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+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f1e32f78d016d65eaf62d619ce157e62f2d402e55bfc70a9a046c3e8d919d007"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 : Pumping Machinery"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.1 Page No : 223"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialisation of variables\n",
+ "\n",
+ "import math \n",
+ "h= 75. \t#ft\n",
+ "e= 0.75\n",
+ "k= 0.01\n",
+ "Q= 3000. \t#gal/min\n",
+ "k1= 1.2\n",
+ "N= 1500.\n",
+ "g= 32.2 \t#ft/sec**2\n",
+ "D= 0.836 \t#ft\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "W= h/e\n",
+ "v1= math.sqrt((W-h)/k)\n",
+ "Q1= Q/374.06\n",
+ "f1= Q1/(k1*D**2)\n",
+ "u1= math.pi*D*N/60\n",
+ "w1= W*g/u1\n",
+ "B= math.degrees(math.atan((f1/(u1-w1))))\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'Diameter of impeller = %.3f ft '%(D)\n",
+ "print ' Blade angle at outlet edge of impeller = %.f degrees '%(B)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of impeller = 0.836 ft \n",
+ " Blade angle at outlet edge of impeller = 30 degrees \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.3 Page No : 226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "V= 150. \t#ft**3/sec\n",
+ "A1= 750. \t#r.p.m\n",
+ "di= 21. \t#in\n",
+ "do= 30. \t#in\n",
+ "v= 50. \t#ft/sec\n",
+ "A= 70. \t#degrees\n",
+ "w= 4.\t#in\n",
+ "p= 30. \t#per cent\n",
+ "p1= 25. \t#per cent\n",
+ "sv= 12.8 \t#ft**3/lb\n",
+ "g= 32.2 \t#ft/sec**2\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "u= A1*2*math.pi*di/(24*60)\n",
+ "u1= A1*2*math.pi*do/(24*60)\n",
+ "f1= V/(math.pi*(do/12)*(1./3))\n",
+ "w1= u1-f1*1/math.tan(math.radians((A)))\n",
+ "v1= math.sqrt(f1**2+w1**2)\n",
+ "P= (u1**2+v**2-(f1**2/(math.sin(math.radians(A)))**2))/(2*g)\n",
+ "h= 30*v1**2/(100*2*g)\n",
+ "Nh= v1**2/(20*2*g)\n",
+ "Prt= P+Nh\n",
+ "W= u1*w1/g\n",
+ "e= Prt*100/W\n",
+ "Power= Prt*V/(sv*550)\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'Total pressure rise = %.1f ft of air'%(Prt)\n",
+ "print ' manometric efficiency = %.1f percent'%(e)\n",
+ "print ' Power = %.2f hp '%(Power)\n",
+ "\n",
+ "#The answer is a bit different due to rounding off error in textbook\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total pressure rise = 137.9 ft of air\n",
+ " manometric efficiency = 58.5 percent\n",
+ " Power = 2.94 hp \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.4 Page No : 228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "g= 32.2 \t#ft/sec**2\n",
+ "u1= 90. \t#ft/sec\n",
+ "w1= 70. \t#ft\n",
+ "e= 0.8\n",
+ "h1= 10. \t#ft\n",
+ "h2= 16. \t#ft\n",
+ "h3= 5. \t#ft\n",
+ "k= 2./5\n",
+ "f1= 20. \t#ft/sec\n",
+ "f= 18. \t#ft/sec\n",
+ "a= 45. \t #degrees\n",
+ "x1= 164.4 \t#ft\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Hm= u1*w1/g\n",
+ "Hm1= e*Hm\n",
+ "lh= Hm-Hm1-h1-h2-h3\n",
+ "vg= k*math.sqrt(f1**2+w1**2)\n",
+ "pr= ((f**2+u1**2-f1**2/(math.sin(math.radians(a)))**2)/(2*g))-h2\n",
+ "pr1= x1-pr\n",
+ "ge= pr1*g*2*100/(vg/k)**2\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'manometer Head = %.1f ft '%(Hm1)\n",
+ "print ' outlet velocity from guides = %.1f ft/sec '%(vg)\n",
