diff options
Diffstat (limited to 'mechanics_of_fluid/Chapter8-_1.ipynb')
| -rwxr-xr-x | mechanics_of_fluid/Chapter8-_1.ipynb | 350 |
1 files changed, 350 insertions, 0 deletions
diff --git a/mechanics_of_fluid/Chapter8-_1.ipynb b/mechanics_of_fluid/Chapter8-_1.ipynb new file mode 100755 index 00000000..a66b62b6 --- /dev/null +++ b/mechanics_of_fluid/Chapter8-_1.ipynb @@ -0,0 +1,350 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0fec1a915dc11972a0b348fe3db6dc5cdc775b2840fa9a3ae3580b08a05d480b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter8-Boundary Layers, Wakes and Other Shear Layers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Displacement thickness and Momentum thickness \n",
+ "delta=0.6; ## mm\n",
+ "\n",
+ "delta1=delta/3.;\n",
+ "\n",
+ "theta=2./15*delta;\n",
+ "\n",
+ "print'%s %.1f %s'%(\"Displacement thickness =\",delta1,\"mm\")\n",
+ "\n",
+ "print'%s %.2f %s'%(\"Momentum thickness =\",theta,\"mm\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Displacement thickness = 0.2 mm\n",
+ "Momentum thickness = 0.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg309"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Evaluate the constants A and B\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "## To determine the values of a1 & a2 following conditions must be satisfied\n",
+ "\n",
+ "## Condition I - When n=0, u/um=0\n",
+ "## Condition II - When n=1, u/um=a1+a2=1\n",
+ "## Condition III - When n=1, d(u/um)/dn = a1+2a2=0\n",
+ "\n",
+ "## By satisfying these conditions, we have\n",
+ "## a1+a2=1;\n",
+ "## a1+2a2=0;\n",
+ "\n",
+ "A=([[1,1],[1,2]]);\n",
+ "B=([[1],[0]]);\n",
+ "X=numpy.linalg.inv(A)*B;\n",
+ "\n",
+ "a1=X[0];\n",
+ "a2=X[1];\n",
+ "\n",
+ "print(\"a1=\",a1[0])\n",
+ "\n",
+ "print(\"a2=\",a1[1])\n",
+ "\n",
+ "\n",
+ "print(\"Evaluate the constants A and B\")\n",
+ "\n",
+ "## A = integrate('(1-f(n))*f(n)','n',0,1)\n",
+ "\n",
+ "def function(n):\n",
+ "\tfun=(1-(2*n-n**2))*(2*n-n**2)\n",
+ "\treturn fun\n",
+ "\n",
+ "A=scipy.integrate.quad(function,0,1)\n",
+ "print(\"A = \",A[0],\"\")\n",
+ "\n",
+ "\n",
+ "## B = differentiation of (2*n-n^2) at n=0, we get \n",
+ "B=2;\n",
+ "\n",
+ "print(\"B =\",B,\"\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "('a1=', 2.0)\n",
+ "('a2=', -1.0)\n",
+ "Evaluate the constants A and B\n",
+ "('A = ', 0.13333333333333333, '')\n",
+ "('B =', 2, '')\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg310"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Evaluate the constants A and B\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "## To determine the values of a1 & a2 following conditions must be satisfied\n",
+ "\n",
+ "## Condition I - When n=0, u/um=0\n",
+ "## Condition II - When n=1, u/um=a1+a2=1\n",
+ "## Condition III - When n=1, d(u/um)/dn = a1+2a2=0\n",
+ "\n",
+ "## By satisfying these conditions, we have\n",
+ "## a1+a2=1;\n",
+ "## a1+2a2=0;\n",
+ "\n",
+ "A=([[1,1],[1,2]]);\n",
+ "B=([[1],[0]]);\n",
+ "X=numpy.linalg.inv(A)*B;\n",
+ "\n",
+ "a1=X[0];\n",
+ "a2=X[1];\n",
+ "\n",
+ "print(\"a1=\",a1[0])\n",
+ "\n",
+ "print(\"a2=\",a1[1])\n",
+ "\n",
+ "\n",
+ "print(\"Evaluate the constants A and B\")\n",
+ "\n",
+ "## A = integrate('(1-f(n))*f(n)','n',0,1)\n",
