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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:dcc8f42948d54861c1b3938fe3127fa1812368cca6f5d25f547cacd56a9fddb2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Fluid Statics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page No : 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#Initialization of variables\n",
+      "z = 20000.       \t#ft\n",
+      "rate = -0.00356 \t#F/ft\n",
+      "T = 59.          \t#F\n",
+      "P = 14.7        \t#psia\n",
+      "gamma = 0.076   \t#lb/ft**3\n",
+      "\t\n",
+      "#calculations\n",
+      "P2 = P*144 - gamma*(z)\n",
+      "P2f = P2/144\n",
+      "P3 = P*math.exp(-gamma*z/(P*144))\n",
+      "P4 = ((P*144)**0.285 -0.285*gamma*z*(P*144)**(-0.715))**(1/0.285)\n",
+      "P4f = P4/144.\n",
+      "P5 = P*((460+T)/(460+T+rate*z))**(gamma*(T+460)/(P*144*rate))\n",
+      "\t\n",
+      "#Results\n",
+      "print ' Constant density'\n",
+      "print ' Final pressure  =  %.2f psia'%(P2f)\n",
+      "print ' \\nIsothermal'\n",
+      "print ' Final pressure  =  %.2f psia'%(P3)\n",
+      "print ' \\nIsentropic'\n",
+      "print ' Final pressure  =  %.1f psia'%(P4f)\n",
+      "print ' \\nLinear decrease'\n",
+      "print ' Final pressure  =  %.1f psia'%(P5)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Constant density\n",
+        " Final pressure  =  4.14 psia\n",
+        " \n",
+        "Isothermal\n",
+        " Final pressure  =  7.17 psia\n",
+        " \n",
+        "Isentropic\n",
+        " Final pressure  =  6.6 psia\n",
+        " \n",
+        "Linear decrease\n",
+        " Final pressure  =  6.8 psia\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 Page No : 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#Initialization of variables\n",
+      "wA = 53.5 \t#weight of A - lb/ft**3\n",
+      "wA2 = 8.4 \t#kN/m**3\n",
+      "wB = 78.8 \t#weight of B - lb/ft**3\n",
+      "wB2 = 12.3 \t#kN/m**3\n",
+      "PB = 30. \t#pressure at B - psi\n",
+      "PB2 = 200. \t#kN/m**2\n",
+      "AB = 1.3 \t#ft\n",
+      "AB2 = 40./100 \t#m\n",
+      "BC = 6.5 \t#ft\n",
+      "BC2 = 2. \t#m\n",
+      "CD = 10. \t#ft\n",
+      "CD2 = 3. \t#m\n",
+      "\t\n",
+      "#calculations\n",
+      "PAbyGB = PB*144/wB - AB*13.55*62.4/wB - (BC+CD) + (AB+BC)*wA/wB\n",
+      "PA = PAbyGB*wB/144.\n",
+      "PAbyGB2 = PB2/wB2 - AB2*13.55*9.81/wB2 - (BC2+CD2) + (AB2+BC2)*wA2/wB2\n",
+      "PA2 = PAbyGB2*wB2\n",
+      "\t\n",
+      "#Results\n",
+      "print ' English units'\n",
+      "print \" Final pressure  =  %.1f psi\"%(PA)\n",
+      "print ' \\n SI Units'\n",
+      "print \" Final pressure  =  %d kPa\"%(PA2+1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " English units\n",
+        " Final pressure  =  16.2 psi\n",
+        " \n",
+        " SI Units\n",
+        " Final pressure  =  106 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page No : 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import *\n",
+      "\n",
+      "#Initialization of variables\n",
+      "W = 500. \t#weight of gate - lb\n",
+      "width = 2. \t#ft\n",
+      "len1 = 4. \t#ft\n",
+      "CGx = 1.2 \t#ft\n",
+      "CGy = 0.9 \t#ft\n",
+      "theta = 30. \t#degrees\n",
+      "gam = 62.4 \t#lb/ft**3\n",
+      "\n",
+      "#calculations\n",
+      "Fv = width*len1 \t#multiply by gam*x\n",
+      "F = width/(2*math.cos(math.radians(theta))) \t#multiply by gam*x*x\n",
+      "vector = roots([F*gam*0.770/2,0, - Fv*gam*width,W*CGx])\n",
+      "\n",
+      "#Result\n",
+      "print 'The gate will remain closed between %.2f ft and %.2f ft'%(vector[2],vector[1])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The gate will remain closed between 0.61 ft and 5.67 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page No : 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#Initialization of variables\n",
+      "z1 = 1. \t#in\n",
+      "z2 = 2. \t#in\n",
+      "z3 = 2. \t#in\n",
+      "sOil = 0.8 \n",
+      "sWater = 1.\n",
+      "Pa = 3. \t#psi\n",
+      "\t\n",
+      "#calculations\n",
+      "Pd = (Pa) + (z2+z1)*sOil*62.4/144 + 62.4*z3/144\n",
+      "Fa = Pa*144*math.pi*z3**2\n",
+      "Fb = sOil*62.4*(z2+z1-(z2+z3)*z2/((z2+z1)*math.pi))*(math.pi*z3**2 /2)\n",
+      "Fc = sOil*62.4*(z2+z1)*(math.pi*z3**2 /2)\n",
