removing problem statements

9 files changed, 2127 insertions, 250 deletions

diff --git a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_1_1.ipynb b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_1_1.ipynb
index 43fa32ce..c3d00505 100644
--- a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_1_1.ipynb
+++ b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_1_1.ipynb
@@ -1,6 +1,7 @@
 {
  "metadata": {
-  "name": "Chapter 1"
+  "name": "",
+  "signature": "sha256:f74ebb3df14e61a18adced8ec1d013b274b4ed30f643d40b946b0969c2ba80e2"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -11,25 +12,46 @@
      "cell_type": "heading",
      "level": 1,
      "metadata": {},
-     "source": "Chapter1: Fluid Statics"
+     "source": [
+      "Chapter1: Fluid Statics"
+     ]
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.1,Page 3"
+     "source": [
+      "Example 1.1,Page 3"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Force\n\n# Variable Decleration\nrho=924; #density\ng=9.81; #grivity\nH=2; #height\nd=2; # depth \n\n#Calculation\np=rho*g*H;\na=d*H;\nF=p*a/2;\n\n#result\nprint\" Total force exerted over the wall in(N) =\",round(F,3)",
+     "input": [
+      "\n",
+      "# Variable Decleration\n",
+      "rho=924; #density\n",
+      "g=9.81; #grivity\n",
+      "H=2; #height\n",
+      "d=2; # depth \n",
+      "\n",
+      "#Calculation\n",
+      "p=rho*g*H;\n",
+      "a=d*H;\n",
+      "F=p*a/2;\n",
+      "\n",
+      "#result\n",
+      "print\" Total force exerted over the wall in(N) =\",round(F,3)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Total force exerted over the wall in(N) = 36257.76\n"
+       "text": [
+        " Total force exerted over the wall in(N) = 36257.76\n"
+       ]
       }
      ],
      "prompt_number": 2
@@ -38,19 +60,47 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.2,Page 5"
+     "source": [
+      "Example 1.2,Page 5"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Force and Pressure\nimport math\n\n#Variable declaration\np_v =50*1000; #pressure\nr =1; #m\np_atm =101.3*1000; #atmospheric pressure\nrho =1000; #density\nH =2.5; #m\ng =9.81; #m/s^2\n\n\n#Calculation\nF= p_v*math.pi*r*r;\np= p_atm + p_v + rho *g*H;\nFd =( p_v+ rho *g*H)*math.pi*r*r+rho*g*2*math.pi*r*r/3;\n\n#result\nprint \" Total vertical force tending to lift the dome (N)\", round(F,3)\nprint \" Absolute pressure at the bottom o f the vessel (Pa)\",round(p,3)\nprint \"Downward force imposed by the gas and liquid (N)\", round(Fd,3)",
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "p_v =50*1000; #pressure\n",
+      "r =1; #m\n",
+      "p_atm =101.3*1000; #atmospheric pressure\n",
+      "rho =1000; #density\n",
+      "H =2.5; #m\n",
+      "g =9.81; #m/s^2\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "F= p_v*math.pi*r*r;\n",
+      "p= p_atm + p_v + rho *g*H;\n",
+      "Fd =( p_v+ rho *g*H)*math.pi*r*r+rho*g*2*math.pi*r*r/3;\n",
+      "\n",
+      "#result\n",
+      "print \" Total vertical force tending to lift the dome (N)\", round(F,3)\n",
+      "print \" Absolute pressure at the bottom o f the vessel (Pa)\",round(p,3)\n",
+      "print \"Downward force imposed by the gas and liquid (N)\", round(Fd,3)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Total vertical force tending to lift the dome (N) 157079.633\n Absolute pressure at the bottom o f the vessel (Pa) 175825.0\nDownward force imposed by the gas and liquid (N) 254673.208\n"
+       "text": [
+        " Total vertical force tending to lift the dome (N) 157079.633\n",
+        " Absolute pressure at the bottom o f the vessel (Pa) 175825.0\n",
+        "Downward force imposed by the gas and liquid (N) 254673.208\n"
+       ]
       }
      ],
      "prompt_number": 5
@@ -59,19 +109,39 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.3,Page 7"
+     "source": [
+      "Example 1.3,Page 7"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Force\n\n#Variable Decleration\na1 =0.3; #m^2\nm =1000; # kg\na2 =0.003; #m^2\nrho_oil =750; #kg /m^3\nH =2; #m\ng =9.81; #m/ s ^2\n\n# Calcualtion\nF1=m*g;\nF2=a2 *( F1/a1 - rho_oil *g*H);\n\n#result\nprint \"The force on the plunger(N)\",round(F2,3)",
+     "input": [
+      "\n",
+      "#Variable Decleration\n",
+      "a1 =0.3; #m^2\n",
+      "m =1000; # kg\n",
+      "a2 =0.003; #m^2\n",
+      "rho_oil =750; #kg /m^3\n",
+      "H =2; #m\n",
+      "g =9.81; #m/ s ^2\n",
+      "\n",
+      "# Calcualtion\n",
+      "F1=m*g;\n",
+      "F2=a2 *( F1/a1 - rho_oil *g*H);\n",
+      "\n",
+      "#result\n",
+      "print \"The force on the plunger(N)\",round(F2,3)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The force on the plunger(N) 53.955\n"
+       "text": [
+        "The force on the plunger(N) 53.955\n"
+       ]
       }
      ],
      "prompt_number": 3
@@ -80,19 +150,44 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.4,Page 8"
+     "source": [
+      "Example 1.4,Page 8"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Height\n\n#variable Decleration\nrho_0 =800; # kg /m^3\nrho_aq =1100; # density of aqueous solution\n\n#calculation\nH =0.5* rho_aq /( rho_aq - rho_0 );\n# For a fixed length of chamber o f 3 m, the\n#interface between the two phases is determined\n#from the p r e s s u r e i n the chamber and d i s c h a r g e\n#p o i n t .\n#r h o 0 \u0003g\u0003H1+rho a q \u0003g\u0003H2=rho a q \u0003g \u0003(H\udbc0\udc000.5) ;\n#H=H1+H2\nrho_0 =600; #kg /m^3\nH1 =0.5* rho_aq /( rho_aq - rho_0 );\n\n#result\nprint \"H(m)\",round(H,3)\nprint \"The lowest possible position of the interface in the chamber below the overflow (m)\",round(H1,2)",
+     "input": [
+      "\n",
+      "#variable Decleration\n",
+      "rho_0 =800; # kg /m^3\n",
+      "rho_aq =1100; # density of aqueous solution\n",
+      "\n",
+      "#calculation\n",
+      "H =0.5* rho_aq /( rho_aq - rho_0 );\n",
+      "# For a fixed length of chamber o f 3 m, the\n",
+      "#interface between the two phases is determined\n",
+      "#from the p r e s s u r e i n the chamber and d i s c h a r g e\n",
+      "#p o i n t .\n",
+      "#r h o 0 \u0003g\u0003H1+rho a q \u0003g\u0003H2=rho a q \u0003g \u0003(H\udbc0\udc000.5) ;\n",
+      "#H=H1+H2\n",
+      "rho_0 =600; #kg /m^3\n",
+      "H1 =0.5* rho_aq /( rho_aq - rho_0 );\n",
+      "\n",
+      "#result\n",
+      "print \"H(m)\",round(H,3)\n",
+      "print \"The lowest possible position of the interface in the chamber below the overflow (m)\",round(H1,2)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "H(m) 1.833\nThe lowest possible position of the interface in the chamber below the overflow (m) 1.1\n"
+       "text": [
+        "H(m) 1.833\n",
+        "The lowest possible position of the interface in the chamber below the overflow (m) 1.1\n"
+       ]
       }
      ],
      "prompt_number": 2
@@ -101,19 +196,39 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.5, Page11"
+     "source": [
+      "Example 1.5, Page11"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#position of interface between legs\n\n#variable decleration\nrho_o =900; #kg /m^3\nrho_n =1070; #kg /m^3\nH =1; #m\ng =9.81; #m/ s ^2\ndp =10*10**3; #change in pressure\n\n#H=H1+H2\n#calculation\nH1 =(dp - rho_n *g*H)/( rho_o - rho_n )/g;\n\n#result\nprint \"The position of the interface between the legs (m)\", round(H1,2)",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "rho_o =900; #kg /m^3\n",
+      "rho_n =1070; #kg /m^3\n",
+      "H =1; #m\n",
+      "g =9.81; #m/ s ^2\n",
+      "dp =10*10**3; #change in pressure\n",
+      "\n",
+      "#H=H1+H2\n",
+      "#calculation\n",
+      "H1 =(dp - rho_n *g*H)/( rho_o - rho_n )/g;\n",
+      "\n",
+      "#result\n",
+      "print \"The position of the interface between the legs (m)\", round(H1,2)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The position of the interface between the legs (m) 0.3\n"
+       "text": [
+        "The position of the interface between the legs (m) 0.3\n"
+       ]
       }
      ],
      "prompt_number": 2
@@ -122,19 +237,42 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.6,Page 13"
+     "source": [
+      "Example 1.6,Page 13"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Density and Fraction\n\n# Variable defining\ndp =22*10**3; #N/m^2\ng =9.81;#m/ s ^2\nH =1.5;#m\nrho =1495.00;#kg /m^3\nrho_s =1270;#kg /m^3\nrho_c =2698;#kg /m^3\n\n# Calculation\np=dp/g/H;\n#rho=f 1 \u0003 r h o s+f 2 \u0003 r h o c\n#f 1+f 2=1\nf2 =( rho - rho_s )/( rho_c - rho_s );\n\n#result\nprint \" the density of the solution with crystals (kg/m^3)\",round(p,3)\nprint \"The fraction of crystals \", round(f2,3)",
+     "input": [
+      "# Variable defining\n",
+      "dp =22*10**3; #N/m^2\n",
+      "g =9.81;#m/ s ^2\n",
+      "H =1.5;#m\n",
+      "rho =1495.00;#kg /m^3\n",
+      "rho_s =1270;#kg /m^3\n",
+      "rho_c =2698;#kg /m^3\n",
+      "\n",
+      "# Calculation\n",
+      "p=dp/g/H;\n",
+      "#rho=f 1 \u0003 r h o s+f 2 \u0003 r h o c\n",
+      "#f 1+f 2=1\n",
+      "f2 =( rho - rho_s )/( rho_c - rho_s );\n",
+      "\n",
+      "#result\n",
+      "print \" the density of the solution with crystals (kg/m^3)\",round(p,3)\n",
+      "print \"The fraction of crystals \", round(f2,3)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " the density of the solution with crystals (kg/m^3) 1495.073\nThe fraction of crystals  0.158\n"
+       "text": [
+        " the density of the solution with crystals (kg/m^3) 1495.073\n",
+        "The fraction of crystals  0.158\n"
+       ]
       }
      ],
      "prompt_number": 12
@@ -143,19 +281,43 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.7, Page 15"
+     "source": [
+      "Example 1.7, Page 15"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding radius\n\n#Variable defining\np_atm =101.3*10**3; #N/m^2\nrho =1000; #kg /m^3\ng =9.81; #m/ s ^2\nH1 =3; #m\na =0.073; #N/m\nr1 =5*10**( -4) ; #m\np1= p_atm + rho *g*H1 +2* a/r1;\n\n#calculation\n# p2=p atm+rho \u0003g\u0003H2+2\u0003a / r2 ;\n# p1 \u00034/3\u0003%pi\u0003 r1 ^3=p2 \u00034/3\u0003%pi\u0003 r2 ^3\n# Solving above two equations we get\nr2 =0.053; #//mm\n\n#result\nprint\"Radius of the bubble(mm)\",round(r2,3)",
+     "input": [
+      "\n",
+      "\n",
+      "#Variable defining\n",
+      "p_atm =101.3*10**3; #N/m^2\n",
+      "rho =1000; #kg /m^3\n",
+      "g =9.81; #m/ s ^2\n",
+      "H1 =3; #m\n",
+      "a =0.073; #N/m\n",
+      "r1 =5*10**( -4) ; #m\n",
+      "p1= p_atm + rho *g*H1 +2* a/r1;\n",
+      "\n",
+      "#calculation\n",
+      "# p2=p atm+rho \u0003g\u0003H2+2\u0003a / r2 ;\n",
+      "# p1 \u00034/3\u0003%pi\u0003 r1 ^3=p2 \u00034/3\u0003%pi\u0003 r2 ^3\n",
+      "# Solving above two equations we get\n",
+      "r2 =0.053; #//mm\n",
+      "\n",
+      "#result\n",
+      "print\"Radius of the bubble(mm)\",round(r2,3)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Radius of the bubble(mm) 0.053\n"
+       "text": [
+        "Radius of the bubble(mm) 0.053\n"
+       ]
       }
      ],
      "prompt_number": 3
@@ -164,19 +326,36 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.8,Page 17"
+     "source": [
+      "Example 1.8,Page 17"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Pressure Difference\n\n#Variable defining\nH =0.2; #m\nrho =1000; #kg /m^3\nrho_Hg =13600; #kg /m^3\ng =9.81; #m/ s ^2\n\n#calculation\ndp =( rho_Hg -rho)*g*H;\n\n#result\nprint \" Differential pressure(Pa)\",round(dp,2)",
+     "input": [
+      "\n",
+      "#Variable defining\n",
+      "H =0.2; #m\n",
+      "rho =1000; #kg /m^3\n",
+      "rho_Hg =13600; #kg /m^3\n",
+      "g =9.81; #m/ s ^2\n",
+      "\n",
+      "#calculation\n",
+      "dp =( rho_Hg -rho)*g*H;\n",
+      "\n",
+      "#result\n",
+      "print \" Differential pressure(Pa)\",round(dp,2)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Differential pressure(Pa) 24721.2\n"
+       "text": [
+        " Differential pressure(Pa) 24721.2\n"
+       ]
       }
      ],
      "prompt_number": 9
@@ -185,19 +364,36 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.9,Page 19"
+     "source": [
+      "Example 1.9,Page 19"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding pressure drop\n\n#variable defining\nrho =1000;\ng =9.81; # m/ s ^2\nH =0.4; #m\n\n#calculation\ndp=rho*g*H;\n\n#result\nprint\" Pressure drop in the pipe(Pa)\",round(dp)",
+     "input": [
+      "\n",
+      "\n",
+      "#variable defining\n",
+      "rho =1000;\n",
+      "g =9.81; # m/ s ^2\n",
+      "H =0.4; #m\n",
+      "\n",
+      "#calculation\n",
+      "dp=rho*g*H;\n",
+      "\n",
+      "#result\n",
+      "print\" Pressure drop in the pipe(Pa)\",round(dp)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Pressure drop in the pipe(Pa) 3924.0\n"
+       "text": [
+        " Pressure drop in the pipe(Pa) 3924.0\n"
+       ]
       }
      ],
      "prompt_number": 1
@@ -206,19 +402,41 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.10,Page 21"
+     "source": [
+      "Example 1.10,Page 21"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#finding quantity\nimport math\n\n#variable defining\ndp =20*10**3; #N/m^2\nrho_Hg =13600; # kg /m^3\nrho =700; #kg /m^3\ng =9.81; #m/ s ^2\nd =0.02; #m\n\n\n#calculation\nH=dp*(rho_Hg - rho)**-1/g;\nV= math.pi /4*d*d*H;\n\n#result\nprint\"Quantity of mercury to be removed (m^3)\",round(V,9)",
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#variable defining\n",
+      "dp =20*10**3; #N/m^2\n",
+      "rho_Hg =13600; # kg /m^3\n",
+      "rho =700; #kg /m^3\n",
+      "g =9.81; #m/ s ^2\n",
+      "d =0.02; #m\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "H=dp*(rho_Hg - rho)**-1/g;\n",
+      "V= math.pi /4*d*d*H;\n",
+      "\n",
+      "#result\n",
+      "print\"Quantity of mercury to be removed (m^3)\",round(V,9)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Quantity of mercury to be removed (m^3) 4.965e-05\n"
+       "text": [
+        "Quantity of mercury to be removed (m^3) 4.965e-05\n"
+       ]
       }
      ],
      "prompt_number": 11
@@ -227,19 +445,39 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.11,Page 23"
+     "source": [
+      "Example 1.11,Page 23"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Pressure\n\n#Variable defining\nimport math\nfrom math import sin\nrho =800; #kg /m^3\ng =9.81; #m/ s ^2\nL =0.12;\n\n\n#calculations\ntheta = math.pi /180*20; #radians\ndp=rho*g*L*(sin(theta));\n\n#result\nprint \"The gauge pressure across the filter(Pa)\",round(dp,2)",
+     "input": [
+      "\n",
+      "#Variable defining\n",
+      "import math\n",
+      "from math import sin\n",
+      "rho =800; #kg /m^3\n",
+      "g =9.81; #m/ s ^2\n",
+      "L =0.12;\n",
+      "\n",
+      "\n",
+      "#calculations\n",
+      "theta = math.pi /180*20; #radians\n",
+      "dp=rho*g*L*(sin(theta));\n",
+      "\n",
+      "#result\n",
+      "print \"The gauge pressure across the filter(Pa)\",round(dp,2)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The gauge pressure across the filter(Pa) 322.100890178\n"
+       "text": [
+        "The gauge pressure across the filter(Pa) 322.100890178\n"
+       ]
       }
      ],
      "prompt_number": 1
@@ -248,19 +486,37 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.12,Page 25"
+     "source": [
+      "Example 1.12,Page 25"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding tension\n\n#Variable defining\nmc =100; #kg\ng =9.81; #m/ s ^2\nrho =1000; #kg /m^3\nrho_c =7930; #kg /m^3\n\n#calculation\nm=mc*rho/ rho_c ;\nF=mc*g-m*g; #tension\n\n#result\nprint(\"The tension in the cable(N)\",round(F,2))\n",
+     "input": [
+      "\n",
+      "#Variable defining\n",
+      "mc =100; #kg\n",
+      "g =9.81; #m/ s ^2\n",
+      "rho =1000; #kg /m^3\n",
+      "rho_c =7930; #kg /m^3\n",
+      "\n",
+      "#calculation\n",
+      "m=mc*rho/ rho_c ;\n",
+      "F=mc*g-m*g; #tension\n",
+      "\n",
+      "#result\n",
+      "print(\"The tension in the cable(N)\",round(F,2))\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "('The tension in the cable(N)', 863.28)\n"
+       "text": [
+        "('The tension in the cable(N)', 863.28)\n"
+       ]
       }
      ],
      "prompt_number": 13