+ "print ' Pressure rise through impeller only = %.1f ft '%(pr)\n",
+ "print ' Guide balde efficiency = %.f per cent '%(ge)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "manometer Head = 156.5 ft \n",
+ " outlet velocity from guides = 29.1 ft/sec \n",
+ " Pressure rise through impeller only = 102.4 ft \n",
+ " Guide balde efficiency = 75 per cent \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.6 Page No : 231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "D1= 7.5 \t#in\n",
+ "Q1= 850. \t#gal/min\n",
+ "p1= 62.4 \t#lb/ft**3\n",
+ "N1= 1800.\n",
+ "D2= 15. \t#in\n",
+ "Q2= 12000. \t#gal/min\n",
+ "p2= 64. \t#lb/ft**3\n",
+ "N1= 1800. \t#r.p.m \n",
+ "H1= 14. \t#lb/ft**2\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "N2= Q2*N1*(D1)**3/(Q1*D2**3)\n",
+ "P1= p1*H1/144\n",
+ "P2= P1*N2**2*D2**2*p2/(N1**2*p1*D1**2)\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'N2 = %.f r.p.m '%(N2+4)\n",
+ "print ' P2 = %.f lb/in**2 '%(P2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "N2 = 3180 r.p.m \n",
+ " P2 = 78 lb/in**2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.8 Page No : 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#initialisation of variables\n",
+ "r= 5.\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "sr= r**2\n",
+ "sr1= r**2/r\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'Corresponding ratio = %.f '%(sr)\n",
+ "print ' Corresponding ratio = %.f '%(sr1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Corresponding ratio = 25 \n",
+ " Corresponding ratio = 5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.9 Page No : 236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "e= 0.88\n",
+ "w= 1.25 \t#in\n",
+ "d= 10. \t#in\n",
+ "q= 630. \t#gal/min\n",
+ "a= 40. \t#degrees\n",
+ "g= 32.2 \t#ft/sec**2\n",
+ "e1= 0.83\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Q= q/(6.24*60)\n",
+ "f1= Q/(e*math.pi*(d/12)*(w/12))\n",
+ "u1= 1000*(w*4/12)*2*math.pi/60\n",
+ "w1= u1-f1*1/math.tan(math.radians(a))\n",
+ "W= u1*w1/g\n",
+ "lr= (f1**2+u1**2-f1**2/(math.sin(math.radians(a)))**2)/(2*g)\n",
+ "mh= e1*W\n",
+ "p= mh-lr\n",
+ "v1= math.sqrt(f1**2+w1**2)\n",
+ "ke= v1**2/(2*g)\n",
+ "pke= p*100/ke\n",
+ "me= 100*lr/W\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'Velocity of flow = %.f ft/sec'%(f1)\n",
+ "print ' Work done = %.1f ft-lb/lb'%(W)\n",
+ "print ' manometric efficiency = %.1f ft'%(mh)\n",
+ "print ' Pressure recovered = %.1f ft head'%(p)\n",
+ "print ' Kinetic energy discharge = %.f ft-lb/lb'%(ke)\n",
+ "print ' Percentage of kinetic energy recovered = %.1f per cent'%(pke)\n",
+ "print ' manometric efficiency = %d percent'%(me)\n",
+ "\n",
+ "#The answer is a bit different due to rounding off error in textbook\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Velocity of flow = 7 ft/sec\n",
+ " Work done = 47.8 ft-lb/lb\n",
+ " manometric efficiency = 39.7 ft\n",
+ " Pressure recovered = 11.2 ft head\n",
+ " Kinetic energy discharge = 20 ft-lb/lb\n",
+ " Percentage of kinetic energy recovered = 55.7 per cent\n",
