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Diffstat (limited to 'Fluid_Mechanics_/Chapter9.ipynb')
| -rw-r--r-- | Fluid_Mechanics_/Chapter9.ipynb | 50 |
1 files changed, 0 insertions, 50 deletions
diff --git a/Fluid_Mechanics_/Chapter9.ipynb b/Fluid_Mechanics_/Chapter9.ipynb index 6c911620..503ee70d 100644 --- a/Fluid_Mechanics_/Chapter9.ipynb +++ b/Fluid_Mechanics_/Chapter9.ipynb @@ -28,23 +28,19 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Determine Head loss\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "S = 1.26 # specific gravity\n", "\n", "mu = 0.826 # kinematic viscosity in Ns/m**2\n", "\n", - "# for water\n", "\n", "rho = 998 # density of water in kg/m**3\n", "\n", "mu1 = 1.005*10**-3 # viscosity in Ns/m**2\n", "\n", - "# for glycerine\n", "\n", "rho1 = S*rho # density of glycerine in kg/m**3\n", "\n", @@ -58,7 +54,6 @@ "\n", "l =100 # length of the pipe\n", "\n", - "# Solution\n", "\n", "V = Q/A\n", "\n", @@ -114,13 +109,10 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Discharge of water\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", - "# for water\n", "\n", "nu = 1.007*10**-6 # viscosity in m**2/s\n", "\n", @@ -136,7 +128,6 @@ "\n", "g = 9.81 # acceleration due to gravity in m/s**2\n", "\n", - "# Solution\n", "\n", "A = sqrt(2*g*D*hf/L)\n", "\n", @@ -183,11 +174,9 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Size of the case iron\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "Q =0.1 # discharge in m**3/s\n", "\n", @@ -201,13 +190,11 @@ "\n", "g = 9.81 # acceleration due to gravity in m/s**2\n", "\n", - "# Solution\n", "\n", "A = 8*L*Q**2/(hf*g*pi**2)\n", "\n", "B = 4*Q/(pi*nu)\n", "\n", - "# for D = 0.172 ; f=0.01\n", "D = 0.172\n", "\n", "r = e/D\n", @@ -216,7 +203,6 @@ "\n", "f = 0.022 # for Re and r\n", "\n", - "# for D1=0.199 ; f=0.021\n", "\n", "D1 = 0.199\n", "\n", @@ -253,11 +239,9 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Head loss \n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "L = 500 # length of the pipe in ft\n", "\n", @@ -267,7 +251,6 @@ "\n", "S = 0.004\n", "\n", - "# Solution\n", "\n", "Hf = S*L\n", "\n", @@ -298,11 +281,9 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Head loss of water\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "Q = 0.1 # water flow rate in m**3/s\n", "\n", @@ -314,7 +295,6 @@ "\n", "g = 9.81 # acceleration due to gravity in m/s**2\n", "\n", - "# Solution\n", "\n", "r = log(d/e,10)\n", "\n", @@ -351,11 +331,9 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Head loss by conveyance method\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "Q = 0.1 # water flow rate in m**3/s\n", "\n", @@ -369,7 +347,6 @@ "\n", "S = 5.43 \n", "\n", - "# Solution\n", "\n", "r = log(d/e,10)\n", "\n", @@ -404,24 +381,18 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Solve using the conveyence method\n", "\n", "from math import *\n", "\n", - "# given\n", "\n", "eps = 0.025*10**-2 # for cast iron epsilon = 0.0025 cm\n", "\n", - "# we get the value of K = 0.432 m**2/s\n", - "# we need to do trial and error to find the value of D\n", "\n", - "# we use the value of D = 0.2 m\n", "\n", "D = 0.2 # value in m\n", "\n", "g = 9.81\n", "\n", - "# Solution\n", "\n", "K = (pi/4)*sqrt(2*g)*(2*log10(D/(eps))+1.14)*D**(2.5)\n", "\n", @@ -452,11 +423,9 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Determine head loss\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "d = 0.1 # diameter of the pipe\n", "\n", @@ -468,7 +437,6 @@ "\n", "g = 9.81 # acceleration due to gravity in m/s**2\n", "\n", - "# for water\n", "\n", "nu = 1.007*10**-6 # viscosity in m**2/s\n", "\n", @@ -476,7 +444,6 @@ "\n", "r = e/(10*d)\n", "\n", - "# Solution\n", "\n", "V = Q/A\n", "\n", @@ -523,13 +490,10 @@ "cell_type": "code", "collapsed": false, "input": [ - "# discharge through the pipe\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", - "# for water\n", "\n", "nu = 1.007*10**-6 # viscosity in m**2/s\n", "\n", @@ -547,7 +511,6 @@ "\n", "f2 = 0.022 # foe e/d2\n", "\n", - "# Solution\n", "\n", "V3 = sqrt(2*g*100/((8.4*(f1)+268.85*(f2)+4.85)))\n", "\n", @@ -555,7 +518,6 @@ "\n", "V2 = (d3/d2)**2*V3\n", "\n", - "# reynolds number for pipe BC\n", "\n", "R1 = V1*d1/nu\n", "\n", @@ -590,11 +552,9 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Replace the flow system\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "D = 0.2 # diameter of pipe 1\n", "\n", @@ -608,7 +568,6 @@ "\n", "r = e/(100*D) \n", "\n", - "# Solution\n", "\n", "V = Q/(pi*(0.2)**2/4)\n", "\n", @@ -653,11 +612,9 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Discharge through each branch\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "e = 0.025 # in cm\n", "\n", @@ -671,7 +628,6 @@ "\n", "g = 9.81\n", "\n", - "# Pipe 1\n", "\n", "r1 = e/D # r1 for pipe 1\n", "\n", @@ -683,7 +639,6 @@ "\n", "hf1 = f*L1*V1**2/(2*g*D1)\n", "\n", - "# pipe 2\n", "\n", "hf2 = hf1\n", "\n", @@ -701,7 +656,6 @@ "\n", "Q2 = V2*(pi*D2**2/4)\n", "\n", - "#pipe 3\n", "\n", "hf3=hf1\n", "\n", @@ -733,7 +687,6 @@ "\n", "print \"Discharge through branch 3 =\",round(q3,3),\"m**3/s\"\n", "\n", - "# Actual head loss\n", "\n", "d = 0.5\n", "\n", @@ -777,11 +730,9 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Find minimum depth below the ridge\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "e = 0.00015 # from moody's chart\n", "\n", @@ -801,7 +752,6 @@ "\n", "L = 1000 # length in ft\n", "\n", - "# Solution\n", "\n", "f = 0.011 # assume\n", "\n", |