removing problem statements

1 files changed, 0 insertions, 129 deletions

diff --git a/Introduction_To_Chemical_Engineering/ch3.ipynb b/Introduction_To_Chemical_Engineering/ch3.ipynb
index e6461254..0b6043d2 100644
--- a/Introduction_To_Chemical_Engineering/ch3.ipynb
+++ b/Introduction_To_Chemical_Engineering/ch3.ipynb
@@ -27,15 +27,12 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the coal consumption\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "w_C = 0.6;   #amount of carbon in coal\n",
       "N2_content = 40.   #in m3 per 100m3 air\n",
       "\n",
-      "# Calculations\n",
       "air_consumed = N2_content/0.79;\n",
       "weight_air = air_consumed*(28.8/22.4);\n",
       "O2_content = air_consumed*32*(0.21/22.4);    #in kg\n",
@@ -50,7 +47,6 @@
       "total_consumption = C_consumption1+C_consumption2;\n",
       "coal_consumption = total_consumption/w_C;\n",
       "\n",
-      "# Results\n",
       "print \"coal consumption = %f kg\"%(coal_consumption)\n"
      ],
      "language": "python",
@@ -78,16 +74,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find amount of ammonia and air consumed\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "NH3_required = (17./63)*1000;    #NH3 required for 1 ton of nitric acid\n",
       "NO_consumption = 0.96;\n",
       "HNO3_consumption = 0.92;\n",
       "\n",
-      "# Calculations and Results\n",
       "NH3_consumed = NH3_required/(NO_consumption*HNO3_consumption);\n",
       "volume_NH3 = NH3_consumed*(22.4/17);\n",
       "print \"volume of ammonia consumed= %f cubic metre/h\"%(volume_NH3)\n",
@@ -122,17 +115,14 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the consumption of NaCl and H2SO4 in HCl consumption\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "HCl_production = 500.   #required to be produced in kg\n",
       "NaCl_required = (117./73)*HCl_production;\n",
       "yield_ = 0.92;\n",
       "purity_NaCl= 0.96;\n",
       "\n",
-      "# Calculations and Results\n",
       "actual_NaCl = NaCl_required/(purity_NaCl*yield_);\n",
       "print \"amount of NaCl required = %f kg\"%(actual_NaCl)\n",
       "\n",
@@ -170,20 +160,15 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the period of service\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "C2H2_produced = (1./64)*0.86;   #in kmol\n",
       "volume_C2H2 = C2H2_produced*22.4*1000;    #in l\n",
       "\n",
-      "# Calculations\n",
-      "#assuming ideal behaviour,\n",
       "volume = (100/101.3)*(273./(273+30));\n",
       "time = (volume_C2H2/volume)*(1./60);\n",
       "\n",
-      "# Results\n",
       "print \"time of service = %f hr\"%(time)\n",
       "\n"
      ],
@@ -212,17 +197,14 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the screen effectiveness\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "xv = 0.88;\n",
       "xf = 0.46;\n",
       "xl = 0.32;\n",
       "F= 100.   #in kg\n",
       "\n",
-      "# Calculations and Results\n",
       "L = (F*(xf-xv))/(xl-xv);\n",
       "V = F-L;\n",
       "print \"L = %f Kg V = %f Kg\"%(L,V)\n",
@@ -264,16 +246,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the flow rate and concentration\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "G1 = 3600.   #in m3/h\n",
       "P = 106.6    #in kPa\n",
       "T = 40   #in degree C\n",
       "\n",
-      "# Calculations and Results\n",
       "q = G1*(P/101.3)*(273./((273+T)));   #in m3/s\n",
       "m = q/22.4;   #in kmol/h\n",
       "y1 = 0.02;\n",