+ " manometric efficiency = 59 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.10 Page No : 239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "W1= 7640. \t#gal/min\n",
+ "W2= 11400. \t#gal/min\n",
+ "Hm= 63. \t#ft\n",
+ "Hm1= 80. \t#ft\n",
+ "ep1= 72. \t#per cent\n",
+ "ep2= 76. \t#per cent\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "whp1= W1*Hm/(60*550)\n",
+ "whp2= W2*Hm1/(60*550)\n",
+ "bhp1= whp1*100/ep1\n",
+ "bhp2= whp2*100/ep2\n",
+ "w1= W2/10\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'For both pumps discharge = %.f gal/min against an 80-ft head'%(W2)\n",
+ "print ' delivery from one pump = %.1f h.p '%(bhp1)\n",
+ "print ' delivery from two pumps = %.1f h.p '%(bhp2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For both pumps discharge = 11400 gal/min against an 80-ft head\n",
+ " delivery from one pump = 20.3 h.p \n",
+ " delivery from two pumps = 36.4 h.p \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.11 Page No : 241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "h= 94. \t#ft\n",
+ "w= 62.4 \t#lb/ft**3\n",
+ "e= 0.58\n",
+ "p= 73.5 \t#per cent\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "WHP= h*e*w/550\n",
+ "BHP= WHP/(p/100)\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'W.H.P= %.2f h.p'%(WHP)\n",
+ "print ' Brake horse power= %.1f'%(BHP)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "W.H.P= 6.19 h.p\n",
+ " Brake horse power= 8.4\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.12 Page No : 243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "sl= 12. \t#ft\n",
+ "l= 20. \t#ft\n",
+ "d= 4. \t#in\n",
+ "dp= 6. \t#in\n",
+ "lst= 18. \t#in\n",
+ "k= 0.025\n",
+ "H= 32. \t#ft\n",
+ "g= 32.2 \t#ft/sec**2\n",
+ "pf= 6. \t#ft\n",
+ "a= 33.83 \n",
+ "a1= 9.53\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "A= math.sqrt((H-sl-d)*g/a)*a1\n",
+ "Q= 2*math.pi*(dp/12)**2*lst/(12*4*60)\n",
+ "v= Q/(math.pi*(d/12)**2/4)\n",
+ "kh= v**2/(2*g)\n",
+ "fh= k*l*v**2*12/(2*g*d)\n",
+ "N= math.sqrt((H-sl-pf)/(kh+fh))\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'premissible speed = %.1f r.p.m'%(A)\n",
+ "print ' maximum premissible speed = %.1f r.p.m'%(N)\n",
+ "\n",
+ "#The answer is a bit different due to rounding off error in textbook\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "premissible speed = 37.2 r.p.m\n",
+ " maximum premissible speed = 168.8 r.p.m\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.13 Page No : 245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "b= 6. \t#in\n",
+ "s= 12. \t #in\n",
+ "d= 4. \t #in\n",
+ "a1= 30. \t#degrees\n",
+ "a2= 90. \t#degrees\n",
+ "a3= 120. \t#degrees\n",
+ "N= 120. \t#r.p.m\n",
+ "n= 4.\n",
+ "#calculations\n",
+ "A= 2*math.pi*N/60\n",
+ "V= math.pi*(b/12)**2*n/4\n",
+ "v= (b/12)**2*A*(b/12)/(d/12)**2\n",
+ "Q1= v*math.pi*(d/12)**2*math.sin(math.radians(a1))/4\n",
+ "Q2= v*math.pi*(d/12)**2*math.sin(math.radians(a2))/4\n",
+ "Q3= v*math.pi*(d/12)**2*math.sin(math.radians(a3))/4\n",
+ "Q4= V-Q1\n",
+ "Q5= Q2-V\n",
+ "Q6= Q3-V\n",
+ "a4= math.degrees(math.asin(V/(v*math.pi*(d/12)**2)))+a1\n",
+ "A= 180-a4\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'rate of flow at a1 = %.3f cuses'%(Q4)\n",
+ "print ' rate of flow at a2 = %.3f cuses'%(Q5)\n",
