{
"metadata": {
"name": "",
"signature": "sha256:97f449fad397b42ecb1501f787392dc6e0950fafed8997a70fb14e2a9c19a198"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3 : Flow Through Orifices Mouthpieces Nozzles"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.1 page no : 70"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"\n",
"import math \n",
"Q = 16. \t\t\t#gpm\n",
"w = 62.4 \t\t\t#lb/ft**3\n",
"d = 1. \t\t\t#in\n",
"h = 2+(5./12) \t\t\t#ft\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"x = 11.5 \t\t\t#ft\n",
"h1 = 1.2 \t\t\t#in\n",
"#CALCULATIONS\n",
"Cd = Q*10/(60*w*(math.pi*(d/12)**2/4)*math.sqrt(2*g*h))\n",
"Cv = math.sqrt(x**2/(4*(h1/12)*h*12**2))\n",
"Cc = Cd/Cv\n",
"Cr = (1-Cv**2)/Cv**2\n",
"#RESULTS\n",
"print 'Cc = %.3f '%(Cc)\n",
"print 'Cv = %.3f '%(Cv)\n",
"print 'Cd = %.3f '%(Cd)\n",
"print 'Cr = %.3f '%(Cr)\n",
"\n",
"# note : answers are slightly different because of rounding error."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Cc = 0.644 \n",
"Cv = 0.975 \n",
"Cd = 0.628 \n",
"Cr = 0.053 \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.2 page no : 71"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"Ww = 261. \t\t\t#lb/min\n",
"a = 1. \t\t\t#in**2\n",
"h = 4. \t\t\t#ft\n",
"y = 5. \t\t\t#ft\n",
"W1 = 10.65 \t\t\t#lb\n",
"l = 1. \t\t\t#ft\n",
"Q = 261. \t\t\t#lb/min\n",
"w = 62.4 \t\t\t#lb/ft**3\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"#CALCULATIONS\n",
"v = Q*144./(w*60)\n",
"F = W1*l/y\n",
"v = F*g*60./Q\n",
"vth = math.sqrt(2*g*h)\n",
"Cv = v/vth\n",
"Q1 = Ww/w\n",
"Qth = vth*60./144\n",
"Cd = Q1/Qth\n",
"Cc = Cd/Cv\n",
"#RESULTS\n",
"print 'Cv = %.3f '%(Cv)\n",
"print 'Cd = %.3f '%(Cd)\n",
"print 'Cc = %.3f '%(Cc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Cv = 0.982 \n",
"Cd = 0.625 \n",
"Cc = 0.637 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.3 page no : 74"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"\n",
"import math \n",
"Q = 10. \t\t\t#ft**3/sec\n",
"a1 = 1. \t\t\t#ft**2\n",
"a2 = 4. \t\t\t#ft**2\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"p1 = 12. \t\t\t#lb/in**2\n",
"v1 = 10. \t\t\t#ft/sec\n",
"w = 62.4 \t\t\t#lb/ft**3\n",
"#RESULTS\n",
"v2 = v1*a1/a2\n",
"Hl = (v1-v2)**2/(2*g)\n",
"p2 = ((p1*144/w)+(v1**2/(2*g))-(v2**2/(2*g))-Hl)*(w/144)\n",
"W = Hl*v1*w/550.\n",
"\n",
"#RESULTS\n",
"print 'Head lost = %.3f ft of water '%(Hl)\n",
"print 'Pressure in larger part of pipe = %.2f lb/in**2 '%(p2)\n",
"print 'Work done = %.3f HP '%(W)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Head lost = 0.873 ft of water \n",
"Pressure in larger part of pipe = 12.25 lb/in**2 \n",
"Work done = 0.991 HP \n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.4 page no : 78"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"\n",
"import math \n",
"Cc = 1.\n",
"Cv = 0.833\n",
"d = 2. \t\t\t#in\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"H = 12. \t\t\t#ft\n",
"Pa = 34. \t\t\t#lb/in**2\n",
"#/CALCULATIONS\n",
"Q = Cc*Cv*math.pi*(d/12)**2*math.sqrt(2*g*H)/4\n",
"Cd = Cc*Cv\n",
"Pc = Pa-0.92*H\n",
"#RESULTS\n",
"#RESULTS\n",
"print 'Discharge = %.3f cu ft/sec '%(Q)\n",
"print 'Coefficient of discharge = %.3f '%(Cd)\n",
"print 'Pressure at Vent-contraction = %.2f ft of water '%(Pc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Discharge = 0.505 cu ft/sec \n",
"Coefficient of discharge = 0.833 \n",
"Pressure at Vent-contraction = 22.96 ft of water \n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.5 page no : 81"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"H = 4. \t\t\t#ft\n",
