ] } ], "prompt_number": 36 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.7, Page Number: 23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "V=94 #volume of water in m^3/sec\n", "d=1000 #density of water in kg/m^3\n", "H=39 #head of water in m\n", "nt=0.80 #overall efficiency\n", "\n", "\n", "#Calculations:\n", "W=V*d #weight of water in kg/sec\n", "w=W*H*9.81/1000 #work done per sec in kW\n", "FC=nt*w #firm capacity in kW\n", "YGO=FC*8760 #yearly gross capacity in kWh\n", "\n", "\n", "#Results:\n", "print \"Firm capacity=\",FC,\"kW\"\n", "print \"Yearly gross output\",round(YGO/10**6),\"* 10^6 kWh\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Firm capacity= 28770.768 kW\n", "Yearly gross output 252.0 * 10^6 kWh\n" ] } ], "prompt_number": 37 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.8, Page Number: 23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable Declaration:\n", "H=100 #Water head in m\n", "Q=1 #discharge, m^3/sec\n", "nh=0.86 #hydraulc efficiency\n", "nelec=0.92 #electrical efficiency\n", "d=1000 #density of water, kg/m^3\n", "\n", "#Calculations:\n", "W=Q*d*9.81 #weight of water in N\n", "Po=W*H*nh*nelec/1000 #power produced, kW\n", "E=Po*1 #in kWh\n", "\n", "#Results:\n", "print \"Electrical energy generated per hr=\",round(E),\"kWh\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Electrical energy generated per hr= 776.0 kWh\n" ] } ], "prompt_number": 38 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.9, Page Number: 23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "CA=5*10**9 #Catchment area in m^2\n", "H=30 #head in m\n", "F=1.25 #Annual rainfall in m\n", "K=0.80 #yeild factor\n", "n=0.70 #overall efficiency\n", "LF=0.40 #Load factor\n", "d=1000 #density of water(kg/m^3)\n", "\n", "#Calculations:\n", "V=CA*F*K #volume of water utilised per annum(m^3)\n", "W=V*d*9.81 #Weight of water available per annum (N)\n", "E=round(W*H*n/(10**11*3600),2)*10**8 #Electrical energy available per annum(kWh)\n", "Pav=E/8760 #average power(kW)\n", "Dmax=Pav/LF #Maximum demand\n", "\n", "#Results:\n", "print \"Average power generated is \",round(Pav),\"kW\"\n", "print \"Rating of generators is\",round(Dmax),\"kW\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Average power generated is 32648.0 kW\n", "Rating of generators is 81621.0 kW\n" ] } ], "prompt_number": 39 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.10, Page Number: 24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable Declaration:\n", "A=2.4 #Area of reservoir(km^2)\n", "V=5*10**6 #Capacity of reservoir(m^3)\n", "H=100 #in m\n", "np=0.95 #penstock efficiency\n", "nt=0.90 #turbine efficiency\n", "ng=0.85 #generation efficiency\n", "L=15000 #load supplied in kW\n", "\n", "#Calculations:\n", "W=V*1000*9.81 #in Newton\n", "n=int(np*nt*ng*1000)/1000 #overall efficiency\n", "E=W*H*n/(1000*3600) #Elecctrical energy generated(kWh)\n", "x=L*3*3600/(A*10**6*9.81*H*n)\n", "\n", "#Results:\n", "print \"Total electrical energy generated is \",round(E),\"kWh\"\n", "print \"Fall in reservoir level is\",round(x*100,3),\"cm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Total electrical energy generated is 989175.0 kWh\n", "Fall in reservoir level is 9.478 cm\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.11, Page Number: 25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration\n", "H=25 #head of reservoir in m\n", "Pr=400 #power required by factory(kW)\n", "n=0.80 #overall efficiency of