+ "print ' rate of flow at a3 = %.3f cuses'%(Q6)\n",
+ "print ' crak angle = %.1f degrees'%(a4)\n",
+ "print ' crak angle = %.1f degrees'%(A)\n",
+ "\n",
+ "#The answer is a bit different due to rounding off error in textbook\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "rate of flow at a1 = 0.169 cuses\n",
+ " rate of flow at a2 = 0.448 cuses\n",
+ " rate of flow at a3 = 0.283 cuses\n",
+ " crak angle = 39.2 degrees\n",
+ " crak angle = 140.8 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.14 Page No : 247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "n= 2. \t#strokes/sec\n",
+ "dp= 6. \t#in\n",
+ "ds= 18. \t#in\n",
+ "ds1=4. \t#in\n",
+ "l= 20. \t#ft\n",
+ "l1= 20. \t#ft\n",
+ "f= 0.008\n",
+ "la= 5. \t#ft\n",
+ "A= 60. \t#r.p.m\n",
+ "f= 0.008\n",
+ "w= 62.4 \t#lb/ft**3\n",
+ "g=32.2\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "V= math.pi*(ds/12)*n*(dp/12)**2/4\n",
+ "vmp= 2*math.pi*A*(ds/24)/60\n",
+ "vmp1= vmp*(dp**2/ds1**2)\n",
+ "hfmax= 4*f*(l-la)*vmp1**2/(2*g*ds1/12)\n",
+ "H1= round(2*hfmax/3,1)\n",
+ "H2= H1*13\n",
+ "Wls= (H1+H2)*w*math.pi/16*1.5*2\n",
+ "mv= V/(math.pi*(ds1/12)**2/4)\n",
+ "lh= round(4*f*(l-la)*mv**2/(2*g*(ds1/12)),2)\n",
+ "lhf= 12*lh\n",
+ "Wls1= (lh+13.21)*w*math.pi*1.5/16 *2 \n",
+ "WS= Wls-Wls1\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'Work lost per second= %.f ft lb/sec'%(Wls)\n",
+ "print ' Work saved per second = %.f ft-lb/sec'%(WS)\n",
+ "\n",
+ "#The answer is a bit different due to rounding off error in textbook\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work lost per second= 875 ft lb/sec\n",
+ " Work saved per second = 352 ft-lb/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.15 Page No : 248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#initialisation of variables\n",
+ "d= 7.5 \t#in\n",
+ "s= 15. \t#in\n",
+ "l= 36. \t#ft\n",
+ "h1= 34. \t#ft\n",
+ "h2= 12. \t#ft\n",
+ "L= 10. \t #ft\n",
+ "g= 32.2 \t#ft/sec**2\n",
+ "f= 0.008\n",
+ "l1= 20. \t#ft\n",
+ "d1= 4. \t#in\n",
+ "h3= 110. \t#ft\n",
+ "w= 62.4 \t#lb/ft**3\n",
+ "l2= 180. \t#ft\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Q= (math.pi/4)*(d)**2*(s/12)*2*(l/60)/144\n",
+ "v= Q/((math.pi/4)*(d1/12)**2)\n",
+ "a= (d/4)**2*(d/12)*(l*2*math.pi/60)**2\n",
+ "H= h1-h2-(L*a/g)-(v**2*0.5/g)-(4*f*l1*v**2/(2*g*(d1/12)))\n",
+ "H1= h1+h3+(L*a/g)+(v**2*0.5/g)+(4*f*l2*v**2/(2*g*(d1/12)))\n",
+ "dh= (H1-H)*w/144\n",
+ "NP= dh*(math.pi/4)*d**2\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'Head at piston = %.2f ft of water absolute'%(H)\n",
+ "print ' Head at piston = %.2f ft of water absolute'%(H1)\n",
+ "print ' Difference on head of piston = %.f lb/in**2'%(dh)\n",
+ "print ' Net load on piston = %.f lb'%(NP)\n",
+ "\n",
+ "#The answer is a bit different due to rounding off error in textbook\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Head at piston = 11.04 ft of water absolute\n",
+ " Head at piston = 161.59 ft of water absolute\n",
+ " Difference on head of piston = 65 lb/in**2\n",
+ " Net load on piston = 2882 lb\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.16 Page No : 250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math \n",
