{
"metadata": {
"name": "",
"signature": "sha256:99b4244f7dba40fbe6b1f9647ffae37a043c932f7cb3e888c2ad7aac3ff9563e"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5 : Flow in Channels"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.1 Page No : 67"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"h= 2.5 \t#ft depth of water\n",
"a= 45. \t#degrees side slope\n",
"x= 5. \t#ft\n",
"Q= 45. \t#cuses\n",
"v= 2.6 \t#ft/sec velocity\n",
"w= 6.92 \t#ft \n",
"C= 120.\n",
"\n",
"#CALCULATIONS\n",
"b= (Q/(v*h))-h\n",
"p= b+2*(h+math.sqrt(2))\n",
"A= h*w\n",
"m= A/p\n",
"i= (v/(C*math.sqrt(m)))**2\n",
"\n",
"#RESULTS\n",
"print 'Width = %.2f ft'%(b) \n",
"print ' Slope = %.6f '%(i) \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Width = 4.42 ft\n",
" Slope = 0.000332 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.2 Page No : 69"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"a= 60. \t#degrees sides inclined\n",
"i= 1./1600\n",
"Q= 8.*10**6 \t#gal/hr discharge\n",
"M= 110.\n",
"w= 6.24 \t#lb/ft**3\n",
"\n",
"#CALCULATIOS\n",
"d= ((Q*2**(2./3)*math.sqrt(1./i))/(w*3600*math.sqrt(3)*M))**(3./8)\n",
"b=6.93 \t#ft\n",
"\n",
"#RESULTS\n",
"print 'Diameter = %.f ft'%(d) \n",
"print ' breadth = %.2f ft'%(b)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter = 6 ft\n",
" breadth = 6.93 ft\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.3 Page No : 71"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"g= 32.2 \t#ft/swc**2\n",
"Q= 40. \t#cuses rate\n",
"w= 5.5 \t#ft\n",
"h= 9. \t#in depth\n",
"d= 0.75 \t#ft\n",
"V= 3. \t#ft/sec\n",
"\n",
"#CALCULATIONS\n",
"D= ((Q*2)**2/(g*(w*2)**2))**(1./3)\n",
"v= Q*d/w\n",
"D1= math.sqrt((2*v**2*d/g)+h/64)-(d/2)\n",
"dD= D1-d\n",
"El= -dD+((v**2*(1-(V/v)**2))/(2*g))\n",
"Els= Q*El*62.4/550\n",
"\n",
"#RESULTS\n",
"print 'Critical depth = %.2f ft'%(D)\n",
"print ' Rise in level = %.f ft'%(D1)\n",
"print ' Horse-power lost = %.3f hp'%(Els) \n",
"\n",
"#The answer is a bit different due to rounding off error in textbook\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Critical depth = 1.18 ft\n",
" Rise in level = 1 ft\n",
" Horse-power lost = 0.961 hp\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.6 Page No : 77"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"b= 3.5 \t#ft\n",
"H= 2.5 \t#ft\n",
"w= 3. \t#ft depth\n",
"h= 6. \t#ft wide\n",
"g= 32.2 \t#ft/sec**2\n",
"\n",
"#CALCULATIONS\n",
"Q= 3.09*b*H**1.5\n",
"v= Q/(w*h)\n",
"H1= H+(v**2/(2*g))\n",
"Q1= 3.09*b*H1**1.5\n",
"hc= (Q1**2/(b**2*g))**(1./3)\n",
"h2= 0.5*(math.sqrt(hc**2+8*hc**2)-hc)\n",
"dh= h2+b-w\n",
"\n",
"#RESULTS\n",
"print \"Flow rate = %.1f cusecs\"%(Q)\n",
"print \" Flow rate = %d cusecs\"%(Q1)\n",
"print ' maximum depth of water downstream = %.3f ft'%(dh) \n",
"print ' Shooting flow depth at hump = %.3f ft'%(h2) \n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Flow rate = 42.8 cusecs\n",
" Flow rate = 45 cusecs\n",
