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diff --git a/Principles_of_Power_System/chapter2_1.ipynb b/Principles_of_Power_System/chapter2_1.ipynb deleted file mode 100644 index daf8974b..00000000 --- a/Principles_of_Power_System/chapter2_1.ipynb +++ /dev/null @@ -1,857 +0,0 @@ -{
- "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 |