@@ -284,16 +263,12 @@
       "Gs = m*(1-y1);\n",
       "print \"moles of benzene free gas = %f kmol drygas/h\"%(Gs)\n",
       "\n",
-      "#for 95% removal\n",
       "Y2 = Y1*(1-0.95);\n",
       "print \"final mole ratio of benzene = %f kmol benzene/kmol dry gas\"%(Y2)\n",
       "\n",
       "x2 = 0.002\n",
       "X2 = 0.002/(1-0.002);\n",
       "\n",
-      "#at equilibrium y* = 0.2406X\n",
-      "#part 1\n",
-      "#for oil rate to be minimum the wash oil leaving the absorber must be in equilibrium with the entering gas\n",
       "\n",
       "y1 = 0.02;\n",
       "x1 = y1/(0.2406);\n",
@@ -301,7 +276,6 @@
       "min_Ls = Gs*((Y1-Y2)/(X1-X2));\n",
       "print \"minimum Ls required = %f kg/h\"%(min_Ls*260)\n",
       "\n",
-      "#for 1.5 times of the minimum\n",
       "Ls = 1.5*min_Ls;\n",
       "print \"flow rate of wash oil = %f kg/h\"%(Ls*260)\n",
       "X1 = X2 + (Gs*((Y1-Y2)/Ls));\n",
@@ -337,47 +311,38 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the extraction of nicotine\n",
       "\n",
       "from scipy.optimize import fsolve \n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "xf = 0.01\n",
       "Xf = xf/(1-xf);\n",
       "Feed = 100   #feed in kg\n",
       "\n",
-      "# Calculations and Results\n",
       "c_nicotine = Feed*Xf;   #nicotine conc in feed\n",
       "c_water = Feed*(1-Xf)    #water conc in feed\n",
       "\n",
-      "#part 1\n",
       "def  F1(x):\n",
       "    return  (x/150.)-0.9*((1-x)/99.);\n",
       "\n",
-      "#initial guess\n",
       "x = 10.;\n",
       "y = fsolve(F1,x)\n",
       "print \"amount of nicotine removed N = %f kg\"%(y)\n",
-      "#part 2\n",
       "def F2(x):\n",
       "    return  (x/50.)-0.9*((1-x)/99.);\n",
       "\n",
-      "#initial guess\n",
       "x = 10.;\n",
       "N1 = fsolve(F2,x)\n",
       "print \"amount of nicotine removed in stage 1, N1 = %f kg\"%(N1)\n",
       "def F3(x):\n",
       "    return  (x/50.)-0.9*((1-x-N1)/99.);\n",
       "\n",
-      "#initial guess\n",
       "x = 10.;\n",
       "N2 = fsolve(F3,x)\n",
       "print \"amount of nicotine removed in stage 2, N2 = %f kg\"%(N2)\n",
       "def F4(x):\n",
       "    return (x/50.)-0.9*((1-x-N2-N1)/99.);\n",
       "\n",
-      "#initial guess\n",
       "x = 10.;\n",
       "N3 = fsolve(F4,x)\n",
       "\n",
@@ -414,11 +379,9 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the amount of water in residue\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
       "vp_water = 31.06   #in kPa\n",
       "vp_benzene = 72.92   #in kPa\n",
       "\n",
@@ -426,13 +389,11 @@
       "x_benzene = vp_benzene/P;\n",
       "x_water = vp_water/P;\n",
       "\n",
-      "# Calculations\n",
       "initial_water = 50./18;   #in kmol of water\n",
       "initial_benzene = 50./78   #in kmol of benzene\n",
       "water_evaporated = initial_benzene*(x_water/x_benzene);\n",
       "water_left = (initial_water - water_evaporated);\n",
       "\n",
-      "# Results\n",
       "print \"amount of water left in residue = %f kg\"%(water_left*18)\n"
      ],
      "language": "python",
@@ -460,15 +421,12 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the vapor content of dimethylanaline\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "po_D = 4.93   #in kPa\n",
       "po_W = 96.3   #in kPa\n",
       "n = 0.75   #vaporization efficiency\n",
       "\n",
-      "# Calculations and Results\n",
       "P = n*po_D+po_W;\n",
       "print \"P = %f kPa\"%(P)\n",