@@ -269,19 +525,39 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.13,Page 27"
+     "source": [
+      "Example 1.13,Page 27"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding length\n\n#Variale defining\nrho =1000.0;\nx =0.06;\nrho_0 =800.0;\nx_0 =0.04;\nrho_L =900.0;\n\n#calculation\nL=( rho*x- rho_0 *x_0)/( rho - rho_0 );\nx_L =L-rho/ rho_L *(L-x);\n\n#result\nprint \" Length of the stem above the liquid of SG 0.9 is (m)\",round(x_L,3)",
+     "input": [
+      "\n",
+      "\n",
+      "#Variale defining\n",
+      "rho =1000.0;\n",
+      "x =0.06;\n",
+      "rho_0 =800.0;\n",
+      "x_0 =0.04;\n",
+      "rho_L =900.0;\n",
+      "\n",
+      "#calculation\n",
+      "L=( rho*x- rho_0 *x_0)/( rho - rho_0 );\n",
+      "x_L =L-rho/ rho_L *(L-x);\n",
+      "\n",
+      "#result\n",
+      "print \" Length of the stem above the liquid of SG 0.9 is (m)\",round(x_L,3)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Length of the stem above the liquid of SG 0.9 is (m) 0.05\n"
+       "text": [
+        " Length of the stem above the liquid of SG 0.9 is (m) 0.05\n"
+       ]
       }
      ],
      "prompt_number": 5
@@ -290,19 +566,41 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 1.14,Page 29\n"
+     "source": [
+      "Example 1.14,Page 29\n"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Time and Quantity\n\n#variable defining\nm_s =5*10**6; #kg\nT2 =4.5; #m\nT1 =3; #m\nrho_hc =950; #kg /m^3\nQ =125; #m^3/h\n\n#calculation\nm_hc =m_s *( T2/T1 -1);\nt= m_hc / rho_hc /Q;\n\n#result\nprint\"Quantity delivered(kg)\",round(m_hc,2)\nprint \"Time taken(hours)\",round(t,3)",
+     "input": [
+      "\n",
+      "\n",
+      "#variable defining\n",
+      "m_s =5*10**6; #kg\n",
+      "T2 =4.5; #m\n",
+      "T1 =3; #m\n",
+      "rho_hc =950; #kg /m^3\n",
+      "Q =125; #m^3/h\n",
+      "\n",
+      "#calculation\n",
+      "m_hc =m_s *( T2/T1 -1);\n",
+      "t= m_hc / rho_hc /Q;\n",
+      "\n",
+      "#result\n",
+      "print\"Quantity delivered(kg)\",round(m_hc,2)\n",
+      "print \"Time taken(hours)\",round(t,3)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Quantity delivered(kg) 2500000.0\nTime taken(hours) 21.053\n"
+       "text": [
+        "Quantity delivered(kg) 2500000.0\n",
+        "Time taken(hours) 21.053\n"
+       ]
       }
      ],
      "prompt_number": 1
@@ -310,7 +608,7 @@
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "",
+     "input": [],
      "language": "python",
      "metadata": {},
      "outputs": []
diff --git a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_2_1.ipynb b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_2_1.ipynb
index a567da1d..72318a60 100644
--- a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_2_1.ipynb
+++ b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_2_1.ipynb
@@ -1,6 +1,7 @@
 {
  "metadata": {
-  "name": "Chapter 2"
+  "name": "",
+  "signature": "sha256:785f37cfaf9aeabbac4169c1fd95bd67ae8666fc83631f2d394e32747e65fd3b"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -11,25 +12,63 @@
      "cell_type": "heading",
      "level": 1,
      "metadata": {},
-     "source": "Chapter 2,Continuity Momentum and Energy"
+     "source": [
+      "Chapter 2,Continuity Momentum and Energy"
+     ]
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 2.1,Page 36"
+     "source": [
+      "Example 2.1,Page 36"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Velocity and Flow rate\nfrom math import pi\n\n\n#variable decleration\nQ1 =0.02; #m^3/ s\nd1 =0.15; #m\nd2 =0.05; #m\nd3 =0.1; #m\nv2 =3; #m/ s\n\n\n#calculation\nQ2=pi*d2 **2/4* v2;\nv3 =(4* Q1/pi -d2 **2* v2)/d3 **2;\nQ3=pi*d3 **2/4* v3;\nv1 =4*( Q2+Q3)/pi/d1 **2;\n\n\n#result\nprint\" Flow rate at pipe 3(m^3/s) =\",round(Q3,4)\nprint\" Flow rate at pipe 2(m^3/s)= \",round(Q2,4)\nprint\" Flow rate at pipe 1(m^3/s)= \",round(Q1,4)\nprint\" Velocity at pipe 1(m/s) =\",round(v1,3)\nprint\" Velocity at pipe 2(m/s) =\",round(v2,3)\nprint\" Velocity at pipe 3(m/s) =\",round(v3,3)\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "Q1 =0.02; #m^3/ s\n",
+      "d1 =0.15; #m\n",
+      "d2 =0.05; #m\n",
+      "d3 =0.1; #m\n",
+      "v2 =3; #m/ s\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "Q2=pi*d2 **2/4* v2;\n",
+      "v3 =(4* Q1/pi -d2 **2* v2)/d3 **2;\n",
+      "Q3=pi*d3 **2/4* v3;\n",
+      "v1 =4*( Q2+Q3)/pi/d1 **2;\n",
+      "\n",
+      "\n",
+      "#result\n",
+      "print\" Flow rate at pipe 3(m^3/s) =\",round(Q3,4)\n",
+      "print\" Flow rate at pipe 2(m^3/s)= \",round(Q2,4)\n",
+      "print\" Flow rate at pipe 1(m^3/s)= \",round(Q1,4)\n",
+      "print\" Velocity at pipe 1(m/s) =\",round(v1,3)\n",
+      "print\" Velocity at pipe 2(m/s) =\",round(v2,3)\n",
+      "print\" Velocity at pipe 3(m/s) =\",round(v3,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Flow rate at pipe 3(m^3/s) = 0.0141\n Flow rate at pipe 2(m^3/s)=  0.0059\n Flow rate at pipe 1(m^3/s)=  0.02\n Velocity at pipe 1(m/s) = 1.132\n Velocity at pipe 2(m/s) = 3.0\n Velocity at pipe 3(m/s) = 1.796\n"
+       "text": [
+        " Flow rate at pipe 3(m^3/s) = 0.0141\n",
+        " Flow rate at pipe 2(m^3/s)=  0.0059\n",
+        " Flow rate at pipe 1(m^3/s)=  0.02\n",
+        " Velocity at pipe 1(m/s) = 1.132\n",
+        " Velocity at pipe 2(m/s) = 3.0\n",
+        " Velocity at pipe 3(m/s) = 1.796\n"
+       ]
       }
      ],
      "prompt_number": 1
@@ -38,19 +77,53 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 2.2,Page 38"
+     "source": [
+      "Example 2.2,Page 38"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Force exerted\n#variable decleration\nimport math\nfrom math import pi\nd1 =0.2; #m\nd2=d1;\np1 =1*10**5; #N/m^2\np2 =80*10**3; #N/m^2\nQ =150.0; #m^3/h\nrho =900; #kg /m^3\ntheta1 =0; # r a d i a n s\ntheta2 =pi; # r a d i a n s\n\n#calculation\na1=pi*d1 **2/4;\na2=pi*d2 **2/4;\nF1=p1*a1; # Upstream f o r c e\nF2=p2*a2; # Downstream f o r c e\nv1 =4*Q /3600/ pi/d1 **2;\nv2=v1;\nflux =rho*Q /3600* v2; #Momentum f l u x\nFx=F1* math.cos ( theta1 )-F2* math.cos ( theta2 )+ flux *( math.cos ( theta2 ) -\nmath.cos ( theta1 ));\nFy=F1* math.sin ( theta1 )-F2* math.sin ( theta2 )-flux *( math.sin ( theta2 )-\nmath.sin ( theta1 ));\nF= math.sqrt (Fx **2+ Fy **2);\n\n#result\nprint\" Force exerted by liquid (kN) =\",round(F,2)\n",
+     "input": [
+      "\n",
+      "#variable decleration\n",
+      "import math\n",
+      "from math import pi\n",
+      "d1 =0.2; #m\n",
+      "d2=d1;\n",
+      "p1 =1*10**5; #N/m^2\n",
+      "p2 =80*10**3; #N/m^2\n",
+      "Q =150.0; #m^3/h\n",
+      "rho =900; #kg /m^3\n",
+      "theta1 =0; # r a d i a n s\n",
+      "theta2 =pi; # r a d i a n s\n",
+      "\n",
+      "#calculation\n",
+      "a1=pi*d1 **2/4;\n",
+      "a2=pi*d2 **2/4;\n",
+      "F1=p1*a1; # Upstream f o r c e\n",
+      "F2=p2*a2; # Downstream f o r c e\n",
+      "v1 =4*Q /3600/ pi/d1 **2;\n",
+      "v2=v1;\n",
+      "flux =rho*Q /3600* v2; #Momentum f l u x\n",
+      "Fx=F1* math.cos ( theta1 )-F2* math.cos ( theta2 )+ flux *( math.cos ( theta2 ) -\n",
+      "math.cos ( theta1 ));\n",
+      "Fy=F1* math.sin ( theta1 )-F2* math.sin ( theta2 )-flux *( math.sin ( theta2 )-\n",
+      "math.sin ( theta1 ));\n",
+      "F= math.sqrt (Fx **2+ Fy **2);\n",
+      "\n",
+      "#result\n",
+      "print\" Force exerted by liquid (kN) =\",round(F,2)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Force exerted by liquid (kN) = 5552.47608322\n"
+       "text": [
+        " Force exerted by liquid (kN) = 5552.47608322\n"
+       ]
       }
      ],
      "prompt_number": 12
@@ -59,19 +132,40 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 2.3,Page 40"
+     "source": [
+      "Example 2.3,Page 40"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding pressure\nfrom math import pi\n\n#variable decleration\nrho =1000; # kg /m^3\nd =0.05; #m\nL =500; #m\nv =1.7; #m/ s\n\n#calculation\na= pi *d **2/4;\nF= rho *a*L*v;\nP=F/a /10**3;\n \n#result    \nprint\"Average pressure(kN/m^2) =\",round(P)\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#variable decleration\n",
+      "rho =1000; # kg /m^3\n",
+      "d =0.05; #m\n",
+      "L =500; #m\n",
+      "v =1.7; #m/ s\n",
+      "\n",
+      "#calculation\n",
+      "a= pi *d **2/4;\n",
+      "F= rho *a*L*v;\n",
+      "P=F/a /10**3;\n",
+      " \n",
+      "#result    \n",
+      "print\"Average pressure(kN/m^2) =\",round(P)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Average pressure(kN/m^2) = 850.0\n"
+       "text": [
+        "Average pressure(kN/m^2) = 850.0\n"
+       ]
       }
      ],
      "prompt_number": 6
@@ -80,19 +174,56 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 2.4,Page 42"
+     "source": [
+      "Example 2.4,Page 42"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding pressure and velocity\n\n#variable decleration\nimport math\ng =9.8; #m/ s ^2\ndz =0.2; #m ; dz1=z1-z2=z1-z2\nrho =1000; # kg /m^3\ndz1 =2; #m ; dz1=z1-z A\ndz2 =0; #m ; dz2=z1-z B\ndz3 = -1.5; #m ; dz3=z1-z C\n\n#calculation\nv2= math.sqrt (2* g*dz);\nv_A =v2;\nv_B =v2;\nv_C =v2;\np_A = rho *g*( dz1 - v_A **2/2/ g);\np_B = rho *g*( dz2 - v_B **2/2/ g);\np_C = rho *g*( dz3 - v_C **2/2/ g);\n\n#result\nprint\" Velocity at pt . A(m/s) =\",round(v_A,2);\nprint\" Velocity at pt . B(m/s) =\",round(v_B,2);\nprint\" Velocity at pt . C(m/s) =\",round(v_C,2);\nprint\" Pressure at pt . A(kN/m^2) =\",round(p_A)\nprint\" Pressure at pt . B (kN/m^2)=\",round(p_B)\nprint\" Pressure at pt . C (kN/m^2)=\",round(p_C)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "import math\n",
+      "g =9.8; #m/ s ^2\n",
+      "dz =0.2; #m ; dz1=z1-z2=z1-z2\n",
+      "rho =1000; # kg /m^3\n",
+      "dz1 =2; #m ; dz1=z1-z A\n",
+      "dz2 =0; #m ; dz2=z1-z B\n",
+      "dz3 = -1.5; #m ; dz3=z1-z C\n",
+      "\n",
+      "#calculation\n",
+      "v2= math.sqrt (2* g*dz);\n",
+      "v_A =v2;\n",
+      "v_B =v2;\n",
+      "v_C =v2;\n",
+      "p_A = rho *g*( dz1 - v_A **2/2/ g);\n",
+      "p_B = rho *g*( dz2 - v_B **2/2/ g);\n",
+      "p_C = rho *g*( dz3 - v_C **2/2/ g);\n",
+      "\n",
+      "#result\n",
+      "print\" Velocity at pt . A(m/s) =\",round(v_A,2);\n",
+      "print\" Velocity at pt . B(m/s) =\",round(v_B,2);\n",
+      "print\" Velocity at pt . C(m/s) =\",round(v_C,2);\n",
+      "print\" Pressure at pt . A(kN/m^2) =\",round(p_A)\n",
+      "print\" Pressure at pt . B (kN/m^2)=\",round(p_B)\n",
+      "print\" Pressure at pt . C (kN/m^2)=\",round(p_C)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Velocity at pt . A(m/s) = 1.98\n Velocity at pt . B(m/s) = 1.98\n Velocity at pt . C(m/s) = 1.98\n Pressure at pt . A(kN/m^2) = 17640.0\n Pressure at pt . B (kN/m^2)= -1960.0\n Pressure at pt . C (kN/m^2)= -16660.0\n"
+       "text": [
+        " Velocity at pt . A(m/s) = 1.98\n",
+        " Velocity at pt . B(m/s) = 1.98\n",
+        " Velocity at pt . C(m/s) = 1.98\n",
+        " Pressure at pt . A(kN/m^2) = 17640.0\n",
+        " Pressure at pt . B (kN/m^2)= -1960.0\n",
+        " Pressure at pt . C (kN/m^2)= -16660.0\n"
+       ]
       }
      ],
      "prompt_number": 2
@@ -101,19 +232,44 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 2.5,Page 44"
+     "source": [
+      "Example 2.5,Page 44"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding pressure drop\nfrom math import pi\n\n#variable decleration\nQ =10.0; # m^3/ hr\nd1 =0.05; #m\nd2 =0.1; #m\nrho =1000.0; #kg /m^3\ng=9.81;\n\n\n#calculation\na1=pi*d1 **2/4;\na2=pi*d2 **2/4;\nv1=Q /3600/ a1;\nv2 =( d1/d2) **2* v1;\nPD=rho*0.057*g;\n\n#results\nprint\" Presure drop (N/m^2)=\",round(PD,3)\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#variable decleration\n",
+      "Q =10.0; # m^3/ hr\n",
+      "d1 =0.05; #m\n",
+      "d2 =0.1; #m\n",
+      "rho =1000.0; #kg /m^3\n",
+      "g=9.81;\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "a1=pi*d1 **2/4;\n",
+      "a2=pi*d2 **2/4;\n",
+      "v1=Q /3600/ a1;\n",
+      "v2 =( d1/d2) **2* v1;\n",
+      "PD=rho*0.057*g;\n",
+      "\n",
+      "#results\n",
+      "print\" Presure drop (N/m^2)=\",round(PD,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Presure drop (N/m^2)= 375.26364312\n"
+       "text": [
+        " Presure drop (N/m^2)= 375.26364312\n"
+       ]
       }
      ],
      "prompt_number": 7
@@ -122,19 +278,48 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 2.7,Page 48"
+     "source": [
+      "Example 2.7,Page 48"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding force required\n\n#variable decleration\nQ =100; #m^3/ hr\nd1 =0.2; #m\nd2 =0.15; #m\np1 =80*10**3; #N/m^2\nrho =1000; # kg /m^3\ng =9.8; #m/ s ^2\npi=3.14;\n\n#calculation\na1=pi*d1 **2/4;\na2=pi*d2 **2/4;\nv1=Q /3600/ a1;\nv2=Q /3600/ a2;\nH_L =0.2* v2 **2/2/ g;\np2=p1+ rho /2*( v1 **2- v2 **2) -rho*g* H_L ;\nF_u =p1*a1; # Upstream f o r c e\nF_d =p2*a2; # Downstream f o r c e\nF_x = rho *Q /3600*( v2 -v1)-F_u +F_d;\n\n#results\nprint\" Force required (N) =\",round(F_x)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "Q =100; #m^3/ hr\n",
+      "d1 =0.2; #m\n",
+      "d2 =0.15; #m\n",
+      "p1 =80*10**3; #N/m^2\n",
+      "rho =1000; # kg /m^3\n",
+      "g =9.8; #m/ s ^2\n",
+      "pi=3.14;\n",
+      "\n",
+      "#calculation\n",
+      "a1=pi*d1 **2/4;\n",
+      "a2=pi*d2 **2/4;\n",
+      "v1=Q /3600/ a1;\n",
+      "v2=Q /3600/ a2;\n",
+      "H_L =0.2* v2 **2/2/ g;\n",
+      "p2=p1+ rho /2*( v1 **2- v2 **2) -rho*g* H_L ;\n",
+      "F_u =p1*a1; # Upstream f o r c e\n",
+      "F_d =p2*a2; # Downstream f o r c e\n",
+      "F_x = rho *Q /3600*( v2 -v1)-F_u +F_d;\n",
+      "\n",
+      "#results\n",
+      "print\" Force required (N) =\",round(F_x)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Force required (N) = -1099.0\n"
+       "text": [
+        " Force required (N) = -1099.0\n"
+       ]
       }
      ],
      "prompt_number": 17
@@ -143,19 +328,39 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 2.9,Page 53"
+     "source": [
+      "Example 2.9,Page 53"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding depression\n\n#variable decleration\nN =60; #rpm\nr2 =0.25; #m\ng =9.8; #m/ s ^2\npi=3.14;\n\nw =2* pi*N /60;\ndz_12 =(w*r2) **2/2/ g; # dz 1 2=z2-z1\nc=w*r2 **2;\ndz_23 =c **2/2/ g/r2 **2; # dz 2 3=z3-z2\ndz_13 = dz_23 + dz_12 ;\n\n#results\nprint\" Total depression(m) =\",round(dz_13,3)\n",
+     "input": [
+      "\n",
+      "#variable decleration\n",
+      "N =60; #rpm\n",
+      "r2 =0.25; #m\n",
+      "g =9.8; #m/ s ^2\n",
+      "pi=3.14;\n",
+      "\n",
+      "w =2* pi*N /60;\n",
+      "dz_12 =(w*r2) **2/2/ g; # dz 1 2=z2-z1\n",
+      "c=w*r2 **2;\n",
+      "dz_23 =c **2/2/ g/r2 **2; # dz 2 3=z3-z2\n",
+      "dz_13 = dz_23 + dz_12 ;\n",
+      "\n",
+      "#results\n",
+      "print\" Total depression(m) =\",round(dz_13,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Total depression(m) = 0.252\n"
+       "text": [
+        " Total depression(m) = 0.252\n"
+       ]
       }
      ],
      "prompt_number": 3
@@ -163,7 +368,7 @@
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "",
+     "input": [],
      "language": "python",
      "metadata": {},
      "outputs": []
diff --git a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_3_1.ipynb b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_3_1.ipynb
index 538dce80..047caec2 100644
--- a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_3_1.ipynb
+++ b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_3_1.ipynb
@@ -1,6 +1,7 @@
 {
  "metadata": {
-  "name": "Chapter 3"
+  "name": "",
+  "signature": "sha256:f2eaea1b025b2ddec764d2fab1514700e255b4e2c5860b0d8b1ba9dafae0ccc9"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -11,25 +12,42 @@
      "cell_type": "heading",
      "level": 1,
      "metadata": {},
-     "source": "Chapter 3,Laminar Flow and Lubrication"
+     "source": [
+      "Chapter 3,Laminar Flow and Lubrication"
+     ]
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.2,Page 63"
+     "source": [
+      "Example 3.2,Page 63"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding fartherest distance\n\n#variable decleration\nRe =2000.0;\nd =0.008; #m\n\n#calculation\nL1 =0.058* Re*d;\n\n#result\nprint\"The furtherest distance the fluid can flow into the 8 mm inside diameter pipe (m)=\",round(L1,3);\n",