+ "from numpy import *\n",
+ "from numpy.linalg import *\n",
+ "\n",
+ "#initialisation of variables\n",
+ "f= 0.009\n",
+ "dc= 3.5 \t#in\n",
+ "ds= 6. \t#in\n",
+ "r= 0.25\n",
+ "sl= 8. \t#ft\n",
+ "d= 2.5 \t#in\n",
+ "l= 14. \t#ft\n",
+ "el= 8. \t#ft\n",
+ "ed= 22.5 \t#in\n",
+ "ph= 4. \t#ft\n",
+ "g= 32.2 \t#ft/sec**2\n",
+ "f= 0.009\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "BC= el+l\n",
+ "v= math.sqrt(BC*g/(l*(d/2)*(r)*(dc/d)**2))*9.55\n",
+ "vec=roots([2,1/r,-1])\n",
+ "H1= 77\n",
+ "MV= math.sqrt(BC*g/(l*(d/2)*(r)*(dc/d)**2))*r*(math.sin(math.radians(H1))+(math.sin(math.radians(2*H1))/8))\n",
+ "mvp= MV*dc**2/d**2\n",
+ "hf= 4*f*(sl+l)*mvp**2/(2*g*(d/12))\n",
+ "\n",
+ "#RESULTS\n",
+ "print 'pump speed = %.1f r.p.m'%(v)\n",
+ "print ' Friction head = %.3f ft'%(hf)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "pump speed = 86.8 r.p.m\n",
+ " Friction head = 1.240 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/MaulikRathod/ch11_1.ipynb b/sample_notebooks/MaulikRathod/ch11_1.ipynb
new file mode 100755
index 00000000..f4d8bc9d
--- /dev/null
+++ b/sample_notebooks/MaulikRathod/ch11_1.ipynb
@@ -0,0 +1,573 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:af2e84ceb74c3bc3148f31759e2a692dba979671e7232787c666481af8571437"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 : SPILLWAYS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.1 pg : 538"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#Given\n",
+ "h = 1.2; \t\t\t\t#head of water\n",
+ "Cd = 2.2; \t\t\t\t#coefficient of discharge\n",
+ "rho = 1; \t\t\t\t#density of water\n",
+ "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n",
+ "\n",
+ "q = Cd*h**1.5;\n",
+ "\n",
+ "#applying bernaulli's equation at u/s water surface at section A and B\n",
+ "#solving it by error and trial method we get\n",
+ "v1 = 13.7;v2 = 14.7;\n",
+ "d1 = 0.212;d2 = 0.197;\n",
+ "\n",
+ "F1 = gamma_w*d1**2*math.cos(math.radians(60))/2;\n",
+ "F2 = gamma_w*d2**2/2;\n",
+ "W = gamma_w*60*2*math.pi*3*((d1+d2)/2)/360;\n",
+ "Fx = rho*q*(v2-v1*math.cos(math.radians(60)))-F1/2+F2;\n",
+ "Fy = rho*q*(v1*math.sin(math.radians(60)))+F1*math.sin(math.radians(60))+W;\n",
+ "F = (Fx**2+Fy**2)**0.5;\n",
+ "F = round(F*100)/100;\n",
+ "\n",
+ "# Results\n",
+ "print \"Resultant force = %.2f kN/m.\"%(F);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant force = 46.68 kN/m.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.2 pg : 539"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\t\t\t\t\n",
+ "#Given\n",
+ "C = 2.4; \t\t\t\t#coefficient of discharge\n",
+ "H = 2; \t\t\t\t#head\n",
+ "L = 100; \t\t\t\t#length of spillway\n",
+ "wc = 8; \t\t\t\t#heigth of weir crest above bottom\n",
+ "g = 9.81; \t\t\t\t#acceleration due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "h = H+wc;\n",
+ "Q1 = C*L*H**(1.5); \t\t\t\t#neglecting approach velocity and end contractions\n",
+ "va = Q1/(h*L);\n",
+ "ha = va**2/(2*g);\n",
+ "Ha = ha+H;\n",
+ "Q = C*L*Ha**1.5;\n",
+ "Q = round(Q*10)/10;\n",
+ "\n",
+ "# Results\n",
+ "print \"discharge over oggy weir = %.2f cumecs.\"%(Q);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "discharge over oggy weir = 690.80 cumecs.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.3 pg : 540"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#capacity of siphon\n",
+ "#head required in oggy spillway\n",
+ "#length of oggy weir required\n",
+ "\n",
+ "#Given\n",
+ "t = 6; \t\t\t\t#tail water elevation\n",
+ "h = 1; \t\t\t\t#heigth of siphon spillway\n",
+ "w = 4; \t\t\t\t#width of siphon spillway\n",
+ "hw = 1.5; \t\t\t\t#head water elevation\n",
+ "C = 0.6; \t\t\t\t#coefficient of discharge\n",
+ "Co = 2.25; \t\t\t\t#coefficient of discharge of oggy spillway\n",
+ "lo = 4; \t\t\t\t#length of oggy spillway\n",
+ "hc = 1.5; \t\t\t\t#head on weir crest\n",
+ "g = 9.81; \t\t\t\t#acceleration due to gravity\n",
+ "\n",
+ "# Calculations and Results\n",
+ "#part (a)\n",
+ "Q = C*h*w*(2*g*(t+hw))**0.5;\n",
+ "Q = round(Q*10)/10;\n",
+ "print \"capacity of siphon = %.2f cumecs.\"%(Q);\n",
+ "\n",
+ "#part (b)\n",
+ "h1 = (Q/(Co*lo))**(2./3);\n",
+ "h1 = round(h1*100)/100;\n",
+ "print \"head required in oggy spillway = %.2f m\"%(h1);\n",
+ "\n",
+ "#part (c)\n",
+ "L = Q/(Co*(hc)**1.5);\n",
+ "L = round(L*100)/100;\n",
+ "print \"length of oggy weir required = %.2f m.\"%(L);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "capacity of siphon = 29.10 cumecs.\n",
+ "head required in oggy spillway = 2.19 m\n",
+ "length of oggy weir required = 7.04 m.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.4 pg : 540"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "\n",
+ "\t\t\t\t\n",
+ "#Given\n",
+ "rl = 435; \t\t\t\t#full reservior level\n",
+ "cl = 429.6; \t\t\t\t#level of centre of siphon\n",
+ "hfl = 435.85; \t\t\t\t#high flood level\n",
+ "hfd = 600; \t\t\t\t#high flood discharge\n",
+ "w = 4; \t\t\t\t#width of throat\n",
+ "h = 2; \t\t\t\t#heigth of throat\n",
+ "C = 0.65; \t\t\t\t#coefficient of discharge\n",
+ "g = 9.81; \t\t\t\t#acceleration due to gravity\n",
+ "\n",
+ "# Calculations\n",
+ "H = hfl-cl;\n",
+ "Q = C*w*h*(2*g*H)**0.5;\n",
+ "n = hfd/Q;\n",
+ "n = round(n*100)/100;\n",
+ "\n",
+ "# Results\n",
+ "print \" number of siphons units required = %.2f.hence provide 11 siphons units.\"%(n);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " number of siphons units required = 10.42.hence provide 11 siphons units.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.5 pg : 541"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "from numpy import arange,zeros\n",
+ "\n",
+ "#design oggy spillway for concrete gravity dam\n",
+ "\n",
+ "#Given\n",
+ "rbl = 250; \t\t\t\t#avarage river bed level\n",
+ "rlc = 350; \t\t\t\t#R.L of spillway crest\n",
+ "s = 0.75; \t\t\t\t#slope on downstream side\n",
+ "Q = 6500; \t\t\t\t#discharge\n",
+ "L = 5*9; \t\t\t\t#length of spillway\n",
+ "Cd = 2.2; \t\t\t\t#coefficient of discharge\n",
+ "t = 2; \t\t\t\t#thickness of each pier\n",
+ "\n",
+ "#step 1. computation of design head\n",
+ "H = (Q/(Cd*L))**(2./3);\n",
+ "P = rlc-rbl;\n",
+ "\n",
+ "#P/H = 6.15,which is<1.33;it is a high overflow spillway\n",
+ "\n",
+ "#H+P/H = 7.15>1.7; hence discharge coefficient is not affected by downstream apron interface\n",