       "\n",
@@ -477,7 +435,6 @@
       "wt_dimethylanaline = (x_dimethylanaline*121)/(x_dimethylanaline*121+x_water*18);\n",
       "print \"weight of dimethylanaline in water = %f\"%(wt_dimethylanaline*100)\n",
       "\n",
-      "#part 1\n",
       "n = 0.8;\n",
       "po_D = 32   #in kPa\n",
       "actual_vp = n*po_D;\n",
@@ -485,7 +442,6 @@
       "steam_required = (p_water*18)/(actual_vp*121);\n",
       "print \"amount of steam required = %f kg steam/kg dimethylanaline\"%(steam_required)\n",
       "\n",
-      "#part 2\n",
       "x_water = p_water/100.;\n",
       "wt_water = x_water*18./(x_water*18+(1-x_water)*121.);\n",
       "print \"weight of water vapor = %f weight of dimethylanaline =%f\"%(wt_water*100,100*(1-wt_water))\n"
@@ -518,22 +474,18 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the amount of water evaporated\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "xf = 0.15;\n",
       "xl = (114.7)/(114.7+1000);\n",
       "xc = 1;\n",
       "\n",
-      "# Calculations\n",
       "K2Cr2O7_feed = 1000*0.15;   #in kg\n",
       "\n",
       "n = 0.8;\n",
       "C = n*K2Cr2O7_feed;\n",
       "V = (K2Cr2O7_feed-120 - 880*0.103)/(-0.103);\n",
       "\n",
-      "# Results\n",
       "print \"amount of water evaporated = %f kg\"%(V)\n"
      ],
      "language": "python",
@@ -561,15 +513,12 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the yield of crystals\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "xc = round(106./286,3);\n",
       "xf = 0.25;\n",
       "xl = round(27.5/127.5,3);\n",
       "\n",
-      "# Calculations\n",
       "water_present = 100*(1-xf);    #in kg\n",
       "V = 0.15*75;   #in kg\n",
       "C = (100*xf - 88.7*xl)/(xc-xl);\n",
@@ -577,7 +526,6 @@
       "\n",
       "yield_ = (C/Na2CO3_feed)*100;\n",
       "\n",
-      "# Results\n",
       "print \"yield = %.1f %%\"%(yield_)\n"
      ],
      "language": "python",
@@ -605,24 +553,20 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the fraction of air recirculated\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "r = 50.   #weight of dry air passing through drier\n",
       "w1 = 1.60   #in kg per kg dry solid\n",
       "w2 = 0.1    #in kg/kg dry solid\n",
       "H0 = 0.016  #in kg water vapor/kg dry air\n",
       "H2 = 0.055  #in kg water vapor/kg dry air\n",
       "\n",
-      "# Calculations and Results\n",
       "y = 1 - (w1-w2)/(r*(H2-H0));\n",
       "print \"fraction of air recirculated = %f\"%(y)\n",
       "\n",
       "H1 = H2 - (w1-w2)/r;\n",
       "print \"humidity of air entering the drier = %f kg water vapor/kg kg dry air\"%(H1)\n",
       "\n",
-      "#check\n",
       "H11 = H2*y+H0*(1-y);\n",
       "if H1 == H11:\n",
       "    print \"fraction of air recirculated = %f  verified\"%(y)\n"
@@ -654,15 +598,12 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the volumetric flow rate and fraction of air passing through the cooler\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "Hf = 0.012;\n",
       "Hi = 0.033;\n",
       "H1 = 0.0075;\n",
       "\n",
-      "# Calculations and Results\n",
       "water_vapor = Hf/18.;   #in kmol of water vapor\n",
       "dry_air = 1/28.9;  #in kmol\n",
       "total_mass = water_vapor+dry_air;\n",
@@ -671,12 +612,10 @@
       "weight = 60/volume;\n",
       "print \"weight of dry air handled per hr = %f kg\"%(weight)\n",
       "\n",
-      "#part 1\n",
       "inlet_watervapor = 0.033/18;    #in kmol of water vapor\n",