+     "input": [
+      "\n",
+      "#variable decleration\n",
+      "Re =2000.0;\n",
+      "d =0.008; #m\n",
+      "\n",
+      "#calculation\n",
+      "L1 =0.058* Re*d;\n",
+      "\n",
+      "#result\n",
+      "print\"The furtherest distance the fluid can flow into the 8 mm inside diameter pipe (m)=\",round(L1,3);\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The furtherest distance the fluid can flow into the 8 mm inside diameter pipe (m)= 0.928\n"
+       "text": [
+        "The furtherest distance the fluid can flow into the 8 mm inside diameter pipe (m)= 0.928\n"
+       ]
       }
      ],
      "prompt_number": 1
@@ -38,19 +56,46 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.4,Page 67"
+     "source": [
+      "Example 3.4,Page 67"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding flow rate\nfrom math import pi\n\n#variable decleration\ndel_p =90*10**3; # N/m^2\nd =0.126; # m\nR =0.126/2; # m\nu =1.2;\nL =60; # m\nRho =1260;\n\n\n#calculation\nQ= pi * del_p * R**4 / (8*u*L);\nRe =4* Rho *Q/(u*pi*d);\n\n#result\nprint \"The glycerol delivery rate is (m^3/s) \",round(Q,4);\nprint\"The Reynolds number is \",round(Re,3);\nprint\"As Re is below 2000 , so, laminar flow\"",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#variable decleration\n",
+      "del_p =90*10**3; # N/m^2\n",
+      "d =0.126; # m\n",
+      "R =0.126/2; # m\n",
+      "u =1.2;\n",
+      "L =60; # m\n",
+      "Rho =1260;\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "Q= pi * del_p * R**4 / (8*u*L);\n",
+      "Re =4* Rho *Q/(u*pi*d);\n",
+      "\n",
+      "#result\n",
+      "print \"The glycerol delivery rate is (m^3/s) \",round(Q,4);\n",
+      "print\"The Reynolds number is \",round(Re,3);\n",
+      "print\"As Re is below 2000 , so, laminar flow\""
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The glycerol delivery rate is (m^3/s)  0.0077\nThe Reynolds number is  82.047\nAs Re is below 2000 , so, laminar flow\n"
+       "text": [
+        "The glycerol delivery rate is (m^3/s)  0.0077\n",
+        "The Reynolds number is  82.047\n",
+        "As Re is below 2000 , so, laminar flow\n"
+       ]
       }
      ],
      "prompt_number": 2
@@ -59,19 +104,42 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.5,Page 69"
+     "source": [
+      "Example 3.5,Page 69"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding diameter\nfrom math import pi\n\n#variable decleration\nu =0.015; #Ns/m^2\nQ =0.004/60; #m^3/ s\ndp =100;\nrho =1100.0; #kg /m^3\n\n\n#calculation\nR =(8* u*Q/( pi *dp)) **(0.25) ;\nRe =(4* rho *Q/( pi*u *(2* R)));\n\n#result\nprint\"Diameter of the pipe(m) =\",round(R,3);\nprint\"Reynolds number =\",round(Re,2)\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#variable decleration\n",
+      "u =0.015; #Ns/m^2\n",
+      "Q =0.004/60; #m^3/ s\n",
+      "dp =100;\n",
+      "rho =1100.0; #kg /m^3\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "R =(8* u*Q/( pi *dp)) **(0.25) ;\n",
+      "Re =(4* rho *Q/( pi*u *(2* R)));\n",
+      "\n",
+      "#result\n",
+      "print\"Diameter of the pipe(m) =\",round(R,3);\n",
+      "print\"Reynolds number =\",round(Re,2)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Diameter of the pipe(m) = 0.013\nReynolds number = 246.38\n"
+       "text": [
+        "Diameter of the pipe(m) = 0.013\n",
+        "Reynolds number = 246.38\n"
+       ]
       }
      ],
      "prompt_number": 6
@@ -80,19 +148,37 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.6,Page 71"
+     "source": [
+      "Example 3.6,Page 71"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding pressure difference\nfrom math import pi\n\n#variable decleration\nmu =0.03; #Ns/m^2\nQ =10.0**( -7) ; #m^3/ s\nu=(7.0);\n\n#calculation\ndp= 8*mu*Q*u/3/pi/0.005**4;\n\n#result\nprint\" pressure difference (N/m^2)=\",round(dp,2)\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#variable decleration\n",
+      "mu =0.03; #Ns/m^2\n",
+      "Q =10.0**( -7) ; #m^3/ s\n",
+      "u=(7.0);\n",
+      "\n",
+      "#calculation\n",
+      "dp= 8*mu*Q*u/3/pi/0.005**4;\n",
+      "\n",
+      "#result\n",
+      "print\" pressure difference (N/m^2)=\",round(dp,2)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " pressure difference (N/m^2)= 28.52\n"
+       "text": [
+        " pressure difference (N/m^2)= 28.52\n"
+       ]
       }
      ],
      "prompt_number": 5
@@ -101,19 +187,43 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.8,Page 75"
+     "source": [
+      "Example 3.8,Page 75"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding pressure difference\nfrom math import pi\n\n#variable decleration\nu =0.1; #Ns/m^2\nd =0.1; #m\nR =0.05; # m\nRho =900; #kg /m^3\nv_max =2; # m/ s\nv= v_max /2; # m/ s\n\n#calculation\nTw = 2*u* v_max /R;\ndel_p =4* u* v_max /R **2;\n\n#result\nprint \"At the pipe wa l l ( r =R) , shear stress (N/m^2)= \",Tw;\nprint\" pressure drop per metre lengthh of pipe is (N/m^2)\",del_p;\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#variable decleration\n",
+      "u =0.1; #Ns/m^2\n",
+      "d =0.1; #m\n",
+      "R =0.05; # m\n",
+      "Rho =900; #kg /m^3\n",
+      "v_max =2; # m/ s\n",
+      "v= v_max /2; # m/ s\n",
+      "\n",
+      "#calculation\n",
+      "Tw = 2*u* v_max /R;\n",
+      "del_p =4* u* v_max /R **2;\n",
+      "\n",
+      "#result\n",
+      "print \"At the pipe wa l l ( r =R) , shear stress (N/m^2)= \",Tw;\n",
+      "print\" pressure drop per metre lengthh of pipe is (N/m^2)\",del_p;\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "At the pipe wa l l ( r =R) , shear stress (N/m^2)=  8.0\n pressure drop per metre lengthh of pipe is (N/m^2) 320.0\n"
+       "text": [
+        "At the pipe wa l l ( r =R) , shear stress (N/m^2)=  8.0\n",
+        " pressure drop per metre lengthh of pipe is (N/m^2) 320.0\n"
+       ]
       }
      ],
      "prompt_number": 9
@@ -122,19 +232,36 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.9,Page 77"
+     "source": [
+      "Example 3.9,Page 77"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding inside diameter\n\n#variable decleration\nimport math\nu =0.032; #Ns/m^2\nRe =2000.0; #maximum v a l u e\nRho =854.0;\ndel_p =150.0; #N/m^2\n\n#calculation\nd =(32* u **2* Re /( Rho* del_p ))**(0.33)  ;\n\nprint \"The maximum inside diameter is found to be (m) \",round(d,3);\n",
+     "input": [
+      "\n",
+      "#variable decleration\n",
+      "import math\n",
+      "u =0.032; #Ns/m^2\n",
+      "Re =2000.0; #maximum v a l u e\n",
+      "Rho =854.0;\n",
+      "del_p =150.0; #N/m^2\n",
+      "\n",
+      "#calculation\n",
+      "d =(32* u **2* Re /( Rho* del_p ))**(0.33)  ;\n",
+      "\n",
+      "print \"The maximum inside diameter is found to be (m) \",round(d,3);\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The maximum inside diameter is found to be (m)  0.082\n"
+       "text": [
+        "The maximum inside diameter is found to be (m)  0.082\n"
+       ]
       }
      ],
      "prompt_number": 7
@@ -143,19 +270,40 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.10,Page 79"
+     "source": [
+      "Example 3.10,Page 79"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding pressure drop\n\nfrom math import pi\n#variable decleration\nrho =1000; #kg /m^3\nu =0.1; #Ns/m^2\ng =9.81; #m/ s ^2\nL =10; #m\nH =2; #m\nQ =14.0/3600; #m^3/ s\nd =0.05; #m\n\n#calculation\ndp=rho*g*(L+H) - (128* Q*u*L/pi /d**4) ;\n\n#result\nprint\" Pressure drop across the valve (N/m^2)=\",round(dp,2)\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "#variable decleration\n",
+      "rho =1000; #kg /m^3\n",
+      "u =0.1; #Ns/m^2\n",
+      "g =9.81; #m/ s ^2\n",
+      "L =10; #m\n",
+      "H =2; #m\n",
+      "Q =14.0/3600; #m^3/ s\n",
+      "d =0.05; #m\n",
+      "\n",
+      "#calculation\n",
+      "dp=rho*g*(L+H) - (128* Q*u*L/pi /d**4) ;\n",
+      "\n",
+      "#result\n",
+      "print\" Pressure drop across the valve (N/m^2)=\",round(dp,2)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Pressure drop across the valve (N/m^2)= 92368.39\n"
+       "text": [
+        " Pressure drop across the valve (N/m^2)= 92368.39\n"
+       ]
       }
      ],
      "prompt_number": 4
@@ -164,19 +312,46 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.12,Page 83"
+     "source": [
+      "Example 3.12,Page 83"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding inclination and viscosity\n\n#variable decleration\nfrom math import pi\nimport math\nQ =3*10**( -6) ; #m^3/ s\nu =0.001; #Ns/m^2\nW =1;\nrho =1000; #kg /m^3\ng =9.81; #m/ s ^2\nd =1.016*10**( -4) ; #m\nd1 =1.25*10**( -4) ; # m\n\n\n#calculation\ntheta = math.asin (3*Q*u/W/rho/g/d **3) ;\nu1=W*rho *g* math.sin ( theta )*( d1 **3) /(3* Q);\n\n#result\nprint\"Exact angle of inclination (radians) =\",round(theta,3)\nprint\" Viscosity of the second liquid (Ns/m^2)=\",round(u1,4)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "from math import pi\n",
+      "import math\n",
+      "Q =3*10**( -6) ; #m^3/ s\n",
+      "u =0.001; #Ns/m^2\n",
+      "W =1;\n",
+      "rho =1000; #kg /m^3\n",
+      "g =9.81; #m/ s ^2\n",
+      "d =1.016*10**( -4) ; #m\n",
+      "d1 =1.25*10**( -4) ; # m\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "theta = math.asin (3*Q*u/W/rho/g/d **3) ;\n",
+      "u1=W*rho *g* math.sin ( theta )*( d1 **3) /(3* Q);\n",
+      "\n",
+      "#result\n",
+      "print\"Exact angle of inclination (radians) =\",round(theta,3)\n",
+      "print\" Viscosity of the second liquid (Ns/m^2)=\",round(u1,4)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Exact angle of inclination (radians) = 1.065\n Viscosity of the second liquid (Ns/m^2)= 0.0019\n"
+       "text": [
+        "Exact angle of inclination (radians) = 1.065\n",
+        " Viscosity of the second liquid (Ns/m^2)= 0.0019\n"
+       ]
       }
      ],
      "prompt_number": 3
@@ -185,19 +360,38 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.17,Page 93"
+     "source": [
+      "Example 3.17,Page 93"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding shear stress\n\n#variable decleration\nu =1.5; #Ns/m^2\nv =0.5; # m/ s\nH =0.02/2; # m\n\n#calculation\nt=-u*3*v/H;\n\n#result\nprint \"The shear stress (N/m^2)=\",t\nprint\"acting on opp. direction of flow\" \n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "u =1.5; #Ns/m^2\n",
+      "v =0.5; # m/ s\n",
+      "H =0.02/2; # m\n",
+      "\n",
+      "#calculation\n",
+      "t=-u*3*v/H;\n",
+      "\n",
+      "#result\n",
+      "print \"The shear stress (N/m^2)=\",t\n",
+      "print\"acting on opp. direction of flow\" \n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The shear stress (N/m^2)= -225.0\nacting on opp. direction of flow\n"
+       "text": [
+        "The shear stress (N/m^2)= -225.0\n",
+        "acting on opp. direction of flow\n"
+       ]
       }
      ],
      "prompt_number": 34
@@ -206,19 +400,48 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.18,Page 95"
+     "source": [
+      "Example 3.18,Page 95"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding torque and viscosity\nfrom math import pi\nfrom scipy import integrate\n\n#variable decleration\nN =600.0/60; #r e v o l u t i o n s pe r s e c\nr =0.025; #m\nt =400; # N/m^2\nl =0.002; # m\ntime=2.0/1000; #s\n\n#calculation\nw =2* pi*N;\nu=t*l/w/r;\ndef integrand (R,w,u,time):\n    return 2*pi*u*w/time*R**3\n\n#a=lambda r:2*pi*0.509*62.8/2*r**3*1000;\nT=integrate.quad(integrand, 0,0.025,args=(w,u,time));\n\n#result\nprint\" Viscosity,(Ns/m^2)=\",round(u,3)\nprint\"Torque(Nm) =\",round(T[0],4)\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "from scipy import integrate\n",
+      "\n",
+      "#variable decleration\n",
+      "N =600.0/60; #r e v o l u t i o n s pe r s e c\n",
+      "r =0.025; #m\n",
+      "t =400; # N/m^2\n",
+      "l =0.002; # m\n",
+      "time=2.0/1000; #s\n",
+      "\n",
+      "#calculation\n",
+      "w =2* pi*N;\n",
+      "u=t*l/w/r;\n",
+      "def integrand (R,w,u,time):\n",
+      "    return 2*pi*u*w/time*R**3\n",
+      "\n",
+      "#a=lambda r:2*pi*0.509*62.8/2*r**3*1000;\n",
+      "T=integrate.quad(integrand, 0,0.025,args=(w,u,time));\n",
+      "\n",
+      "#result\n",
+      "print\" Viscosity,(Ns/m^2)=\",round(u,3)\n",
+      "print\"Torque(Nm) =\",round(T[0],4)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Viscosity,(Ns/m^2)= 0.509\nTorque(Nm) = 0.0098\n"
+       "text": [
+        " Viscosity,(Ns/m^2)= 0.509\n",
+        "Torque(Nm) = 0.0098\n"
+       ]
       }
      ],
      "prompt_number": 18
@@ -227,19 +450,47 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.19,Page 97"
+     "source": [
+      "Example 3.19,Page 97"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding torque and frictional resistance\nfrom math import pi\n\n#variable decleration\nu =0.153; #Ns/m^2\nr =0.05; # m\nN =30; # rps\nt =2.0/10**5; # s\nL =0.2; #m\n\n\n#calculation\ntau =u *(2* pi *N*r/t);\nF= tau *2* pi *r*L;\nT=F*r;\nw =2* pi*N;\nP=T*w;\n\n#result\nprint \"The torque on the bearing is found to be (Nm)= \",round(T,3);\nprint \" and the power required to overcome the resistance is (W)=\",round(P,3);\n\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#variable decleration\n",
+      "u =0.153; #Ns/m^2\n",
+      "r =0.05; # m\n",
+      "N =30; # rps\n",
+      "t =2.0/10**5; # s\n",
+      "L =0.2; #m\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "tau =u *(2* pi *N*r/t);\n",
+      "F= tau *2* pi *r*L;\n",
+      "T=F*r;\n",
+      "w =2* pi*N;\n",
+      "P=T*w;\n",
+      "\n",
+      "#result\n",
+      "print \"The torque on the bearing is found to be (Nm)= \",round(T,3);\n",
+      "print \" and the power required to overcome the resistance is (W)=\",round(P,3);\n",
+      "\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The torque on the bearing is found to be (Nm)=  226.507\n and the power required to overcome the resistance is (W)= 42695.643\n"
+       "text": [
+        "The torque on the bearing is found to be (Nm)=  226.507\n",
+        " and the power required to overcome the resistance is (W)= 42695.643\n"
+       ]
       }
      ],
      "prompt_number": 19
@@ -248,19 +499,39 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 3.20,Page 99"
+     "source": [
+      "Example 3.20,Page 99"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding viscosity of oil\n\n#variable decleration\nt =0.0005;#s\nP =22; \nr =300.0/60; \nR_1 =0.1; \nR_2 =0.0625; \npi=3.14;\n\n#calculation\nw =2* pi*r;\nu =2* t*P/( pi *w **2*(( R_1 )**4 -( R_2) **4) );\n\n#result\nprint\"The viscosity of the oil is (Ns/m^2)= \",round(u,4);\n",
+     "input": [
+      "\n",
+      "#variable decleration\n",
+      "t =0.0005;#s\n",
+      "P =22; \n",
+      "r =300.0/60; \n",
+      "R_1 =0.1; \n",
+      "R_2 =0.0625; \n",
+      "pi=3.14;\n",
+      "\n",
+      "#calculation\n",
+      "w =2* pi*r;\n",
+      "u =2* t*P/( pi *w **2*(( R_1 )**4 -( R_2) **4) );\n",
+      "\n",
+      "#result\n",
+      "print\"The viscosity of the oil is (Ns/m^2)= \",round(u,4);\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The viscosity of the oil is (Ns/m^2)=  0.0839\n"
+       "text": [
+        "The viscosity of the oil is (Ns/m^2)=  0.0839\n"
+       ]
       }
      ],
      "prompt_number": 20
@@ -268,7 +539,7 @@
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "",
+     "input": [],
      "language": "python",
      "metadata": {},
      "outputs": []
diff --git a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_4_1.ipynb b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_4_1.ipynb
index 9b4bb41f..7c5078a5 100644
--- a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_4_1.ipynb
+++ b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_4_1.ipynb
@@ -1,6 +1,7 @@
 {
  "metadata": {
-  "name": "Chapter 4"
+  "name": "",
+  "signature": "sha256:3e9255609d7264d7167f865de1dbf27ba92b4a62cb35c2df65b3f59b0ee5b06e"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -11,25 +12,50 @@
      "cell_type": "heading",
      "level": 1,
      "metadata": {},
-     "source": "Chapter 4, Dimensional Analysis"
+     "source": [
+      "Chapter 4, Dimensional Analysis"
