path: root/Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter4.ipynb

{
 "metadata": {
  "name": "",
  "signature": "sha256:b8e850534ebe1c09700aafa043440ac4a2221bb7e166ce915a3d59a2e23cca27"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter4-Heat transfer theory\n"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex1-pg88"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.1\n",
      "print('Example 4.1\\n\\n');\n",
      "print('Page No. 88\\n\\n');\n",
      "\n",
      "## given\n",
      "K = 45.## Thermal Conductivity in W/m-K\n",
      "L = 5.*10**-3;## thickness in metre\n",
      "T1 = 100.;## in degree celcius\n",
      "T2 = 99.9;## in degree celcius\n",
      "A = 1.;## Area in m^2\n",
      "\n",
      "##By Fourier law of conduction\n",
      "Q = ((K*A*(T1-T2))/L);## in Watts\n",
      "print'%s %.2f %s'%('The rate of conductive heat transfer is ',Q,' W \\n')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.1\n",
        "\n",
        "\n",
        "Page No. 88\n",
        "\n",
        "\n",
        "The rate of conductive heat transfer is  900.00  W \n",
        "\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2-pg89"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.2\n",
      "print('Example 4.2\\n\\n');\n",
      "print('Page No. 89\\n\\n');\n",
      "## given\n",
      "K1 = 45.## Thermal Conductivity  of mild steel in W/m-K\n",
      "K2 = 0.040## Thermal Conductivity  of insulaton in W/m-K\n",
      "L1 = 5.*10**-3;## thickness of mild steel in metre\n",
      "L2 = 50.*10**-3;## thickness of insulation in metre\n",
      "T1 = 100.;## in degree celcius\n",
      "T2 = 25.;## in degree celcius\n",
      "A = 1.;## Area in m^2\n",
      "\n",
      "##By Fourier law of conduction\n",
      "Q = (((T1-T2)/((L1/(K1*A))+(L2/(K2*A)))))## in Watts\n",
      "print'%s %.2f %s'%('The rate of conductive heat transfer is ',Q,' W \\n')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.2\n",
        "\n",
        "\n",
        "Page No. 89\n",
        "\n",
        "\n",
        "The rate of conductive heat transfer is  59.99  W \n",
        "\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex3-pg90"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.3\n",
      "print('Example 4.3\\n\\n');\n",
      "print('Page No. 90\\n\\n');\n",
      "\n",
      "## given\n",
      "K1 = 26.;## Thermal Conductivity  of stainless steel in W/m-K\n",
      "K2 = 0.038;## Thermal Conductivity  of insulaton in W/m-K\n",
      "L1 = 3.*10**-3;## thickness of stainless steel in metre\n",
      "L2 = 40.*10**-3;## thickness of insulation in metre\n",
      "T1 = 105.;## in degree celcius\n",
      "T2 = 25.;## in degree celcius\n",
      "L = 15.;## Length of pipe in metre\n",
      "d1 = 50.*10**-3;## Internal diameter of pipe in metre\n",
      "d2 = 56.*10**-3;## External diameter of pipe in metre\n",
      "\n",
      "r1 = d1/2.;## in metre\n",
      "r2 = d2/2.;## in metre\n",
      "\n",
      "rm_p = ((r2-r1)/math.log(r2/r1));## logarithmic mean radius of pipe in m\n",
      "rm_i = (((r2+L2)-r2)/math.log((r2+L2)/r2));## logarithmic mean radius of insulation in m\n",
      "\n",
      "##By Fourier law of conduction\n",
      "Q = (((T1-T2)/((L1/(K1*2.*math.pi*rm_p))+(L2/(K2*2.*math.pi*rm_i)))));## in W/m\n",
      "Q_L = Q*L;\n",
      "print'%s %.2f %s'%('The rate of conductive heat transfer per 15 m length of pie is ',Q_L,'W')## Deviation in answer due to direct substitution\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.3\n",
