path: root/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter9_1.ipynb

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   "source": [
    "# Chapter 9 - Heat Transfer"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 1: pg 256"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
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   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 9.1\n",
      " The heat lost per hour is (kJ) =  10815.0\n",
      " The interface temperature is (C) =  8.2\n"
     ]
    }
   ],
   "source": [
    "#pg 256\n",
    "print('Example 9.1');\n",
    "\n",
    "#  aim : To determine \n",
    "#  the heat loss per hour through the wall and interface temperature\n",
    "\n",
    "#  Given values\n",
    "x1 = .25;# thickness of brick,[m]\n",
    "x2 = .05;# thickness of concrete,[m]\n",
    "t1 = 30.;# brick face temperature,[C]\n",
    "t3 = 5.;# concrete face temperature,[C]\n",
    "l = 10.;# length of the wall, [m]\n",
    "h = 5.;# height of the wall, [m]\n",
    "k1 = .69;# thermal conductivity of brick,[W/m/K]\n",
    "k2 = .93;# thermal conductivity of concrete,[W/m/K]\n",
    "\n",
    "#  solution\n",
    "A = l*h;# area of heat transfer,[m**2]\n",
    "Q_dot = A*(t1-t3)/(x1/k1+x2/k2);# heat transferred, [J/s]\n",
    "\n",
    "#  so heat loss per hour is\n",
    "Q = Q_dot*3600*10**-3;# [kJ]\n",
    "print ' The heat lost per hour is (kJ) = ',round(Q)\n",
    "\n",
    "#  interface temperature calculation\n",
    "#   for  the brick wall, Q_dot=k1*A*(t1-t2)/x1;\n",
    "#  hence\n",
    "t2 = t1-Q_dot*x1/k1/A;# [C]\n",
    "print ' The interface temperature is (C) = ',round(t2,1)\n",
    "\n",
    "#  End\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2: pg 258"
   ]
  },
  {
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   "execution_count": 2,
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   "outputs": [
    {
     "name": "stdout",
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     "text": [
      "Example 9.2\n",
      " The minimum thickness of the lagging required is (mm) =  38.8\n"
     ]
    }
   ],
   "source": [
    "#pg 258\n",
    "print('Example 9.2');\n",
    "\n",
    "#  aim : To determine\n",
    "#  the minimum \n",
    "#  thickness of the lagging required\n",
    "import math\n",
    "#  Given values\n",
    "r1 = 75./2;# external radious of the pipe,[mm]\n",
    "L = 80.;# length of the pipe,[m]\n",
    "m_dot = 1000.;# flow of steam, [kg/h]\n",
    "P = 2.;# pressure, [MN/m**2]\n",
    "x1 = .98;# inlet dryness fraction\n",
    "x2 = .96;# outlet dryness fraction\n",
    "k = .08;# thermal conductivity of of pipe, [W/m/K]\n",
    "t2 = 27.;# outside temperature,[C]\n",
    "\n",
    "#  solution\n",
    "#  using steam table  at 2 MN/m**2 the enthalpy of evaporation of steam is,\n",
    "hfg = 1888.6;# [kJ/kg]\n",
    "#  so heat loss through the pipe is\n",
    "Q_dot = m_dot*(x1-x2)*hfg/3600;# [kJ]\n",
    "\n",
    "# also from steam table saturation temperature of steam at 2 MN/m**2 is,\n",
    "t1 = 212.4;# [C]\n",
    "# and for thick pipe, Q_dot=k*2*%pi*L*(t1-t2)/log(r2/r1)\n",
    "# hence\n",
    "r2 = r1*math.exp(k*2*math.pi*L*(t1-t2)*10**-3/Q_dot);# [mm]\n",
    "\n",
    "t = r2-r1;# thickness, [mm]\n",
    "#results\n",
    "print ' The minimum thickness of the lagging required is (mm) = ',round(t,1)\n",
    "\n",
    "#  End\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 3: pg 260"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
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   "outputs": [
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     "text": [
      "Example 9.3\n",
      " (a) The heat lost per hour is (kJ) =  8770.0\n",
      " (b) The interface temperature of the lagging is (C) =  71.5\n",
      "There is some rounding off error in the book, so answer is not matching\n"
     ]
    }
   ],
   "source": [
    "#pg 260\n",
    "print('Example 9.3');\n",
    "\n",
    "#  aim : To determine the\n",
    "#  (a) heat loss per hour\n",
    "#   (b) interface temperature og lagging\n",
    "import math\n",
    "# Given values\n",
    "r1 = 50.; # radious of steam main,[mm]\n",
    "r2 = 90.;# radious with first lagging,[mm]\n",
    "r3 = 115.;# outside radious os steam main with lagging,[mm]\n",
    "k1 = .07;# thermal conductivity of 1st lagging,[W/m/K]\n",
    "k2 = .1;# thermal conductivity of 2nd lagging, [W/m/K]\n",
    "P = 1.7;# steam pressure,[MN/m^2]\n",
    "t_superheat = 30.;# superheat of steam, [K]\n",
    "t3 = 24.;# outside temperature of the lagging,[C]\n",
    "L = 20.;# length of the steam main,[m]\n",
    "\n",
    "#  solution\n",
    "#  (a)\n",
    "#  using steam table saturation temperature of steam at 1.7 MN/m^2 is\n",
    "t_sat = 204.3;# [C]\n",
    "# hence\n",
    "t1 = t_sat+t_superheat;# temperature of steam,[C]\n",
    "\n",
    "Q_dot = 2*math.pi*L*(t1-t3)/(math.log(r2/r1)/k1+math.log(r3/r2)/k2);# heat loss,[W]\n",
    "#  heat loss in hour is\n",
    "Q = Q_dot*3600*10**-3;# [kJ]\n",
    "\n",
    "print ' (a) The heat lost per hour is (kJ) = ',round(Q)\n",
    "\n",
    "# (b)\n",
    "#  using Q_dot=2*%pi*k1*(t1-t1)/log(r2/r1) \n",
    "t2 = t1-Q_dot*math.log(r2/r1)/(2*math.pi*k1*L);# interface temperature of lagging,[C]\n",
    "\n",
    "print ' (b) The interface temperature of the lagging is (C) = ',round(t2,1)\n",
    "\n",
    "print 'There is some rounding off error in the book, so answer is not matching'\n",
    "\n",
    "#  End\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4: pg 265"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 9.4\n",
      " The energy emitted from the surface is (kW) =  355.7\n"
     ]
    }
   ],
   "source": [
    "#pg 265\n",
    "print('Example 9.4');\n",
    "\n",
    "# aim : To determine \n",
    "#  the energy emetted from the surface\n",
    "\n",
    "#  Given values\n",
    "h = 3.;# height of surface, [m]\n",
    "b = 4.;# width of surface, [m]\n",
    "epsilon_s = .9;# emissivity of the surface\n",
    "T = 273.+600;# surface temperature ,[K]\n",
    "sigma = 5.67*10**-8;# [W/m^2/K^4]\n",
    "\n",
    "#  solution\n",
    "As = h*b;# area of the surface, [m^2]\n",
    "\n",
    "Q_dot = epsilon_s*sigma*As*T**4*10**-3;# energy emitted, [kW]\n",
    "#results\n",
    "print ' The energy emitted from the surface is (kW) = ',round(Q_dot,1)\n",
    "\n",
    "#  End\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5: pg 265"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 9.5\n",
      " The transfer of energy will be from furnace to sphere and transfer rate is (kW) =  703.0\n",
      " There is some calculation mistake in the book, so answer is not matching\n"
     ]
    }
   ],
   "source": [
    "#pg 265\n",
    "print('Example 9.5');\n",
    "\n",
    "#  aim : To determine \n",
    "#  the rate of energy transfer between furnace and the sphere and its direction\n",
    "import math\n",
    "#  Given values\n",
    "l = 1.25;# internal side of cubical furnace, [m]\n",
    "ti = 800.+273;# internal surface temperature of the furnace,[K]\n",
    "r = .2;# sphere radious, [m]\n",
    "epsilon = .6;# emissivity of sphere\n",
    "ts = 300.+273;# surface temperature of sphere, [K]\n",
    "sigma = 5.67*10**-8;# [W/m**2/K**4]\n",
    "\n",
    "#  Solution\n",
    "Af = 6*l**2;# internal surface area of furnace, [m**2]\n",
    "As =4 *math.pi*r**2;# surface area of sphere, [m**2]\n",
    "\n",
    "#  considering internal furnace to be black\n",
    "Qf = sigma*Af*ti**4*10**-3;# [kW]\n",
    "\n",
    "#  radiation emitted by sphere is\n",
    "Qs = epsilon*sigma*As*ts**4*10**-3; # [kW]\n",
    "\n",
    "#  Hence transfer of energy is\n",
    "Q = Qf-Qs;# [kW]\n",
    "#results\n",
    "print ' The transfer of energy will be from furnace to sphere and transfer rate is (kW) = ',round(Q)\n",
    "print' There is some calculation mistake in the book, so answer is not matching'\n",
    "\n",
