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Diffstat (limited to 'Fluid_mechanics/Chapter_11.ipynb')
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1 files changed, 485 insertions, 0 deletions
diff --git a/Fluid_mechanics/Chapter_11.ipynb b/Fluid_mechanics/Chapter_11.ipynb new file mode 100755 index 00000000..ec13c1a6 --- /dev/null +++ b/Fluid_mechanics/Chapter_11.ipynb @@ -0,0 +1,485 @@ +{
+ "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 |