From 41f1f72e9502f5c3de6ca16b303803dfcf1df594 Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 4 Sep 2015 22:04:10 +0530 Subject: add/remove/update books --- Fluid_mechanics/Chapter_11.ipynb | 485 --------------------------------------- 1 file changed, 485 deletions(-) delete mode 100755 Fluid_mechanics/Chapter_11.ipynb (limited to 'Fluid_mechanics/Chapter_11.ipynb') diff --git a/Fluid_mechanics/Chapter_11.ipynb b/Fluid_mechanics/Chapter_11.ipynb deleted file mode 100755 index ec13c1a6..00000000 --- a/Fluid_mechanics/Chapter_11.ipynb +++ /dev/null @@ -1,485 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:525b51e826bc42c9a4490de8234c350df60c3a09744a8e35af1f480f12f531bc" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 11 - Flow of liquids in open channels" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1a - Pg 454" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the discharge using darcy equation\n", - "#Initialization of variables\n", - "import math\n", - "rho=1.94 #slugs/ft^3\n", - "mu=2.34e-5 #lb-sec/ft^2\n", - "y=5 #ft\n", - "T=25 #ft\n", - "d=10 #ft\n", - "slope=3./2. \n", - "g=32.2 #ft/s^2\n", - "S=0.001\n", - "#calculations\n", - "A=y*d+ 2*0.5*y*(slope*y)\n", - "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n", - "R=A/WP\n", - "e=0.01 #ft\n", - "rr=2*R/e\n", - "f=0.019\n", - "C=math.sqrt(8*g/f)\n", - "V=C*math.sqrt(R*S)\n", - "Q=V*A\n", - "#results\n", - "print '%s %.1f %s' %(\"Discharge using Darcy equation =\",Q,\"ft^3/s\")\n", - "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge using Darcy equation = 569.3 ft^3/s\n", - "The answer is a bit different due to rounding off error in textbook\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1b - Pg 453" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the discharge using kutter ganguillet method\n", - "#Initialization of variables\n", - "import math\n", - "rho=1.94 #slugs/ft^3\n", - "mu=2.34e-5 #lb-sec/ft^2\n", - "y=5 #ft\n", - "T=25 #ft\n", - "d=10 #ft\n", - "slope=3./2. \n", - "g=32.2 #ft/s^2\n", - "S=0.001\n", - "n=0.017\n", - "#calculations\n", - "A=y*d+ 2*0.5*y*(slope*y)\n", - "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n", - "R=A/WP\n", - "e=0.01 #ft\n", - "rr=2*R/e\n", - "f=0.019\n", - "C=(41.65 + 0.00281/S + 1.811/n)/(1+( 41.65 + 0.00281/S)*n/math.sqrt(R))\n", - "V=C*math.sqrt(R*S)\n", - "Q=V*A\n", - "#results\n", - "print '%s %.1f %s' %(\"Discharge using kutter ganguillet formula =\",Q,\" ft^3/s\")\n", - "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge using kutter ganguillet formula = 517.0 ft^3/s\n", - "The answer is a bit different due to rounding off error in textbook\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1c - Pg 455" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the discharge using bazin formula\n", - "#Initialization of variables\n", - "import math\n", - "rho=1.94 #slugs/ft^3\n", - "mu=2.34e-5 #lb-sec/ft^2\n", - "y=5. #ft\n", - "T=25. #ft\n", - "d=10. #ft\n", - "slope=3./2.\n", - "g=32.2 #ft/s^2\n", - "S=0.001\n", - "m=0.21\n", - "#calculations\n", - "A=y*d+ 2*0.5*y*(slope*y)\n", - "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n", - "R=A/WP\n", - "e=0.01 #ft\n", - "rr=2*R/e\n", - "f=0.019\n", - "C=157.6 /(1+ m/math.sqrt(R))\n", - "V=C*math.sqrt(R*S)\n", - "Q=V*A\n", - "#results\n", - "print '%s %.1f %s' %(\"Discharge using bazin formula =\",Q,\" ft^3/s\")\n", - "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge using bazin formula = 688.7 ft^3/s\n", - "The answer is a bit different due to rounding off error in textbook\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1d - Pg 456" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the discharge using darcy equation\n", - "#Initialization of variables\n", - "import math\n", - "rho=1.94 #slugs/ft^3\n", - "mu=2.34e-5 #lb-sec/ft^2\n", - "y=5. #ft\n", - "T=25. #ft\n", - "d=10. #ft\n", - "slope=3./2. \n", - "g=32.2 #ft/s^2\n", - "S=0.001\n", - "n=0.017\n", - "#calculations\n", - "A=y*d+ 2*0.5*y*(slope*y)\n", - "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n", - "R=A/WP\n", - "e=0.01 #ft\n", - "rr=2*R/e\n", - "f=0.019\n", - "C=1.486*math.pow(R,(1./6.)) /n\n", - "V=C*math.sqrt(R*S)\n", - "Q=V*A\n", - "#results\n", - "print '%s %.1f %s' %(\"Discharge using Darcy equation =\",Q,\"ft^3/s\")\n", - "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge using Darcy equation = 516.6 ft^3/s\n", - "The answer is a bit different due to rounding off error in textbook\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1e - Pg 456" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the froude number\n", - "#Initialization of