       "volume_inlet = 22.4*(308./273)*(inlet_watervapor+dry_air);\n",
       "print \"volumetric flow rate of inlet air = %f cubic meter\"%(volume_inlet*weight)\n",
       "\n",
-      "#part 2\n",
       "y = (Hf - Hi)/(H1 - Hi);\n",
       "print \"fraction of inlet air passing through cooler = %f\"%(y)\n"
      ],
@@ -707,25 +646,18 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the fraction of purged recycle and total yield\n",
       "\n",
       "import math \n",
       "\n",
-      "# Variables\n",
-      "#x- moles of N2 and H2 recycled; y - moles of N2 H2 purged\n",
       "Ar_freshfeed = 0.2;\n",
-      "#argon in fresh feed is equal to argon in purge \n",
       "\n",
-      "# Calculations and Results\n",
       "y = 0.2/0.0633;   #argon in purge = 0.0633y\n",
       "x = (0.79*100 - y)/(1-0.79);\n",
       "print \"y = %f kmolx = %f kmol\"%(y,x)\n",
       "\n",
-      "#part 1\n",
       "fraction = y/x;\n",
       "print \"fration of recycle that is purged = %f\"%(fraction)\n",
       "\n",
-      "#part 2\n",
       "yield_ = 0.105*(100+x);\n",
       "print \"overall yield of ammonia = %f kmol\"%(yield_)\n"
      ],
@@ -756,18 +688,14 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find change in enthalpy\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "H0_CH4 = -74.9   #in kJ\n",
       "H0_CO2 = -393.5   #in kJ\n",
       "H0_H2O = -241.8   #in kJ\n",
       "\n",
-      "# Calculations\n",
       "delta_H0 = H0_CO2+2*H0_H2O-H0_CH4;\n",
       "\n",
-      "# Results\n",
       "print \"change in enthalpy = %f kJ\"%(delta_H0)\n"
      ],
      "language": "python",
@@ -795,21 +723,16 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to compare the enthalpy change in two reactions\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "H0_glucose = -1273  #in kJ\n",
       "H0_ethanol = -277.6  #in kJ\n",
       "H0_CO2 = -393.5  #in kJ\n",
       "H0_H2O = -285.8   #in kJ\n",
       "\n",
-      "# Calculations and Results\n",
-      "#for reaction 1\n",
       "delta_H1 = 2*H0_ethanol+2*H0_CO2-H0_glucose;\n",
       "print \"enthalpy change in reaction 1 = %f KJ\"%(delta_H1)\n",
       "\n",
-      "#for reaction 2\n",
       "delta_H2 = 6*H0_H2O+6*H0_CO2-H0_glucose;\n",
       "print \"enthalpy change in reaction 2 = %f kJ\"%(delta_H2)\n",
       "\n",
@@ -845,19 +768,15 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find enthalpy of formation of CuSO4.5H2O\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "delta_H2 = 11.7   #in kJ/mol\n",
       "m_CuSO4 = 16  #in gm\n",
       "m_H2O = 384   #in gm\n",
       "\n",
-      "# Calculations\n",
       "delta_H3 = -((m_CuSO4+m_H2O)*4.18*3.95*159.6)/(16*10**3)\n",
       "delta_H1 = delta_H3 - delta_H2;\n",
       "\n",
-      "# Results\n",
       "print \"enthalpy of formation = %f kJ/mol\"%(delta_H1)\n"
      ],
      "language": "python",
@@ -885,20 +804,16 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the temperature of combustion\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "H_combustion = 1560000    #in kJ/kmol \n",
       "H0_CO2 = 54.56   #in kJ/kmol\n",
       "H0_O2 = 35.2   #in kJ/kmol\n",
       "H0_steam = 43.38   #in kJ/kmol\n",
       "H0_N2 = 33.32   #in kJ/kmol\n",
       "\n",
-      "# Calculations\n",
       "t = H_combustion/(2*H0_CO2+3*H0_steam+0.875*H0_O2+16.46*H0_N2);\n",
       "\n",
-      "# Results\n",
       "print \"theoritical temperature of combustion = %f degree C\"%(t)\n"
      ],
      "language": "python",
