path: root/Transport_Phenomena:_A_Unified_Approach/ch7.ipynb

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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 7 : Integral methods of analysis"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.2 - Page No :273\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "\n",
      "# Variables\n",
      "# given\n",
      "id_ = 4.;  \t\t\t #[m] - insid_e diameter\n",
      "h = 2.;  \t\t\t #[m] - water level\n",
      "ro = 0.03;  \t\t\t #[m] - radius of exit hole\n",
      "rt = id_/2.;  \t\t\t #[m] - insid_e radius\n",
      "g = 9.80665;  \t\t\t #[m/sec**2] - gravitational acceleration\n",
      "\n",
      "# Calculations\n",
      "# using the equation dh/h**(1/2) = -((ro**2)/(rt**2))*(2*g)**(1/2)dt and integrating between h = 2 and h = 1\n",
      "def f6(h): \n",
      "\t return (1./h**(1./2))*(1./(-((ro**2)/(rt**2))*(2*g)**(1./2)))\n",
      "\n",
      "t1 =  quad(f6,2,1)[0]\n",
      "\n",
      "# Results\n",
      "print \" Time required to remove half of the contents of the tank is  t = %.2f sec = %.2f min\"%(t1,t1/60);\n",
      "\t\t\t #integrating between h = 2 and h = 0\n",
      "\n",
      "def f7(h): \n",
      "\t return (1./h**(1./2))*(1./(-((ro**2)/(rt**2))*(2*g)**(1./2)))\n",
      "\n",
      "t2 =  quad(f7,2,0)[0]\n",
      "\n",
      "print \" Time required to empty the tank fully is  t = %.1f sec = %.1f min\"%(t2,t2/60);\n",
      "\n",
      "ro2 = (h**2)*math.sqrt(h*h/g)/(10*60)\n",
      "print \" Ro**2 = %.6f m**2\"%ro2\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Time required to remove half of the contents of the tank is  t = 831.37 sec = 13.86 min\n",
        " Time required to empty the tank fully is  t = 2838.5 sec = 47.3 min\n",
        " Ro**2 = 0.004258 m**2\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.3 - Page No :274\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "# composition of fuel gas\n",
      "nH2 = 24.;\n",
      "nN2 = 0.5;\n",
      "nCO = 5.9;\n",
      "nH2S = 1.5;\n",
      "nC2H4 = 0.1;\n",
      "nC2H6 = 1;\n",
      "nCH4 = 64;\n",
      "nCO2 = 3.0;\n",
      "\n",
      "# Calculations\n",
      "# calculating the theoritical amount of O2 required\n",
      "nO2theoreq = 12+2.95+2.25+0.30+3.50+128;\n",
      "# since fuel gas is burned with 40% excess O2,then O2 required is\n",
      "nO2req = 1.4*nO2theoreq;\n",
      "nair = nO2req/0.21;  \t\t\t # as amount of O2 in air is 21%\n",
      "nN2air = nair*(0.79);  \t\t\t # as amount of N2 in air is 79%\n",
      "nN2 = nN2+nN2air;\n",
      "nO2 = nO2req-nO2theoreq;\n",
      "nH2O = 24+1.5+0.2+3.0+128;\n",
      "nCO2formed = 72.1;\n",
      "nCO2 = nCO2+nCO2formed;\n",
      "nSO2 = 1.5;\n",
      "ntotal = nSO2+nCO2+nO2+nN2+nH2O;\n",
      "mpSO2 = (nSO2/ntotal)*100;\n",
      "mpCO2 = (nCO2/ntotal)*100;\n",
      "mpO2 = (nO2/ntotal)*100;\n",
      "mpN2 = (nN2/ntotal)*100;\n",
      "mpH2O = (nH2O/ntotal)*100;\n",
      "\n",
      "\n",
      "# Results\n",
      "print \" gas                 N2       O2         H2O         CO2       SO2\";\n",
      "print \" moles              %.1f     %.1f      %.1f        %.1f      %.1f\"%(nN2,nO2,nH2O,nCO2,nSO2);\n",
      "print \" mole percent       %.1f      %.1f       %.1f          %.1f      %.1f\"%(mpN2,mpO2,mpH2O,mpCO2,mpSO2);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " gas                 N2       O2         H2O         CO2       SO2\n",
        " moles              785.2     59.6      156.7        75.1      1.5\n",
        " mole percent       72.8      5.5       14.5          7.0      0.1\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.4 - Page No :280\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "# given\n",
      "id_ = 6.;  \t\t\t #[inch] - inlet diameter\n",
      "od = 4.;  \t\t\t #[inch] - outlet diameter\n",
      "Q = 10.;  \t\t\t #[ft**3/sec] - water flow rate\n",
      "alpha2 = math.pi/3;  #[radians] - angle of reduction of elbow\n",
      "alpha1 = 0.;\n",
      "p1 = 100.;  \t\t #[psi] - absolute inlet pressure\n",
      "p2 = 29.;  \t\t\t #[psi] - absolute outlet pressure\n",
      "\n",
      "# Calculations\n",
      "S1 = (math.pi*((id_/12)**2))/4.;\n",
      "S2 = (math.pi*((od/12)**2))/4.;\n",
      "U1 = Q/S1;\n",
      "U2 = Q/S2;\n",
      "mu = 6.72*10**-4;  \t #[lb*ft**-1*sec**-1]\n",
      "p = 62.4;  \t\t\t #[lb/ft**3]\n",
      "Nrei = ((id_/12.)*U1*p)/(mu);\n",
      "\n",
      "# Results\n",
      "print \"Nre(inlet) = %.1e\"%Nrei\n",
      "Nreo = round(((od/12)*U2*p)/(mu),-4);\n",
      "print \"Nre(outlet) = %.1e \"%Nreo\n",
      "\n",
      "# thus\n",
      "b = 1.;\n",
      "w1 = p*Q;  \t\t\t #[lb/sec] - mass flow rate\n",
      "w2 = w1;\n",
      "gc = 32.174;\n",
      "\n",
      "# using the equation (w/gc)*((U1)*(math.cos(alpha1))-(U2)*(math.cos(alpha2)))+p1*S1*math.cos(alpha1)-p2*S2*math.cos(alpha2)+Fextx = 0;\n",
      "Fextx = -(w1/gc)*((U1)*(math.cos(alpha1))-(U2)*(math.cos(alpha2)))-p1*144*S1*math.cos(alpha1)+p2*144*S2*math.cos(alpha2);\n",
      "print \"Fext,x = %.0f lb\"%Fextx\n",
      "Fexty = -(w1/gc)*((U1)*(math.sin(alpha1))-(U2)*(math.sin(alpha2)))-p1*144*S1*math.sin(alpha1)+p2*144*S2*math.sin(alpha2);\n",
      "print \"Fext,y = %.0f lb \"%Fexty\n",
      "print \"The forces Fext,x and Fext,y are the forces exerted on the fluid_ by the elbow.\\\n",
      "\\nFext,x acts to the left and Fext,y acts in the positive y direction.\\\n",
      "\\nNote that the elbow is  horizantal,and gravity acts in the z direction\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Nre(inlet) = 2.4e+06\n",
        "Nre(outlet) = 3.6e+06 \n",
        "Fext,x = -2522 lb\n",
        "Fext,y = 2240 lb \n",
        "The forces Fext,x and Fext,y are the forces exerted on the fluid_ by the elbow.\n",
        "Fext,x acts to the left and Fext,y acts in the positive y direction.\n",
        "Note that the elbow is  horizantal,and gravity acts in the z direction\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.5 - Page No : 282\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import numpy\n",
      "\n",
      "# Variables\n",
      "Fextx = -2522.;  \t\t\t #[lb] - force in x direction\n",
      "Fexty = 2240.;  \t\t\t #[lb] - force in y direction\n",
      "\n",
      "# Calculations\n",
      "# the force exerted by the elbow on the fluid is the resolution of Fext,x and Fext,y , therefore\n",
      "Fext = ((Fextx)**2+(Fexty)**2)**(1./2);\n",
      "alpha = 180. - 41.6\n",
      "\n",
      "# Results\n",
      "print \" the force has a magnitude of %.0f lb and a direction of %.1f from \\\n",
      "\\nthe positive x directionin the second quadrant\"%(Fext,alpha);\n",
      "\n",
      "# Note : answers in book is wrong. they have used wrong values everywhere. Please check it manually."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " the force has a magnitude of 3373 lb and a direction of 138.4 from \n",
        "the positive x directionin the second quadrant\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.6 - Page No :283\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "id_ = 6.;  \t\t\t #[inch] - inlet diameter\n",
      "od = 4.;  \t\t\t #[inch] - outlet diameter\n",
      "Q = 10.;  \t\t\t #[ft**3/sec] - water flow rate\n",
      "alpha2 = math.pi/3;  #[radians] - angle of reduction of elbow\n",
      "alpha1 = 0.;\n",
      "p1 = 100.;  \t\t #[psi] - absolute inlet pressure\n",
      "p2 = 29.;  \t\t\t #[psi] - absolute outlet pressure\n",
      "patm = 14.7;  \t\t #[psi] - atmospheric pressure\n",
      "p1gauge = p1-patm;\n",
      "p2gauge = p2-patm;\n",
      "S1 = (math.pi*((id_/12.)**2))/4.;\n",
      "S2 = (math.pi*((od/12.)**2))/4.;\n",
      "U1 = Q/S1;\n",
      "U2 = Q/S2;\n",
      "p = 62.4;  \t\t\t #[lb/ft**3]\n",
      "b = 1.;\n",
      "w1 = p*Q;  \t\t\t #[lb/sec] - mass flow rate\n",
      "w2 = w1;\n",
      "gc = 32.174;\n",
      "\n",
      "# Calculations\n",
      "# using the equation Fpress = p1gauge*S1-p2gauge*S2*math.cos(alpha2);\n",
      "Fpressx = p1gauge*144*S1-p2gauge*144*S2*math.cos(alpha2);\n",
      "Fpressy = p1gauge*144*S1*math.sin(alpha1)-p2gauge*144*S2*math.sin(alpha2);\n",
      "wdeltaUx = (w1/gc)*((U2)*(math.cos(alpha2))-(U1)*(math.cos(alpha1)));\n",
      "wdeltaUy = (w1/gc)*((U2)*(math.sin(alpha2))-(U1)*(math.sin(alpha1)));\n",
      "Fextx = wdeltaUx-Fpressx;\n",
      "Fexty = wdeltaUy-Fpressy;\n",
      "Fext = ((Fextx)**2+(Fexty)**2)**(1./2);\n",
      "alpha = 180 - 43.4\n",
      "\n",
      "# Results\n",
      "print \" The force has a magnitude of %.0f lb and a direction of %.1f from the positive x directionin the second \\\n",
      "quadrant\"%(round(Fext,-1),alpha);\n",
      "print \" Also there is a force on the elbow in the z direction owing to the weight of the elbow \\\n",
      " plus the weight of the fluid inside\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The force has a magnitude of 3030 lb and a direction of 136.6 from the positive x directionin the second quadrant\n",
        " Also there is a force on the elbow in the z direction owing to the weight of the elbow  plus the weight of the fluid inside\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.7 - Page No :293\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from scipy.integrate import quad \n",
      "# Variables\n",
      "Uo = 1.;  \t\t\t #[m/sec]\n",
      "# using Ux/Uo = y/yo\n",
      "# assuming any particular value of yo will not change the answer,therefore\n",
      "yo = 1;\n",
      "\n",
      "# Calculations\n",
      "def f2(y): \n",
      "\t return (Uo*y)/yo\n",
      "\n",
      "Uxavg =  quad(f2,0,yo)[0]\n",
      "\n",
      "\n",
      "def f3(y): \n",
      "\t return ((Uo*y)/yo)**3\n",
      "\n",
      "Ux3avg =  quad(f3,0,yo)[0]\n",
      "\n",
      "# using the formula alpha = (Uxavg)**3/Ux3avg\n",
      "alpha = (Uxavg)**3/Ux3avg;\n",
      "\n",
      "# Results\n",
      "print \"alpha = \",alpha\n",
      "print \" Note that the kinetic correction factor  alpha has the same final value for \\\n",
      " laminar pipe flow as it has for laminar flow \\nbetween parallel plates.\"\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "alpha =  0.5\n",
        " Note that the kinetic correction factor  alpha has the same final value for  laminar pipe flow as it has for laminar flow \n",
        "between parallel plates.\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.8 - Page No :293\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "Q = 0.03;  \t\t\t #[m**3/sec] - volumetric flow rate\n",
      "id_ = 7.;  \t\t\t #[cm] - insid_e diameter\n",
      "deltaz = -7.;  \t\t #[m] - length of pipe\n",
      "T1 = 25.;  \t\t\t #[degC] - lowere sid_e temperature\n",
      "T2 = 45.;  \t\t\t #[degC] - higher sid_e temperature\n",
      "g = 9.81;  \t\t\t #[m/sec**2] - acceleration due to gravity\n",
      "deltaP = 4.*10**4;   #[N/m**2] - pressure loss due to friction\n",
      "p = 1000.;  \t\t #[kg/m**3] - density of water \n",
      "w = Q*p;\n",
      "C = 4184.;  \t\t #[J/kg*K) - heat capacity of water\n",
      "\n",
      "# Calculations\n",
      "deltaH = w*C*(T2-T1);\n",
      "# using the formula Qh = deltaH+w*g*deltaz\n",
      "Qh = deltaH+w*g*deltaz;\n",
      "\n",
      "# Results\n",
      "print \" the duty on heat exchanger is  Q = %.2e J/sec\"%(Qh);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " the duty on heat exchanger is  Q = 2.51e+06 J/sec\n"
       ]
      }
     ],
     "prompt_number": 56
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.10 - Page No :298\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "d = 0.03;  \t\t\t #[m] - diameter\n",
      "g = 9.784;  \t\t #[m/sec] - acceleration due to gravity\n",
      "deltaz = -1.;\n",
      "\n",
      "# using the equation (1/2)*(U3**2/alpha3-U1**2/alpha1)+g*deltaz = 0\n",
      "# assuming\n",
      "alpha1 = 1.;\n",
      "alpha3 = 1.;\n",
      "\n",
      "U1 = 0.;\n",
      "\n",
      "# Calculations\n",
      "U3 = (-2*g*deltaz+(U1**2)/alpha1)**(1/2.);\n",
      "p = 1000.;  \t\t\t #[kg/m**3] - density of water\n",
      "S3 = (math.pi/4)*(d)**2\n",
      "w = p*U3*S3;\n",
      "\n",
      "# Results\n",
      "print \" the mass flow rate is  w = %.2f kg/sec\"%(w);\n",
      "\n",
      "\n",
      "# using deltap = p*((U3**2)/2+g*deltaz)\n",
      "deltap = p*((U3**2)/2+g*deltaz);\n",
      "p1 = 1.01325*10**5;  \t\t\t #[N/m**2] - is equal to atmospheric pressure\n",
      "p2 = p1+deltap;\n",
      "vp = 0.02336*10**5;\n",
      "if p2>vp:\n",
      "    print \" the siphon can operate since the pressure at point 2 is greater than the value at which the liquid boils\";\n",
      "else:\n",
      "    print \" the siphon cant operate since the pressuer at point 2 is less than \\\n",
      "     the value at which the liquid_ boils\";\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " the mass flow rate is  w = 3.13 kg/sec\n",
        " the siphon can operate since the pressure at point 2 is greater than the value at which the liquid boils\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.11 - Page No :300\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "sp = 1.45;  \t\t\t # specific gravity of trichloroethylene\n",
      "pwater = 62.4;  \t\t\t #[lb/ft**3] - density of water\n",
      "p = sp*pwater;\n",
      "d1 = 1.049;  \t\t\t #[inch] - density of pipe at point 1\n",
      "d2 = 0.6;  \t\t\t #[inch] - density of pipe at point 2\n",
      "d3 = 1.049;  \t\t\t #[inch] - density of pipe at point 3\n",
      "\n",
      "# Calculations\n",
      "# using the formula U1*S1 = U2*S2; we get U1 = U2*(d2/d1);\n",
      "# then using the bernoulli equation deltap/p = (1/2)*(U2**2-U1**2);\n",
      "deltap = 4.2*(144);  \t\t\t #[lb/ft**2] - pressure difference\n",
      "U2 = ((2*(deltap/p)*(1./(1.-(d2/d1)**4)))**(1./2))*(32.174)**(1./2);\n",
      "\n",
      "# umath.sing the formula w = p*U2*S\n",
      "w = p*U2*((math.pi/4)*(0.6/12)**2);\n",
      "w1 = w/(2.20462);\n",
      "\n",
      "# Results\n",
      "print \" the mass flow rate is  w = %.1f lb/sec or in SI units  w = %.2f kg/sec\"%(w,w1);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " the mass flow rate is  w = 3.9 lb/sec or in SI units  w = 1.77 kg/sec\n"
       ]
      }
     ],
     "prompt_number": 60
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.12 - Page No :301\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "Q = 50/(7.48*60);  \t #[ft/sec] - volumetric flow rate of water\n",
      "d1 = 1.;  \t\t\t #[inch] - diameter of pipe\n",
      "deltaz = -5;  \t\t #[ft] - distance between end of pipe and math.tank\n",
      "g = 32.1;  \t\t\t #[ft/sec] - acceleration due to gravity\n",
      "Cp = 1.;  \t\t\t #[Btu/lb*F] - heat capacity of water\n",
      "p = 62.4;  \t\t\t #[lb/ft**3] - density of water\n",
      "S1 = (math.pi/4)*(d1/12.)**2;\n",
      "U1 = Q/S1;\n",
      "w = p*Q;\n",
      "U2 = 0.;\n",
      "gc = 32.174;\n",
      "\n",
      "# Calculations\n",
      "# using the formula deltaH = (w/2)*((U2)**2-(U1)**2)+w*g*deltaz\n",
      "deltaH = -(w/(2*gc))*((U2)**2-(U1)**2)-w*(g/gc)*deltaz;\n",
      "deltaH = deltaH/778;  \t\t\t # converting from ftlb/sec to Btu/sec\n",
      "deltaT = deltaH/(w*Cp);\n",
      "\n",
      "# Results\n",
      "print \" The rise in temperature is %.5f degF\"%(deltaT);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The rise in temperature is 0.01475 degF\n"
       ]
      }
     ],
     "prompt_number": 62
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.13 - Page No :303\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "deltaz = 30.;  \t\t #[ft] - distance between process and the holding math.tank\n",
      "Q = 100.;  \t\t\t #[gpm] - volumetric flow rate of water\n",
      "p1 = 100.;  \t\t #[psig]\n",
      "p2 = 0.;  \t\t\t #[psig]\n",
      "g = 32.1;  \t\t\t #[ft/sec] - acceleration due to gravity\n",
      "sv = 0.0161;  \t\t #[ft**3/lb] - specific volume of water\n",
      "p = 1./sv;  \t\t #[lb/ft**3] - density of water\n",
      "e = 0.77;  \t\t\t # efficiency of centrifugal pump\n",
      "deltap = (p1-p2)*(144);  \t\t\t #[lbf/ft**2]\n",
      "gc = 32.174;\n",
      "\n",
      "# Calculations\n",
      "# using the equation deltap/p+g*(deltaz)+Ws = 0;\n",
      "Wst = -deltap/p-(g/gc)*(deltaz);\n",
      "# using the formula for efficiency e = Ws(theoritical)/Ws(actual)\n",
      "# therefore\n",
      "Wsa = Wst/e;\n",
      "# the calulated shaft work is for a unit mass flow rate of water,therfore for given flow rate multiply it by the flow rate\n",
      "w = (Q*p)/(7.48*60);\n",
      "Wsactual = Wsa*w;\n",
      "power = -Wsactual/(778*0.7070);\n",
      "\n",
      "# Results\n",
      "print \" the required horsepower is %.2f hp\"%(power);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " the required horsepower is 8.55 hp\n"
       ]
      }
     ],
     "prompt_number": 64
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.14 - Page No :304\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "p1 = 5.;  \t\t\t #[atm] - initial pressure\n",
      "p2 = 0.75;  \t\t #[atm] - final pressure after expansion through turbine\n",
      "T = 450.;  \t\t\t #[K] - temperature\n",
      "y = 1.4;  \t\t\t # cp/cv for nitrogen\n",
      "# using the equation Ws = -(y/(y-1))*(p1/density1)*((p2/p1)**((y-1)/y)-1)\n",
      "R = 8314.;  \t\t # gas constant\n",
      "\n",
      "# Calculations\n",
      "p1bydensity = R*T;\n",
      "Ws = -(y/(y-1))*(p1bydensity)*((p2/p1)**((y-1)/y)-1);\n",
      "\n",
      "# Results\n",
      "print \" the shaft work of the gas as it expands through the turbine and transmits its molecular \\\n",
      " energy to the rotating blades is \\n Ws = %.2e J/kmol\"%(Ws);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " the shaft work of the gas as it expands through the turbine and transmits its molecular  energy to the rotating blades is \n",
        " Ws = 5.48e+06 J/kmol\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example  7.15 - Page No :311\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T = 273.15+25;  \t\t #[K] - temperature\n",
      "R = 8.314;  \t\t\t #[kPa*m**3/kmol*K] - gas constant\n",
      "p = 101.325;  \t\t\t #[kPa] - pressure\n",
      "M = 29.;  \t\t\t     # molecular weight of gas\n",
      "pa = (p*M)/(R*T);\n",
      "sg = 13.45;  \t\t\t # specific gravity\n",
      "pm = sg*1000;\n",
      "g = 9.807;  \t\t\t #[m/sec**2] - acceleration due to gravity\n",
      "deltaz = 15./100;  \t\t #[m]\n",
      "\n",
      "# Calculations\n",
      "# using the equation p2-p1 = deltap = (pm-pa)*g*deltaz\n",
      "deltap = -(pm-pa)*g*deltaz;\n",
      "\n",
      "# Results\n",
      "print \" the pressure drop is %.2e N/m**2\"%(deltap);\n",
      "print \" the minus sign means the upstream pressure p1 is greater than p2, i.e ther is a pressure drop.\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " the pressure drop is -1.98e+04 N/m**2\n",
        " the minus sign means the upstream pressure p1 is greater than p2, i.e ther is a pressure drop.\n"
       ]
      }
     ],
     "prompt_number": 68
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.16 - Page No :312\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T = 536.67;  \t\t\t #[degR]; - temperature\n",
      "R = 10.73;  \t\t\t #[(lbf/in**2*ft**3)*lb*mol**-1*degR] - gas constant\n",
      "p = 14.696;  \t\t\t #[lbf/in**2];\n",
      "g = 9.807*3.2808;  \t\t #[ft/sec**2] - acceleration due to gravity\n",
      "M = 29.;  \t\t\t     # molecular weight of gas\n",
      "\n",
      "# Calculations\n",
      "pa = (p*M)/(R*T);\n",
      "sg = 13.45;  \t\t\t # specific gravity\n",
      "pm = sg*62.4;\n",
      "deltaz = 15/(2.54*12);   #[ft]\n",
      "gc = 32.174;\n",
      "\n",
      "# Results\n",
      "# using the equation p2-p1 = deltap = (pm-pa)*g*deltaz\n",
      "deltap = (pm-pa)*(g/gc)*deltaz;\n",
      "print \"the pressure drop is %.0f lbf/ft**2\"%(deltap);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the pressure drop is 413 lbf/ft**2\n"
       ]
      }
     ],
     "prompt_number": 71
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.18 - Page No :315\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "at = 0.049;  \t\t\t #[in**2] - cross sectional area of the manometer tubing\n",
      "aw = 15.5;  \t\t\t #[in**2] - cross sectional area of the well\n",
      "g = 32.174;  \t\t\t #[ft/sec**2] - acceleration due to gravity\n",
      "gc = 32.174;\n",
      "sg = 13.45;  \t\t\t #[ specific garvity of mercury\n",
      "p = 62.4;  \t\t\t     #[lb/ft**3] - density of water;\n",
      "pm = sg*p;\n",
      "deltaz_waterleg = 45.2213;\n",
      "\n",
      "# Calculations\n",
      "# using the equation A(well)*deltaz(well) = A(tube)*deltaz(tube)\n",
      "deltazt = 70.;  \t\t\t #[cm]\n",
      "deltazw = deltazt*(at/aw);\n",
      "deltaz = deltazt+deltazw;\n",
      "deltap_Hg = -pm*(g/gc)*(deltaz/(2.54*12));\n",
      "\n",
      "# Results\n",
      "deltazw = 45.2213;  \t\t\t #[cm]\n",
      "deltap_tap = deltap_Hg+p*(g/gc)*(deltazw/(12*2.54));\n",
      "print \"deltap_tap = %.0f lbf/ft**2\"%(deltap_tap);\n",
      "print \"deltap is negative and therefore p1 is greater than p2\";\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "deltap_tap = -1841 lbf/ft**2\n",
        "deltap is negative and therefore p1 is greater than p2\n"
       ]
      }
     ],
     "prompt_number": 73
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.19 - Page No :317\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "p = 749./760;  \t\t\t #[atm]\n",
      "T = 21.+273.15;  \t\t #[K]\n",
      "R = 82.06;  \t\t\t #[atm*cm**3/K] - gas constant\n",
      "v = (R*T)/p;  \t\t\t #[cm**3/mole] - molar volume\n",
      "M = 29.;  \t\t\t     #[g/mole] - molecular weight\n",
      "pair = M/v;\n",
      "m_air = 53.32;  \t\t\t #[g]\n",
      "m_h2o = 50.22;  \t\t\t #[g]\n",
      "ph2o = 0.998;  \t\t\t     #[g/cm**3] - density of water\n",
      "\n",
      "# Calculations\n",
      "V = (m_air-m_h2o)/(ph2o-pair);  \t\t\t #[cm**3]\n",
      "density = m_air/V;\n",
      "\n",
      "# Results\n",
      "print \" The density of coin is  density = %.2f g/cm**3\"%(density);\n",
      "print \" Consulting a handbook it is seen that this result is correct density for gold\";\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The density of coin is  density = 17.15 g/cm**3\n",
        " Consulting a handbook it is seen that this result is correct density for gold\n"
       ]
      }
     ],
     "prompt_number": 75
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.20 - Page No :318\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "P = 749./760;  \t\t\t #[atm] - pressure\n",
      "T = 21+273.15;  \t\t #[K] - temperature\n",
      "poak = 38*(1./62.4);  \t #[g/cm**3] - density of oak\n",
      "pbrass = 534/62.4;  \t #[g/cm**3] - density of brass\n",
      "m_brass = 6.7348;  \t\t #[g]\n",
      "pair = 0.001184;  \t\t #[g/cm**3] - density of air\n",
      "\n",
      "# Calculations\n",
      "# using the formula m_oak = m_brass*((1-(pair/pbrass))/(1-(pair/poak)))\n",
      "m_oak = m_brass*((1-(pair/pbrass))/(1-(pair/poak)));\n",
      "\n",
      "# Results\n",
      "print \" True mass of wood = %.3f g\"%(m_oak);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " True mass of wood = 6.747 g\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.21 - Page No :320\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T = 545.67;  \t\t\t #[degR] - temperature\n",
      "R = 1545.;  \t\t\t #[Torr*ft**3/degR*mole] - gas constant\n",
      "M = 29.;  \t\t\t     #[g/mole] - molecular weight\n",
      "g = 9.807;  \t\t\t #[m/sec**2] - acceleration due to gravity\n",
      "gc = 9.807;  \n",
      "po = 760.;  \t\t\t #[Torr] - pressure\n",
      "deltaz = 50.;  \t\t\t #[ft]\n",
      "\n",
      "# Calculations\n",
      "# using the equation p = po*exp(-(g/gc)*M*(deltaz/R*T))\n",
      "p = po*math.e**(-(g/gc)*M*(deltaz/(R*T)));\n",
      "\n",
      "# Results\n",
      "print \" p = %.1f Torr \\nThus, the pressure decrease for an elevation of 50ft is very small\"%(p);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " p = 758.7 Torr \n",
        "Thus, the pressure decrease for an elevation of 50ft is very small\n"
       ]
      }
     ],
     "prompt_number": 78
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 7.22 - Page No :321\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "To = 545.67;  \t\t\t #[degR] -  air temperature at beach level\n",
      "betaa = -0.00357;  \t\t #[degR/ft] - constant\n",
      "R = 1545.;  \t\t\t #[Torr*ft**3/degR*mole] - gas constant\n",
      "M = 29.;\n",
      "deltaz = 25000.;  \t\t #[ft]\n",
      "po =  760.  \n",
      "\n",
      "# Calculations\n",
      "# using the equation ln(p/po) = ((M)/(R*betaa))*ln(To/(To+betaa*deltaz)\n",
      "p = po*math.exp(((M)/(R*betaa))*math.log(To/(To+betaa*deltaz)));\n",
      "\n",
      "# Results\n",
      "print \" Pressure = %.2f Torr\"%(p);\n",
      "# using the equation T = To+betaa*deltaz\n",
      "T = To+betaa*deltaz;\n",
      "print \" Temperature = %.2f degR\"%(T);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Pressure = 297.16 Torr\n",
        " Temperature = 456.42 degR\n"
       ]
      }
     ],
     "prompt_number": 19
    }
   ],
   "metadata": {}
  }
 ]
}