updated books

1 files changed, 291 insertions, 0 deletions

diff --git a/Problems_In_Hydraulics/ch4.ipynb b/Problems_In_Hydraulics/ch4.ipynb
new file mode 100755
index 00000000..aefd97a7
--- /dev/null
+++ b/Problems_In_Hydraulics/ch4.ipynb
@@ -0,0 +1,291 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f02a7c0105dc5d8192eae252809dedbf748fa59e0b9b78e4af3b1079610b7a69"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Orifices and Notches"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 Page No : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Cd= 0.98    # velocity\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "H= 2.    \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v= math.sqrt(2*g*H)\n",
+      "t= H/v\n",
+      "h= 0.5*g*t**2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Vertical distance fallen in this ttime = %.3f ft'%(h) \n",
+      "\n",
+      "#Note : The answer given in textbook is wrong.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vertical distance fallen in this ttime = 0.500 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 Page No : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "r= 53.4\n",
+      "T= 60. \t    #F pressure of air\n",
+      "h= 29.7 \t#in of mercury\n",
+      "sm= 13.6\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "d= 1.5  \t#in diameter\n",
+      "Qin= 2. \t#cuses air\n",
+      "g=32.2 \t    #ft/s**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "W= h*sm*w/(r*(460+T)*12)\n",
+      "dP= 0.75*w/(12*W)\n",
+      "Q= math.sqrt(2*g*dP)*math.pi*d**2/(4*144)\n",
+      "W= Q*W*60\n",
+      "Cd= Qin/W\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'coefficient of discharge = %.2f '%(Cd) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "coefficient of discharge = 0.62 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 Page No : 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "H1= 34. \t#ft height\n",
+      "H2= 8. \t    #ft head\n",
+      "H3= 7.   \t#ft pressure head\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "d= 1.5 \t    #in\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v2= math.sqrt(2*g*(H1+H2-H3))\n",
+      "Q= v2*math.pi*d**2/(4*144)\n",
+      "v3= (2*v2+math.sqrt(4*v2**2-4*6*(v2**2-H2*2*5*g)))/12\n",
+      "dr= math.sqrt(v2/v3)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'ratio of diameteres = %.1f '%(dr) \n",
+      "print \" Maximum discharge = %.3f cusec\"%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio of diameteres = 1.6 \n",
+        " Maximum discharge = 0.583 cusec\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 Page No : 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Q1= 8./15 \t#cuses\n",
+      "Q2= 2./15 \t#cuses\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A= math.degrees(math.atan(Q2/Q1))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Angle of inclination = %.2f degrees'%(A) \n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle of inclination = 14.04 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 Page No : 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "r= g**2/((math.sqrt(2))**2*g**2)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'coefficient of contraction = %.1f '%(r) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "coefficient of contraction = 0.5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 Page No : 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "B= 3. \t    #ft long\n",
+      "H= 2.   \t#ft depth of water\n",
+      "H1= 3.75 \t#ft \n",
+      "w= 4.    \t#ft wide\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= 3.33*(B-(H1/5))*H**1.5\n",
+      "v= Q/(H*w)\n",
+      "kh= v**2/(2*g)\n",
+      "Q1= 3.33*(B-(H1/5)-kh)*(((H1/5)+kh)**1.5-kh**1.5)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.2f cuses'%(Q1) \n",
+      "\n",
+      "\n",
+      "# NOte : ANSWER IN THE TEXTBOOK IS WRONG\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 5.42 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file