"d = 1. \t\t\t#in\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"Cc = 0.5\n",
"#CALCULATIONS\n",
"Q = Cc*math.pi*(d/12)**2*math.sqrt(2*g*H)/4\n",
"#RESULTS\n",
"print 'Actual Discharge = %.4f cu ft/sec '%(Q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Actual Discharge = 0.0438 cu ft/sec \n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.6 page no : 82"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"\n",
"D = 4. \t\t\t#ft\n",
"d = 2. \t\t\t#in\n",
"H1 = 6. \t\t\t#ft\n",
"H2 = 2. \t\t\t#ft\n",
"t = 4. \t\t\t#min\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"w = 62.4 \t\t\t#lb/ft**3\n",
"H = 5. \t\t\t#ft\n",
"#CALCULATIONS\n",
"Cd = (2.*(math.pi/4.)*D**2*(math.sqrt(H1)-math.sqrt(H2)))/(t*60*(math.pi/4)*(d/12)**2*math.sqrt(2*g))\n",
"Q = Cd*(math.pi/4)*(d/12)**2*math.sqrt(2*g*H)*w*60/10\n",
"#RESULTS\n",
"print 'Cd = %.3f '%(Cd)\n",
"print 'Discharge = %.1f gpm'%(Q)\n",
"\n",
"# note : answers are slightly different because of rounding error."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Cd = 0.619 \n",
"Discharge = 90.8 gpm\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.7 page no : 84"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"H1 = 10. \t\t\t#ft\n",
"H2 = 2. \t\t\t#ft\n",
"Cd = 0.61\n",
"d1 = 8. \t\t\t#ft\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"d2 = 3. \t\t\t#ft\n",
"#CALCULATIONS\n",
"a = d2**2./144\n",
"H0 = H1*d2/(d1-d2)\n",
"t = math.pi*(d1/2)**2*((2/5.)*(H1**(5./2)-H2**(5./2))+2*H0**2*(math.sqrt(H1)- \\\n",
"math.sqrt(H2))+(4./3)*H0*(H1**(3./2)-H2**(3./2)))/(60*Cd*a*math.sqrt(2*g)*(H1+H0)**2)\n",
"\n",
"#RESULTS\n",
"print 'time required to lower the water level = %.2f min'%(t)\n",
"\n",
"# Note : answer is different because of rounding error."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time required to lower the water level = 5.14 min\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.8 page no : 85"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"#initialisation of variables\n",
"D = 10. \t\t\t#ft\n",
"H1 = 17. \t\t\t#ft\n",
"H2 = 5. \t\t\t#ft\n",
"d = 3. \t\t\t#in\n",
"Cd = 0.62\n",
"g =32.2 \t\t\t#ft/s**2\n",
"#CALCULATIONS\n",
"t1 = (2*math.pi*D**2/4)*(math.sqrt(H1)-math.sqrt(H2))/(Cd*math.sqrt(2*g)*math.pi*(d/12)**2/4)\n",
"t2 = math.pi*(14./15)*H2**(5./2)*4/(Cd*math.pi*(d/12.)**2*math.sqrt(2*g))\n",
"t = t1+t2\n",
"#RESULTS\n",
"print 'time required to empty the vessel = %.f sec'%(t)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time required to empty the vessel = 1885 sec\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.9 page no : 86"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"Cd = 0.8\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"d = 3. \t\t\t#in\n",
"#CALCULATIONS\n",
"t = (60*2/(math.pi*(d/12)**2*math.sqrt(2*g)/4*Cd))*(6-d)**(3./2)/(3*60./2)\n",
"#RESULTS\n",
"print 'time to emptify biler = %.2f min'%(t)\n",
"\n",
"# note : answer is different because of rounding error."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time to emptify biler = 21.98 min\n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.10 page no : 87"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"#initialisation of variables\n",
"A = 100 * 27 # sq ft\n",
"dif = 8 - 3. # ft\n",
"a = 2. # sq ft\n",
"Cd = 0.8 # Co-efficient\n",
"\n",
"#CALCULATIONS\n",
"t1 = round(A*((a/3 * 22.7) - a/3 * 5.19 - (a/3*11.2))/(Cd*2*8.02*dif))\n",
"t2 = round(A*(2./3)*11.2/(Cd*a*8.02*dif))\n",
"t = t1 + t2\n",
