1 files changed, 141 insertions, 0 deletions
diff --git a/mechanics_of_fluid/Chapter12-.ipynb b/mechanics_of_fluid/Chapter12-.ipynb
new file mode 100755
index 00000000..2905ba7e
--- /dev/null
+++ b/mechanics_of_fluid/Chapter12-.ipynb
@@ -0,0 +1,141 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1f6973f8c8d9996abb4c23a2c5352b16bfcd60086af954d2e49808504d3171d0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "CHapter12-Unsteady Flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg557"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#calculate Time for which flow into the tank continues after the power failure \n",
+      "import scipy\n",
+      "from scipy import integrate\n",
+      "Q=0.05; ## m^3/s\n",
+      "d=0.15; ## m^2\n",
+      "h=8.; ## m\n",
+      "g=9.81; ## m/s^2\n",
+      "l=90.; ## m\n",
+      "f=0.007;\n",
+      "\n",
+      "u1=Q/(math.pi/4.*d**2.);\n",
+      "\n",
+      "def function(u):\n",
+      "\tfun=(1./((h*g/l)+(2.*f/d)*u**2))\n",
+      "\treturn fun\n",
+      "\n",
+      "t=scipy.integrate.quad(function,u1,0)\n",
+      "\n",
+      "\n",
+      "print(\"Time for which flow into the tank continues after the power failure is\" )\n",
+      "print'%s %.1f %s'%(\" \",-t[0],\"s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time for which flow into the tank continues after the power failure is\n",
+        "  2.6 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4-pg588"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#calculate Estimate the height of tank required\n",
+      "\n",
+      "print(\"Estimate the height of tank required\")\n",
+      "\n",
+      "f=0.006;\n",
+      "l=1400.; ## m\n",
+      "g=9.81; ## m/s^2\n",
+      "d1=0.75; ## m\n",
+      "d2=3.; ## m\n",
+      "Q=1.2; ## m^3/s\n",
+      "a=20.; ## m\n",
+      "\n",
+      "K=4*f*l/(2*g*d1);\n",
+      "\n",
+      "## 2*K*Y = l*a/(g*A) = 8.919 s^2\n",
+      "\n",
+      "## Y=2*K*Y/2*K\n",
+      "\n",
+      "Y=8.919/(2*K);\n",
+      "## When t=0\n",
+      "\n",
+      "u0=Q/(math.pi/4*d1**2);\n",
+      "\n",
+      "y0=K*u0**2;\n",
+      "\n",
+      "C=-Y/K/math.exp(y0/Y);\n",
+      "\n",
+      "## To determine the height of the surge tank, we consider the condition y = y_max when u = 0. \n",
+      "\n",
+      "## 0 = 1/K*(y_max+Y) + C*exp(y_max/Y)\n",
+      "\n",
+      "## From the above eqn we get\n",
+      "\n",
+      "y_max=-Y;\n",
+      "\n",
+      "H=a-y_max;\n",
+      "print'%s %.1f %s'%(\"The minimum height of the surge tank =\",H,\"m\")\n",
+      "\n",
+      "\n",
+      "print(\"The actual design height should exceed the minimum required, say 23 m\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Estimate the height of tank required\n",
+        "The minimum height of the surge tank = 22.0 m\n",
+        "The actual design height should exceed the minimum required, say 23 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file