plant\n", "\n", "#Calculations:\n", "#part (i):\n", "#(a):\n", "d1 = 10 #discharge in m^3/sec\n", "w1 = d1*1000*9.81 #weight of water in N\n", "P1 = w1*H*n/1000 #power developed(kW)\n", "\n", "#(b)\n", "d2 = 6 #in m^3/sec\n", "P2 = P1*d2/d1 #kW\n", "\n", "#(c)\n", "d3 = 1.5 #in m^3/sec\n", "P3 = P1*d3/d1 #kW\n", "\n", "Ps = Pr-P3 #standby power(kW)\n", "\n", "#part(ii):\n", "Dav = (d1*4+d2*2+d3*6)/12 #avg discharge(m^3/sec)\n", "P = P1*Dav/d1 #power developed(kW)\n", "Pex = P-Pr #Excess power available(kW)\n", "\n", "\n", "#Results:\n", "print \"(i) Standby power is \",round(Ps),\"kW\"\n", "print \"(ii) Excess power available is \",round(Pex,1),\"kW\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Standby power is 106.0 kW\n", "(ii) Excess power available is 597.4 kW\n" ] } ], "prompt_number": 41 }, { "cell_type": "heading", "level": 3, "metadata": { "slideshow": { "slide_type": "-" } }, "source": [ "Example 2.12, Page Number: 25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "#for part (i):\n", "C = 10 #Installed capacity(MW)\n", "H = 20 #head of reservoir(m)\n", "n = 0.80 #overall efficiency\n", "LF = 0.40 #load factor\n", "#for part (ii):\n", "Q2 = 20 #discharge\n", "\n", "\n", "#Calculations:\n", "#for part(i):\n", "U = C*LF*24*7*10**3 #units generated per week(kWh)\n", "Q = U/(H*n*9.81*24*7) #Discharge in m^3/sec\n", "\n", "#for part(ii):\n", "U2 = Q2*9.81*1000*n*H*24/1000 #units generated per day(kWh)\n", "LF2 = U2/(C*10**3*24)\n", "\n", "#Results:\n", "print \"(i) The river discharge is \",round(Q,2),\"m^3/sec\"\n", "print \"(ii) The load factor is \", round(LF2*100,1),\"%\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) The river discharge is 25.48 m^3/sec\n", "(ii) The load factor is 31.4 %\n" ] } ], "prompt_number": 42 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.13, Page Number: 26" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration:\n", "H = 15 #head of reservoir(m)\n", "n = 0.85 #efficiency\n", "L = 0.40 #load factor\n", "\n", "\n", "#Calculations:\n", "Qavg = (500+520+850+800+875+900+546)/7 #in m^3/sec\n", "\n", "#It is clear from graph that on three dyas \n", "#(viz., Sun, Mon. and Sat.), the discharge is less than\n", "#the average discharge.\n", "\n", "V1 = (500+520+546)*24*3600 #Actual volume available in these 3 days(m^3/s)\n", "V2 = 3*Qavg*24*3600 #Vol. of water required in these 3 days(m^3/s)\n", "Pr = V2-V1 #Pondage required(m^3/sec)\n", "Po = Qavg*9.81*1000*H*n #Avg output produced(W)\n", "C = Po/L #Capacity of plant(W)\n", "\n", "#Results:\n", "print \"(i) The average daily discharge is \",Qavg,\"m^3/sec\"\n", "print \"(ii) Pondage required is (\",round(Pr/10**5),\"* 10^5) m^3\"\n", "print \"(iii)Installed capacity of plant is \",round(C/10**6),\"MW\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) The average daily discharge is 713.0 m^3/sec\n", "(ii) Pondage required is ( 495.0 * 10^5) m^3\n", "(iii)Installed capacity of plant is 223.0 MW\n" ] } ], "prompt_number": 43 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.14, Page Number: 29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "w = 0.28 #fuel consumption in kg/kWh\n", "C = 10000 #calorific value of fuel(kcal/kWh)\n", "na = 0.95 #efficiency of alternator\n", "\n", "\n", "#Calculations:\n", "H = w*C/860 #heat produced by 0.28 kg/kWh of fuel\n", "\n", "no = 1/H #Overall efficiency\n", "ne = no/na #Efficiency of engine\n", "\n", "#Results:\n", "print \"(i) The overall efficiency is \",round(no*100,1),\"%\"\n", "print \"(ii)The efficiency of the engine\",round(ne*100,1),\"%\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) The overall efficiency is 30.7 %\n", "(ii)The efficiency of the engine 32.3 %\n" ] } ], "prompt_number": 44 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.15, Page Number: 30" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "\n", "#Variable declaration:\n", "w = 1000 #fuel consumption in kg/day\n", "E = 4000 #Units generated in kWh/day\n", "C = 10000 #calorific value in kcal/kg\n", "na = 0.96 #Alternator efficiency\n", "nem = 0.95 #engine mechanical efficiency\n", "\n", "\n", "#Calculations:\n", "s = w/E #specific fuel consumption(kg/kWh)\n", "E2 = w*C #energy input per day(kcal/day)\n", "no = E*860/E2 #overall efficiency\n", "ne = no/na #engine efficiency\n", "net = ne/nem #engine thermal efficiency\n", "\n", "#Results:\n", "print \"Specific fuel consumption is \",s,\"kg/kWh\"\n", "print \"Overall efficiency is \",round(no*100,1),\"%\"\n", "print \"Thermal efficiency of the engine is \",round(net*100,2),\"%\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Specific fuel consumption is 0.25 kg/kWh\n", "Overall efficiency is 34.4 %\n", "Thermal efficiency of the engine is 37.72 %\n" ] } ], "prompt_number": 45 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.16, Page Number: 30" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "C1 = 700 #capacity of plant 1(kW)\n", "C2 = 2*500 #capacity of plant 2(kW)\n", "pcf = 0.40 #plant capacity factor\n", "w = 0.28 #fuel cunsumption in kg/kWh\n", "H = 10200 #specific heat of fuel in kcal/kg\n", "\n", "#Calculatios:\n", "M = (C1+C2)*30*24 #max energy can be produced in 30 days(kWh)\n", "E = pcf*M #Actual energy produced in 30 days(kWh)\n", "W = E*w #actual fuel consumption in kg\n", "\n", "Po = E*860 #output energy in kWh\n", "Pin = W*H #Input energy in kWh\n", "n = Po/Pin #Overall efficiency\n", "\n", "#Results:\n", "print \"The fuel oil required is \",W,\"kg\"\n", "print \"Ovreall efficiency is\",round(n*100),\"%\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The fuel oil required is 137088.0 kg\n", "Ovreall efficiency is 30.0 %\n" ] } ], "prompt_number": 46 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.17, Page Number: 34" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration\n", "M = 300 #Energy received from reactor(MW)\n", "E1 = 200 #Energy released fron each atom(MeV)\n", "\n", "\n", "#Calculations:\n", "E2 = M*10**6*3600 #Energy released per hour(J)\n", "E3 = E1*1.6*10**-19*10**6 #Energy released per fission(J)\n", "N = E2/E3 #No of atoms fissioned\n", "m = N*235/(6.022*10**23) #mass of uranium fissioned per hr(g)\n", "\n", "#Results:\n", "print \"Mass of Uranium fissioned per hour is\",round(m,2),\"g\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass of Uranium fissioned per hour is 13.17 g\n" ] } ], "prompt_number": 47 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 2.18, Page Number: 35" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "t = 30 #days\n", "w = 2 #weight of uranium(kg)\n", "Eo = 200 #energy released per fission(MeV)\n", "\n", "\n", "#Calculations:\n", "N = 2*1000*6.022*10**23/235 #No of atoms fissioned in 2kg of fuel\n", "Po = N*Eo*(1.6*10**-19)*10**6/(24*60*60*30) #Watt\n", "\n", "\n", "#Results:\n", "print \"Power output is\",round(Po*10**-6,1),\"MW\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Power output is 63.3 MW\n" ] } ], "prompt_number": 48 } ], "metadata": {} } ] }