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diff --git a/backup/mechanics_of_fluid_version_backup/Chapter10.ipynb b/backup/mechanics_of_fluid_version_backup/Chapter10.ipynb deleted file mode 100755 index 9502b293..00000000 --- a/backup/mechanics_of_fluid_version_backup/Chapter10.ipynb +++ /dev/null @@ -1,355 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:0abcc423492072a54fb770a61520bf727ac4ace788fa5f2dbb0a6db0caff8ec2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter10-Flow with free surfaces"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex1-pg436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#calculate The depth of the water under the brigde and the depth of water upstream\n",
- "Q=400.; ## m^3/s\n",
- "b2=20.; ## m\n",
- "g=9.81; ## m/s^2\n",
- "b1=25.; ## m\n",
- "\n",
- "h2=(Q/b2/math.sqrt(g))**(2./3.);\n",
- "## Since energy is conserved\n",
- "## h1 + u1^2/2g = h2 +u2^2/2g = h2 + h2/2 = 3h2/2\n",
- "\n",
- "## h1 + 1/2*g*(Q/(b1h1))^2 = 3*h2/2;\n",
- "\n",
- "## h1^3-5.16*h1^2+13.05 = 0;\n",
- "\n",
- "## By solving this cubic equation\n",
- "\n",
- "h1=4.52; ## m\n",
- "\n",
- "print'%s %.1f %s'%(\" The depth of the water under the brigde =\",h2,\"m\")\n",
- "\n",
- "\n",
- "print'%s %.2f %s'%(\" the depth of water upstream =\",h1,\"m\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The depth of the water under the brigde = 3.4 m\n",
- " the depth of water upstream = 4.52 m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2-pg447"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#calculate Rate of flow \n",
- "w=0.04; ## thickness of block in m\n",
- "d=0.07; ## depth of liquid in m\n",
- "b=0.4; ## m\n",
- "g=9.81; ## m/s^2\n",
- "\n",
- "H=d-w; \n",
- "\n",
- "Q=1.705*b*H**(3/2.);\n",
- "\n",
- "u1=Q/d/b;\n",
- "h=u1**2/(2.*g);\n",
- "\n",
- "H1=H+h;\n",
- "\n",
- "Q1=1.705*b*H1**(3/2.);\n",
- "\n",
- "print'%s %.4f %s'%(\"Rate of flow =\",Q1,\" m^3/s\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rate of flow = 0.0037 m^3/s\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3-pg455"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#calculate depth of water at the throat and the new flow rate and the Froude number at the throat and flow rate\n",
- "h1=0.45; ## m\n",
- "g=9.81; ## m/s^2\n",
- "b1=0.8; ## m\n",
- "h2=0.35; ## m\n",
- "b2=0.3; ## m\n",
- "print(\"the flow rate\")\n",
- "Q=math.sqrt((h1-h2)*2*g/(-(1./(h1*b1)**2)+(1./(h2*b2)**2)));\n",
- "print'%s %.3f %s'%(\"Flow rate =\",Q,\"m^3/s\")\n",
- "\n",
- "\n",
- "print(\" the Froude number at the throat\")\n",
- "Fr2=Q/(math.sqrt(g)*b2*h2**(3/2.));\n",
- "print'%s %.3f %s'%(\"The Froude number at the throat =\",Fr2,\"\")\n",
- "\n",
- "\n",
- "print(\"the depth of water at the throat\")\n",
- "\n",
- "## (h1/h2)^(3) + 1/2*(b2/b1)^2 = 3/2*(h1/h2)^2\n",
- "\n",
- "## The solution for the above eqn is as follows\n",
- "## (h1/h2) = 0.5 + cos(2arcsin(b2/b1)/3)\n",
- "\n",
- "## h1/h2=1.467\n",
- "\n",
- "h2_new=h1/1.467;\n",
- "print'%s %.3f %s'%(\"Depth of water at the throat =\",h2_new,\"m\")\n",
- "\n",
- "print(\"the new flow rate\")\n",
- "w=math.sqrt(g)*b2*h2_new**(3/2.);\n",
- "print'%s %.3f %s'%(\"New flow rate =\",Q,\"m^3/s\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "the flow rate\n",
- "Flow rate = 0.154 m^3/s\n",
- " the Froude number at the throat\n",
- "The Froude number at the throat = 0.790 \n",
- "the depth of water at the throat\n",
- "Depth of water at the throat = 0.307 m\n",
- "the new flow rate\n",
- "New flow rate = 0.154 m^3/s\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex4-pg460"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#calculate depth\n",
- "Q=8.75; ## m^3/s\n",
- "w=5.; ## m\n",
- "n=0.0015; \n",
- "s=1./5000.;\n",
- "\n",
- "## Q/(w*h0) = u = m^(2/3)*i^(1/2)/n = 1/0.015*(w*h0/(w+2*h0))^(2/3)*sqrt(s);\n",
- "## Solution by trial gives h0\n",
- "h0=1.8; ## m\n",
- "\n",
- "q=1.75;\n",
- "g=9.81;\n",
- "hc=(q**2/g)**(1/3); ## critical depth\n",
- "\n",
- "print'%s %.1f %s'%(\"Depth =\",h0,\"m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Depth = 1.8 m\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5-pg469"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#calculate wave length\n",
- "g=9.81; ## m/s^2\n",
- "T=5.; ## s\n",
- "h=4.; ## m\n",
- "\n",
- "## lambda=g*T^2/(2*%pi)*tanh(2*%pi*h/lambda1);\n",
- "## by trial method , we get \n",
- "lambda1=28.04;\n",
- "\n",
- "D=g*T**2/(2*math.pi)*math.tanh(2*math.pi*h/lambda1);\n",
- "print'%s %.1f %s'%(\"Wavelength =\",D,\"m\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelength = 27.9 m\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "EX6-pg470"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#calculate phase velocity and wave length\n",
- "g=9.81; ## m/s^2\n",
- "T=12; ## s\n",
- "\n",
- "c=g*T/(2*math.pi);\n",
- "\n",
- "D=c*T;\n",
- "\n",
- "print\"%s %.1f %s\"%(\"Phase velocity =\",c,\"m/s\")\n",
- "\n",
- "\n",
- "print\"%s %.1f %s\"%(\"Wavelength =\",D,\"m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Phase velocity = 18.7 m/s\n",
- "Wavelength = 224.8 m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7-pg476"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "#Estimate the time elapsed since the waves were generated in a storm occurring 800 km out to sea and Estimate the depth at which the waves begin to be significantly influenced by the sea bed as they approach the shore\n",
- "c=18.74; ## m/s\n",
- "lambd=225.; ## m\n",
- "\n",
- "print(\"Estimate the time elapsed since the waves were generated in a storm occurring 800 km out to sea. \")\n",
- "\n",
- "x=800.*10**3.; ## m\n",
- "cg=c/2.;\n",
- "\n",
- "t=x/cg;\n",
- "\n",
- "print\"%s %.1f %s\"%(\"time elapsed =\",t/3600.,\"hours\")\n",
- "\n",
- "\n",
- "print(\"Estimate the depth at which the waves begin to be significantly influenced by the sea bed as they approach the shore.\")\n",
- "\n",
- "h1=lambd/2.;\n",
- "\n",
- "h2=lambd/(2.*math.pi)*math.atanh(0.99);\n",
- "\n",
- "print\"%s %.1f %s %.1f %s\"%(\"The answers show that h lies in the range between about\",h2,\"m , \",h1, \"m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Estimate the time elapsed since the waves were generated in a storm occurring 800 km out to sea. \n",
- "time elapsed = 23.7 hours\n",
- "Estimate the depth at which the waves begin to be significantly influenced by the sea bed as they approach the shore.\n",
- "The answers show that h lies in the range between about 94.8 m , 112.5 m\n"
- ]
- }
- ],
- "prompt_number": 7
- }
- ],
- "metadata": {}
- }
- ]
-}
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