+     ]
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 4.5,Page 108"
+     "source": [
+      "Example 4.5,Page 108"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding pressure drop\n\n#Variable Declaration\nRho_full =800.0; # kg /m^3\nv_full =1.8; # m/ s\nu_full =9.0/10000 ; # Nm/ s ^2\nRho_model =1000.0; # kg /m^3\nu_model =1.0/1000 ; # Ns/m^2\nd_full = 2.0;\nd_model =1.0;\ndel_p_fmodel =4000.0; # N/m^2\n\n#Calculation\nv_model = (( Rho_full * v_full / u_full )/( Rho_model /u_model ))*( d_full / d_model );\n#del_p_f = del_p_fmodel * Rho_full *( v_full )**2/ Rho_model /(v_model )**2;\n\n#result\nprint \"The velocity in the full scale pipe is expected to be(m/s)\",round(v_model,2);\n",
+     "input": [
+      "\n",
+      "\n",
+      "#Variable Declaration\n",
+      "Rho_full =800.0; # kg /m^3\n",
+      "v_full =1.8; # m/ s\n",
+      "u_full =9.0/10000 ; # Nm/ s ^2\n",
+      "Rho_model =1000.0; # kg /m^3\n",
+      "u_model =1.0/1000 ; # Ns/m^2\n",
+      "d_full = 2.0;\n",
+      "d_model =1.0;\n",
+      "del_p_fmodel =4000.0; # N/m^2\n",
+      "\n",
+      "#Calculation\n",
+      "v_model = (( Rho_full * v_full / u_full )/( Rho_model /u_model ))*( d_full / d_model );\n",
+      "#del_p_f = del_p_fmodel * Rho_full *( v_full )**2/ Rho_model /(v_model )**2;\n",
+      "\n",
+      "#result\n",
+      "print \"The velocity in the full scale pipe is expected to be(m/s)\",round(v_model,2);\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The velocity in the full scale pipe is expected to be(m/s) 3.2\n"
+       "text": [
+        "The velocity in the full scale pipe is expected to be(m/s) 3.2\n"
+       ]
       }
      ],
      "prompt_number": 1
@@ -37,7 +63,7 @@
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "",
+     "input": [],
      "language": "python",
      "metadata": {},
      "outputs": []
diff --git a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_5_1.ipynb b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_5_1.ipynb
index ef67f9f0..19f3f7be 100644
--- a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_5_1.ipynb
+++ b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_5_1.ipynb
@@ -1,6 +1,7 @@
 {
  "metadata": {
-  "name": "Chapter 5"
+  "name": "",
+  "signature": "sha256:3560754f85c552adcb8e17db6d394e5a7c27026cea8efa4cac19d77a4813e8fe"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -27,7 +28,6 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Finding Velocity\n",
       "\n",
       "#Variable Iniialisation\n",
       "import math\n",
@@ -68,7 +68,6 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Finding average rate of flow and velocity\n",
       "\n",
       "%pylab inline\n",
       "#variable initialisation\n",
@@ -109,6 +108,7 @@
        ]
       },
       {
+       "metadata": {},
        "output_type": "display_data",
        "png": 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R7Tk9ce8sco9mgGPHjpGcnExOTo7rsauuusqtE4tc1Guvwb//DV9+aZatEAk0ebOafWwN\npCKTwtNPP82MGTNo0qQJZc5ruq9atcqrgUmQysmBhx82ySAxES67zOmIRLwjNNQkg8OHfWrSZZFJ\nYd68eezevVvrHYn3/fEHDBtmtsxcuxZq1nQ6IhHvCQk5V1fwoaRQZNWuZcuWHDt2zI5YJJilpZm6\nQfnysHy5EoIEBx8sNhfZUnjyySeJjo4mMjKSin/uaRsSEsKSJUu8HpwEiaQkM9z0ppvglVc0wkiC\nhz8mhf/5n/9h3LhxREZGumoKIT5UFBE/t3mzmYj2+ONm5UiRYBIRAX9uZuYrihyS2r59ezZs2FCq\nFx81ahRLly4lNDSU7du3A5CWlsYtt9zCvn37CAsLY/78+dSoUSN/UBqSGhzi4kwN4Z134L/+y+lo\nROyXkmJG1x0+7JGXs2Xp7JiYGJ544gnWrVvH5s2bXV/FMXLkSOLi4vI9NmHCBHr06MGuXbvo1q0b\nEyZMKF3k4t/eew9GjIBFi5QQJHjVq2c2iDpyxOlIXIpsKcTGxhbYXVTcIalJSUn07dvX1VIIDw8n\nISGBunXrkpqaSmxsLDt37swflFoKgcuy4PnnzT4Iy5bBlVc6HZGIs669Fl5+Gbp0cfulbJm8Fh8f\n79YJ/urQoUPU/XP4Vd26dTl06FCBzxs/frzr+9jYWGJjYz0ahzjg7Fm4+26zMuQ335hPSSLBLq/Y\nXIqkEB8f7/F7dKEthRkzZjB06FDKlSs4b2RlZfHRRx8xcuTIi57gry2FmjVr5hviWqtWLdLS0vIH\npZZC4MnMhMGDzffz54PW1BIxXnsN9uyByZPdfimvthQyMzNp37494eHhtGvXjvr162NZFqmpqWzc\nuJGdO3cyZsyYEp8wr9uoXr16pKSkEBoa6tYfIH4gNRX69IHoaPi//zNzEUTEaNkSPv/c6ShcCi00\n33fffWzevJl7772Xs2fPsnbtWhITE8nOznb97p577inxCfv168fMmTMBmDlzJgMGDCh99OL7fv7Z\nLHvdv7/ZGEcJQSQ/H5ur4NVVUocMGUJCQgJHjx6lbt26PP/88/Tv35/BgweTnJysIamBLjERBg0y\nRbQiuhlFgpZlQfXqZhJnrVpuvZQn7p1aOlu845NP4B//gFmz4IYbnI5GxLddcw1MmmRGIrnBlnkK\nIiX2xhtmdvLy5UoIIsWRt4y2D7hoUsjNzWX+/Pl2xSL+LjcXHn3UFJMTE0F7bogUjw/t13zRpFCm\nTBleeeUVu2IRf3bmDNx2G6xbZxJCWJjTEYn4Dx8qNhfZfdSjRw8mTpzI/v37SUtLc32JuBw7ZrqJ\nsrPhq6/cLpaJBB0fSgpFFprDwsIKXOZi79693gtKhWb/kZxslr3u2dMUyrTstUjJ5ebCJZfAwYNm\nJFIp2bLMRVJSklsnkAC2davZA+Hhh+HBB52ORsR/lSkDLVqY1kLHjs6GUtQTnn76abKzs10/Z2Rk\nFLm0hQSBL780rYPXXlNCEPEEHyk2F5kUsrOz6dChA99//z0rVqygQ4cOXKVRJcHtww9h6FAzF+G/\n/9vpaEQCg4/UFYrsPnr55Zfp1q0b11xzDTVr1iQhIYHmzZvbEZv4Gssys5OnToVVq8ybWEQ8IyLC\n/LtyWJGF5oSEBO6++26GDh3K9u3bSU9P591336VBgwbeC0qFZt+TnQ333QfffQdLl8KllzodkUhg\n2bMHrrsO9u0r9UvYUmh+9NFH+fjjj4n481Php59+yvXXX8/PP//s1onFj5w8CbfeCllZsHo1VKvm\ndEQigScsDI4ehRMnHP03VmhLIW9565ycHMqWLZvvd0ePHqV27dreC0otBd9x+LAZYdSypek20iqn\nIt7Ttq1ZEaBDh1Id7tW1j6KioujevTszZswgPT093++8mRDEh/zyi1n2ulcvs6eyEoKId/lAsbnQ\npHDw4EEeeeQR1qxZw5VXXkn//v2ZO3cuf/zxh53xiVO+/dZsD/j442ZP5QImMIqIh/lyUihXrhy9\nevVixowZJCcnM3LkSBYvXkzjxo257bbb7IxR7LZ4MfTtC+++C6XYXU9ESsmXk8L5KlasSEREBC1a\ntKBatWr89NNP3o5LnPL223D33bBsmdlCU0Ts4wNLaF909FFycjJz585l7ty5ZGZmMmTIED777DPC\nw8Ptik/skpsLTz4JCxfC2rXQpInTEYkEnyZN4NAhM+KvShVHQig0KXTq1IkDBw4wePBgpk2bRtu2\nbe2MS+x05gyMGgV795plrzWQQMQZZcvCFVfAzp1mJJIDCk0KEyZMoHPnzpTRqpeBLSMDBg40KzOu\nXAmVKzsdkUhwy6sr+FpSWLduHV26dGHs2LEX/C4kJIQ33njDq4GJDQ4cgBtvNKOMXn/dfEoREWc5\nXGwuNCnkzWAuqNuooP0VxM9s324KyWPHwiOPaMipiK+IiIAZMxw7fZFrH+XJyMigTJkyVLNh+rVm\nNHvZ11+bZStefx2GDHE6GhE5388/mw9sv/5a4kO9OqM5z4YNG2jVqhWtW7cmMjKSqKgoNm7c6NZJ\nxUGzZ5uEMG+eEoKIL2ra1OzA5tBE4SKTwqhRo3j77bfZt28f+/bt46233mLUqFF2xCaeZFnwyisw\nbpxpKVx3ndMRiUhBypWDZs1Mi8EBRSaFcuXKERMT4/q5c+fOlCtX5OKq4ktyckzt4KOPYN06iIx0\nOiIRuRgHi81F3t27du3KXXfdxZA/uxrmzZtH165d2bx5M4B2YfN1p07BbbeZ5XjXrHFrU3ARsYmD\nM5uLLDTHxsZedLTRKi/sFKRCs4ccPWrWMGra1KxyWqGC0xGJSHF8/LFp2S9cWKLDPHHvLPboIzsp\nKXjA7t3QuzfcfDO8+KKGnIr4kx07zKTSEtYVvD766KeffmLlypVkZmbmezwuLs6tk4qXbdgAMTHw\n4IPw0ktKCCL+plkzSE42S9DYrNCk8MYbbzBgwAAmT55My5YtWbRoket3TzzxhC3BSSl8/rmZpTxl\nilntVET8T4UK0Lgx7Npl+6kLLTRPnTqVTZs2UbVqVZKSkrj55ptJSkrigQcesDM+KYmpU+HZZ01i\nuPpqp6MREXfkFZtbtbL1tIUmBcuyqFq1KgBhYWHEx8czaNAg9u3bp/5+X2NZ8MwzMGcOrF4NzZs7\nHZGIuKtlS0eGpRbafRQaGsrWrVtdP1etWpXPP/+c33//nW3bttkSnBRDVhaMGAErVsA33yghiAQK\nh+YqFDr6aP/+/ZQvX5569erle9yyLBITE+ncubP3gtLoo+I5ftyMLqpUybQSHNqUQ0S8YPt2uOWW\nEiUGDUkNZr/9ZgrKHTvC5MlmaryIBI4zZ6BGDbPnSTHnGNmyIJ74oB07oFMns7Dd228rIYgEoooV\n4bLL4JdfbD2tkoK/Wb3aLGb3wgtmcTvNQRAJXA4Um/UR05/Mnw/33WeWv+7e3eloRMTbHCg2q6Xg\nDywL/v1vePhh+PJLJQSRYKGkIBfIyTHLVbz3nhlyGhXldEQiYhcHkoJGH/my06dh6FD4/XezWmKN\nGk5HJCJ2+uMPqFXLDD8vX77Ip2v0USA7dQr69zeF5Lg4JQSRYFS5MjRsaFY9tomSgi/KzDQbd4eG\nmklpFSs6HZGIOMXmLiQlBV+TkQE9e5qlc2fM0BwEkWCnpBDE0tLMyKKrroJ33oGyZZ2OSEScpqQQ\npI4cgeuvh9hYs2xFGf2vERFs369Zo498QUoKdOtmFrd77jnNUhaRc06dgtq1zQikIrqTPXHvdKzD\nOiwsjEsuuYSyZctSvnx51q9f71Qoztq/3ySE4cPhqaecjkZEfM3f/gb16sHevbYsje9YUggJCSE+\nPp5atWo5FYLz9u41XUZjx8JDDzkdjYj4qry6gg1JwdGO66DqIvqrXbuga1d49FElBBG5OBuLzY62\nFLp3707ZsmW56667GDNmTL7fjx8/3vV9bGwssbGx9gboTT/+aIadvvACjBrldDQi4usiIuCrry54\nOD4+nvj4eI+eyrFCc0pKCvXr1+fIkSP06NGDyZMnExMTY4IK5ELz1q3QuzdMnAi33+50NCLiDzZs\ngLvugs2bL/o0v17mon79+gDUqVOHgQMHBkehecMGuOEGePNNJQQRKb7wcPj5Z7NAppc5khROnTrF\niRMnADh58iQrVqygVatWToRin8REs3TFu+/CoEFORyMi/qRaNTMsdd8+r5/KkZrCoUOHGDhwIADZ\n2dncfvvt9OzZ04lQ7LFqFQweDLNmmZaCiEhJ5RWbmzTx6mk0ec3bli83y18vWGBmK4uIlMbDD5tF\nMh9/vNCn+HVNISgsWQLDhsHixUoIIuIem/ZrVlLwlgUL4M474YsvoFMnp6MREX9n01wFdR95w6xZ\nZlJaXJy2zxQRz8jIgAYNzBpIhSyYqe4jX/Tuu6bPb+VKJQQR8Zzq1c0OjPv3e/U0Sgqe9NZb8Pzz\nEB9vmnoiIp5kwzLaSgqeMmmS+UpIsGXRKhEJQjYUm5UUPOFf/zI7pa1eDY0bOx2NiAQqG4rNSgru\nsCz45z9hzhzTQmjY0OmIRCSQ2ZAUNPqotCwLHnkEvv4aVqyAOnWcjkhEAl1aGoSFmZFIBezQqNFH\nTsnNhfvugzVrzCgjJQQRsUOtWlClChw44LVTKCmUVE6OmZS2dSt8+aX5nyQiYhcvF5uVFEoiO9vs\npbxnj1nTqHp1pyMSkWDj5bqCkkJxZWXBkCFw9CgsXQpVqzodkYgEIyUFH3D6NNx8s0kMixdD5cpO\nRyQiwUpJwWGnTkH//lCpEnz8MVSs6HREIhLM8mY1e2mEppLCxWRmmt3SQkNh9mwoX97piEQk2NWu\nbT6cpqR45eWVFAqTkWF2SWvWDGbMgHKObFInInIhL3YhKSkUJC0NuneH6GizfEXZsk5HJCJyjpKC\njY4cgeuvNzulTZ5c6LrlIiKOUVKwSUoKdO0K/frBq68WOI1cRMRxXlxCW0khz/79JiHcfrvZE0EJ\nQUR8VcuWXhuBpKQAsHcvdOkC//gHPPWU09GIiFxcnTqma/vwYY+/tJLCrl2mhfDoo/DQQ05HIyJS\ntJAQr9UVgjsp7NgB110H48fDPfc4HY2ISPF5KSkE7+D7rVuhd2+YONHUEURE/ImXis3B2VLYsMFM\nTHvzTSUEEfFPXlpCO/haComJMHAgTJ8Offs6HY2ISOmopuABq1bBgAHw4YdKCCLi3+rVg7NnzYRb\nDwqepLB8OdxyCyxYYLqORET8Wd4IpJ9+8ujLBkdSWLIEhg2DRYvM8hUiIoHAC8XmwE8KCxaYPZW/\n+AI6dXI6GhERz/FCsTmwk8KsWXD//abrqF07p6MREfEsLxSbAzcpvPsuPP44rFwJUVFORyMi4nlK\nCsX01ltmUbv4eHPRREQCUYMGcPKk2QPGQwIvKUyaZL4SEqB5c6ejERHxHi+sgRRYSeFf/zI7pa1e\nDY0bOx2NiIj3ebjYHBgzmi0Lnn4aFi40LYT69Z2OSETEHmop/IVlwSOPwNKlpoaghCAiwcTDScG/\nWwq5uTB2rFngbuVKqFXL6YhEROylpPCnnBy46y4zxfvLL6F6dacjEhGxX6NGkJEB6ekeeTn/TArZ\n2TBiBPz2m5mYVrWq0xGJiDijTBlo0cJjayD5X1LIyjJ7IJw4YeoIlSs7HZGIiLM82IXkX0nh9GkY\nPNiMzV28GCpWdDoiERHneTAp+M/oo1OnoH9/qFQJPv5YCUFEJE/QJYXMTOjTB0JDYfZsKF/e6YhE\nRHyHB5fQDrEsy/LIK3lQSEgIrrAyMuDGG80fPWUKlC3rbHAiIr4mNxeqVSPk1CncvaU70lKIi4sj\nPDyc5s2b88orrxT+xLQ06N4doqPN8hVBmBDi4+OdDsFn6Fqco2txjq4FZgRSeLhnXsojr1ICOTk5\n3HfffcTFxbFjxw7mzJnDTwUNpTpyBK6/3uyUNnmy+aODkN7w5+hanKNrcY6uxZ88tCK07Xfa9evX\n06xZM8LCwihfvjy33norixcvvvCJXbtCv37w6qtmtJGIiBTOX5PCwYMHadSokevnhg0bcvDgwQuf\nePvtZk8EJQQRkaJ5KCnYXmj+5JNPiIuLY9q0aQDMmjWL7777jsmTJ58LSolARKRU3L2l2z55rUGD\nBuzfv9+uxyzpAAAHP0lEQVT18/79+2nYsGG+5/jggCgRkaBge/dRu3bt+OWXX0hKSiIrK4t58+bR\nr18/u8MQEZEC2N5SKFeuHG+++SY33HADOTk5jB49mhYtWtgdhoiIFMD2lkJcXBwPPvggubm5jBkz\nhieeeOKC59x///00b96cqKgotmzZku/YYs1v8BPF+XsKuxZhYWG0bt2a6OhoOnToYFfIXlPUtdi5\ncycdO3akUqVKTJo0qUTH+ht3rkWwvS8++ugjoqKiaN26Nddeey3btm0r9rH+xp1rUaL3hWWj7Oxs\nq2nTptbevXutrKwsKyoqytqxY0e+5yxdutTq3bu3ZVmW9e2331pXX311sY/1J+5cC8uyrLCwMOv3\n33+3NWZvKc61OHz4sLVhwwbrqaeesiZOnFiiY/2JO9fCsoLvffHNN99Y6enplmVZ1rJly4L6flHY\ntbCskr0vbG0pFGeOwpIlSxg+fDgAV199Nenp6aSmphZ/foOfKO21OHTokOv3VoAU5ItzLerUqUO7\ndu0o/5d1r4LxfVHYtcgTTO+Ljh07Uv3PDbauvvpqDhw4UOxj/Yk71yJPcd8XtiaF4sxRKOw5v/32\nW/HmN/gJd64FmGG73bt3p127dq7hvf6q2HNXPHysL3L37wnm98X06dO58cYbS3Wsr3PnWkDJ3he2\nFpqLO/8gUD7pXIy712Lt2rVceumlHDlyhB49ehAeHk5MTIwnQ7SNO/NSAm1Oi7t/T2JiIvXr1w+6\n98WqVat47733SExMLPGx/sCdawEle1/Y2lIozhyFvz7nwIEDNGzYsFjH+pPSXosGDRoAcOmllwKm\nK2HgwIGsX7/ehqi9w53/t8H4vriY+vXrA8H1vti2bRtjxoxhyZIl1KxZs0TH+gt3rgWU8H3hkSpI\nMZ09e9Zq0qSJtXfvXuvMmTNFFlfXrVvnKpYU51h/4s61OHnypHX8+HHLsiwrMzPT6tSpk7V8+XJ7\n/wAPKsn/22effTZfcTUY3xd5/notgvF9sW/fPqtp06bWunXrSnysP3HnWpT0fWFrUrAsy/riiy+s\nK664wmratKn10ksvWZZlWVOmTLGmTJnies69995rNW3a1GrdurW1adOmix7rz0p7LXbv3m1FRUVZ\nUVFRVsuWLYPiWqSkpFgNGza0LrnkEqtGjRpWo0aNrBMnThR6rD8r7bUIxvfF6NGjrVq1allt2rSx\n2rRpY7Vv3/6ix/qz0l6Lkr4vfHKTHRERcUZwblIgIiIFUlIQEREXJQUREXFRUhARERclBQl68fHx\n9O3bF4DPPvvMY4un3XLLLezevbvYz9+2bRujR4/2yLlFSsv2pbNF7JA3qK6kM1v79u3rShDu+PXX\nXzl58iRNmzYt9jGtW7dm9+7dHD58mNDQULdjECkNtRQkYCQlJXHllVcyfPhwWrVqxf79+7nnnnto\n3749kZGRjB8/3vXcuLg4WrRoQdu2bVm4cKHr8RkzZjB27FgARowYwSeffOL6XdWqVQFISUmhS5cu\nREdH06pVK9auXXtBLHPnzs23eVTVqlV57LHHiIyMpEePHnz77bd07dqVpk2b8tlnn7me17t3bxYs\nWOCxayJSUkoKElB+/fVX7r33Xn744Qcuu+wyXnzxRTZs2MD3339PQkIC27dv5/Tp09x55518/vnn\nbNq0idTU1AJbFH99LO/n2bNn06tXL7Zs2cK2bdto06bNBccmJibSrl0718+nTp2iW7du/PDDD1Sr\nVo1nnnmGr7/+moULF/LMM8+4ntehQwdWr17tqcshUmLqPpKAcvnll+fbRGTevHlMmzaN7OxsUlJS\n2LFjBzk5OTRu3NjVtTN06FCmTp1a7HN06NCBUaNGcfbsWQYMGEBUVNQFz9m3b59rvRmAChUqcMMN\nNwDQqlUrKlWqRNmyZYmMjCQpKcn1vPr16+f7WcRuailIQKlSpYrr+7179zJp0iS+/vprvv/+e/r0\n6cPp06cvaAEUNqm/XLly5ObmApCbm0tWVhYAMTExrFmzhgYNGjBixAg+/PDDAo8//3XP3/ugTJky\nVKhQwfV9dnZ2vmMCbYVP8S9KChKwjh8/TpUqVbjkkks4dOgQy5YtIyQkhPDwcJKSktizZw8Ac+bM\nKfD4sLAwNm3aBJgNj86ePQtAcnIyderU4Y477uCOO+7It01qnssvv5yUlJQSx5ySksLll19e4uNE\nPEXdRxJQzv+UHRUVRXR0NOHh4TRq1IjOnTsDULFiRaZOnUqfPn3429/+RkxMDCdPnnQdn/caY8aM\noX///rRp04ZevXq5Cs2rVq1i4sSJlC9fnmrVqvHBBx9cEEfnzp3ZuHEjbdu2vSCuv/58/vfr16+n\nS5cunrgUIqWiBfFEvGDPnj2MHTuWpUuXlui42NhY5s+fryGp4hh1H4l4QZMmTahWrVqJJ681a9ZM\nCUEcpZaCiIi4qKUgIiIuSgoiIuKipCAiIi5KCiIi4qKkICIiLkoKIiLi8v8vJedrtH6NpQAAAABJ\nRU5ErkJggg==\n"
       },
@@ -135,7 +135,6 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Finding throat diameter and overall length\n",
       "\n",
       "#variable declaration\n",
       "from math import pi\n",
@@ -187,7 +186,6 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Finding flow through orifice and venturimeter\n",
       "\n",
       "#Variable Decleration\n",
       "Cd_o =0.65;\n",
@@ -235,7 +233,6 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#finding Cd\n",
       "\n",
       "#Variable declaration\n",
       "from math import pi\n",
@@ -283,7 +280,6 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Finding differential pressure\n",
       "\n",
       "#variable declaration\n",
       "from math import pi\n",
@@ -328,7 +324,7 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#finding rate of flow\n",
+      "\n",
       "\n",
       "#variable declaration\n",
       "from math import pi\n",
@@ -373,7 +369,7 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Finding mass flow rate\n",
+      "\n",
       "\n",
       "#variable decleration\n",
       "from math import pi\n",
@@ -422,7 +418,7 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Finding flow rate of water and Cd\n",
+      "\n",
       "\n",
       "#variable decleration\n",
       "from math import pi\n",
diff --git a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_6_1.ipynb b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_6_1.ipynb
index 19e4fb13..28a995b9 100644
--- a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_6_1.ipynb
+++ b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_6_1.ipynb
@@ -1,6 +1,7 @@
 {
  "metadata": {
-  "name": "Chapter 6"
+  "name": "",
+  "signature": "sha256:935baa5cece6ad60c6e41bbc750e73048ac600a6bffe3888ab4bb403a9f445e6"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -11,25 +12,50 @@
      "cell_type": "heading",
      "level": 1,
      "metadata": {},
-     "source": "Chapter 6:Tank drainage and variable head flow"
+     "source": [
+      "Chapter 6:Tank drainage and variable head flow"
+     ]
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 6.1,Page 143"
+     "source": [
+      "Example 6.1,Page 143"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#finding no. of orifice\n\n#variable decleration\nfrom math import pi\nimport math\nQ =5000.0/3600/24; #m^3 per second\nC_d =0.6;\nr =0.01/2; # m\ng =9.8; #m/ s ^2\nH =0.2; # m\n\n\n#calculation\na_o = pi *r**2;\nn=Q/C_d/ a_o / math.sqrt (2*g*H);\n\n#result\nprint\"The number of orifices required are \",round (n)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "from math import pi\n",
+      "import math\n",
+      "Q =5000.0/3600/24; #m^3 per second\n",
+      "C_d =0.6;\n",
+      "r =0.01/2; # m\n",
+      "g =9.8; #m/ s ^2\n",
+      "H =0.2; # m\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "a_o = pi *r**2;\n",
+      "n=Q/C_d/ a_o / math.sqrt (2*g*H);\n",
+      "\n",
+      "#result\n",
+      "print\"The number of orifices required are \",round (n)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The number of orifices required are  621.0\n"
+       "text": [
+        "The number of orifices required are  621.0\n"
+       ]
       }
      ],
      "prompt_number": 10
@@ -38,19 +64,41 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 6.2,Page 145"
+     "source": [
+      "Example 6.2,Page 145"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding coeff. of velocity\n\n#variable decleration\nfrom math import pi\nimport math\nx =0.86; # m\ng =9.8; # m/ s\ny =0.96; # m\nH =0.2; # m\n\n#calculation\nv_act =x* math.sqrt (g /2/ y);\nv= math.sqrt (2* g*H);\nCv= v_act /v;\n\n#result\nprint \"The coeff. of velocity for the orifices found to be \",round(Cv,3)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "from math import pi\n",
+      "import math\n",
+      "x =0.86; # m\n",
+      "g =9.8; # m/ s\n",
+      "y =0.96; # m\n",
+      "H =0.2; # m\n",
+      "\n",
+      "#calculation\n",
+      "v_act =x* math.sqrt (g /2/ y);\n",
+      "v= math.sqrt (2* g*H);\n",
+      "Cv= v_act /v;\n",
+      "\n",
+      "#result\n",
+      "print \"The coeff. of velocity for the orifices found to be \",round(Cv,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The coeff. of velocity for the orifices found to be  0.981\n"
+       "text": [
+        "The coeff. of velocity for the orifices found to be  0.981\n"
+       ]
       }
      ],
      "prompt_number": 1
@@ -59,19 +107,44 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 6.3,Page 147"
+     "source": [
+      "Example 6.3,Page 147"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time taken for drainage\n\n#variable declaration\nfrom math import pi\nimport math\nVt =1; # m^3\nd_t =1; # m\nC_d =0.6;\nd_o =0.02; # m\ng =9.8; # m/ s ^2\n\n\n#calculation\na_o = pi *( d_o ) **2/4;\nA= pi *( d_t ) **2/4;\nH1 =4* Vt/ pi /( d_t) **2;\nt=A/C_d/ a_o * math.sqrt (2* H1/g);\n\n#results\nprint\" Total drainage is found to take (s) \",round(t,2)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "from math import pi\n",
+      "import math\n",
+      "Vt =1; # m^3\n",
+      "d_t =1; # m\n",
+      "C_d =0.6;\n",
+      "d_o =0.02; # m\n",
+      "g =9.8; # m/ s ^2\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "a_o = pi *( d_o ) **2/4;\n",
+      "A= pi *( d_t ) **2/4;\n",
+      "H1 =4* Vt/ pi /( d_t) **2;\n",
+      "t=A/C_d/ a_o * math.sqrt (2* H1/g);\n",
+      "\n",
+      "#results\n",
+      "print\" Total drainage is found to take (s) \",round(t,2)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Total drainage is found to take (s)  2124.49586383\n"
+       "text": [
+        " Total drainage is found to take (s)  2124.49586383\n"
+       ]
       }
      ],
      "prompt_number": 14
@@ -80,19 +153,41 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 6.4,Page 149"
+     "source": [
+      "Example 6.4,Page 149"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time taken to drain tank\n\n#variable declaration\nfrom math import pi\nimport math\nC_d =0.6;\nd_o =0.05; #m\ng =9.81; # m/ s ^ 2 ;\nR =2; \nH1 =1.5; \n\n#calculation\na_o = pi *d_o **2/4;\nt= pi /C_d/ a_o*(1.333* R*H1 **(1.5) -0.4* H1 **(2.5))/math.sqrt(2*g);\n\n#results\nprint\"The time to drain the tank is found to be (s)\",round(t,3);",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "from math import pi\n",
+      "import math\n",
+      "C_d =0.6;\n",
+      "d_o =0.05; #m\n",
+      "g =9.81; # m/ s ^ 2 ;\n",
+      "R =2; \n",
+      "H1 =1.5; \n",
+      "\n",
+      "#calculation\n",
+      "a_o = pi *d_o **2/4;\n",
+      "t= pi /C_d/ a_o*(1.333* R*H1 **(1.5) -0.4* H1 **(2.5))/math.sqrt(2*g);\n",
+      "\n",
+      "#results\n",
+      "print\"The time to drain the tank is found to be (s)\",round(t,3);"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The time to drain the tank is found to be (s) 2285.001\n"
+       "text": [
+        "The time to drain the tank is found to be (s) 2285.001\n"
+       ]
       }
      ],
      "prompt_number": 16
@@ -101,19 +196,46 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 6.6,Page 153"
+     "source": [
+      "Example 6.6,Page 153"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time taken to drain tank and rate of flow\n\n#variable declaration\nimport math\nCd =0.62;\na =0.01;# m^2\ng =9.81; # m/ s ^2\nH =0.3; #m\nA1 =4*2; # m^2\nH1 =0.300; # m\nH2 =0.100; # m\nA2 =2*2; # m^2\n\n#calculation\nQ=Cd*a* math.sqrt (2* g*H);\nt =2* A1 *( math.sqrt(H1) -math.sqrt(H2) )/( Cd*a* math.sqrt (2*g) *(1+ A1/A2));\n\n#results\nprint \"The rate of flow(m^3/s) =\",round(Q,3)\nprint \"The time taken to reduce the difference in levels to 10 cm is (s) \", round(t,3)\n\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "import math\n",
+      "Cd =0.62;\n",
+      "a =0.01;# m^2\n",
+      "g =9.81; # m/ s ^2\n",
+      "H =0.3; #m\n",
+      "A1 =4*2; # m^2\n",
+      "H1 =0.300; # m\n",
+      "H2 =0.100; # m\n",
+      "A2 =2*2; # m^2\n",
+      "\n",
+      "#calculation\n",
+      "Q=Cd*a* math.sqrt (2* g*H);\n",
+      "t =2* A1 *( math.sqrt(H1) -math.sqrt(H2) )/( Cd*a* math.sqrt (2*g) *(1+ A1/A2));\n",
+      "\n",
+      "#results\n",
+      "print \"The rate of flow(m^3/s) =\",round(Q,3)\n",
+      "print \"The time taken to reduce the difference in levels to 10 cm is (s) \", round(t,3)\n",
+      "\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The rate of flow(m^3/s) = 0.01504\nThe time taken to reduce the difference in levels to 10 cm is (s)  44.957\n"
+       "text": [
+        "The rate of flow(m^3/s) = 0.01504\n",
+        "The time taken to reduce the difference in levels to 10 cm is (s)  44.957\n"
+       ]
       }
      ],
      "prompt_number": 4
@@ -122,19 +244,42 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 6.8,Page 157"
+     "source": [
+      "Example 6.8,Page 157"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time taken to level tank \n\n#variable declaration\nimport math\nQs =0.4; #m^3/ s\nH1 =1.5; # m\nQ =0.2; #m^3/ s\nH2 =0.5; # m\nl =15; # m\nb =10; #m\ne=2.69;\n\n#calculation\nA=l*b;\nk=Qs*H1 **( -0.5) ;\nt=-2*A/k**2*(Q*math.log((Q-k*math.sqrt( H2))/(Q-k*math.sqrt(H1)),e)+k*(math.sqrt(H2)-math.sqrt(H1)));\n#result\nprint \"The time reqd. for the level in the tank to fall to 1 m is(s) \",round(t,3)\n\n",
+     "input": [
+      "\n",
+      "#variable declaration\n",
+      "import math\n",
+      "Qs =0.4; #m^3/ s\n",
+      "H1 =1.5; # m\n",
+      "Q =0.2; #m^3/ s\n",
+      "H2 =0.5; # m\n",
+      "l =15; # m\n",
+      "b =10; #m\n",
+      "e=2.69;\n",
+      "\n",
+      "#calculation\n",
+      "A=l*b;\n",
+      "k=Qs*H1 **( -0.5) ;\n",
+      "t=-2*A/k**2*(Q*math.log((Q-k*math.sqrt( H2))/(Q-k*math.sqrt(H1)),e)+k*(math.sqrt(H2)-math.sqrt(H1)));\n",
+      "#result\n",
+      "print \"The time reqd. for the level in the tank to fall to 1 m is(s) \",round(t,3)\n",
+      "\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The time reqd. for the level in the tank to fall to 1 m is(s)  1536.35\n"
+       "text": [
+        "The time reqd. for the level in the tank to fall to 1 m is(s)  1536.35\n"
+       ]
       }
      ],
      "prompt_number": 5
@@ -143,19 +288,49 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 6.9,Page 159"
+     "source": [
+      "Example 6.9,Page 159"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time taken to level tank \n\n#variable declaration\nfrom math import pi\nimport numpy as np\nimport numpy \nimport math\nCd =0.62;\nd =0.05;\na_o = pi *d **2/4;\ng =9.81; #m/ s ^2\n\n#calculation\nk=Cd*a_o* math.sqrt (2*g);\n# We have g o t two s imultaneous equations\n#Q_k \u0003 0 . 6 5 ^ ( 1 / 2 ) =0.1/90\u0003A\n# Q_k \u0003 1 . 2 2 5 ^ ( 1 / 2 ) =0.05/120\u0003A\n\nM=np.array([[1,-0.1/90],[1,-0.05/120]]);\nN=np.array([[k *math.sqrt(0.65)] ,[k *math.sqrt(1.225)]]);\nX=np.linalg.solve(M,N)\nQ=X[0][0];\nA=X[1][0];\nprint \"The Area of the tank(m^2) =\",round(A,3)\nprint \" Flow rate(m^3/s) =\",round(Q,3)\n",
+     "input": [
+      "\n",
+      "#variable declaration\n",
+      "from math import pi\n",
+      "import numpy as np\n",
+      "import numpy \n",
+      "import math\n",
+      "Cd =0.62;\n",
+      "d =0.05;\n",
+      "a_o = pi *d **2/4;\n",
+      "g =9.81; #m/ s ^2\n",
+      "\n",
+      "#calculation\n",
+      "k=Cd*a_o* math.sqrt (2*g);\n",
+      "# We have g o t two s imultaneous equations\n",
+      "#Q_k \u0003 0 . 6 5 ^ ( 1 / 2 ) =0.1/90\u0003A\n",
+      "# Q_k \u0003 1 . 2 2 5 ^ ( 1 / 2 ) =0.05/120\u0003A\n",
+      "\n",
+      "M=np.array([[1,-0.1/90],[1,-0.05/120]]);\n",
+      "N=np.array([[k *math.sqrt(0.65)] ,[k *math.sqrt(1.225)]]);\n",
+      "X=np.linalg.solve(M,N)\n",
+      "Q=X[0][0];\n",
+      "A=X[1][0];\n",
+      "print \"The Area of the tank(m^2) =\",round(A,3)\n",
+      "print \" Flow rate(m^3/s) =\",round(Q,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The Area of the tank(m^2) = 2.334\n Flow rate(m^3/s) = 0.007\n"
+       "text": [
+        "The Area of the tank(m^2) = 2.334\n",
+        " Flow rate(m^3/s) = 0.007\n"
+       ]
       }
      ],
      "prompt_number": 6
@@ -164,19 +339,48 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 6.10,Page 161"
+     "source": [
+      "Example 6.10,Page 161"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time taken to level tank \n\n#variable declaration\nfrom math import pi\nfrom numpy import matrix\nfrom numpy import linalg\nimport math\nH1 =1.5; # m\nV =0.75; # m^3\nd1 =1.2; # m\nu =0.08; # Ns/m^2\nL =3; # m\nrho =1100; # kg /m^3\ng =9.81; # m/ s ^2\nd =0.025; # m\ne=2.71;\n\n#calculation\na= pi*d **2/4;\nA= pi*d1 **2/4;\nH2=H1 -(V/A);\nround(H2,2);\nt= -32*u*L*A/(a*rho*g*d **2) *math.log(H2/H1,e);\n\n#result\nprint\"Time taken(s) =\",round(t,3);",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "from numpy import matrix\n",
+      "from numpy import linalg\n",
+      "import math\n",
+      "H1 =1.5; # m\n",
+      "V =0.75; # m^3\n",
+      "d1 =1.2; # m\n",
+      "u =0.08; # Ns/m^2\n",
+      "L =3; # m\n",
+      "rho =1100; # kg /m^3\n",
+      "g =9.81; # m/ s ^2\n",
+      "d =0.025; # m\n",
+      "e=2.71;\n",
+      "\n",
+      "#calculation\n",
+      "a= pi*d **2/4;\n",
+      "A= pi*d1 **2/4;\n",
+      "H2=H1 -(V/A);\n",
+      "round(H2,2);\n",
+      "t= -32*u*L*A/(a*rho*g*d **2) *math.log(H2/H1,e);\n",
+      "\n",
+      "#result\n",
+      "print\"Time taken(s) =\",round(t,3);"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Time taken(s) = 1535.7565254\n"
+       "text": [
+        "Time taken(s) = 1535.7565254\n"
+       ]
       }
      ],
      "prompt_number": 17
@@ -184,7 +388,7 @@
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "",
+     "input": [],
      "language": "python",
      "metadata": {},
      "outputs": []
diff --git a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_7_1.ipynb b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_7_1.ipynb
index 0942f575..eff449e4 100644
--- a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_7_1.ipynb
+++ b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_7_1.ipynb
@@ -1,6 +1,7 @@
 {
  "metadata": {
-  "name": "Chapter 7"
+  "name": "",
+  "signature": "sha256:9d2bc1b337823a33d27dffeb338a599749ddc3b72fed4860934dfca72c26b017"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -11,25 +12,48 @@
      "cell_type": "heading",
      "level": 1,
      "metadata": {},
-     "source": "Chapter 7,Open channels notches and Weirs"
+     "source": [
+      "Chapter 7,Open channels notches and Weirs"
+     ]
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 7.2,Page 171"
+     "source": [
+      "Example 7.2,Page 171"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Discharge\n\n#variable decleration\nimport math\nl =1; # m\nb =0.3; # m\nn =0.014; # s /m^ ( 1 / 3 )\ni =1.0/1000;\n\n#calculations\nA=l*b;\nP =2* b+l;\nm=A/P;\nQ=A/n*m **(0.67) * math.sqrt (i);\n\n#results\nprint\"The delivery of water through the channel is found to be (m^3/s)= \",round(Q,3)\n",
+     "input": [
+      "\n",
+      "#variable decleration\n",
+      "import math\n",
+      "l =1; # m\n",
+      "b =0.3; # m\n",
+      "n =0.014; # s /m^ ( 1 / 3 )\n",
+      "i =1.0/1000;\n",
+      "\n",
+      "#calculations\n",
+      "A=l*b;\n",
+      "P =2* b+l;\n",
+      "m=A/P;\n",
+      "Q=A/n*m **(0.67) * math.sqrt (i);\n",
+      "\n",
+      "#results\n",
+      "print\"The delivery of water through the channel is found to be (m^3/s)= \",round(Q,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The delivery of water through the channel is found to be (m^3/s)=  0.221\n"
+       "text": [
+        "The delivery of water through the channel is found to be (m^3/s)=  0.221\n"
+       ]
       }
      ],
      "prompt_number": 13
@@ -38,28 +62,81 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 7.3 Page 173"
+     "source": [
+      "Example 7.3 Page 173"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Depth and Chezy's Coeff.\n\n%pylab inline\n#variable decleration\nimport math\nimport numpy as np\nfrom pylab import *\nn =0.015; #m^(-1/3) s\ni =1;\nH =[4.0, 4.1, 4.2, 4.13];\n\n#calculation\nA =12* H[0];\nP =12+2* H[0];\nm=A/P;\nC=m **(0.167) /n;\nQ=C*A* math.sqrt (m*i);\n#An analytical soln for depth H is not possible.It is so necessary to use a graphical soution\n#The corresponding values of A, P, MHD (m) , Q are given below as taken from values of H\n\nA =[48, 49.2, 50.4, 49.56];\nP =[20, 20.2, 20.4, 20.26];\nm =[2.4, 2.44, 2.47, 2.45];\nQ =[57.36, 59.38, 61.39, 59.98];\nfigure()\nplot (H,Q,'g')\nr =[4.13, 4.13];\ns =[57, 60];\nplot (r,s, ' r ' )\nt =[4, 4.13];\nu =[60, 60];\nplot (t,u, ' y ' )\nxlabel('Depth (H)')\nylabel('Flow Rate')\ntitle('Depth vs Flow Rate')\nshow()\n# So, the depth is found to be approx. 4.13\ndepth =4.13; #m\nC1 =(2.45) **(0.167) /n;\n\n#results\nprint\"Depth(m) = \",round(depth,3)\nprint\"Chezy Coeff. =\",round(C1,3)\n",
+     "input": [
+      "\n",
+      "%pylab inline\n",
+      "#variable decleration\n",
+      "import math\n",
+      "import numpy as np\n",
+      "from pylab import *\n",
+      "n =0.015; #m^(-1/3) s\n",
+      "i =1;\n",
+      "H =[4.0, 4.1, 4.2, 4.13];\n",
+      "\n",
+      "#calculation\n",
+      "A =12* H[0];\n",
+      "P =12+2* H[0];\n",
+      "m=A/P;\n",
+      "C=m **(0.167) /n;\n",
+      "Q=C*A* math.sqrt (m*i);\n",
+      "#An analytical soln for depth H is not possible.It is so necessary to use a graphical soution\n",
+      "#The corresponding values of A, P, MHD (m) , Q are given below as taken from values of H\n",
+      "\n",
+      "A =[48, 49.2, 50.4, 49.56];\n",
+      "P =[20, 20.2, 20.4, 20.26];\n",
+      "m =[2.4, 2.44, 2.47, 2.45];\n",
+      "Q =[57.36, 59.38, 61.39, 59.98];\n",
+      "figure()\n",
+      "plot (H,Q,'g')\n",
+      "r =[4.13, 4.13];\n",
+      "s =[57, 60];\n",
+      "plot (r,s, ' r ' )\n",
+      "t =[4, 4.13];\n",
+      "u =[60, 60];\n",
+      "plot (t,u, ' y ' )\n",
+      "xlabel('Depth (H)')\n",
+      "ylabel('Flow Rate')\n",
+      "title('Depth vs Flow Rate')\n",
+      "show()\n",
+      "# So, the depth is found to be approx. 4.13\n",
+      "depth =4.13; #m\n",
+      "C1 =(2.45) **(0.167) /n;\n",
+      "\n",
+      "#results\n",
+      "print\"Depth(m) = \",round(depth,3)\n",
+      "print\"Chezy Coeff. =\",round(C1,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "\nWelcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\nFor more information, type 'help(pylab)'.\n"
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'.\n"
+       ]
       },
       {
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        "output_type": "display_data",
        "png": 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XGYoSNUKZw1M//fQTfn5+JCcnM2XKFLKysggLC2Po0KGVVWMxMjwlqqq7BXeJ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\nAAAAAElFTkSuQmCC\n"
       },
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Depth(m) =  4.13\nChezy Coeff. = 77.428\n"
+       "text": [
+        "Depth(m) =  4.13\n",
+        "Chezy Coeff. = 77.428\n"
+       ]
       }
      ],
      "prompt_number": 15
@@ -68,19 +145,36 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 7.4,Page 175"
+     "source": [
+      "Example 7.4,Page 175"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding flow area\n\n#variable decleration\nimport math\nQ =300.0/60; # m^3/ s\ni =1.0/1600;\n\n#calculation\nH=(Q /140* math.sqrt (2/ i)) **(0.67) ;\nA =2* H**2;\n\n#result\nprint \"The minimum flow area is found to be(m^2)= \",round(A,3)\n",
+     "input": [
+      "\n",
+      "#variable decleration\n",
+      "import math\n",
+      "Q =300.0/60; # m^3/ s\n",
+      "i =1.0/1600;\n",
+      "\n",
+      "#calculation\n",
+      "H=(Q /140* math.sqrt (2/ i)) **(0.67) ;\n",
+      "A =2* H**2;\n",
+      "\n",
+      "#result\n",
+      "print \"The minimum flow area is found to be(m^2)= \",round(A,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The minimum flow area is found to be(m^2)=  5.132\n"
+       "text": [
+        "The minimum flow area is found to be(m^2)=  5.132\n"
+       ]
       }
      ],
      "prompt_number": 16
@@ -89,28 +183,83 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 7.5,Page 177"
+     "source": [
+      "Example 7.5,Page 177"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Depth and Chezy's Coeff.\n\n%pylab inline\n#variable decleration\nfrom math import pi\nimport math\nimport numpy as np\nfrom pylab import *\nd =0.9144; # m\nC =100; # m^ ( 1 / 2 ) s ^(-1)\nR=d/2;\nH =[0.1, 0.15, 0.2, 0.25, 0.201];\n\n\n#calculations\ntheta = math.acos ((R-H[0])/R);\nA=R **2*( theta -math.sin (2* theta )/2);\nP =2* R* theta ;\nm=A/P;\n#An analytical soln for depth H is not possible.It is so necessary to use a graphical soution\n#The corresponding values of A, P, MHD (m) , Q are given below as taken from values of H\n\n\ntheta =[0.674, 0.834, 0.973, 1.101, 0.975];\nA =[0.039, 0.070, 0.106, 0.146, 0.107];\nP =[0.616, 0.763, 0.890, 1.006, 0.891];\nm =[0.063, 0.092, 0.119,0.145,0.120];\nQ =[248.7, 543.2, 932.2 ,1412.9, 940.0];\nfigure()\nplot (H,Q,)\nxlabel('Depth H')\nylabel('Flow Rate')\ntitle('Depth Vs Flow Rate');\ni =[0.201, 0.201];\nj =[0 ,940];\nplot (i,j)\nk =[0, 0.201];\nl  =[940, 940];\nplot (k,l)\nshow()\n# So, the depth is found to be approx. 0.201\nDepth =0.201; # m\n\n#result\nprint\"The depth in the channel(m) =\",round(Depth,3)\n",
+     "input": [
+      "\n",
+      "\n",
+      "%pylab inline\n",
+      "#variable decleration\n",
+      "from math import pi\n",
+      "import math\n",
+      "import numpy as np\n",
+      "from pylab import *\n",
+      "d =0.9144; # m\n",
+      "C =100; # m^ ( 1 / 2 ) s ^(-1)\n",
+      "R=d/2;\n",
+      "H =[0.1, 0.15, 0.2, 0.25, 0.201];\n",
+      "\n",
+      "\n",
+      "#calculations\n",
+      "theta = math.acos ((R-H[0])/R);\n",
+      "A=R **2*( theta -math.sin (2* theta )/2);\n",
+      "P =2* R* theta ;\n",
+      "m=A/P;\n",
+      "#An analytical soln for depth H is not possible.It is so necessary to use a graphical soution\n",
+      "#The corresponding values of A, P, MHD (m) , Q are given below as taken from values of H\n",
+      "\n",
+      "\n",
+      "theta =[0.674, 0.834, 0.973, 1.101, 0.975];\n",
+      "A =[0.039, 0.070, 0.106, 0.146, 0.107];\n",
+      "P =[0.616, 0.763, 0.890, 1.006, 0.891];\n",
+      "m =[0.063, 0.092, 0.119,0.145,0.120];\n",
+      "Q =[248.7, 543.2, 932.2 ,1412.9, 940.0];\n",
+      "figure()\n",
+      "plot (H,Q,)\n",
+      "xlabel('Depth H')\n",
+      "ylabel('Flow Rate')\n",
+      "title('Depth Vs Flow Rate');\n",
+      "i =[0.201, 0.201];\n",
+      "j =[0 ,940];\n",
+      "plot (i,j)\n",
+      "k =[0, 0.201];\n",
+      "l  =[940, 940];\n",
+      "plot (k,l)\n",
+      "show()\n",
+      "# So, the depth is found to be approx. 0.201\n",
+      "Depth =0.201; # m\n",
+      "\n",
+      "#result\n",
+      "print\"The depth in the channel(m) =\",round(Depth,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "\nWelcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\nFor more information, type 'help(pylab)'.\n"
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'.\n"
+       ]
       },
       {
+       "metadata": {},
        "output_type": "display_data",
        "png": 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hvfAABYiISLORn18zPNq29W49ChARkWYgPx9CQhzTvhAeoAAREfF5BQW+Fx6gABER8WmF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\nDQAAAABJRU5ErkJggg==\n"
       },
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The depth in the channel(m) = 0.201\n"
+       "text": [
+        "The depth in the channel(m) = 0.201\n"
+       ]
       }
      ],
      "prompt_number": 9
@@ -119,19 +268,38 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 7.7,Page 180"
+     "source": [
+      "Example 7.7,Page 180"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding flow area\n\n#variable decleration\nimport math\nCd =0.56;\nB =1.2; # m\ng =9.8; # m/ s ^2\nH =0.018; # m\n\n\n#calcualtion\nQ =2.0/3* Cd*B* math.sqrt (2*g)*H **(1.5) ;\n\n#results\nprint \"The rate of flow of liquid over the weir is (m^3/s)=\",round(Q,4)\n",
+     "input": [
+      "\n",
+      "#variable decleration\n",
+      "import math\n",
+      "Cd =0.56;\n",
+      "B =1.2; # m\n",
+      "g =9.8; # m/ s ^2\n",
+      "H =0.018; # m\n",
+      "\n",
+      "\n",
+      "#calcualtion\n",
+      "Q =2.0/3* Cd*B* math.sqrt (2*g)*H **(1.5) ;\n",
+      "\n",
+      "#results\n",
+      "print \"The rate of flow of liquid over the weir is (m^3/s)=\",round(Q,4)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The rate of flow of liquid over the weir is (m^3/s)= 0.0048\n"
+       "text": [
+        "The rate of flow of liquid over the weir is (m^3/s)= 0.0048\n"
+       ]
       }
      ],
      "prompt_number": 10
@@ -140,19 +308,39 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 7.8,Page 182"
+     "source": [
+      "Example 7.8,Page 182"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding height of weir\n\n#variable decleration\nfrom math import pi\nimport math\nH2 =5.5;\nQ1 =217.0;\nQ2 =34.0;\nH1 =8.5;\n\n#calculation\nH0 =( H2 *( Q1/Q2) **(0.67) -H1)/(( Q1/Q2) **(0.67) -1);\n\n#result\nprint\"The height of the weir crest above the surface of the river is found to be(m)\",round(H0,2)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "from math import pi\n",
+      "import math\n",
+      "H2 =5.5;\n",
+      "Q1 =217.0;\n",
+      "Q2 =34.0;\n",
+      "H1 =8.5;\n",
+      "\n",
+      "#calculation\n",
+      "H0 =( H2 *( Q1/Q2) **(0.67) -H1)/(( Q1/Q2) **(0.67) -1);\n",
+      "\n",
+      "#result\n",
+      "print\"The height of the weir crest above the surface of the river is found to be(m)\",round(H0,2)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The height of the weir crest above the surface of the river is found to be(m) 4.28\n"
+       "text": [
+        "The height of the weir crest above the surface of the river is found to be(m) 4.28\n"
+       ]
       }
      ],
      "prompt_number": 5
@@ -161,19 +349,45 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 7.9,Page 184"
+     "source": [
+      "Example 7.9,Page 184"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time taken\n\n#variable decleration\nfrom math import pi\nfrom scipy import integrate\nimport math\nH =0.07; # a v e r a g e head\nrate = -0.02/600; # (dH/ dt )\nH1 =0.08; # m\nH2 =0.01; # m\n\n\n#calculation\nk=- rate /H **(1.5) ;\ndef integrand(H,k):\n    return -1/k*H**(-1.5);\n\nt=integrate.quad(integrand,H1,H2,args=(k))\n\n#result\nprint \"Time taken (s) =\",round(t[0],2);\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "from math import pi\n",
+      "from scipy import integrate\n",
+      "import math\n",
+      "H =0.07; # a v e r a g e head\n",
+      "rate = -0.02/600; # (dH/ dt )\n",
+      "H1 =0.08; # m\n",
+      "H2 =0.01; # m\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "k=- rate /H **(1.5) ;\n",
+      "def integrand(H,k):\n",
+      "    return -1/k*H**(-1.5);\n",
+      "\n",
+      "t=integrate.quad(integrand,H1,H2,args=(k))\n",
+      "\n",
+      "#result\n",
+      "print \"Time taken (s) =\",round(t[0],2);\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Time taken (s) = 7183415.25\n"
+       "text": [
+        "Time taken (s) = 7183415.25\n"
+       ]
       }
      ],
      "prompt_number": 18
@@ -182,19 +396,37 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 7.10,Page 186"
+     "source": [
+      "Example 7.10,Page 186"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding flow area\n\n#variable decleration\nimport math\nCd =0.62;\ng =9.81; # m/ s ^2\nH =0.03; # m\n\n#calculations\nQ =8.0/15* Cd* math.sqrt (2*g)*H **(2.5) ;\n\n#results\nprint\"Rate of flow (m^3/s) =\",round(Q,6)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "import math\n",
+      "Cd =0.62;\n",
+      "g =9.81; # m/ s ^2\n",
+      "H =0.03; # m\n",
+      "\n",
+      "#calculations\n",
+      "Q =8.0/15* Cd* math.sqrt (2*g)*H **(2.5) ;\n",
+      "\n",
+      "#results\n",
+      "print\"Rate of flow (m^3/s) =\",round(Q,6)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Rate of flow (m^3/s) = 0.000228\n"
+       "text": [
+        "Rate of flow (m^3/s) = 0.000228\n"
+       ]
       }
      ],
      "prompt_number": 3
@@ -203,19 +435,43 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 7.11,Page 188"
+     "source": [
+      "Example 7.11,Page 188"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time taken\nfrom scipy import integrate\n#variable decleration\nimport math\nl =4; # m\nb =2; # m\nH1 =0.15; # m\nH2 =0.05; # m\nA=8;\n\n#calculation\nk=3*1.5/A;\ndef integrand (H,A):\n    return -A/1.5*H**(-2.5)\n \nt=integrate.quad(integrand,H1,H2,args=(A))\n\n#results\nprint \"Time taken to reduce the head in the the tank (s)=\",round(t[0])\n",
+     "input": [
+      "\n",
+      "from scipy import integrate\n",
+      "#variable decleration\n",
+      "import math\n",
+      "l =4; # m\n",
+      "b =2; # m\n",
+      "H1 =0.15; # m\n",
+      "H2 =0.05; # m\n",
+      "A=8;\n",
+      "\n",
+      "#calculation\n",
+      "k=3*1.5/A;\n",
+      "def integrand (H,A):\n",
+      "    return -A/1.5*H**(-2.5)\n",
+      " \n",
+      "t=integrate.quad(integrand,H1,H2,args=(A))\n",
+      "\n",
+      "#results\n",
+      "print \"Time taken to reduce the head in the the tank (s)=\",round(t[0])\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Time taken to reduce the head in the the tank (s)= 257.0\n"
+       "text": [
+        "Time taken to reduce the head in the the tank (s)= 257.0\n"
+       ]
       }
      ],
      "prompt_number": 17
@@ -223,7 +479,7 @@
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "",
+     "input": [],
      "language": "python",
      "metadata": {},
      "outputs": []
diff --git a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_8_1.ipynb b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_8_1.ipynb
index 4c96cf24..3d67d746 100644
--- a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_8_1.ipynb
+++ b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_8_1.ipynb
@@ -1,6 +1,7 @@
 {
  "metadata": {
-  "name": "Chapter 8"
+  "name": "",
+  "signature": "sha256:b23dbd5a6884d4ea13e28c9cb79f4fe4cdf3c73f39b3e9719031299037425f06"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -11,25 +12,56 @@
      "cell_type": "heading",
      "level": 1,
      "metadata": {},
-     "source": "Chapter 8,Pipe Friction and Turbulent Flow"
+     "source": [
+      "Chapter 8,Pipe Friction and Turbulent Flow"
+     ]
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.4,Page 199"
+     "source": [
+      "Example 8.4,Page 199"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Internal diameter and reynold's number\n\n#variable declaration\nfrom math import pi\nimport math\nrho =867.0; #kg /m^3\nQ =12.0/3600; #m^3/ s\nu =7.5*10**( -4) ; # Ns/m^2\nL =200.0; # m\nH =10.0; # m\ng =9.81; # m/ s ^2\ne=2.718;\n\n#calculation\nd=(H*2*g /(4*0.079*(4* rho *Q/pi/u)**( -0.25) *L *(4* Q/pi)**2) )**( -0.2105);\nd=round(d,4);\nRe =4* rho *Q/ pi /d/u;\n\n#result\nprint\" Internal diameter of the pipe line (m)=\",round(d,5)\nprint\"Re =\",round(Re,2)\nprint(\"The value o f Reynolds number l i e s between 4000 and 10^5 , confirming the validity of using the Blasius equationfor smooth walled pipes\")",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "from math import pi\n",
+      "import math\n",
+      "rho =867.0; #kg /m^3\n",
+      "Q =12.0/3600; #m^3/ s\n",
+      "u =7.5*10**( -4) ; # Ns/m^2\n",
+      "L =200.0; # m\n",
+      "H =10.0; # m\n",
+      "g =9.81; # m/ s ^2\n",
+      "e=2.718;\n",
+      "\n",
+      "#calculation\n",
+      "d=(H*2*g /(4*0.079*(4* rho *Q/pi/u)**( -0.25) *L *(4* Q/pi)**2) )**( -0.2105);\n",
+      "d=round(d,4);\n",
+      "Re =4* rho *Q/ pi /d/u;\n",
+      "\n",
+      "#result\n",
+      "print\" Internal diameter of the pipe line (m)=\",round(d,5)\n",
+      "print\"Re =\",round(Re,2)\n",
+      "print(\"The value o f Reynolds number l i e s between 4000 and 10^5 , confirming the validity of using the Blasius equationfor smooth walled pipes\")"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Internal diameter of the pipe line (m)= 0.0505\nRe = 97152.8\nThe value o f Reynolds number l i e s between 4000 and 10^5 , confirming the validity of using the Blasius equationfor smooth walled pipes\n"
+       "text": [
+        " Internal diameter of the pipe line (m)= 0.0505\n",
+        "Re = 97152.8\n",
+        "The value o f Reynolds number l i e s between 4000 and 10^5 , confirming the validity of using the Blasius equationfor smooth walled pipes\n"
+       ]
       }
      ],
      "prompt_number": 1
@@ -38,19 +70,46 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.5,Page 202"
+     "source": [
+      "Example 8.5,Page 202"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding minimum allowable height\n\n#variable declaration\nfrom math import pi\nimport math\nm =40.0/60; # kg / s\nrho =873; #kg /m^3\nd =0.025; #m\nu =8.8*10** -4; # Ns/m^2\ndp =55.0*10**3; #N/m^2\nL =18; #m\ng =9.81; # m/ s ^2\n\n#calculation\nv2 =4*m/ rho / pi /d **2;\nRe=rho*v2*d/u;\n#using MOODY's Chart\nf =0.0055;\nH=dp/rho/g + v2 **2/2/ g + v2 **2/2/ g *(4* f*L/d +1.5) ;\n\n#result\nprint \"The minimum allowable height (m) =\",round(H,3)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "from math import pi\n",
+      "import math\n",
+      "m =40.0/60; # kg / s\n",
+      "rho =873; #kg /m^3\n",
+      "d =0.025; #m\n",
+      "u =8.8*10** -4; # Ns/m^2\n",
+      "dp =55.0*10**3; #N/m^2\n",
+      "L =18; #m\n",
+      "g =9.81; # m/ s ^2\n",
+      "\n",
+      "#calculation\n",
+      "v2 =4*m/ rho / pi /d **2;\n",
+      "Re=rho*v2*d/u;\n",
+      "#using MOODY's Chart\n",
+      "f =0.0055;\n",
+      "H=dp/rho/g + v2 **2/2/ g + v2 **2/2/ g *(4* f*L/d +1.5) ;\n",
+      "\n",
+      "#result\n",
+      "print \"The minimum allowable height (m) =\",round(H,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The minimum allowable height (m) = 8.684\n"
+       "text": [
+        "The minimum allowable height (m) = 8.684\n"
+       ]
       }
      ],
      "prompt_number": 2
@@ -59,19 +118,49 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.6,Page 204"
+     "source": [
+      "Example 8.6,Page 204"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding minimum allowable height\n\n#variable declaration\nfrom math import pi\nimport math\nQ =15.0/3600; # m^3/ s\nd =0.05; #m\nRho =780.0;\nu =1.7*10**( -3) ; #Ns/m^2\nf =0.0065;\nL =100; #m\ng =9.8; #m^2/ s\n\n#calculation\nv =4* Q/ pi /d **2;\ndel_pf =2*f* Rho *v**2*L/d;\nH_f =4* f*L*v **2/( d*2*g);\nH_exit =v **2/2/ g;\nH_entrance =v **2/4/ g;\nH= H_f+ H_exit + H_entrance ;\n\n#result\nprint \"The pressure drop due to friction is (kN/m^2)\",round(del_pf,2)\nprint \" and the difference in levels is (m)\",round(H,3)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "from math import pi\n",
+      "import math\n",
+      "Q =15.0/3600; # m^3/ s\n",
+      "d =0.05; #m\n",
+      "Rho =780.0;\n",
+      "u =1.7*10**( -3) ; #Ns/m^2\n",
+      "f =0.0065;\n",
+      "L =100; #m\n",
+      "g =9.8; #m^2/ s\n",
+      "\n",
+      "#calculation\n",
+      "v =4* Q/ pi /d **2;\n",
+      "del_pf =2*f* Rho *v**2*L/d;\n",
+      "H_f =4* f*L*v **2/( d*2*g);\n",
+      "H_exit =v **2/2/ g;\n",
+      "H_entrance =v **2/4/ g;\n",
+      "H= H_f+ H_exit + H_entrance ;\n",
+      "\n",
+      "#result\n",
+      "print \"The pressure drop due to friction is (kN/m^2)\",round(del_pf,2)\n",
+      "print \" and the difference in levels is (m)\",round(H,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The pressure drop due to friction is (kN/m^2) 91324.16\n and the difference in levels is (m) 12.292\n"
+       "text": [
+        "The pressure drop due to friction is (kN/m^2) 91324.16\n",
+        " and the difference in levels is (m) 12.292\n"
+       ]
       }
      ],
      "prompt_number": 3
@@ -80,19 +169,49 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.7,Page 206"
+     "source": [
+      "Example 8.7,Page 206"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding rate of flow\n\n#variable declaration\nimport math\nf =0.005;\nL =10; # m\nd =0.025; # m\ng =9.81; # m/ s ^2\n\n#Calculations\n# H L=4\u0003 f \u0003L/d\u0003v ^2/2/ g+0.5\u0003 v ^2/2/ g\n# H L=8.5\u0003 v ^2/2/ g\n# By Be r n o u l l i e q u a t i o n we g e t\n#H=2.62+9.5\u0003 v2 ^2/2/ g\n# Applying the Be r n o u l l i e q u a t i o n between the liquid surface and discharge point\n# H L=33.5\u0003 v2 ^2/2/ g\n#Solving above two we g e t\nH =2.62+9.5* v2 **2/2/ g;\nv2 =1.9; # m/ s\nQ= pi*d **2/4* v2;\n\n#result\nprint\"Rate of flow (m^3/s)=\",round(Q,5)\nprint\"The minimum allowable height(m) =\",round(H,3)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "import math\n",