variables\n", - "import math\n", - "rho=1.94 #slugs/ft^3\n", - "mu=2.34e-5 #lb-sec/ft^2\n", - "y=5. #ft\n", - "T=25. #ft\n", - "d=10. #ft\n", - "slope=3./2. \n", - "g=32.2 #ft/s^2\n", - "S=0.001\n", - "n=0.017\n", - "#calculations\n", - "A=y*d+ 2*0.5*y*(slope*y)\n", - "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n", - "R=A/WP\n", - "e=0.01 #ft\n", - "rr=2*R/e\n", - "f=0.019\n", - "C=(41.65 + 0.00281/S + 1.811/n)/(1+( 41.65 + 0.00281/S)*n/math.sqrt(R))\n", - "V=C*math.sqrt(R*S)\n", - "T=d+ 2*(slope*y)\n", - "yh=A/T\n", - "Nf=V/(math.sqrt(g*yh))\n", - "#results\n", - "print '%s %.2f' %(\"froude number = \",Nf)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "froude number = 0.56\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1f - Pg 456" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the critical depth\n", - "#Initialization of variables\n", - "import math\n", - "rho=1.94 #slugs/ft^3\n", - "mu=2.34e-5 #lb-sec/ft^2\n", - "y=5. #ft\n", - "T=25. #ft\n", - "d=10. #ft\n", - "slope=3./2. \n", - "g=32.2 #ft/s^2\n", - "S=0.001\n", - "n=0.017\n", - "#calculations\n", - "A=y*d+ 2*0.5*y*(slope*y)\n", - "WP=d+ 2*math.sqrt(3*3 +2*2) /2 *y\n", - "R=A/WP\n", - "e=0.01 #ft\n", - "rr=2*R/e\n", - "f=0.019\n", - "C=(41.65 + 0.00281/S + 1.811/n)/(1+( 41.65 + 0.00281/S)*n/math.sqrt(R))\n", - "V=C*math.sqrt(R*S)\n", - "Q=V*A\n", - "T=d+ 2*(slope*y)\n", - "yh=A/T\n", - "yc=2.88 #ft\n", - "#results\n", - "print '%s' %(\"yc is obtained using trial and error method\")\n", - "print '%s %.2f %s' %(\"Critical depth =\",yc,\"ft\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "yc is obtained using trial and error method\n", - "Critical depth = 2.88 ft\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2 - Pg 459" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the minimum scale ratio\n", - "#Initialization of variables\n", - "import math\n", - "Re=4000.\n", - "rho=1.94 #slugs/ft^3\n", - "vm=5.91 #ft/s\n", - "mu=3.24e-5 #ft-lb/s^2\n", - "Rm=3.12 #ft\n", - "#calculations\n", - "lam3=Re*mu/(vm*4*Rm*rho)\n", - "lam=math.pow(lam3,(2./3.))\n", - "#results\n", - "print '%s %.2e' %(\"Minimum scale ratio = \",lam)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Minimum scale ratio = 9.36e-03\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3 - Pg 462" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the discharge, depth of the channel, froude numbers and also the force applied\n", - "#Initialization of variables\n", - "import math\n", - "yc=2. #ft\n", - "g=32.2 #ft/s^2\n", - "d=10. #ft\n", - "gam=62.4\n", - "rho=1.94\n", - "B=10. #ft\n", - "#calculations\n", - "Vc=math.sqrt(g*yc)\n", - "Ac=yc*d\n", - "Q=Vc*Ac\n", - "y1=5.88 #ft\n", - "y2=0.88 #ft\n", - "V1=2.73 #ft/s\n", - "V2=18.25 #ft/s\n", - "Nf1=0.198\n", - "Nf2=3.43\n", - "F= 0.5*gam*y1*y1 *B - 0.5*gam*y2*y2 *B - Q*rho*V2 +Q*rho*V1\n", - "#results\n", - "print '%s %.1f %s' %(\"Discharge in the channel =\",Q,\"ft^3/s\")\n", - "print '%s %.2f %s %.2f %s' %(\"\\n Depth of the channel at upstream and downstream =\",y1,\"ft and\",y2, \"ft\")\n", - "print '%s %.3f %s %.3f' %(\"\\n froude numbers at upstream and downstream =\",Nf1,\" and \",Nf2)\n", - "print '%s %d %s' %(\"\\n Force applied =\",F,\"lb\")\n", - "print '%s' %(\"The answers are a bit different from textbook due to rounding off error\")" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge in the channel = 160.5 ft^3/s\n", - "\n", - " Depth of the channel at upstream and downstream = 5.88 ft and 0.88 ft\n", - "\n", - " froude numbers at upstream and downstream = 0.198 and 3.430\n", - "\n", - " Force applied = 5713 lb\n", - "The answers are a bit different from textbook due to rounding off error\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4 - Pg 470" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the distance from vena contracta and also the total distance\n", - "#Initialization of variables\n", - "import math\n", - "S0=0.0009\n", - "n=0.018\n", - "w=20 #ft\n", - "d=0.5 #ft\n", - "Q=400 #ft^3/s\n", - "g=32.2 #ft/s^2\n", - "#calculations\n", - "y2=4 #ft\n", - "V2=Q/(w*y2)\n", - "Nf2=V2/math.sqrt(g*y2)\n", - "yr=0.5*(math.sqrt(1+ 8*Nf2*Nf2) -1)\n", - "y1=yr*y2\n", - "L1=32.5\n", - "L2=37.1 \n", - "L3=51.4\n", - "L=L1+L2+L3\n", - "#results\n", - "print '%s %.1f %s %.2f %s' %(\"distance from vena contracta =\",y2,\"ft and\",y1,\"ft\")\n", - "print '%s %.1f %s' %(\"\\n Total distance =\",L,\" ft\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "distance from vena contracta = 4.0 ft and 1.20 ft\n", - "\n", - " Total distance = 121.0 ft\n" - ] - } - ], - "prompt_number": 9 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit