From 6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- .../Chapter12-.ipynb | 141 --------------------- 1 file changed, 141 deletions(-) delete mode 100755 backup/mechanics_of_fluid_version_backup/Chapter12-.ipynb (limited to 'backup/mechanics_of_fluid_version_backup/Chapter12-.ipynb') diff --git a/backup/mechanics_of_fluid_version_backup/Chapter12-.ipynb b/backup/mechanics_of_fluid_version_backup/Chapter12-.ipynb deleted file mode 100755 index 2905ba7e..00000000 --- a/backup/mechanics_of_fluid_version_backup/Chapter12-.ipynb +++ /dev/null @@ -1,141 +0,0 @@ -{ - "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 -- cgit