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diff --git a/Hydraulics/Chapter_5.ipynb b/Hydraulics/Chapter_5.ipynb new file mode 100644 index 00000000..7fdc90dc --- /dev/null +++ b/Hydraulics/Chapter_5.ipynb @@ -0,0 +1,1214 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : Flow Through Pipes" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.1 pageno : 131" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "find\n", + "a) a channel having bottom width\n", + "b) channel whose section is an arc of a circle\n", + "'''\n", + "import math\n", + "import numpy \n", + "#initialisation of variables\n", + "h = 4. \t\t\t#ft\n", + "h1 = 3. \t\t\t#ft\n", + "r = 3. \t\t\t#ft\n", + "h2 = 1.5 \t\t\t#ft\n", + "#CALCULATIONS\n", + "m = (h*h1+(h1**2/2))/(h+(h/2)*math.sqrt(h1**2+(h1/2)**2))\n", + "a = 2*numpy.degrees(numpy.arccos(h2/r))\n", + "P = 2*math.pi*r*a/360.\n", + "A = r**2*((2*math.pi/3.)-math.sin(math.radians(a)))/2.\n", + "H = A/(2*math.pi)\n", + "#RESULTS\n", + "print 'Hydraulic mean depth m = %.2f ft'%m\n", + "print 'hydraulic mean depth = %.2f ft '%(H)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hydraulic mean depth m = 1.54 ft\n", + "hydraulic mean depth = 0.88 ft \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.2 pageno : 133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the head lost\n", + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "d = 3. \t\t\t#ft\n", + "l = 5280. \t\t\t#ft\n", + "v = 3. \t\t\t#ft/sec\n", + "f = 0.005\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "C = 115.\n", + "#CALCULATIONS\n", + "hf = 4*f*l*v**2/(2*g*v)\n", + "m = d/4\n", + "hf1 = (v/C)**2*4*l/3\n", + "#RESULTS\n", + "print 'hf = %.2f ft '%(hf)\n", + "print 'hf = %.1f ft '%(hf1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "hf = 4.92 ft \n", + "hf = 4.8 ft \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.3 page no : 134" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the pressure difference between two points\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "d = 6. \t\t\t#in\n", + "Q = 2. \t\t\t#cfs\n", + "l = 1000. \t\t\t#ft\n", + "f = 0.0055\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "h = 70. \t\t\t#ft\n", + "#CALCULATIONS\n", + "v = Q/(math.pi*(d/12)**2/4)\n", + "hf = 4*f*l*w*(Q/(math.pi*(d/12)**2/4))**2/((d/12)*2*144*g)\n", + "ft_water = round(hf*144/w + 70)\n", + "P = ft_water*w/144\n", + "\n", + "#RESULTS\n", + "print 'pressure = %.1f lb/in**2 '%(hf)\n", + "print 'presure difference = %.0f lb/in**2 '%(P)\n", + "\n", + "# Answers are slightly different because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pressure = 30.7 lb/in**2 \n", + "presure difference = 61 lb/in**2 \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.4 pageno : 135" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the discharge\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "d = 6. \t\t\t#in\n", + "hf = 7.7 \t\t\t#ft\n", + "f = 0.005\n", + "l = 1000. \t\t\t#ft\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "#CALCULATIONS\n", + "C = math.sqrt(2*g/f)\n", + "Q = math.pi*C*(d/12)**2.5*(hf/1000)**0.5 /8\n", + "#RESULTS\n", + "print 'Discharge = %.2f cfs '%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Discharge = 0.69 cfs \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.5 page no : 136" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the size of the supply main \n", + "\n", + "import math \n", + "#initialisation of variables\n", + "Q = 400000.\n", + "d = 4. \t\t\t#miles\n", + "h = 50. \t\t\t#ft\n", + "q = 40. \t\t\t#gallons of water\n", + "t = 8. \t\t\t#hr\n", + "f = 0.0075\n", + "w = 6.24 \t\t\t#lb/ft**3\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "#CALCULATIONS\n", + "Q1 = round(Q*q*0.5/(t*60*60*w),1)\n", + "d = (4*f*(d*5280)*Q1**2*16/(math.pi**2*h*2*g))**0.2*12\n", + "\n", + "#RESULTS\n", + "print 'size of the supply = %.3f in '%(d)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "size of the supply = 43.579 in \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.6 pageno : 138" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the critical velocities \n", + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "Q = 0.7 \t\t\t#cfs\n", + "d = 6. \t\t\t#in\n", + "v1 = 1.084*10**-5 \t\t\t#ft**2/sec\n", + "v2 = 0.394*10**-5 \t\t\t#ft**2/sec\n", + "R = 2320.