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| author | nice | 2014-09-16 17:48:17 +0530 |
|---|---|---|
| committer | nice | 2014-09-16 17:48:17 +0530 |
| commit | b8bb8bbfa81499ad7fc3f3508be257da65f543af (patch) | |
| tree | 204976d3209b79a52e8518c65fa27a4ca48f8489 /Principles_of_Power_System/chapter2.ipynb | |
| parent | 2792e8d6ecab454e3cb8fb1ea1f26f1613bc1e1c (diff) | |
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diff --git a/Principles_of_Power_System/chapter2.ipynb b/Principles_of_Power_System/chapter2.ipynb new file mode 100755 index 00000000..daf8974b --- /dev/null +++ b/Principles_of_Power_System/chapter2.ipynb @@ -0,0 +1,857 @@ +{
+ "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": {}
+ }
+ ]
+}
\ No newline at end of file |