path: root/Principles_of_Power_System/chapter2.ipynb

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 "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": {}
  }
 ]
}