+ "\n",
+ "Kp = 0.01;\n",
+ "Ka = 0.1;\n",
+ "N = 4;\n",
+ "He = 17.5; \t\t\t\t#assumed\n",
+ "Le = L-2*(N*Kp+Ka)*He;\n",
+ "He1 = (Q/(Cd*Le))**(2./3);\n",
+ "He1 = round(He1*100)/100;\n",
+ "#He1 is almost equal to He\n",
+ "print \"crest profile will be designed for Hd = %.2f m.\"%(He1);\n",
+ "\n",
+ "#step 2. determination of d/s profile\n",
+ "\n",
+ "#equating the slope of d/s side and derivative of profile equation suggested by WES\n",
+ "x = 27.03;\n",
+ "y = 0.04372*x**1.85;\n",
+ "print \"downstream profile:\";\n",
+ "x = arange(1,27)\n",
+ "y = zeros(26)\n",
+ "for i in range(26):\n",
+ " y[i] = 0.04372*x[i]**1.85;\n",
+ " y[i] = round(y[i]*1000)/1000;\n",
+ "\n",
+ "print \"x y\";\n",
+ "for i in range(26):\n",
+ " print \"%i %.2f\"%(x[i],y[i]);\n",
+ "\n",
+ "print \"27.03 19.48\";\n",
+ "#step 3. determination of u/s profile\n",
+ "# math.cosidering equation for vertical u/s face and Hd = 17.58\n",
+ "\n",
+ "print \"upstream profile:\";\n",
+ "x = [-0.5, -0.1, -1.5, -2.0, -3.0, -4.0, -4.75];\n",
+ "y = zeros(7)\n",
+ "for i in range(7):\n",
+ " if i==6:\n",
+ " y[i] = 0.0633*(x[i]+4.75)**1.85+2.2151-1.2643*(x[i]+4.75)**0.625;\n",
+ " y[i] = round(y[i]*1000)/1000;\n",
+ " continue\n",
+ " \n",
+ " y[i] = 0.0633*(x[i]+4.7466)**1.85+2.2151-1.2643*(x[i]+4.7466)**0.625;\n",
+ " y[i] = round(y[i]*1000)/1000;\n",
+ "\n",
+ "print \"x y\";\n",
+ "for i in range(7):\n",
+ " print \"%.2f %.2f\"%(x[i],y[i]);\n",
+ "\n",
+ "\n",
+ "#step 4.design of d/s bucket\n",
+ "\n",
+ "R = P/4;\n",
+ "print \"radius of bucket = %i m.\"%(R);\n",
+ "print \"bucket will subtend angle of 60 degree at the centre.\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "crest profile will be designed for Hd = 17.58 m.\n",
+ "downstream profile:\n",
+ "x y\n",
+ "1 0.04\n",
+ "2 0.16\n",
+ "3 0.33\n",
+ "4 0.57\n",
+ "5 0.86\n",
+ "6 1.20\n",
+ "7 1.60\n",
+ "8 2.05\n",
+ "9 2.55\n",
+ "10 3.10\n",
+ "11 3.69\n",
+ "12 4.34\n",
+ "13 5.03\n",
+ "14 5.77\n",
+ "15 6.55\n",
+ "16 7.38\n",
+ "17 8.26\n",
+ "18 9.18\n",
+ "19 10.15\n",
+ "20 11.16\n",
+ "21 12.21\n",
+ "22 13.31\n",
+ "23 14.45\n",
+ "24 15.63\n",
+ "25 16.86\n",
+ "26 18.13\n",
+ "27.03 19.48\n",
+ "upstream profile:\n",
+ "x y\n",
+ "-0.50 0.01\n",
+ "-0.10 -0.00\n",
+ "-1.50 0.14\n",
+ "-2.00 0.25\n",
+ "-3.00 0.60\n",
+ "-4.00 1.20\n",
+ "-4.75 2.21\n",
+ "radius of bucket = 25 m.\n",
+ "bucket will subtend angle of 60 degree at the centre.\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.6 pg : 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "\t\t\t\t#design length and depth of stilling bamath.sin\n",
+ "\t\t\t\t\n",
+ "#Given\n",
+ "q = 1; \t\t\t\t#discharge of spillway\n",
+ "Cd = 0.7; \t\t\t\t#coefficient of discharge\n",
+ "h1 = 10; \t\t\t\t#heigth of crest above downstream silting bamath.sin\n",
+ "g = 9.81; \t\t\t\t#acceleration due to gravity\n",
+ "Cv = 0.9; \t\t\t\t#coefficient of velocity\n",
+ "\n",
+ "# Calculations\n",
+ "h = (3*q/(2*Cd*(2*g)**0.5))**(2./3);\n",
+ "H = h1+h/2;\n",
+ "vt = (2*g*H)**0.5;\n",
+ "v1 = Cv*vt;\n",
+ "y1 = q/v1;\n",
+ "F1 = v1/(g*y1)**0.5;\n",
+ "\t\t\t\t#F>1, flow is super-critical\n",
