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diff --git a/Fundamentals_of_Fluid_Mechanics/ch_12.ipynb b/Fundamentals_of_Fluid_Mechanics/ch_12.ipynb new file mode 100644 index 00000000..9c63bf6a --- /dev/null +++ b/Fundamentals_of_Fluid_Mechanics/ch_12.ipynb @@ -0,0 +1,328 @@ +{ + "metadata": { + "name": "ch 12" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12 :Turbomachines" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 12.2 Page no.697" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 12.2\n", + "#Find The ideal head rise\n", + "#and Tangential velocity component and The power transferred to the fluid.\n", + "\n", + "#Given\n", + "Q=1400.0 #gpm, water rate\n", + "N=1750.0 #rpm speed\n", + "b=2.0 #in height of blade\n", + "r1=1.9 #in inner radius\n", + "r2=7.0 #in outer radius\n", + "beta2=23.0 #degrees exit blade angle\n", + "beta2=beta2*3.14/180 #Radian\n", + "alpha1=90.0 #degrees water entering angle\n", + "\n", + "#Calculation\n", + "import math\n", + "U2=r2*2*math.pi*N/(60*12) #ft/sec\n", + "Vr2=(1400/(7.48*60*2*math.pi*(r2/12)*(b/12))) #ft/sec\n", + "V2tangential=round(U2,0)-(Vr2*1/(math.tan(beta2))) #ft/sec\n", + "hi=U2*V2tangential/32.2 #ft\n", + "print \"The ideal head rise=\",round(hi,2),\"ft\"\n", + "d=1.94 #slugs/(ft**3)\n", + "Wshaft=(d*Q*U2*V2tangential/(7.58*60))/550 #hp\n", + " \n", + "#result\n", + "print \"Tangential velocity component is \",round(V2tangential,0),\"ft/s\"\n", + "print \"The power transferred to the fluid=\",round(Wshaft,1),\"hp\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ideal head rise= 315.27 ft\n", + "Tangential velocity component is 95.0 ft/s\n", + "The power transferred to the fluid= 110.226 hp\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 12.3 Page no.702\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 12.3\n", + "#Find the maximum height at which the pump can be located.\n", + "\n", + "#Given\n", + "Q=0.5 #(ft**3)/sec\n", + "NPSHr=15 #ft\n", + "T=80 #degree F\n", + "patm=14.7 #psi\n", + "KL=20\n", + "D=4.0 #in\n", + "\n", + "#Calculation\n", + "import math\n", + "V=Q/((math.pi/4)*(D/12.0)**2.0) #ft/sec\n", + "hL=KL*(V**2)/(2*32.2) #ft\n", + "#from value of T\n", + "pv=0.5069 #psi\n", + "sw=62.22 #lb/(ft**3) sw =specific weight\n", + "z1max=(patm*144/sw)-hL-(pv*144/sw)-NPSHr #ft\n", + "\n", + "#result\n", + "print \"The maximum height at which the pump can be located=\",round(z1max,2),\"feet\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum height at which the pump can be located= 7.65 feet\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 12.5 Page no.708" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 12.5\n", + "#For peak efficiency predict the discharge actual head rise and shaft horsepower.\n", + "\n", + "#Given\n", + "D1=8.0 #in\n", + "N1=1200.0 #rpm\n", + "D2=12.0 #in\n", + "N2=1000.0 #rpm\n", + "T=60.0 #degree F\n", + "CQ=0.0625\n", + "\n", + "#Calculation\n", + "import math\n", + "Q1=CQ*(N1/60)*(2*math.pi)*(D1/12.0)**3.0 #(ft**3)/sec\n", + "print \"The discharge=\",round(Q1*7.48*60,0),\"gpm\"\n", + "CH=0.19\n", + "ha=CH*((N1*2*math.pi/60)**2.0)*((D1/12)**2.0)/32.2 #ft\n", + "print \"The actual headrise=\",round(ha,1),\"ft\"\n", + "CP=0.014\n", + "Wshaft=(CP*(1.94)*((N1*2*math.pi/60)**3.0)*((D1/12.0)**5.0))/550.0 #hp\n", + "\n", + "#Result\n", + "print \"The shaft horsepower=\",round(Wshaft,0),\"hp\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The discharge= 1044.0 gpm\n", + "The actual headrise= 41.4 ft\n", + "The shaft horsepower= 13.0 hp\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 12.6 Page no.719" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 12.6\n", + "#determine The nozzle diameter for maximum power outpu.