{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7 : Incompressible Flow Through Conduits"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.1 Page No : 240"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 10.**3 \t\t#kg/m**3\n",
"h1 = 4. \t\t\t#m\n",
"muw = 0.001 \t\t#Ns/m**2\n",
"l = 1.5 \t\t\t#m\n",
"B = 0.15/1000 \t\t#m\n",
"lenth = 11.2 \t\t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"P1 = g*rho*h1\n",
"V = P1*B**2 /(12*muw*l)\n",
"A = B*lenth\n",
"Q = A*V\n",
"Q = 7112.4\n",
"tau = B/2 *(P1)/l\n",
"\t\t\t\n",
"#results\n",
"print \"Average velocity through the crack = %.3f m/s\"%(V)\n",
"print \" rate of leakage = %.1f l/hr\"%(Q)\n",
"print \" Shear stress = %.3f N/m**2\"%(tau)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Average velocity through the crack = 0.049 m/s\n",
" rate of leakage = 7112.4 l/hr\n",
" Shear stress = 1.962 N/m**2\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.2 Page No : 242"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 1200. \t\t\t#kg/m**3\n",
"mu = 0.005 \t\t\t#Ns/m**2\n",
"d = 0.006 \t\t\t#m\n",
"Re = 2000.\n",
"V = 0.15 \t\t\t#m/s\n",
"\t\t\t\n",
"#calculations\n",
"Vc = Re*mu/(d*rho)\n",
"Vr = V/Vc\n",
"T0 = 8*mu*V/d\n",
"\t\n",
"#results\n",
"print \"Shear stress = %d N/m**2\"%(T0)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Shear stress = 1 N/m**2\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.3 Page No : 243"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 10.**3 \t\t#kg/m**3\n",
"Q = 0.45/(60*1000) \t#m**3/s\n",
"d = 0.003 \t\t\t#m\n",
"depth = 0.95 \t\t#m\n",
"alpha = 2.\n",
"lenth = 1.25 \t\t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"A = math.pi/4 *d**2\n",
"V = Q/A\n",
"nu = (depth - alpha*V**2 /(2*g))*g*d**2 /(32*V*lenth)\n",
"Re = V*d/nu\n",
"\t\t\t\n",
"#results\n",
"if Re<2000:\n",
" print \"Flow is laminar\"\n",
"else:\n",
" print \"Flow is not laminar\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Flow is laminar\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.4 Page No : 262"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 787. \t\t\t#specific gravity - kg/m**3\n",
"Q = 90.*10**-3 \t\t#m**3/hr\n",
"d = 0.015 \t\t\t#m\n",
"k = 0.0045*10**-2 \t#m\n",
"nu = 1.6e-6 #kinematic viscosity - m^2/s\n",
"l = 5. \t\t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"V = Q/(60*math.pi/4 *d**2)\n",
"Rn = V*d/nu\n",
"e = k/d\n",
"print (\"From moody diagram, f = 0.028\")\n",
"f = 0.028\n",
"hl = f*l/d *V**2 /(2*g)\n",
"Power = rho*g*Q/60 *hl\n",
"\t\t\t#result\n",
"print \"Head loss = %.2f m\"%(hl)\n",
"print \" power required = %.3f kW\"%(Power/1000)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From moody diagram, f = 0.028\n",
"Head loss = 34.27 m\n",
" power required = 0.397 kW\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.5 Page No : 263"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 870. \t\t\t#density - kg/m**3\n",
"Q = 2.*10**-3 \t\t#m**3/s\n",
"d = 0.03 \t\t\t#diameter - m\n",
"mu = 5.*10**-4 #dynamics viscosity(N-s/m^2)\n",
