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diff --git a/mechanics_of_fluid/Chapter10_1.ipynb b/mechanics_of_fluid/Chapter10_1.ipynb new file mode 100755 index 00000000..9502b293 --- /dev/null +++ b/mechanics_of_fluid/Chapter10_1.ipynb @@ -0,0 +1,355 @@ +{
+ "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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