    "#  End\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6: pg 271"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 9.6\n",
      " The overall heat transfer coefficient for the wall is (W/m**2 K) =  0.313\n",
      " The heat loss per hour through the wall is (kJ) =  1148.0\n"
     ]
    }
   ],
   "source": [
    "#pg 271\n",
    "print('Example 9.6');\n",
    "\n",
    "#  aim : To determine\n",
    "#  the overall transfer coefficient and the heat loss per hour\n",
    "\n",
    "#  Given values\n",
    "x1 = 25*10**-3;# Thickness of insulating board, [m]\n",
    "x2 = 75*10**-3;# Thickness of fibreglass, [m]\n",
    "x3 = 110*10**-3;# Thickness of brickwork, [m]\n",
    "k1 = .06;# Thermal conductivity of insulating board, [W/m K]\n",
    "k2 = .04;# Thermal conductivity of fibreglass, [W/m K]\n",
    "k3 = .6;# Thermal conductivity of brickwork, [W/m K]\n",
    "Us1 = 2.5;#  surface heat transfer coefficient of the inside wall,[W/m**2 K]\n",
    "Us2 = 3.1;#  surface heat transfer coefficient of the outside wall,[W/m**2 K]\n",
    "ta1 = 27.;# internal ambient temperature, [C]\n",
    "ta2 = 10.;# external ambient temperature, [C]\n",
    "h = 6.;# height of the wall, [m]\n",
    "l = 10.;# length of the wall, [m]\n",
    "\n",
    "#  solution\n",
    "U = 1/(1/Us1+x1/k1+x2/k2+x3/k3+1/Us2);# overall heta transfer coefficient,[W/m**2 K]\n",
    "\n",
    "A = l*h;# area ,[m**2]\n",
    "\n",
    "Q_dot = U*A*(ta1-ta2);# heat loss [W]\n",
    "\n",
    "#  so heat loss per hour is\n",
    "Q = Q_dot*3600*10**-3;# [kJ]\n",
    "#results\n",
    "print ' The overall heat transfer coefficient for the wall is (W/m**2 K) = ',round(U,3)\n",
    "print ' The heat loss per hour through the wall is (kJ) = ',round(Q)\n",
    "\n",
    "#  End\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 7: pg 272"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example 9.7\n",
      " The heat loss per hour is (kJ) =  24533.0\n",
      " The surface temperature of the lagging is (C) =  46.99\n",
      "there is minor variation in the answer due to rounding off error in textbook\n"
     ]
    }
   ],
   "source": [
    "#pg 272\n",
    "print('Example 9.7');\n",
    "\n",
    "#  aim : To determine  \n",
    "#  the heat loss per hour and the surface temperature of the lagging\n",
    "import math\n",
    "#  Given values\n",
    "r1 = 75.*10**-3;# External radiou of the pipe, [m]\n",
    "t_l1 = 40.*10**-3;# Thickness of lagging1, [m]\n",
    "t_l2 = t_l1;\n",
    "k1 = .07;# thermal conductivity of lagging1, [W/m K]\n",
    "k2 = .1;# thermal conductivity of lagging2, [W/m K]\n",
    "Us = 7;# surface transfer coefficient for outer surface, [W/m**2 K]\n",
    "L = 50.;# length of the pipe, [m]\n",
    "ta = 27.;# ambient temperature, [C]\n",
    "P = 3.6;# wet steam pressure, [MN/m**2]\n",
    "\n",
    "#  solution\n",
    "#  from steam table saturation temperature of the steam at given pressure is,\n",
    "t1 =  244.2;# [C]\n",
    "r2 = r1+t_l1;# radious of pipe with lagging1,[m]\n",
    "r3 = r2+t_l2;# radious of pipe with both the lagging, [m]\n",
    "\n",
    "R1 = math.log(r2/r1)/(2*math.pi*L*k1);# resistance due to lagging1,[C/W]\n",
    "R2 = math.log(r3/r2)/(2*math.pi*L*k2);# resistance due to lagging2,[C/W]\n",
    "R3 = 1/(Us*2*math.pi*r3*L);# ambient resistance, [C/W]\n",
    "\n",
    "#  hence overall resistance is,\n",
    "Req = R1+R2+R3;# [C/W]\n",
    "tdf = t1-ta;# temperature driving force, [C]\n",
    "Q_dot = tdf/Req;# rate of heat loss, [W]\n",
    "#  so heat loss per hour is,\n",
    "Q = Q_dot*3600*10**-3;# heat loss per hour, [kJ]\n",
    "\n",
    "#  using eqn [3]\n",
    "t3 = ta+Q_dot*R3;# surface temperature of the lagging, [C]\n",
    "#results\n",
    "print ' The heat loss per hour is (kJ) = ',round(Q,0)\n",
    "print ' The surface temperature of the lagging is (C) = ',round(t3,2)\n",
    "\n",
    "print 'there is minor variation in the answer due to rounding off error in textbook'\n",
    "\n",
    "#  End\n"
   ]
  }
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