{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "## Chapter 10 : Open Channel Flow " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 10.1 Page no 363" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "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" ] } ], "source": [ "# Top width, area of lfow, hydraulic radius\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", "# Given\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", "# Solution\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\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 10.2 Page no 366" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Discharge for the channel = 16.1 m**3/s\n" ] } ], "source": [ "# Discharge for the trapezoidal channel\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", "# Given\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", "# Solution\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", "# from mannings eqquation\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\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 10.4 Page no 373" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "depth(y) = 5.05 ft\n", "base width(b) = 4.18 ft\n" ] } ], "source": [ "# Determine cross sectional area\n", "\n", "from __future__ import division\n", "\n", "from math import *\n", "\n", "# Given\n", "\n", "z = 1\n", "\n", "Q = 10000/60 # discharge of water in ft**#/s\n", "\n", "# Solution\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\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 10.5 Page no 378" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Critical depth = 3.44 m\n" ] } ], "source": [ "# Calculate criticcal depth\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", "# Given\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", "# Solution\n", "\n", "Yc = (20**2/g)**(1/3)\n", "\n", "print \"Critical depth =\",round(Yc,2),\"m\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 10.6 Page no 380" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "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" ] } ], "source": [ "# determine normal depth, flow regine, critical depth\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", "# Given\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", "# Solution\n", "\n", "# We use a trial and error method here to find the value of y i.e. normal depth\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", "# we use trail and error to find the value of yc for critical depth\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" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 10.8 Page no 390" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "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" ] } ], "source": [ "# Height, type and length of jump and loss of energy\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", "# Given\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", "# Solution\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" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 10.12 Page no 409" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Discharge = 503.0 cfs\n" ] } ], "source": [ "# Determine flow rate\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", "# Given\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", "# Solution\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\"" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": false }, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.3" } }, "nbformat": 4, "nbformat_minor": 0 }