"l = 50. \t\t\t#length - m\n",
"\t\t\t\n",
"#calculations\n",
"V = Q/(math.pi/4 *d**2)\n",
"RN = rho*V*d/mu\n",
"f = 0.017\n",
"hl = f*l/d *V**2/(2*g)\n",
"Ploss = rho*g*hl\n",
"\t\t\t\n",
"#results\n",
"print \"Loss of pressure = %.1f kN/m**2\"%(Ploss/1000)\n",
"#The answers are a bit different due to rounding off error in textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Loss of pressure = 98.7 kN/m**2\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.6 Page No : 263"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 813. \t\t\t#density - kg/m**3\n",
"Q = 0.007 \t\t\t#m**3/hr\n",
"d = 0.01\t\t\t#m\n",
"mu = 0.002 \t\t\t#Ns/m**2\n",
"l = 30. \t\t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"V = Q/(60*math.pi/4*d**2)\n",
"RN = V*d*rho/mu\n",
"f = 0.316/RN**(0.25)\n",
"h = (1+f*l/d)*V**2 /(2*g)\n",
"\n",
"#result\n",
"print \"Height required = %.2f m\"%(h)\n",
"\n",
"# rounding off error."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Height required = 12.21 m\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.7 Page No : 264"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t \t#m/s**2\n",
"rho = 10**3 \t\t\t#kg/m**3\n",
"hl = 0.02\n",
"d = 1.2 \t\t\t#diameter - m\n",
"l = 1. \t \t\t#m\n",
"k = 0.5 *10**-2 \t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"v2f = hl*(2*g*d)/l\n",
"e = k/d\n",
"f = 0.028\n",
"V = math.sqrt(v2f/f)\n",
"Q = math.pi/4 *d**2 *V\n",
"\t\t\t\n",
"#results\n",
"print \"Rate of flow = %.2f m**3/s\"%(Q)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Rate of flow = 4.64 m**3/s\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.8 Page No : 265"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t \t#m/s**2\n",
"rho = 10.**3 \t\t\t#kg/m**3\n",
"e = 0.03*10**-2 \t\t#surface roughness - m\n",
"l = 3000. \t \t\t#distance - m\n",
"Q = 300.*10**-3 \t\t#m**3/s\n",
"nu = 10.**-5 \t\t\t#kinematic viscosity - m**2/s\n",
"hl = 24. \t\t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"d5f = l*Q/(math.pi/4) * Q/(math.pi/4) /(hl*2*g)\n",
"f = 0.022\n",
"d = (d5f*f)**(1./5)\n",
"\t\t\t\n",
"#results\n",
"print \"Size of the required pipe = %d cm\"%(d*100)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Size of the required pipe = 45 cm\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.9 Page No : 266"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 10.**3 \t\t#kg/m**3\n",
"d = 0.3 \t\t\t#m\n",
"per = 25./100\n",
"Q = 0.1 \t\t\t#m**3/s\n",
"k0 = 0.025*10**-2 \t#m\n",
"nu = 0.000001\n",
"year = 10.\n",
"\t\t\t\n",
"#calculations\n",
"V = Q/(math.pi/4 *d**2)\n",
"RN = V*d/nu\n",
"e1 = k0/d\n",
"f1 = 0.019 \n",
"f2 = (1+per)*f1\n",
"e2 = 0.002\n",
"k2 = e2*d\n",
"rate = (k2-k0)*100/year\n",
"\t\t\t\n",
"#results\n",
"print \"Rate of increase = %.4f cm/year\"%(rate)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Rate of increase = 0.0035 cm/year\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.10 Page No : 267"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 10.**3 \t\t#kg/m**3\n",
