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| author | debashisdeb | 2014-06-20 15:42:42 +0530 |
|---|---|---|
| committer | debashisdeb | 2014-06-20 15:42:42 +0530 |
| commit | 83c1bfceb1b681b4bb7253b47491be2d8b2014a1 (patch) | |
| tree | f54eab21dd3d725d64a495fcd47c00d37abed004 /Fluid_Mechanics_/Chapter10.ipynb | |
| parent | a78126bbe4443e9526a64df9d8245c4af8843044 (diff) | |
| download | Python-Textbook-Companions-83c1bfceb1b681b4bb7253b47491be2d8b2014a1.tar.gz Python-Textbook-Companions-83c1bfceb1b681b4bb7253b47491be2d8b2014a1.tar.bz2 Python-Textbook-Companions-83c1bfceb1b681b4bb7253b47491be2d8b2014a1.zip | |
removing problem statements
Diffstat (limited to 'Fluid_Mechanics_/Chapter10.ipynb')
| -rw-r--r-- | Fluid_Mechanics_/Chapter10.ipynb | 24 |
1 files changed, 0 insertions, 24 deletions
diff --git a/Fluid_Mechanics_/Chapter10.ipynb b/Fluid_Mechanics_/Chapter10.ipynb index fc9fd648..679a7dd2 100644 --- a/Fluid_Mechanics_/Chapter10.ipynb +++ b/Fluid_Mechanics_/Chapter10.ipynb @@ -29,13 +29,11 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Top width, area of lfow, hydraulic radius\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", - "# Given\n", "\n", "b = 3 # base of the channel\n", "\n", @@ -43,7 +41,6 @@ "\n", "y = 2 # depth of the channel\n", "\n", - "# Solution\n", "\n", "T = b + 2*z*y\n", "\n", @@ -99,13 +96,11 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Discharge for the trapezoidal channel\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", - "# Given\n", "\n", "z = 1.0 # slide slope\n", "\n", @@ -117,7 +112,6 @@ "\n", "n = 0.012 # for concrete\n", "\n", - "# Solution\n", "\n", "A = (b+z*y)*y\n", "\n", @@ -125,7 +119,6 @@ "\n", "R = A/P\n", "\n", - "# from mannings eqquation\n", "\n", "Q = A*(1/n)*(R**(2/3)*S**(1/2))\n", "\n", @@ -156,19 +149,16 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Determine cross sectional area\n", "\n", "from __future__ import division\n", "\n", "from math import *\n", "\n", - "# Given\n", "\n", "z = 1\n", "\n", "Q = 10000/60 # discharge of water in ft**#/s\n", "\n", - "# Solution\n", "\n", "y = (Q/(1.828*2.25*sqrt(0.5)))**(2/5)\n", "\n", @@ -204,13 +194,11 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Calculate criticcal depth\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", - "# Given\n", "\n", "y = 2.5 # depth\n", "\n", @@ -218,7 +206,6 @@ "\n", "g = 9.81 # acceleration due to gravity in m/s**2\n", "\n", - "# Solution\n", "\n", "Yc = (20**2/g)**(1/3)\n", "\n", @@ -249,13 +236,11 @@ "cell_type": "code", "collapsed": false, "input": [ - "# determine normal depth, flow regine, critical depth\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", - "# Given\n", "\n", "Q = 15 # flow rate in m**3/s\n", "\n", @@ -267,9 +252,7 @@ "\n", "z = 0.5 # slope\n", "\n", - "# Solution\n", "\n", - "# We use a trial and error method here to find the value of y i.e. normal depth\n", "\n", "y = 2.22 # we take the value of y as 2.2 m\n", "\n", @@ -289,7 +272,6 @@ "\n", "print \"b )F = \",round(F,2),\" Since the Froude number is less than 1, the flow is subcritical\"\n", "\n", - "# we use trail and error to find the value of yc for critical depth\n", "\n", "yc = 1.08\n", "\n", @@ -325,13 +307,11 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Height, type and length of jump and loss of energy\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", - "# Given\n", "\n", "b = 60 # base width in ft\n", "\n", @@ -341,7 +321,6 @@ "\n", "g = 32.2\n", "\n", - "# Solution\n", "\n", "V1 = Q/(b*y1)\n", "\n", @@ -394,13 +373,11 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Determine flow rate\n", "\n", "from math import *\n", "\n", "from __future__ import division\n", "\n", - "# Given\n", "\n", "d = 6 # depth of the channel\n", "\n", @@ -412,7 +389,6 @@ "\n", "g = 32.2\n", "\n", - "# Solution\n", "\n", "y2 = y1 - h - 0.75\n", "\n", |