+ "y2 = 1;\n",
+ "v2 = q/y2;\n",
+ "F2 = v2/(g*y2)**0.5; \t\t\t\t#<1\n",
+ "y2 = (y1/2)*((1+8*F1**2)**0.5-1);\n",
+ "de = y2-1;\n",
+ "le = 5*(y2-y1);\n",
+ "de = round(de*1000)/1000;\n",
+ "le = round(le*10)/10;\n",
+ "\n",
+ "# Results\n",
+ "print \"stilling bamath.sin should be depressed by %.2f m.\"%(de);\n",
+ "print \"length of stilling bamath.sin = %.2f m.\"%(le);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "stilling bamath.sin should be depressed by 0.58 m.\n",
+ "length of stilling bamath.sin = 7.50 m.\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.7 pg : 563"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "\n",
+ "\t\t\t\t\n",
+ "#Given\n",
+ "q = 7.83; \t\t\t\t#discharge through spillway\n",
+ "w = 12.5; \t\t\t\t#width of fall\n",
+ "d = 2.; \t\t\t\t#depth of water in downstream\n",
+ "g = 9.8;\n",
+ "\n",
+ "y1 = 0.5;\n",
+ "v1 = q/y1;\n",
+ "F1 = v1/(g*y1)**0.5;\n",
+ "\n",
+ "#F>1,flow is super-critical\n",
+ "\n",
+ "# Calculations\n",
+ "v2 = q/d;\n",
+ "F2 = v2/(g*d)**0.5;\n",
+ "y2 = (y1/2)*((1+8*F1**2)**0.5-1);\n",
+ "de = y2-d;\n",
+ "le = 5*(y2-y1);\n",
+ "de = round(de*100)/100;\n",
+ "le = round(le*10)/10;\n",
+ "\n",
+ "# Results\n",
+ "print \"stilling bamath.sin should be depressed by %.2f m.\"%(de);\n",
+ "print \"length of stilling bamath.sin = %.2f m.\"%(le); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "stilling bamath.sin should be depressed by 2.76 m.\n",
+ "length of stilling bamath.sin = 21.30 m.\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.8 pg : 564"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "\n",
+ "\t\t\t\t\n",
+ "#Given\n",
+ "Ag = 5*2.5; \t\t\t\t#area of gate\n",
+ "miu = 0.25; \t\t\t\t#coefficient of friction\n",
+ "w = 0.5; \t\t\t\t#weigth of gate\n",
+ "h = 2; \t\t\t\t#head of water over crest\n",
+ "g = 9.81; \t\t\t\t#acceleration due to gravity\n",
+ "gamma_w = 1000; \t\t\t\t#unit weigth of water\n",
+ "\n",
+ "\n",
+ "# Calculations\n",
+ "m = w*g*1000;\n",
+ "F = gamma_w*Ag*h*h*g/10;\n",
+ "ff = miu*F;\n",
+ "tf = (m+ff)/1000;\n",
+ "\n",
+ "# Results\n",
+ "print \"force to be exerted to lift the gate = %.2f kN.\"%(tf);\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "force to be exerted to lift the gate = 17.17 kN.\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.9 pg : 564"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "\n",
+ "\t\t\t\t\n",
+ "#Given\n",
+ "q = 19; \t\t\t\t#dischrge through spillway\n",
+ "E = 1; \t\t\t\t#energy loss\n",
+ "\n",
+ "\t\t\t\t#from energy loss equation;E = (y2-y1)**3/4y2y1; and solving it we get\n",
+ "\t\t\t\t#x = 0.5*(-1+(1+294.39*(x-1)**9/64*x**3))\n",
+ "\t\t\t\t#by trial and error method x = 2.806\n",
+ "x = 2.806;\n",
+ "y1 = 4*x/(x-1)**3;\n",
+ "y2 = x*y1;\n",
+ "y1 = round(y1*1000)/1000;\n",
+ "y2 = round(y2*1000)/1000;\n",
+ "print \"depth of flow at both end of jumps = %.2f m and %.2f m. respectively.\"%(y1,y2);\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "depth of flow at both end of jumps = 1.91 m and 5.35 m. respectively.\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
generated by cgit (git 2.34.1) at 2025-01-23 21:17:45 +0000