" maximum depth of water downstream = 2.226 ft\n",
" Shooting flow depth at hump = 1.726 ft\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.7 Page No : 79"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"m= 60./26\n",
"i= 1./2000\n",
"h1= 3. \t#ft depth\n",
"h2= 5. \t#ft depth\n",
"m1= 10./3\n",
"C= 90. # constant\n",
"l= 500. \t#ft depth\n",
"H= 20. \t#ft broad\n",
"H1= 29.62 \t#ft\n",
"g= 32.2 \t#ft/s**2\n",
"\n",
"#CALCULATIONS\n",
"v= 90*math.sqrt(m*i)\n",
"v1= v*h1/h2\n",
"dh= (i-(v1**2/(C**2*m1)))*l/(1-v1**2/(g*h2))\n",
"h3= h2-dh\n",
"V= h1*v/h3\n",
"\n",
"#RESULTS\n",
"print 'Height of water 1000 ft upstream = %.3f ft'%(h3) \n",
"print ' Height of water upstream = %.3f ft'%(h3) \n",
"\n",
"#The answer is a bit different due to rounding off error in textbook\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Height of water 1000 ft upstream = 4.808 ft\n",
" Height of water upstream = 4.808 ft\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.8 Page No : 80"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"v= 5. \t #ft/sec\n",
"m= 60./26\n",
"i= 1./2000\n",
"h= 5.5 \t#ft\n",
"m1= 110./31\n",
"d= 3. \t #ft\n",
"g= 32.2 \t#ft/sec**2\n",
"\n",
"#CALCULATIONS\n",
"C= v/(math.sqrt(m*i))\n",
"v1= v*d/h\n",
"r= (i-(v1**2/(C**2*m1)))/(1-(v1**2/(g*h)))\n",
"x= 1/r\n",
"\n",
"#RESULTS\n",
"print 'Distance upstream = %.f ft'%(round(x,-1)) \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Distance upstream = 2380 ft\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.9 Page No : 81"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"from numpy import *\n",
"from numpy.linalg import *\n",
"\n",
"#initialisation of variables\n",
"g= 32.2 \t#ft/sec**2\n",
"Q= 12 \t#cuses\n",
"\n",
"#CALCULATIONS\n",
"hc= (Q/(3*math.sqrt(g)))**(2./3)\n",
"vec=roots([1,6,12,8,0,-8.95,-8.95])\n",
"H=vec[2]\n",
"\n",
"#RESULTS\n",
"print 'Critical depth = %.2f ft'%(hc) \n",
"print ' Critical depth = %.2f ft'%(H) \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Critical depth = 0.79 ft\n",
" Critical depth = 0.89 ft\n"
]
},
{
"output_type": "stream",
"stream": "stderr",
"text": [
"-c:17: ComplexWarning: Casting complex values to real discards the imaginary part\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.11 Page No : 85"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"Cd= 0.64 # coefficient\n",
"g= 32.2 \t#ft/sec**2\n",
"A= 12.5 \t#ft**2\n",
"H= 24.8 \t#ft\n",
"Q= 3200. \t#cuses\n",
"b= 150. \t#ft wide\n",
"A1= 5.*10**6 # avg surface area\n",
"h= 9. \t#ft\n",
"h1= 6. \t #in\n",
"\n",
"#CALCULATIONS\n",
"N= Q/(Cd*A*math.sqrt(2*g*H))\n",
"H1= (Q/(3.2*b))**(2./3)\n",
"ES= (H1-(h1/12))*A1*h\n",
"\n",
"#RESULTS\n",
"print 'number of siphons = %.f '%(N) \n",
"print ' Extra Storage = %.2e ft**3'%(ES) \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"number of siphons = 10 \n",
" Extra Storage = 1.37e+08 ft**3\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}