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diff --git a/backup/mechanics_of_fluid_version_backup/Chapter2-.ipynb b/backup/mechanics_of_fluid_version_backup/Chapter2-.ipynb new file mode 100755 index 00000000..7cf5ac34 --- /dev/null +++ b/backup/mechanics_of_fluid_version_backup/Chapter2-.ipynb @@ -0,0 +1,326 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f587ddef55b28db4a216d9527633644d2f15945b69280f84d70b5536f38a83a6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2- Fluid Statics\n"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##calculate hieght above sea level to which ballon will rise\n",
+ "d=1.5; ##m\n",
+ "m=1.2; ## kg\n",
+ "rate=0.0065; ## K/m\n",
+ "R=287.; ## J/(kg.K)\n",
+ "T_0=288.15; ## K\n",
+ "p_0=101*10**3; ## Pa\n",
+ "g=9.81; ## m/s^2\n",
+ "\n",
+ "rho=m/(math.pi*d**3/6);\n",
+ "rho_0=p_0/R/T_0;\n",
+ "\n",
+ "## log(rho/rho_0)=(g/R*rate - 1)*log((T_0-rate*z)/T_0)\n",
+ "\n",
+ "z=1/rate*(T_0-T_0*math.exp(math.log(rho/rho_0)/(g/R/rate-1)));\n",
+ "print'%s %.f %s'%(\"The height above sea level to which the ballon will rise =\",z,\"m\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The height above sea level to which the ballon will rise = 5708 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg64"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate total force and Distance of line of action of total force from top of cylinder\n",
+ "d=2.; ## m\n",
+ "a=1.; ## radius in m\n",
+ "rho=880.; ## density of oil in kg/m^3\n",
+ "g=9.81; ## m/s^2\n",
+ "rho_w=1000.; ## density of water in kg/m^3\n",
+ "\n",
+ "C_0=4*a/3/math.pi; ## centroid of the upper semicircle\n",
+ "h1=a-C_0; ## distance of the centroid from the top\n",
+ "\n",
+ "P1=rho*g*h1; ## Pressure of the oil at this point\n",
+ "F1=P1*math.pi*a**2/2; ## Force exerted by the oil on the upper half of the wall\n",
+ "\n",
+ "cp1=a**4*(math.pi/8-8/(9*math.pi)); ## (AK^2)_C\n",
+ "\n",
+ "cp2=cp1/(math.pi*a**2/2*h1); ## Centre of Pressure below the centroid\n",
+ "\n",
+ "cp0=cp2+h1; ## Centre of Pressure below the top\n",
+ "\n",
+ "P_w=(rho*g*a)+(rho_w*g*C_0);\n",
+ "F_w=P_w*math.pi*a**2/2;\n",
+ "\n",
+ "h2=C_0+rho/rho_w;\n",
+ "cp2_w=cp1/(math.pi*a**2/2*h2);\n",
+ "cp0_w=a+C_0+cp2_w; ## below the top of cylinder\n",
+ "\n",
+ "F_total=F1+F_w;\n",
+ "\n",
+ "## F1*cp0 + F_w*cp0_w = F_total*x\n",
+ "\n",
+ "x=(F1*cp0 + F_w*cp0_w)/F_total;\n",
+ "\n",
+ "print'%s %.1f %s'%(\"Total force =\",F_total,\"N\")\n",
+ "print'%s %.3f %s'%(\"Distance of line of action of total force from top of cylinder =\",x,\"m\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total force = 27905.5 N\n",
+ "Distance of line of action of total force from top of cylinder = 1.260 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "rho=1000.; ## kg/m**3\n",
+ "g=9.81; ## m/s**2\n",
+ "r=4.; ## m\n",
+ "h=2.; ## m\n",
+ "l=5.; ## m\n",
+ "theta=math.pi/6.;\n",
+ "A=h*l;\n",
+ "\n",
+ "F_h=rho*g*h*A; ## Horizontal force\n",
+ "\n",
+ "C0=(2.**2./(12.*2.))+2.; ## distance of line of action below the free surface\n",
+ "\n",
+ "AB=4.-4.*math.cos(theta);\n",
+ "\n",