        "\n",
        "\n",
        "Page No. 90\n",
        "\n",
        "\n",
        "The rate of conductive heat transfer per 15 m length of pie is  322.84 W\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex4-pg93"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.4\n",
      "print('Example 4.4\\n\\n');\n",
      "print('Page No. 93\\n\\n');\n",
      "\n",
      "## given\n",
      "dH = 12.*10**-3;## Outer diameter of pipe in m\n",
      "dC = 10.*10**-3;## Inner diameter of pipe in m\n",
      "L = 1.*10**-3;## im m\n",
      "h_H = 10.*10**3;## Heat Transfer Coefficient  on vapour side in W/m^2-K\n",
      "h_C = 4.5*10**3;## Heat Transfer Coefficient  on vapour side in W/m^2-K\n",
      "K = 26.;## Thermal Conductivity of metal in W/m-K\n",
      "dM = (dH + dC)/2.;## mean diameter in m\n",
      "h_Hf = 6.*10**3;## Fouling factor for hot side\n",
      "h_Cf = 6.*10**3;## Fouling factor for cold side\n",
      "\n",
      "U = (1./h_H)+((L*dH)/(K*dM))+(dH/(dC*h_C));\n",
      "Uh = (1./U);## in W/m^2-K\n",
      "print'%s %.2f %s'%('The original heat transfer coefficient is ',Uh,' W/sq.m K \\n' )## Deviation in answer due to direct substitution\n",
      "\n",
      "u = (1./h_H)+(1./h_Hf)+((L*dH)/(K*dM))+(dH/(dC*h_C))+(dH/(dC*h_Cf));\n",
      "Uf = (1./u);## in W/m^2-K\n",
      "print'%s %.2f %s'%('The final heat transfer coefficient due to fouling is ',math.ceil(Uf),' W/m^2-K \\n')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.4\n",
        "\n",
        "\n",
        "Page No. 93\n",
        "\n",
        "\n",
        "The original heat transfer coefficient is  2447.23  W/sq.m K \n",
        "\n",
        "The final heat transfer coefficient due to fouling is  1290.00  W/m^2-K \n",
        "\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex5-pg95"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.5\n",
      "print('Example 4.5\\n\\n');\n",
      "print('Page No. 95\\n\\n');\n",
      "\n",
      "## given\n",
      "m_h = 1.05;## Mass flow rate of hot liquid in kg/s\n",
      "Thi = 130.;## Inlet Temperature of hot liquid in degree celcius\n",
      "Tho = 30.;## Outlet Temperature of hot fluid in degree celcius\n",
      "Cph =  2.45*10**3;## Specific heat capacity of hot liquid in J/kg-K\n",
      "\n",
      "m_c = 4.10;## Mass flow rate of cold liquid in kg/s\n",
      "Tci =  20.;## Inlet Temperature of cold liquid in degree celcius\n",
      "Cpc =  4.18*10**3;## Specific heat capacity of cold liquid in J/kg-K\n",
      "\n",
      "A = 6.8;## Area of heat exchanger in m^2\n",
      "Q = m_h*Cph*(Thi-Tho);## in Watts\n",
      "\n",
      "##From heat balance\n",
      "## m_c*Cpc*(Tci-Tco)= m_h*Cph*(Thi-Tho)= UAlTm = Q\n",
      "Tco = ((Q/(m_c*Cpc))+Tci);\n",
      "print'%s %.2f %s'%(' The Outlet Temperature of cold fluid is ',Tco,' degree celcius\\n')\n",
      "## As counter flow  heat exchanger \n",
      "T1 = Thi-Tco;\n",
      "T2 = Tho-Tci;\n",
      "Tm = ((T1-T2)/math.log(T1/T2));\n",
      "\n",
      "U = (Q/(A*Tm));\n",
      "print'%s %.2f %s'%('The overall heat transfer coefficient is ',U,' W/sq.m K \\n')## Deviation in answer due to direct substitution\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.5\n",
        "\n",
        "\n",
        "Page No. 95\n",
        "\n",
        "\n",
        " The Outlet Temperature of cold fluid is  35.01  degree celcius\n",
        "\n",
        "The overall heat transfer coefficient is  1002.06  W/sq.m K \n",
