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+{
+ "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": [
+ "# 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\""
+ ],
+ "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": [
+ "# 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\""
+ ],
+ "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": [
+ "# 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\""
+ ],
+ "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": [
+ "# 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\""
+ ],
+ "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": [
+ "# 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"
+ ],
+ "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": [
+ "# 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"
+ ],
+ "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": [
+ "# 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\""
+ ],
+ "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": {}
+ }
+ ]
+} \ No newline at end of file
generated by cgit (git 2.34.1) at 2024-12-20 11:50:32 +0000