+ "F_v=rho*g*l*(AB*1.+math.pi*r**2.*theta/(2.*math.pi)-1./2.*h*r*math.cos(theta));\n",
+ "BC=0.237; ## m\n",
+ "\n",
+ "F_net=math.sqrt(F_h**2.+F_v**2.);\n",
+ "\n",
+ "phi=math.atan(F_v/F_h);\n",
+ "\n",
+ "print'%s %.1f %s'%(\"Net force =\",F_net,\"N\")\n",
+ "print'%s %.3f %s'%(\"Angle between net force and horizontal =\",phi,\"degrees\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Net force = 205712.4 N\n",
+ "Angle between net force and horizontal = 0.305 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#GM and states that negative or positive give explanation\n",
+ "m=10.; ## kg\n",
+ "M=80.; ## kg\n",
+ "H=1.5; ## m\n",
+ "rho=1026.; ## kg/m**3\n",
+ "g=9.81; ## m/s**2\n",
+ "d=1.; ## m\n",
+ "\n",
+ "## m*H + M*H/2 =(M+m)(OG)\n",
+ "\n",
+ "OG=(m*H + M*H/2)/(M+m);\n",
+ "\n",
+ "## For vertical equilibrium, buoyancy = weight\n",
+ "h=(M+m)/(rho*math.pi/4*d**2);\n",
+ "\n",
+ "BM=(math.pi*d**4./64.)/(math.pi*d**2.*h/4.);\n",
+ "OB=h/2.;\n",
+ "\n",
+ "GM=OB+BM-OG;\n",
+ "\n",
+ "print'%s %.4f %s'%(\"GM =\",GM,\"m\")\n",
+ "\n",
+ "print(\"Since this is negative (i.e. M is below G) buoy is unstable.\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "GM = -0.2179 m\n",
+ "Since this is negative (i.e. M is below G) buoy is unstable.\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Least vertical downward force and Depth of immersion\n",
+ "m=10.; ## kg\n",
+ "M=80.; ## kg\n",
+ "OG=0.8333; ## m\n",
+ "rho=1026.; ## kg/m^3\n",
+ "g=9.81; ## m/s^2\n",
+ "d=1.; ## m\n",
+ "W=(m+M)*g;\n",
+ "\n",
+ "## W(OG) = (W + F)(OB + BM) = rho*g*%pi/4*d^2*h1*(h1/2+d^2/(16*h1))\n",
+ "\n",
+ "h1=math.sqrt(2*(W*OG/(rho*g*math.pi/4.*d**2) - d**2/16.));\n",
+ "\n",
+ "F=rho*g*math.pi/4*d**2*h1 - W;\n",
+ "\n",
+ "print'%s %.f %s'%(\"Least vertical downward force =\",F,\"N\")\n",
+ "print'%s %.3f %s'%(\"Depth of immersion =\",h1,\"m\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Least vertical downward force = 1072 N\n",
+ "Depth of immersion = 0.247 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Volume left in the tank and Pressure at the lowest corners of the tank \n",
+ "a=5.; ## m/s^2\n",
+ "s=0.5; ## m\n",
+ "phi=math.atan(1./4.); ## degrees\n",
+ "g=9.81; ## m/s^2\n",
+ "rho=880.; ## kg/m^3\n",
+ "\n",
+ "a_x=a*math.cos(phi); ## Horizontal component of acceleration\n",
+ "a_z=a*math.sin(phi); ## Vertical component of acceleration\n",
+ "\n",
+ "theta=math.atan(a_x/(a_z+g)); ## b=tan(theta)\n",
+ "\n",
+ "d=(math.tan(phi)+math.tan(theta))/(1-math.tan(phi)*math.tan(theta));\n",
+ "\n",
+ "c=s*d;\n",
+ "\n",
+ "V=s*(s**2-s*c/2.);\n",
+ "Z=V*1000.\n",
+ "print'%s %.1f %s'%(\"Volume left in the tank =\",Z,\"L\")\n",
+ "P=rho*g*s*math.cos(phi);\n",
+ "print'%s %.f %s'%(\"Pressure at the lowest corners of the tank =\",P,\"Pa\")\n",
+ "#apporxmatilty it is correct which is 4190 is given in text book"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Volume left in the tank = 76.5 L\n",
+ "Pressure at the lowest corners of the tank = 4188 Pa\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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