diff options
author | Thomas Stephen Lee | 2015-09-04 22:04:10 +0530 |
---|---|---|
committer | Thomas Stephen Lee | 2015-09-04 22:04:10 +0530 |
commit | 41f1f72e9502f5c3de6ca16b303803dfcf1df594 (patch) | |
tree | f4bf726a3e3ce5d7d9ee3781cbacfe3116115a2c /Fluid_mechanics/Chapter_11.ipynb | |
parent | 9c9779ba21b9bedde88e1e8216f9e3b4f8650b0e (diff) | |
download | Python-Textbook-Companions-41f1f72e9502f5c3de6ca16b303803dfcf1df594.tar.gz Python-Textbook-Companions-41f1f72e9502f5c3de6ca16b303803dfcf1df594.tar.bz2 Python-Textbook-Companions-41f1f72e9502f5c3de6ca16b303803dfcf1df594.zip |
add/remove/update books
Diffstat (limited to 'Fluid_mechanics/Chapter_11.ipynb')
-rwxr-xr-x | Fluid_mechanics/Chapter_11.ipynb | 485 |
1 files changed, 0 insertions, 485 deletions
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 |