+ "\n",
+ "def function(n):\n",
+ "\tfun=(1-(2*n-n**2))*(2*n-n**2)\n",
+ "\treturn fun\n",
+ "\n",
+ "A=scipy.integrate.quad(function,0,1)\n",
+ "print(\"A = \",A[0],\"\")\n",
+ "\n",
+ "\n",
+ "## B = differentiation of (2*n-n^2) at n=0, we get \n",
+ "B=2;\n",
+ "\n",
+ "print(\"B =\",B,\"\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "('a1=', 2.0)\n",
+ "('a2=', -1.0)\n",
+ "Evaluate the constants A and B\n",
+ "('A = ', 0.13333333333333333, '')\n",
+ "('B =', 2, '')\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the velocity of the airstream and frictional drag of the plat\n",
+ "v=1.5*10**(-5); ## m^2/s\n",
+ "Re_t=5.*10**5;\n",
+ "x_t=1.2; ## m\n",
+ "rho=1.21; ## kg/m^3\n",
+ "\n",
+ "u_m=v*Re_t/x_t;\n",
+ "\n",
+ "print'%s %.2f %s'%(\"the velocity of the airstream =\",u_m,\"m/s\")\n",
+ "\n",
+ "\n",
+ "theta=0.646*x_t/math.sqrt(Re_t);\n",
+ "\n",
+ "F=rho*u_m**2*theta;\n",
+ "\n",
+ "D_F=2.*F*x_t;\n",
+ "print'%s %.3f %s'%(\"the frictional drag of the plate, D_F =\",D_F,\"N\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity of the airstream = 6.25 m/s\n",
+ "the frictional drag of the plate, D_F = 0.124 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg316"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the boundary layer thickness at the rear of the train and the frictional drag acting on the train and the power required to overcome the frictional drag\n",
+ "u_m = 50.; ## m/s or 180 km/h\n",
+ "v=1.5*10**(-5); ## m^2/s\n",
+ "l=100.; ## m\n",
+ "rho=1.2; ## kg/m^3\n",
+ "b=8.3; ## m\n",
+ "\n",
+ "delta = 0.37*(v/u_m)**(1/5.)*l**(4/5.);\n",
+ "\n",
+ "print'%s %.3f %s'%(\"the boundary layer thickness at the rear of the train =\",delta,\"m\")\n",
+ "Re_l = u_m*l/v;\n",
+ "C_F=0.074*(Re_l)**(-1/5.);\n",
+ "F=0.037*rho*u_m**2*l*Re_l**(-1/5.);\n",
+ "\n",
+ "D_F = F*b;\n",
+ "\n",
+ "print'%s %.4f %s'%(\"the frictional drag acting on the train, D_F =\",D_F,\"N\")\n",
+ "\n",
+ "\n",
+ "P=D_F*u_m;\n",
+ "print'%s %.1f %s'%(\"the power required to overcome the frictional drag =\",P/1000,\"kW\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the boundary layer thickness at the rear of the train = 0.731 m\n",
+ "the frictional drag acting on the train, D_F = 1818.9691 N\n",
+ "the power required to overcome the frictional drag = 90.9 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the proportion of the plate occupied by the laminar boundary layer and the skin friction coefficient CF evaluated at the trailing edge\n",
+ "Re_t=5.*10**5;\n",
+ "Re_l=5.*10**6;\n",
+ "\n",
+ "r1=Re_t/Re_l; ## r1=x_t/l\n",
+ "r2=1-36.9*(1./Re_t)**(3/8.); ## r2=x_0/x_t\n",
+ "\n",
+ "r=r1*r2; ## r=x_0/l;\n",
+ "\n",
+ "print'%s %.2f %s'%(\"the proportion of the plate occupied by the laminar boundary layer =\",r*100,\"%\")\n",
+ "\n",
+ "C_F = 0.074/Re_l**(1/5.)*(1-r)**(4/5.);\n",
+ "print'%s %.4f %s'%(\"the skin friction coefficient CF evaluated at the trailing edge =\",C_F,\"\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the proportion of the plate occupied by the laminar boundary layer = 7.31 %\n",
+ "the skin friction coefficient CF evaluated at the trailing edge = 0.0032 \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |