{
"metadata": {
"name": "",
"signature": "sha256:8fbfdaa034f011c9304b20c16adbf915d0a8f58f4b89051b6e8db7227dd1440f"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7 : Flow Under Varying Head"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.1 Page No : 139"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"from scipy.integrate import quad\n",
"\n",
"#initialisation of variables\n",
"g= 32.2 \t#ft/sec**2\n",
"d= 6. \t#ft\n",
"di= 2. \t #in\n",
"h= 9. \t #ft\n",
"Cd= 0.6\n",
"\n",
"#CALCULATIONS\n",
"def fun(H):\n",
" return H**-0.5*(d/2)**2*math.pi/(Cd*math.pi*math.sqrt(2*g)/144)\n",
"\n",
"\n",
"vec2=quad(fun,0,h)\n",
"T= vec2[0]\n",
"\n",
"#RESULTS\n",
"print 'Time to emptify = %.f sec'%(T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time to emptify = 1615 sec\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.2 Page No : 140"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"from scipy.integrate import quad\n",
"\n",
"#initialisation of variables\n",
"d1= 4. \t#ft\n",
"d2= 2. \t #in\n",
"l= 300. \t#ft\n",
"P= 5. \t #lb/in**2\n",
"h1= 3. \t #ft\n",
"h2= 6. \t #ft\n",
"f= 0.01\n",
"\n",
"#CALCULATIONS\n",
"X= P*2.31*10*(d2/12)**5/(f*l)\n",
"A= math.pi*d1**2/4\n",
"\n",
"def fun(h):\n",
" return A*math.sqrt((P*2.31*10*(d2/12)**5/(f*l))-(10*(d2/12)**5*h/(f*l)))/(10*(d2/12)**5/(f*l))/7\n",
"\n",
"vec2=quad(fun,h1,h2)\n",
"T= vec2[0]\n",
"\n",
"#RESULTS\n",
"print 'time for the channel to fall = %.2f sec'%(T)\n",
"\n",
"# rounding error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time for the channel to fall = 689.35 sec\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.3 Page No : 141"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from scipy.integrate import quad\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"d= 10. \t#in\n",
"l= 15. \t#ft\n",
"di= 3. \t#in\n",
"Cd= 0.62 \n",
"g=32.2\n",
"\n",
"#CALCULATIONS\n",
"def fun(H):\n",
" return -l*2*math.sqrt((d/2)**2-((d/2)-H)**2)/(Cd*(math.pi*(di/12)**2/4)*H**0.5*math.sqrt(2*g))\n",
"\n",
"vec2=quad(fun,0,d/2)\n",
"T= vec2[0]\n",
"secs = -T%60\n",
"mins = -T/60\n",
"#RESULTS\n",
"print 'time for the channel to fall = %d mins and %d seconds'%(mins,secs)\n",
"\n",
"# rounding error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time for the channel to fall = 27 mins and 54 seconds\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.4 Page No : 142"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"from scipy.integrate import quad\n",
"\n",
"#initialisation of variables\n",
"h= 4. \t#ft\n",
"w= 6. \t#ft\n",
"l= 100. \t#yd\n",
"a= 60. \t#degrees\n",
"h1= 3. \t#ft\n",
"h2= 2. \t#ft\n",
"Cd= 0.6\n",
"g=32.2 \t#ft/s**2\n",
"\n",
"#CALCULATIONS\n",
"A= l*3*w\n",
"def fun(H):\n",
" return -A*H**-2.5/(Cd*(8./15)*(math.tan(math.radians(a/2)))*math.sqrt(2*g))\n",
"\n",
"vec2=quad(fun,h1,(h1-h2))\n",
"T= vec2[0]\n",
"\n",
"#RESULTS\n",
"print 'time for the channel to fall = %.f sec'%(T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time for the channel to fall = 654 sec\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.5 Page No : 143"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"A= 1./16 \t#mile**2\n",