+      "f =0.005;\n",
+      "L =10; # m\n",
+      "d =0.025; # m\n",
+      "g =9.81; # m/ s ^2\n",
+      "\n",
+      "#Calculations\n",
+      "# H L=4\u0003 f \u0003L/d\u0003v ^2/2/ g+0.5\u0003 v ^2/2/ g\n",
+      "# H L=8.5\u0003 v ^2/2/ g\n",
+      "# By Be r n o u l l i e q u a t i o n we g e t\n",
+      "#H=2.62+9.5\u0003 v2 ^2/2/ g\n",
+      "# Applying the Be r n o u l l i e q u a t i o n between the liquid surface and discharge point\n",
+      "# H L=33.5\u0003 v2 ^2/2/ g\n",
+      "#Solving above two we g e t\n",
+      "H =2.62+9.5* v2 **2/2/ g;\n",
+      "v2 =1.9; # m/ s\n",
+      "Q= pi*d **2/4* v2;\n",
+      "\n",
+      "#result\n",
+      "print\"Rate of flow (m^3/s)=\",round(Q,5)\n",
+      "print\"The minimum allowable height(m) =\",round(H,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Rate of flow (m^3/s)= 0.0009321875\nThe minimum allowable height(m) = 4.36796126402\n"
+       "text": [
+        "Rate of flow (m^3/s)= 0.0009321875\n",
+        "The minimum allowable height(m) = 4.36796126402\n"
+       ]
       }
      ],
      "prompt_number": 16
@@ -101,19 +220,42 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.8,Page 208"
+     "source": [
+      "Example 8.8,Page 208"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding rate of flow\n\n#variable declaration\nimport math\nd_A =0.025; # m\nv_A =1.21; # m/ s\nd_B =0.05; # m\nv_B =1.71; # m/ s\n\n#calculation\nQ_A = pi *d_A **2* v_A /4;\nQ_B = pi *d_B **2* v_B /4;\n\n#result\nprint \"The rate of flow through parallel pipes A is \",round(Q_A,5)\nprint\"The rate of flow through parallel pipes B is (m^3/s)=\",round(Q_B,5);\n\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "import math\n",
+      "d_A =0.025; # m\n",
+      "v_A =1.21; # m/ s\n",
+      "d_B =0.05; # m\n",
+      "v_B =1.71; # m/ s\n",
+      "\n",
+      "#calculation\n",
+      "Q_A = pi *d_A **2* v_A /4;\n",
+      "Q_B = pi *d_B **2* v_B /4;\n",
+      "\n",
+      "#result\n",
+      "print \"The rate of flow through parallel pipes A is \",round(Q_A,5)\n",
+      "print\"The rate of flow through parallel pipes B is (m^3/s)=\",round(Q_B,5);\n",
+      "\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The rate of flow through parallel pipes A is  0.00059\nThe rate of flow through parallel pipes B is (m^3/s)= 0.00336\n"
+       "text": [
+        "The rate of flow through parallel pipes A is  0.00059\n",
+        "The rate of flow through parallel pipes B is (m^3/s)= 0.00336\n"
+       ]
       }
      ],
      "prompt_number": 4
@@ -122,19 +264,47 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.9,Page 210"
+     "source": [
+      "Example 8.9,Page 210"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding rate of flow\n\n#variable declaration\nfrom math import pi\nimport math\nd2 =0.06; # m\nd1 =0.12; # m\nk =0.44;\nf =0.05;\nL1 =500; # m\ng =9.81; # m/ s ^2\n\n#Calcualtion\n# v1=d2 ^2/ d1 ^2\u0003 v2\n# H f=4\u0003 f \u0003L1 /16/ d\u0003 v2 ^2/2/ g\n# H c=k\u0003 v2 ^2/2/ g\n# H f=4\u0003 f \u0003L2/d\u0003 v2 ^2/2/ g\n# H_exit=v2 ^2/2/ g\nv2= math.sqrt (30*2* g /173.4);\nQ= pi*d2 **2/4* v2;\n\n#result\nprint\"The rate of flow (m^3/s) =\",round(Q,5)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "from math import pi\n",
+      "import math\n",
+      "d2 =0.06; # m\n",
+      "d1 =0.12; # m\n",
+      "k =0.44;\n",
+      "f =0.05;\n",
+      "L1 =500; # m\n",
+      "g =9.81; # m/ s ^2\n",
+      "\n",
+      "#Calcualtion\n",
+      "# v1=d2 ^2/ d1 ^2\u0003 v2\n",
+      "# H f=4\u0003 f \u0003L1 /16/ d\u0003 v2 ^2/2/ g\n",
+      "# H c=k\u0003 v2 ^2/2/ g\n",
+      "# H f=4\u0003 f \u0003L2/d\u0003 v2 ^2/2/ g\n",
+      "# H_exit=v2 ^2/2/ g\n",
+      "v2= math.sqrt (30*2* g /173.4);\n",
+      "Q= pi*d2 **2/4* v2;\n",
+      "\n",
+      "#result\n",
+      "print\"The rate of flow (m^3/s) =\",round(Q,5)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The rate of flow (m^3/s) = 0.00521\n"
+       "text": [
+        "The rate of flow (m^3/s) = 0.00521\n"
+       ]
       }
      ],
      "prompt_number": 5
@@ -143,19 +313,51 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.10,Page 212"
+     "source": [
+      "Example 8.10,Page 212"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding rate of flow\n\n#variable declaration\nimport math\nm =12.0*10**3/3600; #kg / s\nRho =815; #kg /m^3\nd =0.05; #m\ne =0.02;\nd1 =50; # m\nd2 =0.038; # m\ng =9.8; #m\n\n#calculation\nv =4* m/ Rho / pi /d**2;\nf1 =1/(2* math.log (d1/e,10) +2.28) **2;\nL_eq =d1 +2* d1*d;\nH_50mm =4* f1* L_eq *v **2/( d *2* g);\nv =4* m/( Rho * pi *d2 **2) ;\nf2 =1/(2* math.log (38/ e,10) +2.28) **2;\nL_eq =d1 +2* d1*d2;\nH_38mm =4* f2* L_eq *v **2/( d2 *2* g);\nHr =0.2* v **2/(2* g);\nH_L = H_50mm + H_38mm +Hr;\ndel_p_f =Rho*g* H_L ;\n\n#result\nprint \"The total pressure drop due to friction through the pipe system is (N/m^2) =\",round(del_p_f,3)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "import math\n",
+      "m =12.0*10**3/3600; #kg / s\n",
+      "Rho =815; #kg /m^3\n",
+      "d =0.05; #m\n",
+      "e =0.02;\n",
+      "d1 =50; # m\n",
+      "d2 =0.038; # m\n",
+      "g =9.8; #m\n",
+      "\n",
+      "#calculation\n",
+      "v =4* m/ Rho / pi /d**2;\n",
+      "f1 =1/(2* math.log (d1/e,10) +2.28) **2;\n",
+      "L_eq =d1 +2* d1*d;\n",
+      "H_50mm =4* f1* L_eq *v **2/( d *2* g);\n",
+      "v =4* m/( Rho * pi *d2 **2) ;\n",
+      "f2 =1/(2* math.log (38/ e,10) +2.28) **2;\n",
+      "L_eq =d1 +2* d1*d2;\n",
+      "H_38mm =4* f2* L_eq *v **2/( d2 *2* g);\n",
+      "Hr =0.2* v **2/(2* g);\n",
+      "H_L = H_50mm + H_38mm +Hr;\n",
+      "del_p_f =Rho*g* H_L ;\n",
+      "\n",
+      "#result\n",
+      "print \"The total pressure drop due to friction through the pipe system is (N/m^2) =\",round(del_p_f,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The total pressure drop due to friction through the pipe system is (N/m^2) = 479792.576\n"
+       "text": [
+        "The total pressure drop due to friction through the pipe system is (N/m^2) = 479792.576\n"
+       ]
       }
      ],
      "prompt_number": 6
@@ -164,19 +366,36 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.11,Page 214"
+     "source": [
+      "Example 8.11,Page 214"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding rate of flow\n\n#variable declaration\nimport math\n#H L=1.2\u0003 v ^2/2/ g\n# H L=4\u0003 f \u0003 L eq /d\u0003V^2/2/ g\n# L eq=60\u0003d\n# H L=240\u0003 f \u0003v ^2/2/ g\n# Combining the two e q u a t i o n s f o r head l o s s\n# 1.2\u0003v^2/2/ g=240\u0003 f_\u0003v^2/2/ g\n\nf =1.2/240;\nprint \" Friction factor =\",round(f,3)\n",
+     "input": [
+      "\n",
+      "#variable declaration\n",
+      "import math\n",
+      "#H L=1.2\u0003 v ^2/2/ g\n",
+      "# H L=4\u0003 f \u0003 L eq /d\u0003V^2/2/ g\n",
+      "# L eq=60\u0003d\n",
+      "# H L=240\u0003 f \u0003v ^2/2/ g\n",
+      "# Combining the two e q u a t i o n s f o r head l o s s\n",
+      "# 1.2\u0003v^2/2/ g=240\u0003 f_\u0003v^2/2/ g\n",
+      "\n",
+      "f =1.2/240;\n",
+      "print \" Friction factor =\",round(f,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Friction factor = 0.005\n"
+       "text": [
+        " Friction factor = 0.005\n"
+       ]
       }
      ],
      "prompt_number": 7
@@ -185,19 +404,54 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.12,Page 216"
+     "source": [
+      "Example 8.12,Page 216"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding rate of flow\n\n#variable declaration\nimport math\n# dp AB+dp BC=dp AD+dp DC\n# dp AD=2\u0003 f \u0003 rho \u0003v ^2\u0003L/d\n# dp AD=16600\u0003(3_Q) ^2\n# Li k ewi s e\n# dp AB=16600\u0003Q^2\n# dp BC=16600\u0003(Q+0.5) ^2\n# dp DC=16600\u0003(2.1_Q) ^2\n#By s o l v i n g above 5 e qu a t i o ns , we g e t\nQ =1.175; # litres per second\n\n#Calculation\ndp_AD =16600*(3 - Q) **2;\ndp_AB =16600* Q**2;\ndp_BC =16600*( Q +0.5) **2;\ndp_DC =16600*(2.1 - Q) **2;\n\nprint\"The rate of flow from B to C (l/s)=\",round(Q+0.5,3)\nprint \"dp_AD(kN/m^2 =\",round(dp_AD/1000,3);\nprint\"dp AB (kN/m^2)=\",round(dp_AB/1000,3);\nprint\"dp BC(kN/m^2) =\",round(dp_BC/1000,3);\nprint\"dp DC (kN/m^2)=\",round(dp_DC/1000,3);\nprint\"The lowest pressure is in the pipe coneecting C and D\"",
+     "input": [
+      "\n",
+      "#variable declaration\n",
+      "import math\n",
+      "# dp AB+dp BC=dp AD+dp DC\n",
+      "# dp AD=2\u0003 f \u0003 rho \u0003v ^2\u0003L/d\n",
+      "# dp AD=16600\u0003(3_Q) ^2\n",
+      "# Li k ewi s e\n",
+      "# dp AB=16600\u0003Q^2\n",
+      "# dp BC=16600\u0003(Q+0.5) ^2\n",
+      "# dp DC=16600\u0003(2.1_Q) ^2\n",
+      "#By s o l v i n g above 5 e qu a t i o ns , we g e t\n",
+      "Q =1.175; # litres per second\n",
+      "\n",
+      "#Calculation\n",
+      "dp_AD =16600*(3 - Q) **2;\n",
+      "dp_AB =16600* Q**2;\n",
+      "dp_BC =16600*( Q +0.5) **2;\n",
+      "dp_DC =16600*(2.1 - Q) **2;\n",
+      "\n",
+      "print\"The rate of flow from B to C (l/s)=\",round(Q+0.5,3)\n",
+      "print \"dp_AD(kN/m^2 =\",round(dp_AD/1000,3);\n",
+      "print\"dp AB (kN/m^2)=\",round(dp_AB/1000,3);\n",
+      "print\"dp BC(kN/m^2) =\",round(dp_BC/1000,3);\n",
+      "print\"dp DC (kN/m^2)=\",round(dp_DC/1000,3);\n",
+      "print\"The lowest pressure is in the pipe coneecting C and D\""
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The rate of flow from B to C (l/s)= 1.675\ndp_AD(kN/m^2 = 55.288\ndp AB (kN/m^2)= 22.918\ndp BC(kN/m^2) = 46.573\ndp DC (kN/m^2)= 14.203\nThe lowest pressure is in the pipe coneecting C and D\n"
+       "text": [
+        "The rate of flow from B to C (l/s)= 1.675\n",
+        "dp_AD(kN/m^2 = 55.288\n",
+        "dp AB (kN/m^2)= 22.918\n",
+        "dp BC(kN/m^2) = 46.573\n",
+        "dp DC (kN/m^2)= 14.203\n",
+        "The lowest pressure is in the pipe coneecting C and D\n"
+       ]
       }
      ],
      "prompt_number": 8
@@ -206,19 +460,51 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.13,Page 221"
+     "source": [
+      "Example 8.13,Page 221"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time taken\n\n#variable declaration\nfrom math import pi\nimport math\nfrom scipy import integrate\nH2 =0.5; #m\nH1 =2.0; #m\nA =4.0; #m^2\nf =0.005;\nL =20.0; #m\nd =0.025; #m\ng =9.81; # m/ s ^2\n\n#calculation\na= pi *d **2/4;\nc=A*math.sqrt( 4*f*L/d  +2.5)/a/math.sqrt(2*g); # constant term of integration\ndef integrand(H,c):\n    return -c*(H)**(-0.5)\n#integrate function '-A\u0003*sqrt ( ( 4*f \u0003L/d ) +2.5) / a /( sqrt(2\u0003*g ) )* \u0003(H) ^(-1/2)\n#f=lambda H:-4*1978.18*H**(-0.5);\n\n\nt=integrate.quad(integrand,H1,H2,args=(c))\n                 \n#result\nprint \"Time taken (s)=\",round(t[0],2)",
+     "input": [
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "from math import pi\n",
+      "import math\n",
+      "from scipy import integrate\n",
+      "H2 =0.5; #m\n",
+      "H1 =2.0; #m\n",
+      "A =4.0; #m^2\n",
+      "f =0.005;\n",
+      "L =20.0; #m\n",
+      "d =0.025; #m\n",
+      "g =9.81; # m/ s ^2\n",
+      "\n",
+      "#calculation\n",
+      "a= pi *d **2/4;\n",
+      "c=A*math.sqrt( 4*f*L/d  +2.5)/a/math.sqrt(2*g); # constant term of integration\n",
+      "def integrand(H,c):\n",
+      "    return -c*(H)**(-0.5)\n",
+      "#integrate function '-A\u0003*sqrt ( ( 4*f \u0003L/d ) +2.5) / a /( sqrt(2\u0003*g ) )* \u0003(H) ^(-1/2)\n",
+      "#f=lambda H:-4*1978.18*H**(-0.5);\n",
+      "\n",
+      "\n",
+      "t=integrate.quad(integrand,H1,H2,args=(c))\n",
+      "                 \n",
+      "#result\n",
+      "print \"Time taken (s)=\",round(t[0],2)"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Time taken (s)= 11190.29\n"
+       "text": [
+        "Time taken (s)= 11190.29\n"
+       ]
       }
      ],
      "prompt_number": 39
@@ -227,19 +513,43 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.14,Page 223"
+     "source": [
+      "Example 8.14,Page 223"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding pressure drop\n\n#variable decleration\nfrom math import pi\nd0 =0.15; # m\nd1 =0.1; #m\nQ =50.0/3600; # m^3/ s\nf =0.0052;\nRho =972;\n\n\n#calculation\na= pi /4*(( d0)**2 -( d1) **2) ;\nP= pi *(( d0)+( d1));\nd_eq =4*a/P;\nv=Q/a;\ndel_p_f =2*f* Rho *v**2/ d_eq ;\n\n#result\nprint\" the pressure drop due to friction per metre lenfth of tube is found to be(Nm^2/m)\",round(del_p_f,3)\n",
+     "input": [
+      "\n",
+      "#variable decleration\n",
+      "from math import pi\n",
+      "d0 =0.15; # m\n",
+      "d1 =0.1; #m\n",
+      "Q =50.0/3600; # m^3/ s\n",
+      "f =0.0052;\n",
+      "Rho =972;\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "a= pi /4*(( d0)**2 -( d1) **2) ;\n",
+      "P= pi *(( d0)+( d1));\n",
+      "d_eq =4*a/P;\n",
+      "v=Q/a;\n",
+      "del_p_f =2*f* Rho *v**2/ d_eq ;\n",
+      "\n",
+      "#result\n",
+      "print\" the pressure drop due to friction per metre lenfth of tube is found to be(Nm^2/m)\",round(del_p_f,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " the pressure drop due to friction per metre lenfth of tube is found to be(Nm^2/m) 404.636\n"
+       "text": [
+        " the pressure drop due to friction per metre lenfth of tube is found to be(Nm^2/m) 404.636\n"
+       ]
       }
      ],
      "prompt_number": 12
@@ -248,19 +558,44 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.15,Page 225"
+     "source": [
+      "Example 8.15,Page 225"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Head Loss\n\n#variable decleration\nfrom math import pi\nfrom scipy import integrate\nf =0.005;\nQ =0.07; #m^3/ s\ng =9.81; #m/ s ^2\nL=3;\n\n#calculation\nu=0.3-0.0666*L;\ndef integrand(l,f,Q,g,u):\n    return 32*f*Q**2/pi**2/g*(0.3-0.0666*l)**-5    \n\nH_f=integrate.quad(integrand,0,3,args=(f,Q,g,u));\nH_f=round(H_f[0],1)\n\n#result\nprint\" Fractional head loss(m) =\",H_f\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable decleration\n",
+      "from math import pi\n",
+      "from scipy import integrate\n",
+      "f =0.005;\n",
+      "Q =0.07; #m^3/ s\n",
+      "g =9.81; #m/ s ^2\n",
+      "L=3;\n",
+      "\n",
+      "#calculation\n",
+      "u=0.3-0.0666*L;\n",
+      "def integrand(l,f,Q,g,u):\n",
+      "    return 32*f*Q**2/pi**2/g*(0.3-0.0666*l)**-5    \n",
+      "\n",
+      "H_f=integrate.quad(integrand,0,3,args=(f,Q,g,u));\n",
+      "H_f=round(H_f[0],1)\n",
+      "\n",
+      "#result\n",
+      "print\" Fractional head loss(m) =\",H_f\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Fractional head loss(m) = 0.3\n"
+       "text": [
+        " Fractional head loss(m) = 0.3\n"
+       ]
       }
      ],
      "prompt_number": 3
@@ -269,19 +604,42 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 8.16,Page 228"
+     "source": [
+      "Example 8.16,Page 228"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#variable decleration\nfrom math import pi\nimport math\ng =9.81; # m/ s ^2\nH =4; # m\nf =0.006;\nL =50; # m\nd =0.1; # m\ne=2.718;\n\n\n#calculation\nv1= math.sqrt (2* g*H /(4* f*L/d + 13) );\nv2=0.99*v1;\nt=2/v1*math.log((v1+v2)/(v1-v2),e)\n\n#result\nprint\"Time taken (s)=\",round(t,2)\n",
+     "input": [
+      "\n",
+      "from math import pi\n",
+      "import math\n",
+      "g =9.81; # m/ s ^2\n",
+      "H =4; # m\n",
+      "f =0.006;\n",
+      "L =50; # m\n",
+      "d =0.1; # m\n",
+      "e=2.718;\n",
+      "\n",
+      "\n",
+      "#calculation\n",
+      "v1= math.sqrt (2* g*H /(4* f*L/d + 13) );\n",
+      "v2=0.99*v1;\n",
+      "t=2/v1*math.log((v1+v2)/(v1-v2),e)\n",
+      "\n",
+      "#result\n",
+      "print\"Time taken (s)=\",round(t,2)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Time taken (s)= 5.98\n"
+       "text": [
+        "Time taken (s)= 5.98\n"
+       ]
       }
      ],
      "prompt_number": 14
@@ -289,7 +647,7 @@
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "",
+     "input": [],
      "language": "python",
      "metadata": {},
      "outputs": []
diff --git a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_9_1.ipynb b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_9_1.ipynb
index 3b69a21d..3a6f5367 100644
--- a/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_9_1.ipynb
+++ b/Fluid_Mechanics_-_Worked_Examples_For_Engineers/Chapter_9_1.ipynb
@@ -1,6 +1,7 @@
 {
  "metadata": {
-  "name": "Chapter 9"
+  "name": "",
+  "signature": "sha256:8db0dc01c526205f6578e94b8a174e884a60249a5154cf023d12bae195b0ba58"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -11,25 +12,74 @@
      "cell_type": "heading",
      "level": 1,
      "metadata": {},
-     "source": "Chapter 9,Pumps"
+     "source": [
+      "Chapter 9,Pumps"
+     ]
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 9.2 (part1 and part2), Page 239"
+     "source": [
+      "Example 9.2 (part1 and part2), Page 239"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding head loss \nimport math\n\n#Variable Declaration\nd =0.1; # m\nv_r =2; # m/ s\nf =0.005;\ng =9.81; #m/ s ^2\npi=3.14;\nL_s =2; #m\nL_r =10; # m\nQ1 =1.1*10**( -2) ; #m^3/ s\nz_t =12; #m\nz_s=5 #m\nL1 =20; #m\n\n#Calculation\nQ= math.pi*d**2/4*v_r ;\nH =12-70*Q-4300*Q*Q ;\nk =2*g*H/ v_r/v_r-(4*f*( L_s +L_r)/d)-1;\n\n#result\nprint\"The head loss across the restriction orifice (in terms of velocity head)\",round(k,3) \n\n#Part2\n#Finding System Head and Delivered Head\n# For the case of the valve being fully open\n\n#calculation\nv_t =4*Q1/math.pi/d**2;\nv_r =((2*g*( z_t-z_s )+(4*f*L1/d+1)*v_t**2) /(4*f*L_r/d+k+1))**(0.5) ;\nH1 =4*f* L_r /d*v_r**2/2/ g + 4*f* L_s /d*( v_r**2+ v_t**2) /2/g+k* v_r**2/2/ g + v_r**2/2/ g;\nQ= math.pi*d**2/4*( v_t+ v_r );\nH2 =12 -70*Q -4300* Q**2;\n\n#result\nprint\" System head(m)=\",round(H1,3)\nprint\" Delivered head(m)=\",round(H2,3)\nprint\"The delivered head therefore closely matches the system head at the flow rate of 1.1*\u000310^( -2) m^3/ s\"\n",
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "#Variable Declaration\n",
+      "d =0.1; # m\n",
+      "v_r =2; # m/ s\n",
+      "f =0.005;\n",
+      "g =9.81; #m/ s ^2\n",
+      "pi=3.14;\n",
+      "L_s =2; #m\n",
+      "L_r =10; # m\n",
+      "Q1 =1.1*10**( -2) ; #m^3/ s\n",
+      "z_t =12; #m\n",
+      "z_s=5 #m\n",
+      "L1 =20; #m\n",
+      "\n",
+      "#Calculation\n",
+      "Q= math.pi*d**2/4*v_r ;\n",
+      "H =12-70*Q-4300*Q*Q ;\n",
+      "k =2*g*H/ v_r/v_r-(4*f*( L_s +L_r)/d)-1;\n",
+      "\n",
+      "#result\n",
+      "print\"The head loss across the restriction orifice (in terms of velocity head)\",round(k,3) \n",
+      "\n",
+      "#Part2\n",
+      "#Finding System Head and Delivered Head\n",
+      "# For the case of the valve being fully open\n",
+      "\n",
+      "#calculation\n",
+      "v_t =4*Q1/math.pi/d**2;\n",
+      "v_r =((2*g*( z_t-z_s )+(4*f*L1/d+1)*v_t**2) /(4*f*L_r/d+k+1))**(0.5) ;\n",
+      "H1 =4*f* L_r /d*v_r**2/2/ g + 4*f* L_s /d*( v_r**2+ v_t**2) /2/g+k* v_r**2/2/ g + v_r**2/2/ g;\n",
+      "Q= math.pi*d**2/4*( v_t+ v_r );\n",
+      "H2 =12 -70*Q -4300* Q**2;\n",
+      "\n",
+      "#result\n",
+      "print\" System head(m)=\",round(H1,3)\n",
+      "print\" Delivered head(m)=\",round(H2,3)\n",
+      "print\"The delivered head therefore closely matches the system head at the flow rate of 1.1*\u000310^( -2) m^3/ s\"\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The head loss across the restriction orifice (in terms of velocity head) 44.863\n System head(m)= 7.603\n Delivered head(m)= 7.628\nThe delivered head therefore closely matches the system head at the flow rate of 1.1*\u000310^( -2) m^3/ s\n"
+       "text": [
+        "The head loss across the restriction orifice (in terms of velocity head) 44.863\n",
+        " System head(m)= 7.603\n",
+        " Delivered head(m)= 7.628\n",
+        "The delivered head therefore closely matches the system head at the flow rate of 1.1*\u000310^( -2) m^3/ s\n"
+       ]
       }
      ],
      "prompt_number": 3
@@ -38,19 +88,48 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 9.6, Page 247"
+     "source": [
+      "Example 9.6, Page 247"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding time and volume of liquid\n\nfrom math import pi\n#Variable Declaration\nNPSH =5; # m\np_v =18*10**3; # N/m^2\np_l =0.94*101.3*10**3; # N/m^2\nrho =970;#kg /m^3\ng =9.81; # m/ s ^2\nz_s =3; # m\nH_L =0.5; # m\nd =3; # m\nh =2.5; # m\nQ =5; #m^3/h\n\n#calculation\nz1= NPSH +( p_v -p_l)/ rho /g + z_s + H_L ;\nV= pi /4*d**2*(h-z1);\nt=V/Q;\n\n#result\nprint\"Quantity of liquid delivered (m^3)=\",round(V,2)\nprint\"Time taken (h)=\",round(t,2) \n",
+     "input": [
+      "\n",
+      "\n",
+      "from math import pi\n",
+      "#Variable Declaration\n",
+      "NPSH =5; # m\n",
+      "p_v =18*10**3; # N/m^2\n",
+      "p_l =0.94*101.3*10**3; # N/m^2\n",
+      "rho =970;#kg /m^3\n",
+      "g =9.81; # m/ s ^2\n",
+      "z_s =3; # m\n",
+      "H_L =0.5; # m\n",
+      "d =3; # m\n",
+      "h =2.5; # m\n",
+      "Q =5; #m^3/h\n",
+      "\n",
+      "#calculation\n",
+      "z1= NPSH +( p_v -p_l)/ rho /g + z_s + H_L ;\n",
+      "V= pi /4*d**2*(h-z1);\n",
+      "t=V/Q;\n",
+      "\n",
+      "#result\n",
+      "print\"Quantity of liquid delivered (m^3)=\",round(V,2)\n",
+      "print\"Time taken (h)=\",round(t,2) \n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Quantity of liquid delivered (m^3)= 14.95\nTime taken (h)= 2.99\n"
+       "text": [
+        "Quantity of liquid delivered (m^3)= 14.95\n",
+        "Time taken (h)= 2.99\n"
+       ]
       }
      ],
      "prompt_number": 4
@@ -59,19 +138,41 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 9.8,Page 251"
+     "source": [
+      "Example 9.8,Page 251"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#finding minimum level of alarm\n\n#variable Declaration\nN_s =0.14; #m^ ( 3 / 4 ) s ^(-3/2)\nH =30; #m\np_v =7.38*10**3; #N/m^2\np_l =50*10**3; #N/m^2\nrho =992; #kg /m^3\ng =9.81; #m/ s ^2\nH_L =0.2; #m\n\n#calculation\nNPSH =2.8* N_s **(1.33) *H;\nz1= NPSH +( p_v -p_l)/ rho /g+H_L;\n\n#result\nprint\"The minimum level of the alarm (m)=\",round(z1,3)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#variable Declaration\n",
+      "N_s =0.14; #m^ ( 3 / 4 ) s ^(-3/2)\n",
+      "H =30; #m\n",
+      "p_v =7.38*10**3; #N/m^2\n",
+      "p_l =50*10**3; #N/m^2\n",
+      "rho =992; #kg /m^3\n",
+      "g =9.81; #m/ s ^2\n",
+      "H_L =0.2; #m\n",
+      "\n",
+      "#calculation\n",
+      "NPSH =2.8* N_s **(1.33) *H;\n",
+      "z1= NPSH +( p_v -p_l)/ rho /g+H_L;\n",
+      "\n",
+      "#result\n",
+      "print\"The minimum level of the alarm (m)=\",round(z1,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "The minimum level of the alarm (m)= 1.967\n"
+       "text": [
+        "The minimum level of the alarm (m)= 1.967\n"
+       ]
       }
      ],
      "prompt_number": 5
@@ -80,28 +181,96 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 9.10,Page 255"
+     "source": [
+      "Example 9.10,Page 255"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Final Speed\n\n%pylab inline\n#Vabiable Declaration\nimport numpy as np\nfrom pylab import *\ndz =10; #z2-z1\ng =9.81; #m/ s ^2\nd =0.05; #m\nf =0.005;\nL =100; #m\nN1 =1200; #rpm\n\n# H=z2-z1+16\u0003Q^2/2/ g /%pi ^2/d ^4\u0003(4\u0003 f \u0003L/d+1)\n# H=10+5.42\u000310^5\u0003Q^2\n\n#Calculation\nQ =[0.000, 0.002, 0.004, 0.006, 0.008, 0.010];\nH_p =[40.0, 39.5, 38.0, 35.0, 30.0, 20.0];\nH_s =[10.0, 12.2, 18.7, 29.5, 44.7, 64.2];\nfigure()\nplot (Q,H_p , 'b')\nplot (Q,H_s , 'r')\nxlabel('FLOW (m^3/s)')\nylabel('HEAD (m)')\ntitle('system and pump')\n#xtitle (\" \" , \"Flow\" , \"Head\")\n#legend (\"pump\" , \" system\")\na =[0.0066, 0.0066];\nb =[0, 33.8];\n\nplot (a,b)\ne =[0, 0.0066];\nf =[33.8, 33.8];\nplot (e,f)\ni =[0.0049, 0.0049];\nh =[0, 23];\n\nplot (i,h)\nl =[0, 0.00495];\nm =[23, 23];\nplot (l,m)\nshow()\n\n#From graph\nH1 =34.0; #m\nH2 =23.0; # m\nQ1 =0.0066; #m^3/ s\nQ2 =0.00495; #m^3/ s\nN2=N1 *( H2/H1)**(0.5) ;\n\n#result\nprint \"Duty point(m^3/s) =\",round(Q1,4)\nprint \"The speed of the pump to reduce the flow by 25% =\",round(N2,3)\n\n\n\n\n",
+     "input": [
+      "\n",
+      "%pylab inline\n",
+      "#Vabiable Declaration\n",
+      "import numpy as np\n",
+      "from pylab import *\n",
+      "dz =10; #z2-z1\n",
+      "g =9.81; #m/ s ^2\n",
+      "d =0.05; #m\n",
+      "f =0.005;\n",
+      "L =100; #m\n",
+      "N1 =1200; #rpm\n",
+      "\n",
+      "# H=z2-z1+16\u0003Q^2/2/ g /%pi ^2/d ^4\u0003(4\u0003 f \u0003L/d+1)\n",
+      "# H=10+5.42\u000310^5\u0003Q^2\n",
+      "\n",
+      "#Calculation\n",
+      "Q =[0.000, 0.002, 0.004, 0.006, 0.008, 0.010];\n",
+      "H_p =[40.0, 39.5, 38.0, 35.0, 30.0, 20.0];\n",
+      "H_s =[10.0, 12.2, 18.7, 29.5, 44.7, 64.2];\n",
+      "figure()\n",
+      "plot (Q,H_p , 'b')\n",
+      "plot (Q,H_s , 'r')\n",
+      "xlabel('FLOW (m^3/s)')\n",
+      "ylabel('HEAD (m)')\n",
+      "title('system and pump')\n",
+      "#xtitle (\" \" , \"Flow\" , \"Head\")\n",
+      "#legend (\"pump\" , \" system\")\n",
+      "a =[0.0066, 0.0066];\n",
+      "b =[0, 33.8];\n",
+      "\n",
+      "plot (a,b)\n",
+      "e =[0, 0.0066];\n",
+      "f =[33.8, 33.8];\n",
+      "plot (e,f)\n",
+      "i =[0.0049, 0.0049];\n",
+      "h =[0, 23];\n",
+      "\n",
+      "plot (i,h)\n",
+      "l =[0, 0.00495];\n",
+      "m =[23, 23];\n",
+      "plot (l,m)\n",
+      "show()\n",
+      "\n",
+      "#From graph\n",
+      "H1 =34.0; #m\n",
+      "H2 =23.0; # m\n",
+      "Q1 =0.0066; #m^3/ s\n",
+      "Q2 =0.00495; #m^3/ s\n",
+      "N2=N1 *( H2/H1)**(0.5) ;\n",
+      "\n",
+      "#result\n",
+      "print \"Duty point(m^3/s) =\",round(Q1,4)\n",
+      "print \"The speed of the pump to reduce the flow by 25% =\",round(N2,3)\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "\nWelcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\nFor more information, type 'help(pylab)'.\n"
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'.\n"
+       ]
       },
       {
+       "metadata": {},
        "output_type": "display_data",
        "png": 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YIirD8ePA228D+/bJTvdffgHq1rV2VDahaVNg9mzZ7fTVV8DAgXI84803gcBA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\n"
       },
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "Duty point(m^3/s) = 0.0066\nThe speed of the pump to reduce the flow by 25% = 986.974\n"
+       "text": [
+        "Duty point(m^3/s) = 0.0066\n",
+        "The speed of the pump to reduce the flow by 25% = 986.974\n"
+       ]
       }
      ],
      "prompt_number": 6
@@ -110,19 +279,70 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 9.11,Page 257"
+     "source": [
+      "Example 9.11,Page 257"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding head, rotational frequency, power\n\n#Varable Declaration\nQ =0.05; # m^3/ s\nv =2; # m/ s\nf =0.005;\nL_s =5; # m\nd =0.178; #m\ng =9.81; # m/ s ^2\nL_d =20; # m\np2 =1.5*10**5; # N/m^2\np1 =0.5*10**5; # N/m^2\nrho =1000; # kg /m^3\nz2 =15; # m\nz1 =5; # m\nN1 =1500.0/60; # rps\npi=3.14;\nQ =[0, 5, 10, 15, 20, 25];\nH =[9.25, 8.81, 7.85, 6.48, 4.81, 2.96];\nP=[- 0.96, 1.03, 1.19, 1.26, 1.45];\nn1 =[0, 45, 75, 800, 75, 50];\nH =27.96; #m\nH1 =6.48; #m\nQ1 =0.015; #m^3/ s\nQ =0.05; #m^3/ s\nD1 =0.15; #m\nn =0.80;\n\n#calculation\nd =(4* Q/ pi /v)**(0.5) ;\nH_f_s =4* f* L_s /d*v **2/2/ g;\nH_f_d =4* f* L_d /d*v **2/2/ g;\nH =1/(1 -0.25) *(( p2 -p1)/ rho /g + v **2/2/ g + z2 - z1 +H_f_s + H_f_d );\n# n=rho \u0003g\u0003Q\u0003H/P\nN=N1 *(H/H1) **(0.75) *( Q1/Q) **(0.5) ;\nD=D1 *(Q*N1/Q1/N)**(0.2) ;\nP= rho *g*Q*H/n;\n\n#result\nprint \" Differential Head (m) =\",round(H,3);\nprint \"impeller diameter (m)=\",round(D,3)\nprint \"The rotational speed at maximum efficiency (rps)=\",round(N,3)\nprint \"Power input to the pump (W)=\",round(P,3)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#Varable Declaration\n",
+      "Q =0.05; # m^3/ s\n",
+      "v =2; # m/ s\n",
+      "f =0.005;\n",
+      "L_s =5; # m\n",
+      "d =0.178; #m\n",
+      "g =9.81; # m/ s ^2\n",
+      "L_d =20; # m\n",
+      "p2 =1.5*10**5; # N/m^2\n",
+      "p1 =0.5*10**5; # N/m^2\n",
+      "rho =1000; # kg /m^3\n",
+      "z2 =15; # m\n",
+      "z1 =5; # m\n",
+      "N1 =1500.0/60; # rps\n",
+      "pi=3.14;\n",
+      "Q =[0, 5, 10, 15, 20, 25];\n",
+      "H =[9.25, 8.81, 7.85, 6.48, 4.81, 2.96];\n",
+      "P=[- 0.96, 1.03, 1.19, 1.26, 1.45];\n",
+      "n1 =[0, 45, 75, 800, 75, 50];\n",
+      "H =27.96; #m\n",
+      "H1 =6.48; #m\n",
+      "Q1 =0.015; #m^3/ s\n",
+      "Q =0.05; #m^3/ s\n",
+      "D1 =0.15; #m\n",
+      "n =0.80;\n",
+      "\n",
+      "#calculation\n",
+      "d =(4* Q/ pi /v)**(0.5) ;\n",
+      "H_f_s =4* f* L_s /d*v **2/2/ g;\n",
+      "H_f_d =4* f* L_d /d*v **2/2/ g;\n",
+      "H =1/(1 -0.25) *(( p2 -p1)/ rho /g + v **2/2/ g + z2 - z1 +H_f_s + H_f_d );\n",
+      "# n=rho \u0003g\u0003Q\u0003H/P\n",
+      "N=N1 *(H/H1) **(0.75) *( Q1/Q) **(0.5) ;\n",
+      "D=D1 *(Q*N1/Q1/N)**(0.2) ;\n",
+      "P= rho *g*Q*H/n;\n",
+      "\n",
+      "#result\n",
+      "print \" Differential Head (m) =\",round(H,3);\n",
+      "print \"impeller diameter (m)=\",round(D,3)\n",
+      "print \"The rotational speed at maximum efficiency (rps)=\",round(N,3)\n",
+      "print \"Power input to the pump (W)=\",round(P,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Differential Head (m) = 27.958\nimpeller diameter (m)= 0.173\nThe rotational speed at maximum efficiency (rps)= 40.992\nPower input to the pump (W)= 17141.964\n"
+       "text": [
+        " Differential Head (m) = 27.958\n",
+        "impeller diameter (m)= 0.173\n",
+        "The rotational speed at maximum efficiency (rps)= 40.992\n",
+        "Power input to the pump (W)= 17141.964\n"
+       ]
       }
      ],
      "prompt_number": 10
@@ -131,19 +351,37 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 9.12,Page 259"
+     "source": [
+      "Example 9.12,Page 259"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#finding suction speed\n\n#Variable Declaration\nN =2000.0/60; #rps\nQ =50.0/3600; #m^3/ s\ng =9.81; #m/ s ^2\nH =5.0; #m\n\n#Calculation\nS_n =N*Q **(0.5) /(g)**(0.75)/(H) **(0.75) ;\n\n#result\nprint \" Suction specific speed =\",round(S_n,3)\n",
+     "input": [
+      "\n",
+      "\n",
+      "#Variable Declaration\n",
+      "N =2000.0/60; #rps\n",
+      "Q =50.0/3600; #m^3/ s\n",
+      "g =9.81; #m/ s ^2\n",
+      "H =5.0; #m\n",
+      "\n",
+      "#Calculation\n",
+      "S_n =N*Q **(0.5) /(g)**(0.75)/(H) **(0.75) ;\n",
+      "\n",
+      "#result\n",
+      "print \" Suction specific speed =\",round(S_n,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Suction specific speed = 0.212\n"
+       "text": [
+        " Suction specific speed = 0.212\n"
+       ]
       }
      ],
      "prompt_number": 8
@@ -152,19 +390,44 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": "Example 9.14,Page 264"
+     "source": [
+      "Example 9.14,Page 264"
+     ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#Finding Coeff. of Discharge and Power required\n\n# Variable Declaration\nA =0.01; #m^2\nL =0.3; # m\nN =60/60; # rps\nV_act =10.6/3600; # m^3/ s\nrho =1000; # kg /m^3\ng =9.81; # m/ s ^2\nQ =10.6/3600; # m^3/ s\nH =15; # m\n\n#calculation\nV=A*L*N;\nCd= V_act /V;\nP= rho *g*Q*H;\n\n#result\nprint \" Coeff. of Discharge =\",round(Cd,2)\nprint \"The power required (W)=\",round(P,3)\n",
+     "input": [
+      "\n",
+      "# Variable Declaration\n",
+      "A =0.01; #m^2\n",
+      "L =0.3; # m\n",
+      "N =60/60; # rps\n",
+      "V_act =10.6/3600; # m^3/ s\n",
+      "rho =1000; # kg /m^3\n",
+      "g =9.81; # m/ s ^2\n",
+      "Q =10.6/3600; # m^3/ s\n",
+      "H =15; # m\n",
+      "\n",
+      "#calculation\n",
+      "V=A*L*N;\n",
+      "Cd= V_act /V;\n",
+      "P= rho *g*Q*H;\n",
+      "\n",
+      "#result\n",
+      "print \" Coeff. of Discharge =\",round(Cd,2)\n",
+      "print \"The power required (W)=\",round(P,3)\n"
+     ],
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": " Coeff. of Discharge = 0.98\nThe power required (W)= 433.275\n"
+       "text": [
+        " Coeff. of Discharge = 0.98\n",
+        "The power required (W)= 433.275\n"
+       ]
       }
      ],
      "prompt_number": 9
@@ -172,7 +435,7 @@
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "",
+     "input": [],
      "language": "python",
      "metadata": {},
      "outputs": []