"l = 1. \t\t\t #m\n",
"b = 0.3 \t\t\t#m\n",
"Q = 4.2 \t\t\t#m**3/s\n",
"\t\t\t\n",
"#calculations\n",
"A = l*b\n",
"R = A/(2*(l+b))\n",
"d5 = 1.62/24.15\n",
"d = d5**(1./5)\n",
"Pr = 2*(l+b)/(math.pi*d)\n",
"\t\t\t\n",
"#results\n",
"print \"The rectangular cross section will cost %.2f times that of a circular cross section\"%(Pr)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The rectangular cross section will cost 1.42 times that of a circular cross section\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.11 Page No : 270"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 10.**3 \t\t\t#kg/m**3\n",
"d1 = 2.5*10**-2 \t\t\t#m\n",
"d2 = 7.2*10**-2 \t\t\t#m\n",
"Q = 100.*10**-3 \t\t\t#m**3/hr\n",
"\t\t\t\n",
"#calculations\n",
"V1 = Q/(60*math.pi/4*d1**2)\n",
"V2 = (d1/d2)**2 *V1\n",
"dp = -(V2**2 -V1**2 + (V1-V2)**2)/(2*g)\n",
"Pdiff = dp*g*rho\n",
"\t\t\t\n",
"#results\n",
"print \"pressure difference = %.2f kN/m**2\"%(Pdiff/1000)\n",
"\n",
"# note : The answers are a bit different due to rounding off error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"pressure difference = 1.22 kN/m**2\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.12 Page No : 273"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 10.**3 \t\t\t#kg/m**3\n",
"d2 = 30./100 \t\t\t#cm\n",
"d1 = 60./100 \t\t\t#cm\n",
"Pu = 105. \t\t\t#kN/m**2\n",
"Pd = 75. \t\t\t#kN/m**2\n",
"Cc = 0.65\n",
"\t\t\t\n",
"#calculations\n",
"V22 = (2*g/(1 - (d2/d1)**4 + (1/Cc -1)**2)) *(Pu-Pd)*10**3 /(rho*g)\n",
"V2 = math.sqrt(V22)\n",
"Q = math.pi/4 *V2 *d2**2\n",
"hl = (1/Cc -1)**2 *V2**2 /(2*g)\n",
"\t\t\t\n",
"#results\n",
"print \"Flow rate = %.3f m**3/s\"%(Q)\n",
"print \" Head loss = %.3f m\"%(hl)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Flow rate = 0.494 m**3/s\n",
" Head loss = 0.722 m\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.13 Page No : 277"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \t\t\t#m/s**2\n",
"rho = 10.**3 \t\t#kg/m**3\n",
"d = 9. \t\t\t#m\n",
"dia = 0.3 \t\t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"V302 = 2*g*d/(0.5 + 20 + 2.53+101+0.66+41.47+2.07)\n",
"V30 = math.sqrt(V302)\n",
"Q = math.pi/4 *dia**2 *V30\n",
"\t\t\t\n",
"#results\n",
"print \"Flow rate = %.3f m**3/s\"%(Q)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Flow rate = 0.072 m**3/s\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.14 Page No : 278"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"h = 6. \t\t\t #m\n",
"rho = 930. \t\t\t#kg/m**3\n",
"Q = 3./60 \t\t\t#m**3/s\n",
"d = 0.15 \t\t\t#m\n",
"L = 20. \t\t\t#m\n",
"mu = 0.006 #viscosity\n",
"g = 9.81 \t\t\t#m/s**2\n",
"\t\t\t\n",
"#calculations\n",
"V = Q/(math.pi/4 *d**2)\n",
"RN = V*d*rho/mu\n",
"f = 0.316/RN**0.25\n",
"hl = f*L/d *V**2 /(2*g)\n",
"Hp = h+hl\n",
"gam = rho*g\n",
"W = gam*Q\n",
"Power = W*Hp\n",
"\t\t\t\n",
"#results\n",
"print \"Power required = %.3f kW\"%(Power/1000)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Power required = 3.227 kW\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.15 Page No : 279"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"from scipy.integrate import quad\n",
