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-{
- "metadata": {
- "name": "",
- "signature": "sha256:db7a54602402420ebc03881b70578491b776f0322d83cb209fd1e5faed86cdad"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2 : Fluid Statics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.1 Page no 44"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "\n",
- "d = 10 # diameter of hydraulic press in meters\n",
- "\n",
- "d1 = 1 # diameter of piston in meters\n",
- "\n",
- "W = 1000 # weight in Newtons\n",
- "\n",
- "Ap = math.pi*d1**2/4 # Area of piston in m**2\n",
- "\n",
- "Ar = math.pi*d**2/4 # Area of rram in m**2\n",
- "\n",
- "\n",
- "p = W/Ar # pressure to be supplied by the oil in N/cm**2\n",
- "\n",
- "F = p*Ap # Force applied on the piston\n",
- "\n",
- "print \"Using the pascal's law a weight of 1000N can be lifted by applying a force only of \",round(F,1),\"N\"\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Using the pascal's law a weight of 1000N can be lifted by applying a force only of 10.0 N\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.2 Page no 53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "\n",
- "h = 1 # ocean depth below the surface in km\n",
- "\n",
- "gma = 10070 # Specific weight of sea water\n",
- "\n",
- "\n",
- "P =gma*h # Pressure in kN/m**2\n",
- "\n",
- "print \"Pressure = \",round(P),\"kN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure = 10070.0 kN/m**2\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.3 Page no 53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "\n",
- "p1 = 150*10**3 # Pressure at point 1 in kN/m**2\n",
- "\n",
- "Sg = 0.85 # Specific gravity of oil\n",
- "\n",
- "h = 0.8 # height of oil 2 i tank in meters \n",
- "\n",
- "g = 9810 # specific gravity \n",
- "\n",
- "h1 = 2.0 # height of oil 3 in tank\n",
- "\n",
- "\n",
- "p2 = (p1 + Sg*h*g)\n",
- "\n",
- "p3 = (p2 + g*h1)/1000\n",
- "\n",
- "print \"Pressure at the bottom of the tank is\",round(p3,1),\"kN/m**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure at the bottom of the tank is 176.3 kN/m**3"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.4 Page no 54"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "from __future__ import division\n",
- "\n",
- "Po = 570 # mercury barometer reading in mm\n",
- "\n",
- "T = 273 -5 # temperature in K\n",
- "\n",
- "p = 750 # mercury barometer reading in mm\n",
- "\n",
- "n = 1.2345 # for polytropic atmosphere\n",
- "\n",
- "R =287 # univerasl gas constant in J/Kg-K\n",
- "\n",
- "g = 9.81 \n",
- "\n",
- "r = p/Po\n",
- "\n",
- "\n",
- "y = -(R*T/(g*0.19))*(1 - (r)**((n-1)/n))\n",
- "\n",
- "print \"Height of the mountain is\",round(y,0),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Height of the mountain is 2208.0 m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.5 Page no 57"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "\n",
- "h1 = 500 # height in mm\n",
- "\n",
- "h2 = 950 # height in mm\n",
- "\n",
- "S1 = 1 # specific gravity fo water\n",
- "\n",
- "S2 = 1.5 # specific gravity of liquid 2\n",
- "\n",
- "w = 9810 # specific weight of water\n",
- "\n",
- "\n",
- "ha = ((h2*S2)-(h1*S1))/1000\n",
- "\n",
- "Pa = w*ha/1000 # Pressure in kPa\n",
- "\n",
- "print \"Pressure in the pipe = \",round(Pa,3),\"kPa\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure in the pipe = 9.074 kPa\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.6 Page no 66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "\n",
- "from __future__ import division\n",
- "\n",
- "from math import *\n",
- "\n",
- "d = 1 # diameter of the gate in m\n",
- "\n",
- "w = 9810 # specific weight in N/m**3\n",
- "\n",
- "A = pi*d**2/4 # area of the gate\n",
- "\n",
- "Ig = pi*d**4/64 # mamm moment of inertia\n",
- "\n",
- "theta = pi/2\n",
- "\n",
- "y1= 5+0.5\n",
- "\n",
- "\n",
- "F = w*round(A,3)*(y1) # the answer will come out to be different as they have used the value of Area as 0.78 \n",
- "\n",
- "\n",
- "h1 = y1 + (Ig*math.sin(theta)*math.sin(theta)/(A*y1))\n",
- "\n",
- "\n",
- "F1 = (F*(h1 - 5))/d\n",
- "\n",
- "print \"Magnitude of the force required to open the gate = \",round(F1,0),\"N\" \n",
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of the force required to open the gate = 21659.0 N\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.7 Page no 67"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "\n",
- "l =2 # length of the plate in m\n",
- "\n",
- "b =1 # width of the plate\n",
- "\n",
- "theta = pi/3\n",
- "\n",
- "h = 0.75 # depth of the plate\n",
- "\n",
- "w = 9810 # specific weight of water\n",
- "\n",
- "\n",
- "A = 1*2\n",
- "\n",
- "y1 = h + 1*sin(theta)\n",
- "\n",
- "F = w*A*y1/1000\n",
- "\n",
- "print \"(a) Total force = \",round(F,2),\"kN\"\n",
- "\n",
- "Ig = (b*l**3)/12\n",
- "\n",
- "h1 = y1 + (Ig*sin(theta)*sin(theta)/(2*y1))\n",
- "\n",
- "print \"(b) Position of center of pressure = \",round(h1,3),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Total force = 31.71 kN\n",
- "(b) Position of center of pressure = 1.616 m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.8 Page no 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "\n",
- "d = 6 # diameter of the gate in ft\n",
- "\n",
- "A =pi*d**2/4 # area of the gate\n",
- "\n",
- "p1 = 600 # pressure on top in psia\n",
- "\n",
- "y1 = 10 +2 + 3*sin(pi/6)\n",
- "\n",
- "F = 62.4*A*y1\n",
- "\n",
- "F1 = p1*A\n",
- "\n",
- "\n",
- "Tf = F+F1\n",
- "\n",
- "print \"Total hydrostatic force =\",round(Tf,0),\"lbs\"\n",
- "\n",
- "Ig = pi*d**4/64\n",
- "\n",
- "h1 = y1 + ((Ig*sin(pi/6)*sin(pi/6))/(A*y1))\n",
- "\n",
- "H = ((F*h1)+(F1*y1))/Tf\n",
- "\n",
- "print \"point of location on the center plate = \",round(H,2),\"ft\"\n",
- "\n",
- "\n",
- "Hf = p1/62.4 # equivalent fluid height\n",
- "\n",
- "y2 = Hf+y1\n",
- "\n",
- "Tf1 = 62.4*A*y2\n",
- "\n",
- "h2 = y2 + ((Ig*sin(pi/6)*sin(pi/6))/(A*y2))\n",
- "\n",
- "H1 = y2-Hf\n",
- "\n",
- "print \" OR point of location on the center plate from method 2 = \",round(H1,2),\"ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total hydrostatic force = 40783.0 lbs\n",
- "point of location on the center plate = 13.52 ft\n",
- " OR point of location on the center plate from method 2 = 13.5 ft\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.10 Page no 74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "import math\n",
- "\n",
- "\n",
- "R = 4 # radius of the gate in ft\n",
- "\n",
- "w = 1 # width of the gate in ft\n",
- "\n",
- "gma =62.4 # specific weight of water\n",
- "\n",
- "y1 = 4 # distance of center of the gate\n",
- "\n",
- "xv1 = 2 # distance in ft\n",
- "\n",
- "xv2 = 1.7 # distance in ft\n",
- "\n",
- "\n",
- "Fh = R*y1*gma\n",
- "\n",
- "Ig = w*R**3/12\n",
- "\n",
- "yh = y1 + (Ig/(R*y1))\n",
- "\n",
- "Fv1 = R*2*gma\n",
- "\n",
- "Fv2 = pi*R**2*gma/4\n",
- "\n",
- "Fv = Fv1 + Fv2\n",
- "\n",
- "Xv = (Fv1*xv1+Fv2*xv2)/(Fv)\n",
- "\n",
- "print \"Horizontal component acting on the plate = \",round(Fh,2),\"lbs\"\n",
- "\n",
- "print \"Vertical component acting on the plate = \",round(Fv,2),\"lbs\"\n",
- "\n",
- "print \"location of Xv =\",round(Xv,2),\"ft\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Horizontal component acting on the plate = 998.4 lbs\n",
- "Vertical component acting on the plate = 1283.34 lbs\n",