"d= 2. \t#ft\n",
"h= 18. \t#ft\n",
"h1= 5. \t#ft\n",
"f= 0.006\n",
"l= 200. \t#ft\n",
"h2= 10. \t#ft\n",
"g= 32.2 \t#ft/sec**2\n",
"\n",
"#CALCULATIONS\n",
"X= math.sqrt(1./((1.5+(4*f*l/d))/(2*g)))\n",
"def fun(H):\n",
" return A*5280**2*H**-0.5/(math.pi*d**2*X/4)\n",
"\n",
"vec2=quad(fun,h-h1,h)\n",
"T= vec2[0]\n",
"hours = T/3600\n",
"mins = T%3600/60\n",
"\n",
"#RESULTS\n",
"print 'time for the channel to fall = %d hours and %d mins sec'%(hours,round(mins,-1))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time for the channel to fall = 48 hours and 20 mins sec\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.6 Page No : 144"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"l= 8. \t#ft\n",
"b= 6. \t#ft\n",
"h= 10. \t#ft\n",
"r= 3.\n",
"Cd= 0.6\n",
"A1= 36. \t#ft**2\n",
"A2= 12. \t#ft**2\n",
"l1= 6. \t#ft\n",
"h1= 1. \t#ft\n",
"d= 2. \t#in\n",
"g=32.2 \t#ft/s**2\n",
"\n",
"#CALCULATIONS\n",
"def fun(H):\n",
" return H**-0.5/(Cd*(math.pi*(d/12)**2/4)*math.sqrt(2*g)*((1/A1)+(1/A2)))\n",
"\n",
"vec2=quad(fun,0,(b-h1))\n",
"T= vec2[0]\n",
"\n",
"#RESULTS\n",
"print 'time for the levels to become equal = %.f sec'%(T)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time for the levels to become equal = 383 sec\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.7 Page No : 145"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"h1= 3. \t#ft\n",
"h2= 4. \t#ft diameter\n",
"r= 0.95 \t#m**-1\n",
"k= 27.65 \t#sec\n",
"Cd= 0.95\n",
"\n",
"#CALCULATIONS\n",
"T= k*(math.log(r*math.sqrt(h2)-1)+(r*math.sqrt(h2)-1))-k*(math.log(r*math.sqrt(h1)-1)+(r*math.sqrt(h1)-1))\n",
"h= ((h2-h1)/Cd)**2\n",
"\n",
"#RESULTS\n",
"print 'Time = %.2f sec'%(T)\n",
"print ' Increase in water level = %.2f ft'%(h)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time = 16.23 sec\n",
" Increase in water level = 1.11 ft\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.8 Page No : 146"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"t= 75. \t#sec\n",
"h= 10.5 \t#in constant\n",
"h1= 13.5 \t#in\n",
"\n",
"#CALCULATIONS\n",
"r= t*math.pi*math.sqrt(2*h**2)/math.log((math.sqrt(2*h1**2)+h1)/(math.sqrt(2*h**2)-h))\n",
"t= -r*((1/h1)-(1/h))\n",
"\n",
"#RESULTS\n",
"print 'A/K = %.f '%(r)\n",
"print ' Time taken = %.1f sec'%(t)\n",
"\n",
"# rounding off error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"A/K = 1737 \n",
" Time taken = 36.8 sec\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.9 Page No : 148"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"g= 9.8 \t#m/sec**2\n",
"h1= 10. \t#in\n",
"h2= 12. \t#in\n",
"r1= 1.32\n",
"r2= 1.56\n",
"r3= 1.97\n",
"r4= 4.10\n",
"r5= 2.64\n",
"\n",
"#CALCULATIONS\n",
"Q= math.sqrt(32.2)*(h2/18)**1.5\n",
"T= 10**5*(r1+2*r3+r4+4*(r3+r5))/(6*h2*60*60)\n",
"\n",
"#RESULTS\n",
"print 'Actual discharge = %.2f CBH**1.5 cuses'%(Q)\n",
"print ' Time = %.1f hr'%(T)\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": [
"Actual discharge = 3.09 BH**1.5 cuses\n",
" Time = 10.7 hr\n"
]
}
],
"prompt_number": 20
}
],
"metadata": {}
}
]
}