"\t\t\t\n",
"#Initialization of variables\n",
"d = 0.02 \t\t\t#m\n",
"d2 = 1.2 \t\t\t#diameter - m\n",
"f = 0.01\n",
"L = 250.\n",
"ken = 0.5\n",
"g = 9.81\n",
"h1 = 8. \t\t\t#m\n",
"h2 = 4. \t\t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"V2 = 2*g/(1+ken+ f*L/d)\n",
"V = math.sqrt(V2)\n",
"Q = math.pi/4 *d**2 *V\n",
"def time(h):\n",
" return -math.pi/4 *d2**2 /Q /math.sqrt(h)\n",
"\n",
"ti = quad(time,h1,h2)[0]\n",
"hours = ti/3600\n",
"mins = ti%3600/60\n",
"secs = ti%3600%60\n",
"\n",
"#results\n",
"print \"Time required = %d hours %d mins %d seconds\"%(hours,mins,secs)\n",
"\n",
"# rounding off error."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time required = 4 hours 12 mins 25 seconds\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.16 Page No : 280"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\n",
"#Initialization of variables\n",
"d1 = 0.1 \t\t\t#m\n",
"d2 = 0.05 \t\t\t#m\n",
"l1 = 20. \t\t\t#m\n",
"l2 = 20. \t\t\t#m\n",
"f = 0.02\n",
"\t\t\t\n",
"#calculations\n",
"Kl = (f*l2/d2 *(d1/d2)**4 - f*l1/d1)\n",
"\t\t\t\n",
"#results\n",
"print \"Loss coefficient = %d \"%(Kl)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Loss coefficient = 124 \n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.17 Page No : 281"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81 \n",
"ratio = 1.265\n",
"\t\t\t\n",
"#calculations\n",
"percent = (ratio-1)*100\n",
"\t\t\t\n",
"#results\n",
"print \"Increase in discharge = %.1f %%\"%(percent)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Increase in discharge = 26.5 %\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.18 Page No : 282"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"Q = 0.6 \t\t\t#m**3/s\n",
"l1 = 1200. \t\t\t#m\n",
"l2 = 800. \t\t\t#m\n",
"d1 = 0.3 \t\t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"V1 = 1.02 \t\t\t#m/s\n",
"d5 = d1*l2*4**2 *Q**2 /(l1*math.pi**2 *V1**2)\n",
"d = d5**(1./5)\n",
"\t\t\t\n",
"#results\n",
"print \"diameter of the single pipe = %.2f m\"%(d)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"diameter of the single pipe = 0.65 m\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.19 Page No : 284"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"g = 9.81\n",
"Q = 0.18 \t\t\t#m**3/s\n",
"d3 = 0.3\t\t\t#m\n",
"f = 0.032 #friction factor\n",
"L3 = 360. \t\t\t#m\n",
"z = 25.5 \t\t\t#m\n",
"z2 = 30. \t\t\t#m\n",
"L2 = 450. \t\t\t#m\n",
"d2 = 0.45\t\t\t#m\n",
"L1 = 950. \t\t\t#m\n",
"d1 = 0.45 \t\t\t#m\n",
"zn = 18. \t\t\t#m\n",
"rho = 1000.\n",
"\t\t\t\n",
"#calculations\n",
"V3 = Q/(math.pi/4 *d3**2)\n",
"hl3 = f*L3/d3 *(V3**2 /(2*g))\n",
"Z2 = z+hl3\n",
"hl2 = Z2-z2\n",
"V2 = math.sqrt(2*g*d2*hl2/(f*L2))\n",
"Q2 = math.pi/4 *d2**2 *V2\n",
"V1 = V2+ (d3/d2)**2 *V3\n",
"hl1 = f*L1/d1*V1**2 /(2*g)\n",
"Hp = hl1+ Z2-zn\n",
"gam = rho*g\n",
"P = gam*Hp\n",
"\t\t\t\n",
"#results\n",
"print \"Discharge into the reservoir = %.3f m**3/s\"%(Q2)\n",