+      "Fd = 62.4*(z2+z3)*z2/((z2+z1)*math.pi)*(math.pi*z3**2 /2)\n",
+      "F = Fa+Fb+Fc+Fd\n",
+      "yPa = z2+z1\n",
+      "yCb = z2+z1-(z2+z3)*z2/((z2+z1)*math.pi) \n",
+      "ICb = math.pi*(z2+z3)**4 /128 -0.5*math.pi*z2**2 *((z2+z3)*z2/((z2+z1)*math.pi))**2\n",
+      "yPb = yCb+ICb/(yCb*0.5*math.pi*z2**2)\n",
+      "yPc = z2+z1+ (z2+z3)*z2/((z2+z1)*math.pi) \n",
+      "ICd = ICb\n",
+      "yPd = z2+z1 + (z2+z3)*z2/((z2+z1)*math.pi) + ICb/((z2+z3)*z2/((z2+z1)*math.pi)*0.5*math.pi*z3**2 )\n",
+      "yP = (Fa*yPa+Fb*yPb+Fc*yPc+Fd*yPd)/F\n",
+      "\t\n",
+      "#Results\n",
+      "print ' case 1'\n",
+      "print ' Pressure at the bottom  =  %.1f psi'%(Pd)\n",
+      "print ' \\n case 2'\n",
+      "print ' Net force  =  %d lb'%( F+3)\n",
+      "print ' Location of net force =  %.2f ft'%( yP)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " case 1\n",
+        " Pressure at the bottom  =  4.9 psi\n",
+        " \n",
+        " case 2\n",
+        " Net force  =  7380 lb\n",
+        " Location of net force =  3.10 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 Page No : 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#Initialization of variables\n",
+      "dia = 4. \t#m\n",
+      "P = 35. \t#kN/m**2\n",
+      "theta = 30. #degrees\n",
+      "\t\n",
+      "#calculations\n",
+      "Fx = P*(dia-dia*(1-math.cos(math.radians(theta)))/2.)\n",
+      "Fz = P*dia*math.sin(math.radians(theta))/2\n",
+      "dist = (dia-dia*(1-math.cos(math.radians(theta)))/2.)\n",
+      "Fxb = 9.81*dist*dist/2 \n",
+      "Fzb = 9.81*((180+theta)*math.pi*(dia/2)**2/360 + math.sqrt(3) /2 + dia/2)\n",
+      "\t\n",
+      "#Results\n",
+      "print ' part a'\n",
+      "print ' Horizontal force =  %.1f kN/m to the right'%( Fx)\n",
+      "print ' Vertical force  =  %.1f kN/m upward' %( Fz)\n",
+      "print ' \\n part b'\n",
+      "print ' force by the fluid  =  %.1f kN/m to the right'%(Fxb)\n",
+      "print ' weight of the cross-hatched volume of liquid  = %.1f kN/m Upward'%(Fzb )\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " part a\n",
+        " Horizontal force =  130.6 kN/m to the right\n",
+        " Vertical force  =  35.0 kN/m upward\n",
+        " \n",
+        " part b\n",
+        " force by the fluid  =  68.3 kN/m to the right\n",
+        " weight of the cross-hatched volume of liquid  = 100.0 kN/m Upward\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 Page No : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy.linalg import solve\n",
+      "from numpy import *\n",
+      "\n",
+      "#Initialization of variables\n",
+      "d1 = 4. \t#diameter - in\n",
+      "h1 = 3.75 \t#in\n",
+      "w1 = 0.85 \t#weight of cylinder - lb\n",
+      "gamma = 52. #lb/ft**3\n",
+      "d2 = 5. \t#in\n",
+      "depth = 3. \t#in\n",
+      "\t\n",
+      "#calculations\n",
+      "A = array([[(d1/2)*(d1/2), -(d2/2)*(d2/2)+(d1/2)*(d1/2)],[ 1,1]])\n",
+      "b = array([[0],[w1*12*(12*2/d1)**2 /(gamma*math.pi)]])\n",
+      "C = solve(A,b)\n",
+      "x = C[0]\n",
+      "y = C[1]\n",
+      "height = depth-x\n",
+      "\t\n",
+      "#Results\n",
+      "print 'Bottom of the cylinder will be %.2f in above the bottom of hollow cylinder'%(height)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bottom of the cylinder will be 2.19 in above the bottom of hollow cylinder\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 Page No : 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#Initialization of variables\n",
+      "v = 180. \t#velocity - m/s\n",
+      "angle = 40. #degrees\n",
+      "a = 4.   \t#m/s**2\n",
+      "r = 2600. \t#radius - m\n",
+      "g = 9.81 \t#m/s**2\n",
+      "\t\n",
+      "#calculations\n",
+      "#Assume outward and right as positive\n",
+      "an = round(-v*v/r,1)\n",
+      "at = -a\n",
+      "ax = at*math.cos(math.radians(angle)) +an*math.sin(math.radians(angle))\n",
+      "az = at*math.sin(math.radians(angle)) -an*math.cos(math.radians(angle))\n",
+      "tangent = ax/(az+g)\n",
+      "theta = math.degrees(math.atan(tangent))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print 'Angle made by the free liquid  =  %.1f degrees'%(-theta)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle made by the free liquid  =  33.4 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file