- "location of Xv = 1.82 ft\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.11 Page no 77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "from __future__ import division\n",
- "\n",
- "\n",
- "p = 50 # pressure in psia\n",
- "\n",
- "gma = 62.4 # specific weight in ft\n",
- "\n",
- "h1 = p*144/gma # equivalent height of water surface in ft\n",
- "\n",
- "R = 4 # radius of gate in ft\n",
- "\n",
- "w = 1 # width of the gate in ft\n",
- "\n",
- "A = R*w\n",
- "\n",
- "y1 = h1 + 2.5 + 2 # center of pressure\n",
- "\n",
- "xv1 = 2 # center of pressure1 for x direction force\n",
- "\n",
- "xv2 = 1.7 # center of pressure2 for x direction force\n",
- "\n",
- "\n",
- "Fh = gma*A*y1 # hiorizontal force\n",
- "\n",
- "Ig = 5.33 # moment of inertia\n",
- "\n",
- "yh = y1 + (Ig/(A*y1)) # location of horizontal component\n",
- "\n",
- "y2 = h1+2.5\n",
- "\n",
- "Fv1 = gma*(R*y2) # vertical force component 1\n",
- "\n",
- "Fv2 = gma*(pi*R**2/4) # vrtical force component 2\n",
- "\n",
- "Fv = Fv1 + Fv2 # vertical force component\n",
- "\n",
- "Xv = (Fv1*xv1+Fv2*xv2)/(Fv)\n",
- "\n",
- "print \"(a) Horizontal component acting on the plate = \",round(Fh,0),\"lbs\" # The answer for horizontal force in the book is wrong\n",
- "\n",
- "print \"(b) Vertical component acting on the plate = \",round(Fv,2),\"lbs\"\n",
- "\n",
- "print \"location of vertical component Xv =\",round(Xv,2),\"ft from the left wall\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Horizontal component acting on the plate = 29923.0 lbs\n",
- "(b) Vertical component acting on the plate = 30208.14 lbs\n",
- "location of vertical component Xv = 1.99 ft from the left wall\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.13 Page no 84"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "from __future__ import division\n",
- "\n",
- "\n",
- "l = 3 # length in m\n",
- "\n",
- "b = 4 # breadth in m\n",
- "\n",
- "h = 15 # height in m\n",
- "\n",
- "S = 0.9 # specific gravity of barge\n",
- "\n",
- "Sw = 1.09 # specific gravity of water\n",
- "\n",
- "Wd = 150*10**3 # addditional weight in kN\n",
- "\n",
- "V = l*b*h # volume in m**3\n",
- "\n",
- "Wb = gma*S*V # weight of barge\n",
- "\n",
- "Tw = Wb + Wd # total weight in kN\n",
- "\n",
- "gma = 9810 # specific density \n",
- "\n",
- "\n",
- "Fb = Tw # since barge is floating\n",
- "\n",
- "V1 = Fb/((gma/1000)*Sw) # volume in m**3\n",
- "\n",
- "\n",
- "d = (V1/(h*b))/1000\n",
- "\n",
- "print \"Depth to which water would rise = \",round(d,2),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Depth to which water would rise = 0.25 m\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.14 Page no 85 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "from __future__ import division\n",
- "\n",
- "import numpy as np\n",
- "\n",
- "\n",
- "W = 0.4 * 9.81 # weight of the solid cylinder in N\n",
- "\n",
- "\n",
- "A = np.array([(1,-0.96),(1,1)])\n",
- "\n",
- "b = np.array([0,6.37])\n",
- "\n",
- "x = np.linalg.solve(A,b)\n",
- "\n",
- "X = x[0]\n",
- "\n",
- "Y = x[1]\n",
- "\n",
- "print \"X = \",round(X,2),\"cm\"\n",
- "\n",
- "print \"Y = \",round(Y,2),\"cm\"\n",
- "\n",
- "b = 8 -x[0]\n",
- "\n",
- "print \"The bottom of the solid cylinder will be \",round(b,2),\"cm above the bottom\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "X = 3.12 cm\n",
- "Y = 3.25 cm\n",
- "The bottom of the solid cylinder will be 4.88 cm above the bottom\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.15 Page no 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "\n",
- "l =100 # length of the pan in cm\n",
- "\n",
- "w = 20 # width of the pan in cm\n",
- "\n",