@@ -926,16 +841,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the heat of reaction and consumption of coke\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "H_NaCl = 410.9   #in MJ/kmol\n",
       "H_H2SO4 = 811.3   #in MJ/kmol\n",
       "H_Na2SO4 = 1384   #in MJ/kmol\n",
       "H_HCl = 92.3   #in MJ/kmol\n",
       "\n",
-      "# Calculations and Results\n",
       "Q = H_Na2SO4 + 2*H_HCl -2*H_NaCl-H_H2SO4;\n",
       "print \"heat of reaction = %f MJ\"%(Q)\n",
       "\n",
@@ -969,21 +881,17 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the rate of heat flow\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "cp_water = 146.5   #in kj/kg\n",
       "cp_steam = 3040   #in kJ/kg\n",
       "d = 0.102  #in m\n",
       "u = 1.5  #in m/s\n",
       "density = 1000   #in kg/m3\n",
       "\n",
-      "# Calculations\n",
       "m = (3.14/4)*d**2*u*density;\n",
       "Q = m*(cp_steam-cp_water);\n",
       "\n",
-      "# Results\n",
       "print \"rate of heat flow = %f kW\"%(Q)\n"
      ],
      "language": "python",
@@ -1011,10 +919,8 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the amount of air required for combustion and composition of flue gas\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "wt_C = 0.75   #in kg\n",
       "wt_H2 = 0.05   #in kg\n",
       "wt_O2 = 0.12   #in kg\n",
@@ -1022,7 +928,6 @@
       "wt_S = 0.01    #in kg\n",
       "wt_ash = 0.04  #in kg\n",
       "\n",
-      "# Calculations and Results\n",
       "O2_C = wt_C*(32./12);  #in kg\n",
       "O2_H2 = wt_H2*(16./2);   #in kg\n",
       "O2_S = wt_S*(32./32);   #in kg\n",
@@ -1060,7 +965,6 @@
       "print \"N2 = %f %%\"%(x_N2*100)\n",
       "print \"O2 = %f %%\"%(x_O2*100)\n",
       "\n",
-      "# Note : answers are slightly different because of rounding error."
      ],
      "language": "python",
      "metadata": {},
@@ -1092,17 +996,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the composition of flue gas\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "C = 0.8   #in kg\n",
       "H2 = 0.05   #in kg\n",
       "S = 0.005   #in kg\n",
       "ash = 0.145  #in kg\n",
       "\n",
-      "# Calculations and Results\n",
-      "#required oxygen in kg\n",
       "C_O2 = C*(32./12);  \n",
       "H2_O2 = H2*(16./2);\n",
       "S_O2 = S*(32./32);\n",
@@ -1113,7 +1013,6 @@
       "wt_airsupplied = 1.25*wt_air;\n",
       "print \"amount of air supplied = %f kg\"%(wt_airsupplied)\n",
       "\n",
-      "#flue gas composition\n",
       "m_N2 = wt_airsupplied*0.77;   #in kg\n",
       "mole_N2 = m_N2/28.;\n",
       "\n",
@@ -1131,7 +1030,6 @@
       "\n",
       "m = m_N2+m_O2+m_CO2+m_H2O+m_SO2\n",
       "\n",
-      "#percent by weight\n",
       "w_N2 = m_N2/m;\n",
       "print \"percentage of N2 by weight = %f\"%(w_N2*100)\n",
       "\n",
@@ -1149,7 +1047,6 @@
       "\n",
       "m1 = mole_N2+mole_O2+mole_CO2+mole_H2O+mole_SO2\n",
       "\n",
-      "#percent by mole \n",
       "x_N2 = mole_N2/m1;\n",
       "print \"percentage of N2 by mole = %f\"%(x_N2*100)\n",
       "\n",
@@ -1201,16 +1098,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find volumetric composition of flue glass\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "wt_H2 = 0.15;\n",
       "wt_C = 0.85;\n",
       "O2_H2 = wt_H2*(16./2);\n",
       "O2_C = wt_C*(32./12);\n",
       "\n",
-      "# Calculations and Results\n",
       "total_O2 = O2_H2+O2_C;\n",
       "wt_air = total_O2/0.23;\n",
       "air_supplied = 1.15*(wt_air);\n",
@@ -1258,20 +1152,16 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the excess air supplied\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "N2 = 80.5   #in m3\n",
       "air_supplied = N2/0.79   #in m3\n",
       "volume_O2 = air_supplied*0.21;  #in m3\n",
       "O2_fluegas = 6.1   #in m3\n",
       "\n",
-      "# Calculations\n",
       "O2_used = volume_O2 - O2_fluegas;\n",
       "excess_air_supplied = (O2_fluegas/O2_used)*100;\n",
       "\n",
-      "# Results\n",
       "print \"percentage of excess air supplied = %f\"%(excess_air_supplied)\n"
      ],
      "language": "python",
@@ -1299,15 +1189,12 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the outlet temperature of water\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "q_NTP = 10*(200/101.3)*(273./313);\n",
       "m_CO2 = 44*(q_NTP/22.4);\n",
       "s_CO2 = 0.85   #in kJ/kg K\n",
       "\n",
-      "# Calculations\n",
       "Q = m_CO2*s_CO2*(40-20)   #Q = ms*delta_T\n",
       "\n",
       "d0 = 0.023   #in mm\n",
@@ -1322,7 +1209,6 @@
       "s_water = 4.19    #in kJ/kg K\n",
       "t = 15+(Q/(m_water*s_water));\n",
       "\n",
-      "# Results\n",
       "print \"exit water temperature = %f degree C\"%(t)\n"
      ],
      "language": "python",
@@ -1350,14 +1236,11 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the area of heating surface\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "F = 1000   #in kg\n",
       "xF = 0.01   \n",
       "\n",
-      "# Calculations and Results\n",
       "solid_feed = F*xF;\n",
       "water_feed = F - solid_feed;\n",
       "\n",
@@ -1379,7 +1262,6 @@
       "tc = 108.4  #in degree C\n",
       "hc = 454   #in kJ/kg\n",
       "\n",
-      "#applying heat balance\n",
       "S = (F*hF-V*Hv-L*hL)/(hc-Hs);\n",
       "print \"weight of steam required = %f kg/hr\"%(S)\n",
       "\n",
@@ -1415,10 +1297,8 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the top and bottom product,condenser duty,heat input to rebpoiler\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "hF = 171  #in kJ/kg\n",
       "hD = 67   #in kJ/kg\n",
       "hL = hD;\n",
@@ -1432,7 +1312,6 @@
       "xW = 0.02;\n",
       "xD = 0.97;\n",
       "\n",
-      "# Calculations and Results\n",
       "D = F*(xF-xW)/(xD-xW);\n",
       "W = F-D;\n",
       "\n",
@@ -1480,23 +1359,19 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the rate of crystal formation, cooling water rate, required area\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "F = 1000.;    #in kg\n",
       "V = 0.05*F;   #in kg\n",
       "xF = 0.48;\n",
       "xL = 75./(100+75);\n",
       "xC = 1.;\n",
       "\n",
-      "# Calculations and Results\n",
       "C = (F*xF-950*xL)/(1-0.429);\n",
       "print  \"rate of crystal formation = %f kg\"%(C)\n",
       "\n",
       "L = F-C-V;\n",
       "\n",
-      "#cooling water\n",
       "W = (F*2.97*(85-35)+126.9*75.2-V*2414)/(4.19*11);\n",
       "print \"rate of cooling water = %f kg\"%(W)\n",
       "\n",
@@ -1535,16 +1410,12 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#to find the heat of combustion\n",
       "\n",
       "import math \n",
-      "# Variables\n",
       "delta_n = 10-12.;  #mole per mole napthanlene\n",
       "\n",
-      "#basis 1g\n",
       "moles_napthalene = (1./128);\n",
       "\n",
-      "# Calculations and Results\n",
       "print ('part 1')\n",
       "Qv = 40.28   #in kJ\n",
       "Qp = Qv-(delta_n*moles_napthalene*8.3144*298./1000);\n",