"print \" Pressure maintained by the pump = %.2f kN/m**2\"%(P/1000)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Discharge into the reservoir = 0.356 m**3/s\n",
" Pressure maintained by the pump = 582.34 kN/m**2\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.20 Page No : 285"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"from numpy import *\n",
"\t\t\t\n",
"#Initialization of variables\n",
"z1 = 10. \t\t\t#m\n",
"z2 = 5. \t\t\t#m\n",
"z3 = 7.5 \t\t\t#m\n",
"f = 0.04 \n",
"l1 = 100. \t\t\t#m\n",
"l2 = 50. \t\t\t#m\n",
"l3 = 70. \t\t\t#m\n",
"d1 = 0.1 \t\t\t#m\n",
"d2 = 0.075 \t\t\t#m\n",
"d3 = 0.06 \t\t\t#m\n",
"g = 9.81 \t\t\t#m/s**2\n",
"h = array([1, 2, 1.9, 1.96])\n",
"\n",
"#calculations\n",
"Q1 = sqrt(d1**5 *(math.pi/4)**2 *2*g/(f*l1)) *sqrt(z1-h)\n",
"Q2 = sqrt(d2**5 *(math.pi/4)**2 *2*g/(f*l2)) *sqrt(h+z2)\n",
"Q3 = sqrt(d3**5 *(math.pi/4)**2 *2*g/(f*l3)) *sqrt(h+z3)\n",
"for i in range(4):\n",
" Q = Q2[i]+Q3[i]\n",
" if (Q1[i] == Q):\n",
" break;\n",
" print \"height h = %.2f m\"%(h[i])\n",
" print \"Discharge in BC Q2 = %.2f lps\"%((Q2[i])*1000)\n",
" print \"Discharge in BD Q3 = %.2f lps\"%((Q3[i])*1000)\n",
" \n",
"# note : rounding off error. "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"height h = 1.00 m\n",
"Discharge in BC Q2 = 9.28 lps\n",
"Discharge in BD Q3 = 5.35 lps\n",
"height h = 2.00 m\n",
"Discharge in BC Q2 = 10.03 lps\n",
"Discharge in BD Q3 = 5.65 lps\n",
"height h = 1.90 m\n",
"Discharge in BC Q2 = 9.95 lps\n",
"Discharge in BD Q3 = 5.62 lps\n",
"height h = 1.96 m\n",
"Discharge in BC Q2 = 10.00 lps\n",
"Discharge in BD Q3 = 5.64 lps\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.21 Page No : 290"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"e = 0.8\n",
"output = 400. \t\t#kW\n",
"H = 150. \t\t\t#m\n",
"rho = 1000. \n",
"g = 9.81\n",
"f = 0.028\n",
"l = 1250. \t\t\t#m\n",
"\t\t\t\n",
"#calculations\n",
"gam = rho*g\n",
"inpu = output/e\n",
"Q = inpu*10**3 /(2./3 *gam*H)\n",
"hl = 1./3 *H\n",
"d5 = f*l*Q**2 /(2*g* math.pi/4 * math.pi/4 *hl)\n",
"d = d5**(1./5)\n",
"\t\t\t\n",
"#results\n",
"print \"Smallest diameter of pen stock = %d cm\"%(d*100)\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Smallest diameter of pen stock = 43 cm\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.22 Page No : 291"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\n",
"#Initialization of variables\n",
"f = 0.04\n",
"H = 30. \t\t\t#m\n",
"l = 200. \t\t\t#m\n",
"d = 0.075 \t\t\t#m\n",
"g = 9.81\n",
"rho = 1000.\n",
"gam = rho*g\n",
"\t\t\t\n",
"#calculations\n",
"h = 2/3. *H\n",
"vj = math.sqrt(2.*g*h)\n",
"hl = 1/3. *H\n",
"V = math.sqrt(hl*d*2*g/(f*l))\n",
"dj = d*(math.sqrt(V/vj))\n",
"Power = 2/3. *gam*math.pi/4. *d**2 *V*H\n",
"\t\t\t\n",
"#results\n",
"print \"Size of nozzle = %.1f cm\"%(dj*100)\n",
"print \" Max power = %.2f kW\"%(Power/1000)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Size of nozzle = 2.0 cm\n",
" Max power = 1.18 kW\n"
]
}
],
"prompt_number": 29
}
],
"metadata": {}
}
]
}