- "d = 4 # depth of the pan in cm\n",
- "\n",
- "L = 1.5 # load in N/m\n",
- "\n",
- "gma = 9810 # sepcific weight\n",
- "\n",
- "\n",
- "Fb = gma*(d*w*l/(2*l**3)) # weight on the pan\n",
- "\n",
- "W = Fb-L # weight of the pan\n",
- "\n",
- "X1 = w/3\n",
- "\n",
- "X2 = w/2\n",
- "\n",
- "theta = math.atan(d/w)*180/pi\n",
- "\n",
- "x = ((X2-X1)*cos(theta*pi/180))\n",
- "\n",
- "\n",
- "M = W*x\n",
- "\n",
- "print \"Weight of the pan =\",round(W,1),\"N\"\n",
- "print \"Magnitude of right momentum = \",round(M,1),\"N.cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Weight of the pan = 37.7 N\n",
- "Magnitude of right momentum = 123.4 N.cm\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example no 2.16 Page no 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "from __future__ import division\n",
- "\n",
- "\n",
- "\n",
- "Io = 15*4**3/12 # moment of inertia in m**4\n",
- "\n",
- "V = 15*4*2.71 # Volume in m**3\n",
- "\n",
- "Gb = ((3/2)-(2.71/2)) \n",
- "\n",
- "W = 1739.2 # weight of the barge from the previous example in kN\n",
- "\n",
- "Mg = (Io/V)-Gb # metacentric height in m\n",
- "\n",
- "print \"(a) Metacentric height = \", round(Mg,3),\"m\"\n",
- "\n",
- "M = W*Mg*sin(pi*6/180)\n",
- "\n",
- "print \"(b) Righting moment =\",round(M,1),\"kN.m\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Metacentric height = 0.347 m\n",
- "(b) Righting moment = 63.1 kN.m\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example no 2.17 Page no 92"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "from __future__ import division\n",
- "\n",
- "\n",
- "l=6 # length of the tank\n",
- "\n",
- "w =2 # width of the tank\n",
- "\n",
- "d = 3 # depth of the tank\n",
- "\n",
- "a = 3 # acceleration in m/s**2\n",
- "\n",
- "theta = pi/6\n",
- "\n",
- "W = 9810 # specific weight\n",
- "\n",
- "X = 0\n",
- "\n",
- "po=0 # pressure at the origin\n",
- "\n",
- "\n",
- "A = np.array([(1,-1),(1,1)])\n",
- "\n",
- "b = np.array([-1.38,3.0])\n",
- "\n",
- "x = np.linalg.solve(A,b)\n",
- "\n",
- "Y1 = x[0]\n",
- "\n",
- "Y2 = x[1]\n",
- "\n",
- "\n",
- "P = po - W*(2.61*X/9.81) - W*(1+(1.5/9.81))*(-Y2)\n",
- "\n",
- "print\"Amximum pressure occurs at the bottom of the tank =\",round(P,3),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Amximum pressure occurs at the bottom of the tank = 24768.9 N\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.18 Page no 98"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from math import *\n",
- "\n",
- "from __future__ import division\n",
- "\n",
- "\n",
- "d = 1 # diamter of the jar in ft\n",
- "\n",
- "h =2 # height of the jar in ft\n",
- "\n",
- "H = 1 # height of water in the jar in ft\n",
- "\n",
- "N = 150 # RPM\n",
- "\n",
- "g = 32.2 # acceleration due to gravity in ft/s**2\n",
- "\n",
- "\n",
- "w = 2*pi*N/60\n",
- "\n",
- "ho = H+((w**2*(d/2)**2)/(4*g))\n",
- "print \"(a) Height of paraboliod revolution of base = \", round(ho,2),\"ft\"\n",
- "\n",
- "Pmax = 62.4*ho\n",
- "\n",
- "print \"(b) Maximum pressure corresponding to maximum height = \",round(Pmax,1),\"lbs/ft**2\"\n",
- "\n",
- "z = H - ((w**2*(d/2)**2)/(4*g))\n",
- "\n",
- "r = 0.2 # distance from center\n",
- "\n",
- "y = -(0.52-0.25)\n",
- "\n",
- "P = po + (62.4*w**2*r**2/(2*g))-(62.4*y)\n",
- "\n",
- "print \"(c) Pressure =\",round(P,1),\"lbs/ft**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Height of paraboliod revolution of base = 1.48 ft\n",
- "(b) Maximum pressure corresponding to maximum height = 92.3 lbs/ft**2\n",
- "(c) Pressure = 26.4 lbs/ft**2\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
generated by cgit (git 2.34.1) at 2024-12-20 10:35:33 +0000