\n", + "#The maximum power output and The angular velocity of the rotor.\n", + "\n", + "#Given\n", + "z0=200 #ft height of left section\n", + "l=1000 #ft length of pipe\n", + "f=0.02\n", + "D=8 #in. diameter of pipe\n", + "B=150 #degree \n", + "R=1.5 #ft radius of wheel\n", + "z1=0 #ft height of right section\n", + "#calculation\n", + "import math\n", + "#energy equation between a point on surface of lake and the nozzle outlet\n", + "#z0=(V1**2)/(2*32.2) + hL\n", + "#from continuity equation, V=(A1*V1/A)=((D1/D)**2)*V1\n", + "#neglecting minor losses, \n", + "#z0=(1+(f*l/D)*((D1/D)**4))*(V1**2)/(2*32.2)\n", + "#Wshaft=d*Q*u*(U-V1)*(1-cos(B))\n", + "#The maximum power occurs at U=V1/2 and dWshaft/dD1=0\n", + "a=(2*32.2*z0)**(0.5) \n", + "b=f*(l/(D/12.0))*(1/(D/12.0))**4.0 \n", + "c=a*math.pi*1.94*(1-math.cos(B*math.pi/180))/4.0 \n", + "d=(c*a*a/4) #1.04*(10**6)\n", + "#by the above conditions, and applying Q=(math.pi*(D1**2)*V1/4)\n", + "D1=(1/(2*b))**(0.25) #ft, nozzle diameter\n", + "\n", + "#result\n", + "print \"The nozzle diameter for maximum power output=\",round(D1,2),\"ft\"\n", + "Wshaft=-((d*D1**2)/(1+(b*D1**4))**(1.5))/550.0 #hp\n", + "print \"The maximum power output=\",round(Wshaft,0),\"hp\"\n", + "V1=a/(1+(b*(D1**4)))**0.5 #ft/sec\n", + "omega=(V1/(2*R))*60/(2*math.pi) #rpm\n", + "print \"The angular velocity of the rotor=\",round(omega,0),\"rpm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The nozzle diameter for maximum power output= 0.24 ft\n", + "The maximum power output= -59.0 hp\n", + "The angular velocity of the rotor= 295.0 rpm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 12.8 Page no.723" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 12.8\n", + "#find The shaft energy per unit mass of air.\n", + "#Given\n", + "ri=0.133 #in.\n", + "ro=0.168 #in.\n", + "N=300000 #rpm\n", + "\n", + "#calculation\n", + "import math\n", + "rm=0.5*(ro+ri)/12\n", + "U=(N*2*math.pi/60)*rm #ft/sec\n", + "wshaft=(-U)*(2*U)/32.174 #ft*lb/lbm\n", + "\n", + "#Result\n", + "print \"The shaft energy per unit mass of air=\",round(wshaft,0),\"ft*lb/lbm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The shaft energy per unit mass of air= -9650.0 Ft*lb/lbm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 12.9 Page No.727" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Example 12.9\n", + "#What type of turbine should we selected.\n", + "\n", + "#Given\n", + "w=6.0 #rev/s angular velocity\n", + "q=10.0 #ft**3/s , flow rate\n", + "h=20.0 #ft, head\n", + "gamma=62.4 #lb/ft**3, specific wt\n", + "n=94 # %, assumed efficiency\n", + "#calculation\n", + "import math\n", + "##rev/min\n", + "Wshaft=gamma*q*h*(n*10**-2)/(550.0) #shaft power\n", + "N=w1*math.sqrt(Wshaft)/(h)**(1.25)\n", + "print \"Accordint to information a mixed flow Francis Turbine must be used.\"\n", + "\n", + "g=32.2 #ft/s**2\n", + "V1=math.sqrt(2*g*h) #ft/s velocity\n", + "D=V1/(w*2*math.pi) #ft\n", + "d1=math.sqrt(4*q/(math.pi*V1))\n", + "print \"So a Pelton wheel with a diameter=\",round(D,3),\"ft\" \"supplied with water through nozzle of diameter=\",round(d1,3),\"ft\" \"is not a practical design.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Accordint to information a mixed flow Francis Turbine must be used.\n", + "So a Pelton wheel with a diameter= 0.952 ftsupplied with water through nozzle of diameter= 0.596 ftis not a practical design.\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
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