"\n",
"#RESULTS\n",
"print 'Total time to empty the tank = %d sec'%t"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total time to empty the tank = 491 sec\n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.11 page no : 89"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"#initialisation of variables\n",
"H1 = 9. \t\t\t#ft\n",
"H2 = 4. \t\t\t#ft\n",
"Cd = 0.6\n",
"a = 4. \t\t\t#in**2\n",
"A1 = 72. \t\t\t#ft**2\n",
"A2 = 24. \t\t\t#ft**2\n",
"g =32.2 \t\t\t#ft/s**2\n",
"#CALCULATIONS\n",
"t = (2*A1*A2/(A1+A2))*(math.sqrt(H1)-math.sqrt(H2))*144/(Cd*60*a*math.sqrt(2*g))\n",
"#RESULTS\n",
"print 'time required to reduce the water level difference = %.1f min'%(t)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time required to reduce the water level difference = 4.5 min\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.12 pageno : 91"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"import math \n",
"l = 80. \t\t\t#ft\n",
"w = 12. \t\t\t#ft\n",
"t = 3. \t\t\t#min\n",
"Hl = 12. \t\t\t#ft\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"Cd = 0.6\n",
"#CALCULATIONS\n",
"s = math.sqrt(2*l*w*Hl**(1./2)/(Cd*math.sqrt(2*g)*t*60.))\n",
"#RESULTS\n",
"print 'side of the square orifice = %.2f ft'%(s)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"side of the square orifice = 2.77 ft\n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.13 page no : 92"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#initialisation of variables\n",
"import math \n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"Cd = 0.6\n",
"d = 2. \t\t\t#in\n",
"H1 = 5. \t\t\t#ft\n",
"\n",
"#CALCULATIONS\n",
"K = round(Cd * math.pi/4 * (d/12)**2 * math.sqrt(g*2),3)\n",
"t = d*math.pi*(0.5*math.log(1.89) - 0.235)/K**2 \n",
"v = round(math.sqrt(2*g*H1)/2.)\n",
"q = v*Cd*math.pi*(d/12)**2./4\n",
"\n",
"#RESULTS\n",
"print \"Time required to raise the level is : %.2f sec\"%t\n",
"print 'Total discharge = %.3f cfs'%(q)\n",
"\n",
"# Note : answers may vary because of rounding error. Please calculate manually."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time required to raise the level is : 47.47 sec\n",
"Total discharge = 0.118 cfs\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.14 page no : 95"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"#initialisation of variables\n",
"Cd = 0.62\n",
"H = 9. \t\t\t#in\n",
"l = 3. \t\t\t#ft\n",
"g = 32.2 \t\t\t#t/sec**2\n",
"#CALCULATIONS\n",
"Q1 = Cd*(H*l/12)*math.sqrt(2*g*3*H/24.)\n",
"Q2 = Cd*2*l*math.sqrt(2*g)*((H/6)**(3./2)-(H/12)**(3./2))/3\n",
"#RESULTS\n",
"print 'Discharge by appropriate formula = %.2f cfs'%(Q1)\n",
"print ' Discharge by exact formula = %.2f cfs'%(Q2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Discharge by appropriate formula = 11.87 cfs\n",
" Discharge by exact formula = 11.82 cfs\n"
]
}
],
"prompt_number": 54
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.15 pageno : 95"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"#initialisation of variables\n",
"\n",
"Cd = 0.62\n",
"B = 2.5 \t\t\t#ft\n",
"H2 = 8. \t\t\t#ft\n",
"H1 = 7. \t\t\t#ft\n",
"g = 32.2 \t\t\t#ft/sec**2\n",
"h = 4. \t\t\t#ft\n",
"#CALCULATIONS\n",
"Q1 = round(2*Cd*B*math.sqrt(2*g)*(H2**(3./2)-H1**(3./2))/3)\n",
"Q2 = Cd*math.sqrt(2*g)*math.sqrt(H2)*B*(h-1)\n",
"Q = Q1+Q2\n",
"#RESULTS\n",
"print 'Total discharge = %d cfs'%(Q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total discharge = 139 cfs\n"
]
}
],
"prompt_number": 60
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}