\n", + "#CALCULATIONS\n", + "Re = (4*Q)/(math.pi*.5*v2)\n", + "v3 = R*v1/(d/12.)\n", + "v4 =R*v2/(d/12.) \n", + "v = Q*4/(math.pi*(d/12.)**2)\n", + "\n", + "#RESULTS\n", + "print 'Re at 80 C %.0f'%Re\n", + "print 'crititcal velocity = %.4f ft/sec '%(v4)\n", + "print 'actual velocity = %.2f ft/sec '%(v)\n", + "\n", + "# Note : answer is correct. Please calculate it manually." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Re at 80 C 452420\n", + "crititcal velocity = 0.0183 ft/sec \n", + "actual velocity = 3.57 ft/sec \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.7 pageno : 140" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find pressure \n", + "import math \n", + "#initialisation of variables\n", + "p = 0.91 \t\t\t#units\n", + "u = 0.21 \t\t\t#poise\n", + "q = 200. \t\t\t#gallons\n", + "h = 40. \t\t\t#ft\n", + "l = 200. \t\t\t#ft\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "d = 3./4 \t\t\t#in\n", + "g =32.2 \t\t\t#ft/s**2\n", + "#CALCULATIONS\n", + "v = u/(p*(30.5)**2)\n", + "Q = q*10/(w*3600*p)\n", + "V = Q/(math.pi*(d/12)**2/4)\n", + "Re = V*(d/12)/v\n", + "F = 64/Re\n", + "Hf = F*l*V**2/(2*g*(d/12))\n", + "Ht = Hf+h\n", + "P = w*p*Ht/144\n", + "#RESULTS\n", + "print 'Pressure head = %.1f lb/sq in '%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pressure head = 31.6 lb/sq in \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.8 page no : 142" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the value of n\n", + "from math import log10\n", + "#initialisation of variables\n", + "h = 1.5 # H\n", + "v = 2. # V\n", + "\n", + "#CALCULATIONS\n", + "logh = round(log10(h),4)\n", + "logk = -0.415 # from fig.\n", + "logv = round(log10(v),4)\n", + "n = (logh-logk)/logv\n", + "\n", + "#RESULTS\n", + "print 'n = %.3f '%(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "n = 1.964 \n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.10 page no : 146" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the pressure \n", + "\n", + "import math \n", + "#initialisation of variables\n", + "pb = 20. \t\t\t#lb/in**2\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "Q = 1.96 \t\t\t#cfs\n", + "d1 = 0.5 \t\t\t#ft\n", + "d2 = 1. \t\t\t#ft\n", + "f = 0.005\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "l1 = 300. \t\t\t#ft\n", + "H = 14.015 \t\t\t#ft of water\n", + "#CALCULATIONS\n", + "v1 = Q/(math.pi*d1**2/4.)\n", + "v2 = Q/(math.pi*d2**2./4.)\n", + "hf1 = 4*f*l1*v1**2./(2.*g*d1)\n", + "hf2 = 4*f*l1*v2**2/(2*g*d2)\n", + "h = (v1-v2)**2/(2*g)\n", + "h1 = v1**2/(2*g)\n", + "h2 = v2**2/(2*g)\n", + "P = H*w/144\n", + "#RESULTS\n", + "print 'Loss of head at C = %.2f ft '%(h1)\n", + "print 'Loss of head at C = %.3f ft '%(h2)\n", + "print 'Pressure differnece at discharge end = %.2f lb/in**2 '%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of head at C = 1.55 ft \n", + "Loss of head at C = 0.097 ft \n", + "Pressure differnece at discharge end = 6.07 lb/in**2 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.11 page no : 148" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find rate of flow \n", + "import math \n", + "#initialisation of variables\n", + "d = 8. \t\t\t#in\n", + "l = 6000. \t\t\t#ft\n", + "H = 100. \t\t\t#ft\n", + "H1 = 1000. \t\t\t#ft\n", + "f = 0.008\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "h1 = 24. \t\t\t#ft\n", + "h2 = 34. \t\t\t#ft \n", + "h3 = 25. \t\t\t#ft\n", + "w = 6.24 \t\t\t#lb/ft**3\n", + "#CALCULATIONS\n", + "v = math.sqrt(H*d*2*g/(4*f*l*12))\n", + "h = -h1+(v**2/(2*g))+h3+(4*f*H1*v**2/(2*g*(d/12)))\n", + "Q = round(math.pi*(d/12)**2*v*3600*w/4,-2)\n", + "#RESULTS\n", + "print 'minimum depth = %.f ft '%(h)\n", + "print 'Discharge = %.f gpm'%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "minimum depth = 18 ft \n", + "Discharge = 37100 gpm\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.12 page no : 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find discharge when the syphone is running full\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "h = 25. \t\t\t#ft\n", + "l = 2000. \t\t\t#ft\n", + "d = 12. \t\t\t#in\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "f = 0.005\n", + "dz = 16. \t\t\t#ft\n", + "zb = 25. \t\t\t#ft\n", + "zc = -16. \t\t\t#ft\n", + "#CALCULATIONS\n", + "v = math.sqrt(2*g*h/(1.5+(4*f*l/(d/12))))\n", + "Q = math.pi*(d/12)**2*v/4\n", + "l1 = (34-dz)*l/(zb-zc-dz)\n", + "#RESULTS\n", + "print 'Discharge = %.1f cfs '%(Q)\n", + "print 'length of the inlet = %.f ft of water '%(l1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Discharge = 4.9 cfs \n", + "length of the inlet = 1440 ft of water \n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.13 page no : 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the necessary height of water surface\n", + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "d1 = 2 \t\t\t#in\n", + "l1 = 25 \t\t\t#ft\n", + "d2 = 4 \t\t\t#in\n", + "l2 = 140 \t\t\t#ft\n", + "v = 4 \t\t\t#ft/sec\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "f = 0.0065\n", + "#CALCULATIONS\n", + "v1 = v*(d2/d1)**2\n", + "H = (0.5*v1**2/(2*g))+(4*f*l1*12*v1**2/(d1*2*g))+((v1-v)**2/(2*g))+(4*f*l2*12*v**2/(d2*2*g))+(v**2/(2*g))\n", + "#RESULTS\n", + "print 'necessaey height of water = %.3f ft '%(H)\n", + "\n", + "# Note : answer is slightly different because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "necessaey height of water = 22.688 ft \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.14 page no : 155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#find the diameter of uniform pipe \n", + "\n", + "import math \n", + "#initialisation of variables\n", + "l1 = 3000. \t\t\t#ft\n", + "d1 = 18. \t\t\t#in\n", + "l2 = 1500. \t\t\t#ft\n", + "d2 = 15. \t\t\t#ft\n", + "l3 = 1000. \t\t\t#ft\n", + "d3 = 12. \t\t\t#in\n", + "#CALCULATIONS\n", + "d = ((l1+l2+l3)/((l1/d1**5)+(l2/d2**5)+(l3/d3**5)))**(1./5)\n", + "#RESULTS\n", + "print 'Diameter = %.2f in '%(d)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter = 14.86 in \n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.15 page no : 156" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Find the diameter of the parallel mains\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "D = 9. \t\t\t#in\n", + "#CALCULATIONS\n", + "d = D/(2**0.4)\n", + "#RESULTS\n", + "print 'diameter of paralle mains = %.2f in '%(d)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "diameter of paralle mains = 6.82 in \n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.16 page no : 157" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the increase in discharge\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "d = 2. \t\t\t#ft\n", + "l = 5280. \t\t\t#ft\n", + "f = 0.01\n", + "H = 100. \t\t\t#ft\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "#CALCLATIONS\n", + "v = math.sqrt(H*2*d*g/(4*f*l))\n", + "Q = math.pi*d**2*v/4.\n", + "r = d\n", + "v2 = math.sqrt(H/((r**2+1)*(4*f*l/(2*2*2*g))))\n", + "Q1 = 2*math.pi*d**2*v2/4.\n", + "dQ = Q1-Q\n", + "p = dQ*100./Q\n", + "#RESULTS\n", + "print 'percentage increase in discharge = %.1f %% '%(p)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentage increase in discharge = 26.5 % \n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.17 Page no : 160" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Find the velocity of flow in each pipe\n", + "'''\n", + "\n", + "import math\n", + "\n", + "# Variables\n", + "d1 = 2 # ft\n", + "d2 = 1.5 # ft\n", + "d3 = 1 # ft\n", + "l1 = 2000 # ft\n", + "l2 = 3000 # ft\n", + "l3 = 1500 # ft\n", + "za = 100 # ft\n", + "zb = 70 # ft\n", + "zc = 50 # ft\n", + "zd = 80 # ft\n", + "f = 0.007 # ft\n", + "v3 = 7.93\n", + "\n", + "# Calculations\n", + "\n", + "v1 = round(math.sqrt(111-1.412*(v3**2)),1)\n", + "v2 = round(math.sqrt(-23.4+.745*(v3**2)),2)\n", + "pd_w = round((za - zd) - ( 29 * v1**2/64.4),2)\n", + "Q1 = (math.pi/4)*d1**2*v1\n", + "Q2 = (math.pi/4)*d2**2*v2\n", + "Q3 = Q1 - Q2\n", + "\n", + "# Results\n", + "print \"V1 = %.1f ft/sec\"%v1\n", + "print \"V2 = %.2f ft/sec\"%v2\n", + "print \"V3 = %.2f ft/sec\"%v3\n", + "print \"Pressure at the junction point : %.2f ft of water\"%pd_w\n", + "print \"Discharge in section : %.2f cfs\"%Q3\n", + "\n", + "# Note : Answers may vary because of rounding error. Please check it manually." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V1 = 4.7 ft/sec\n", + "V2 = 4.84 ft/sec\n", + "V3 = 7.93 ft/sec\n", + "Pressure at the junction point : 10.05 ft of water\n", + "Discharge in section : 6.21 cfs\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.18 page no : 164" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the time taken to lower the level of water\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "A = 10000. \t\t\t#ft**2\n", + "H1 = 50. \t\t\t#ft\n", + "H2 = 40. \t\t\t#ft\n", + "l = 1500. \t\t\t#ft\n", + "d = 6. \t\t\t#in\n", + "f = 0.0075\n", + "g = 32.2 \t\t\t#f/sec**2\n", + "#CALCULATIONS\n", + "t = 2.*A*math.sqrt((1.5+(4*f*l/(d/12)))/(2*g))*(math.sqrt(H1)-math.sqrt(H2))/(math.pi*(d/12)**2/4)\n", + "#RESULTS\n", + "print 'Time taken to lower the level of water = %.f hours '%(t/3600)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken to lower the level of water = 25 hours \n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.19 page no : 164" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the time required to empty the overhead\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "l = 24. \t\t\t#ft\n", + "b = 12. \t\t\t#ft\n", + "f = 0.006\n", + "d = 4. \t\t\t#in\n", + "l1 = 25. \t\t\t#ft\n", + "H1 = 6. \t\t\t#ft\n", + "H = 20. \t\t\t#ft\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "Cd = 0.6\n", + "#CALCULATIONS\n", + "a = math.pi*(d/12)**2/4\n", + "A = l*b\n", + "H2 = H1+H\n", + "t = round(2*A*math.sqrt((1.5+(4*f*l1/(d/12)))/(2*g))*(math.sqrt(H2)-math.sqrt(H))/a,-1)\n", + "t1 = 2*A*math.sqrt((1.5+(4*f*l1/(d/12)))/(2*g))*math.sqrt(H1)/a\n", + "t2 = 2*A*math.sqrt(H1)/(Cd*a*math.sqrt(2*g))\n", + "#RESULTS\n", + "print 'Time taken to lower the pipe = %.f sec '%(t)\n", + "print 'Time taken to lower the pipe = %.f sec '%(t1)\n", + "print 'Time taken to lower the pipe = %.f sec '%(t2)\n", + "\n", + "# note : answers may vary becasue of ronding error. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken to lower the pipe = 940 sec \n", + "Time taken to lower the pipe = 3660 sec \n", + "Time taken to lower the pipe = 3358 sec \n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.20 page no : 165\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the time taken \n", + "\n", + "import math \n", + "#initialisation of variables\n", + "d = 2 \t\t\t#ft\n", + "l = 1000 \t\t\t#ft\n", + "f = 0.0075\n", + "H1 = 20 \t\t\t#ft\n", + "A1 = 100000 \t\t\t#ft**2\n", + "A2 = 50000 \t\t\t#ft**2\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "#CALCULATIONS\n", + "h = 2*A1/A2\n", + "H2 = H1-h\n", + "t = 2*A1*A2*math.sqrt(1.5+(4*f*l/2))*0.47/((A1+A2)*(math.pi*d**2/4)*math.sqrt(2*g))/60\n", + "#RESULTS\n", + "print 'Time taken to lower the level of water = %.f min '%(t)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken to lower the level of water = 84 min \n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.21 page no : 167" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find diameter of the pipe and efficiency of transmission.\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "H = 1000. \t\t\t#lb/in**2\n", + "Hf = 100. \t\t\t#lb/in**2\n", + "l = 10. \t\t\t#miles\n", + "HP = 100.\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "w = 64.4 \t\t\t#lb/ft**3\n", + "f = 0.006\n", + "#CALCULATIONS\n", + "n = (H-Hf)*100/H\n", + "v = Hf*550/((math.pi/4)*n*10*144)\n", + "r = Hf*144*2*g/(w*4*f*l*5280)\n", + "d = (v**2/r)**(1./5)\n", + "#RESULTS\n", + "print 'Diameter = %.4f ft '%(d)\n", + "\n", + "# answer may vary because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter = 0.4808 ft \n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.22 page no : 170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the horse-powers of the jet.