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diff --git a/Fluid_Mechanics_/Chapter10.ipynb b/Fluid_Mechanics_/Chapter10.ipynb deleted file mode 100644 index 679a7dd2..00000000 --- a/Fluid_Mechanics_/Chapter10.ipynb +++ /dev/null @@ -1,426 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:3b6a6304ae5564c5b042191e18c023d5d141ad8d64a979c29e54da09ebc8a32e" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Chapter 10 : Open Channel Flow\n", - " " - ] - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 10.1 Page no 363" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "from math import *\n", - "\n", - "from __future__ import division\n", - "\n", - "\n", - "b = 3 # base of the channel\n", - "\n", - "z = 0.5 # slope of the channel\n", - "\n", - "y = 2 # depth of the channel\n", - "\n", - "\n", - "T = b + 2*z*y\n", - "\n", - "print \"Top width =\",round(T,0),\"m\"\n", - "\n", - "A = (b+z*y)*y\n", - "\n", - "print \"Area of flow =\",round(A,0),\"m**2\"\n", - "\n", - "P = b + 2*y*sqrt(1+z**2)\n", - "\n", - "print \"Wetted perimeter =\",round(P,3),\"m\"\n", - "\n", - "R = A/P\n", - "\n", - "print \"Hydraulic radius =\",round(R,2),\"m\"\n", - "\n", - "D = A/T\n", - "\n", - "print \"Hydraulic depth =\",round(D,2),\"m\"\n", - "\n", - "Z = A*sqrt(D)\n", - "\n", - "print \"Secton Factor =\",round(Z,2),\"m**2\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Top width = 5.0 m\n", - "Area of flow = 8.0 m**2\n", - "Wetted perimeter = 7.472 m\n", - "Hydraulic radius = 1.07 m\n", - "Hydraulic depth = 1.6 m\n", - "Secton Factor = 10.12 m**2\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 10.2 Page no 366" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "from math import *\n", - "\n", - "from __future__ import division\n", - "\n", - "\n", - "z = 1.0 # slide slope\n", - "\n", - "b = 3.0 # base width\n", - "\n", - "y = 1.5 # depth\n", - "\n", - "S = 0.0009\n", - "\n", - "n = 0.012 # for concrete\n", - "\n", - "\n", - "A = (b+z*y)*y\n", - "\n", - "P = P = b + 2*y*sqrt(1+z**2)\n", - "\n", - "R = A/P\n", - "\n", - "\n", - "Q = A*(1/n)*(R**(2/3)*S**(1/2))\n", - "\n", - "print \"Discharge for the channel =\",round(Q,2),\"m**3/s\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge for the channel = 16.1 m**3/s\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 10.4 Page no 373" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "from __future__ import division\n", - "\n", - "from math import *\n", - "\n", - "\n", - "z = 1\n", - "\n", - "Q = 10000/60 # discharge of water in ft**#/s\n", - "\n", - "\n", - "y = (Q/(1.828*2.25*sqrt(0.5)))**(2/5)\n", - "\n", - "print \"depth(y) =\",round(y,2),\"ft\"\n", - "\n", - "b = 0.828*y\n", - "\n", - "print \"base width(b) =\",round(b,2),\"ft\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "depth(y) = 5.05 ft\n", - "base width(b) = 4.18 ft\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 10.5 Page no 378" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "from math import *\n", - "\n", - "from __future__ import division\n", - "\n", - "\n", - "y = 2.5 # depth\n", - "\n", - "V = 8 # velocity in m/s\n", - "\n", - "g = 9.81 # acceleration due to gravity in m/s**2\n", - "\n", - "\n", - "Yc = (20**2/g)**(1/3)\n", - "\n", - "print \"Critical depth =\",round(Yc,2),\"m\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Critical depth = 3.44 m\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 10.6 Page no 380" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "from math import *\n", - "\n", - "from __future__ import division\n", - "\n", - "\n", - "Q = 15 # flow rate in m**3/s\n", - "\n", - "w = 4.0 # bottom width\n", - "\n", - "S = 0.0008 # bed slope\n", - "\n", - "n = 0.025 # manning constant\n", - "\n", - "z = 0.5 # slope\n", - "\n", - "\n", - "\n", - "y = 2.22 # we take the value of y as 2.2 m\n", - "\n", - "Q = ((4+0.5*y)*(y/(n))*(((4+0.5*y)*y)/(4+2.236*y))**(0.667)*(S)**(0.5))\n", - "\n", - "print \"a )Normal Depth =\",round(y,2),\"m\"\n", - "\n", - "A = (4+0.5*y)*y\n", - "\n", - "T = (w+2*z*y)\n", - "\n", - "D = A/T\n", - "\n", - "V = (Q/A)\n", - "\n", - "F =V/(sqrt(9.81*D)) \n", - "\n", - "print \"b )F = \",round(F,2),\" Since the Froude number is less than 1, the flow is subcritical\"\n", - "\n", - "\n", - "yc = 1.08\n", - "\n", - "Q1 = (4+z*yc)*yc*sqrt((9.81*(4+0.5*yc)*yc)/(4+2*z*yc)) \n", - "\n", - "print \"c )Critical depth is = \",round(yc,2),\"m\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "a )Normal Depth = 2.22 m\n", - "b )F = 0.31 Since the Froude number is less than 1, the flow is subcritical\n", - "c )Critical depth is = 1.08 m\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "\n", - "Example 10.8 Page no 390" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "from math import *\n", - "\n", - "from __future__ import division\n", - "\n", - "\n", - "b = 60 # base width in ft\n", - "\n", - "y1 = 2.5 # base depth in ft\n", - "\n", - "Q = 2500 # discharge in ft**3/s\n", - "\n", - "g = 32.2\n", - "\n", - "\n", - "V1 = Q/(b*y1)\n", - "\n", - "F1 = V1/sqrt(g*y1)\n", - "\n", - "y2 = y1*0.5*(sqrt(1+8*F1**2)-1)\n", - "\n", - "V2 = Q/(b*y2)\n", - "\n", - "print \"Since F1 =\",round(F1,2),\" It is a weak jump\"\n", - "\n", - "L = y2*4.25\n", - "\n", - "print \"Length of the jump =\",round(L,0),\"ft\"\n", - "\n", - "E1 = y1+(V1**2/(2*g))\n", - "\n", - "E2 = y2+(V2**2/(2*g))\n", - "\n", - "El = E1-E2\n", - "\n", - "Te = El*62.4*Q/543\n", - "\n", - "print \"Total energy loss =\",round(Te,2),\"HP\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Since F1 = 1.86 It is a weak jump\n", - "Length of the jump = 23.0 ft\n", - "Total energy loss = 133.7 HP\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 3, - "metadata": {}, - "source": [ - "Example 10.12 Page no 409" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "from math import *\n", - "\n", - "from __future__ import division\n", - "\n", - "\n", - "d = 6 # depth of the channel\n", - "\n", - "w= 12 # width of the channel\n", - "\n", - "h = 1.0 # height of the channel\n", - "\n", - "p = 9 # pressure drop in m\n", - "\n", - "g = 32.2\n", - "\n", - "\n", - "y2 = y1 - h - 0.75\n", - "\n", - "V1 = sqrt(2*g*0.75/((1.41)**2-1))\n", - "\n", - "Q = w*b*V1/10\n", - "\n", - "print \"Discharge =\",round(Q,0),\"cfs\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge = 503.0 cfs\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
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