        "\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex6-pg98"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.6\n",
      "print('Example 4.6\\n\\n');\n",
      "print('Page No. 98\\n\\n');\n",
      "\n",
      "## given\n",
      "v = 1.23;## velocity in m/s\n",
      "d = 25.*10**-3;## diameter in m\n",
      "p = 980.;## density in kg/m^3\n",
      "u = 0.502*10**-3;## viscosity in Ns/m^2\n",
      "Cp = 3.76*10**3;## Specific heat capacity in J/kg-K\n",
      "K = 0.532;## Thermal conductivity in W/m-K\n",
      "\n",
      "Re = (d*v*p)/u;##Reynolds Number\n",
      "Pr = (Cp*u)/K;## Prandtl Number\n",
      "Re_d = (Re)**0.8;\n",
      "Pr_d = (Pr)**0.4;\n",
      "\n",
      "## By Dittus-Boelter Equation\n",
      "##Nu = 0.0232 * Re^0.8 Pr^0.4 = (hd)/K\n",
      "Nu = 0.0232 * Re_d * Pr_d;## Nusselt Number\n",
      "h = (Nu*K)/d;##W/m^2-K\n",
      "print'%s %.2f %s'%('The film heat transfer coefficient is ',h,' W/sq.m K\\n')## Deviation in answer due to direct substitution\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.6\n",
        "\n",
        "\n",
        "Page No. 98\n",
        "\n",
        "\n",
        "The film heat transfer coefficient is  5446.85  W/sq.m K\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex7-pg99"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.7\n",
      "print('Example 4.7\\n\\n');\n",
      "print('Page No. 99\\n\\n');\n",
      "\n",
      "## (a) without insulation\n",
      "## given\n",
      "d_a = 0.150;## Diameter of pipe in m\n",
      "T1_a = 60.;## Surface temperature in degree celcius\n",
      "T2_a = 10.;## Ambient temperature in degree celcius\n",
      "\n",
      "##For laminar flow in pipe,h= 1.41*((T1-T2)/d)^0.25\n",
      "h_a = 1.41*((T1_a-T2_a)/d_a)**0.25;##W/m^2-K\n",
      "A_a = math.pi * d_a;## Surface Area per unit length in m^2/m\n",
      "Q_a = h_a*A_a*(T1_a - T2_a);## in W/m\n",
      "print'%s %.2f %s'%('The heat loss per unit length without insulation is ',math.ceil(Q_a),' W/m \\n')\n",
      "\n",
      "## (b) with insulation\n",
      "## given\n",
      "d_b = 0.200;## Diameter of pipe in m\n",
      "T1_b = 20.;## Surface temperature in degree celcius\n",
      "T2_b = 10.;## Ambient temperature in degree celcius\n",
      "\n",
      "##For laminar flow in pipe,h= 1.41*((T1-T2)/d)^0.25\n",
      "h_b = 1.41*((T1_b-T2_b)/d_b)**0.25;##W/m^2-K\n",
      "A_b = math.pi * d_b;## Surface Area per unit length in m^2/m\n",
      "Q_b = h_b*A_b*(T1_b - T2_b);## in W/m\n",
      "print'%s %.2f %s'%('the heat loss per unit length with insulation is ',Q_b,' W/m')\n",
      "## Deviation in answer due to direct substitution\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.7\n",
        "\n",
        "\n",
        "Page No. 99\n",
        "\n",
        "\n",
        "The heat loss per unit length without insulation is  142.00  W/m \n",
        "\n",
        "the heat loss per unit length with insulation is  23.56  W/m\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex8-pg103"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.8\n",
      "print('Example 4.8\\n\\n');\n",
      "print('Page No. 103\\n\\n');\n",
      "\n",
      "## given\n",
      "d = 0.100;## Diameter of pipe in m\n",
      "T1 = 383.;## Surface temperature in Kelvin\n",
      "T2 = 288.;## Surrounding air temperature in Kelvin\n",
      "e = 0.9;## Emissivity of pipe\n",
      "A = math.pi * d;## Surface Area per unit length in m^2/m\n",
      "\n",
      "## By Stefan-Blotzmann law, the radiative heat transfer rate is   Q = 5.669*e*A*((T1/100)^4-(T2/100)^4)\n",
      "Q = 5.669*e*A*((T1/100.)**4-(T2/100.)**4);## in W/m\n",
      "print'%s %.2f %s'%('The radiative heat loss per unit length is ',math.ceil(Q),' W/sq.m')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.8\n",
        "\n",
        "\n",
        "Page No. 103\n",
        "\n",
        "\n",
        "The radiative heat loss per unit length is  235.00  W/sq.m\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex9-pg103"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.9\n",
      "print('Example 4.9\\n\\n');\n",
      "print('Page No. 103\\n\\n');\n",
      "\n",
      "## given\n",
      "A = 1.;## Area in m^2\n",
      "T1 = 423.;## Surface temperature in Kelvin\n",
      "T2 = 293.;## Surrounding air temperature in Kelvin\n",
      "T1_c = 150.;## Surface temperature in degree celcius\n",
      "T2_c = 20.;## Ambient temperature in degree celcius\n",
      "e = 0.9;## Emissivity of pipe\n",
      "\n",
      "##(a) Horizontal Pipe\n",
      "d = 0.100;## Diameter of pipe in m\n",
      "##For laminar flow in pipe,Q= (1.41*((T1-T2)/d)^0.25)*(T1-T2)\n",
      "Q_Ca = (1.41*((T1_c-T2_c)/d)**0.25)*(T1_c-T2_c);## Convective heat transfer rate in W/m^2\n",
      "## By Stefan-Blotzmann law, the radiative heat transfer rate is   Q = 5.669*e*((T1/100)^4-(T2/100)^4)\n",
      "Q_Ra = 5.669*e*((T1/100.)**4-(T2/100.)**4);## in W/m^2\n",
      "Q_Ta = Q_Ra + Q_Ca;## IN W/m^2\n",
      "print'%s %.2f %s'%('The total heat loss from per square meter area is ',Q_Ta,' W/sq.m\\n')## Deviation in answer due to direct substitution\n",
      "\n",
      "\n",
      "##(b) Vertical Pipe\n",
      "##For turbulent flow in pipe,Q= (1.24*(T1-T2)^1.33)\n",
      "Q_Cb = (1.24*(T1-T2)**1.33);## Convective heat transfer rate in W/m^2\n",
      "## By Stefan-Blotzmann law, the radiative heat transfer rate is   Q = 5.669*e*((T1/100)^4-(T2/100)^4)\n",
      "Q_Rb = 5.669*e*((T1/100.)**4-(T2/100.)**4);## in W/m^2\n",
      "Q_Tb = Q_Rb + Q_Cb;## IN W/m^2\n",
      "print'%s %.2f %s'%('The total heat loss from per square meter area is ',math.floor(Q_Tb),' W/sq.m\\n')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.9\n",
        "\n",
        "\n",
        "Page No. 103\n",
        "\n",
        "\n",
        "The total heat loss from per square meter area is  2358.09  W/sq.m\n",
        "\n",
        "The total heat loss from per square meter area is  2060.00  W/sq.m\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10-pg106"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 4.10\n",
      "print('Example 4.10\\n\\n');\n",
      "print('Page No. 106\\n\\n');\n",
      "\n",
      "## given\n",
      "T1 = 150.;## Surface temperature in degree celcius\n",
      "T2 = 20.;## Ambient temperature in degree celcius\n",
      "d = 0.100; ##Outside diametr of pipe in m\n",
      "h = 10.;## Outside film coefficient in W/m^2-K\n",
      "t = 25.*10**-3;## thickness of insulation in m\n",
      "K = 0.040;## Thermal conductivity of insulation in W/m-K\n",
      "\n",
      "r2 = d/2.;##in m\n",
      "r1 = r2+t;## in m\n",
      "Q = ((T1-T2)/((1./(2.*math.pi*r1*h))+(math.log(r1/r2)/(2.*math.pi*K))));## in W/m\n",
      "print'%s %.2f %s'%('The heat loss per unit length is ',Q,' W/m')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 4.10\n",
        "\n",
        "\n",
        "Page No. 106\n",
        "\n",
        "\n",
        "The heat loss per unit length is  71.21  W/m\n"
       ]
      }
     ],
     "prompt_number": 13
    }
   ],
   "metadata": {}
  }
 ]
}