\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "h1 = 1640. \t\t\t#ft\n", + "h2 = 40. \t\t\t#ft\n", + "d = 8. \t\t\t#in\n", + "l = 2. \t\t\t#miles\n", + "D = 3. \t\t\t#ft\n", + "f = 0.006\n", + "Cv = 0.98\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "#CALCULATIONS\n", + "r = (d/12)/D\n", + "vact = Cv*math.sqrt(2*g*(h1-h2)/(1+(4*f*l*5280*r**4/D)))\n", + "HP = round(round(w*vact**3*(math.pi*(d/12)**2/4)/(550*2*g),-2),-3)\n", + "#RESULTS\n", + "print 'Horse Power of Jet = %.f HP '%(HP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Horse Power of Jet = 15000 HP \n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.23 page no : 171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find volume of flow \n", + "\n", + "import math \n", + "#initialisation of variables\n", + "p = 60. \t\t\t#lb/in**2\n", + "l = 300. \t\t\t#ft\n", + "D = 2.5 \t\t\t#in\n", + "d = 7./8 \t\t\t#in\n", + "f = 0.018\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "#CALCULATIONS\n", + "r = (D/d)**4\n", + "V = math.sqrt(2*g*144*p/(w*(r+0.5+(4*f*l/(D/12)))))\n", + "Q = V*(math.pi*(D/12)**2)/4\n", + "#RESULTS\n", + "print 'Volume of flow = %.3f cu ft/sec '%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volume of flow = 0.246 cu ft/sec \n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.24 page no : 171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find maximum power transmitted by the jet\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "D = 3. \t\t\t#in\n", + "l = 800. \t\t\t#ft\n", + "H = 120. \t\t\t#ft\n", + "f = 0.01\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "#CALCULATIONS\n", + "d = ((D/12)**5/(8*f*l))**0.25\n", + "hf = H/3\n", + "dh = H-hf\n", + "v = math.sqrt(hf*(D/12)*2*g/(4*f*l))\n", + "HPmax = w*math.pi*((D/48)**2/4)*v*dh/550\n", + "#RESULTS\n", + "print 'HPmax = %.3f HP '%(HPmax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HPmax = 0.125 HP \n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.25 page no : 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find height\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "l = 2. \t\t\t#miles\n", + "Q = 2.*10**6 \t\t\t#gal/day\n", + "d = 12. \t\t\t#in\n", + "t = 16. \t\t\t#sec\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "#CALCULATIO\n", + "Q1 =Q*10/(w*24*60*60)\n", + "hi = l*5280*Q1/((math.pi*(d/12)**2./4)*(g*t))\n", + "#RESULTS\n", + "print 'height = %.1f ft '%(hi)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "height = 96.8 ft \n" + ] + } + ], + "prompt_number": 61 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.26 pageno : 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find rise of pressure in the pipe\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "d = 6. \t\t\t#in\n", + "Q = 0.7854 \t\t\t#cfs\n", + "E = 30*10**6 \t\t\t#lb/in**2\n", + "t = 0.25 \t\t\t#in\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "K = 300000. \t\t\t#lb/in**2\n", + "#CALCULATIONS\n", + "v = Q/(math.pi*(d/12)**2/4)\n", + "p = v/(math.sqrt(144*(g/w)*((1/K)+(d/(t*E)))))\n", + "#RESULTS\n", + "print 'rise of presure in the pipe = %.f lb/in**2 '%(p)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rise of presure in the pipe = 228 lb/in**2 \n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 5.27 page no : 183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find economical diameter of pipe line\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "f = 0.005\n", + "Q = 100. \t\t\t#cuses\n", + "m = 40. \t\t\t#Rs\n", + "n = 0.75\n", + "n1 = 0.065\n", + "K = 15. \t\t\t#Rs\n", + "#CALCULATIONS\n", + "d = ((5*w/(1.5*550*10))*n*f*Q**3*m/(K*n1))**(1/6.5)\n", + "#RESULTS\n", + "print 'economical diameter of pipe line = %.3f ft '%(d)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "economical diameter of pipe line = 3.795 ft \n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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