path: root/Introduction_to_Heat_Transfer_by_S._K._Som/Chapter7.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {
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   },
   "source": [
    "# Chapter 07:Principles of forced convection"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.1:pg-296"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 1\n",
      "First we check from reynolds no. that the flow is laminar or tubulent\n",
      "Reynold number is\n",
      "Re= 20000.0\n",
      "which is less than critical reynolds number,So the flow is laminar.\n",
      "The average nusselt number over the entire length under the situation is given by NuL=0.664*Re**0.5*Pr**(1/3)\n",
      "NuL= 93.9037805416\n",
      "Heat flux in W/(m**2*K) is\n",
      "h= 2.72320963571\n",
      "The rate of heat transfer per unit width in W is\n",
      "Q= 408.481445356\n"
     ]
    }
   ],
   "source": [
    " \n",
    "import math \n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 1\"\n",
    "#Engine oil at temprature,Tinf=60°C with a velocity of Uinf=1m/s flows over plate of length(L)=5m whose temprature(Tw)=30°C\n",
    "Tw=30;\n",
    "L=5;\n",
    "Uinf=1;\n",
    "Tinf=60;\n",
    "#The properties at a film temprature of 45°C are as follows density(rho=870kg/m**3),Prandtl number(Pr=2850),conductivity(k=0.145W/(m*°C)),kinematic viscosity(nu=250*10**-6m**2/s).\n",
    "rho=870;\n",
    "Pr=2850;\n",
    "k=0.145;\n",
    "nu=250*10**-6;\n",
    "print\"First we check from reynolds no. that the flow is laminar or tubulent\"\n",
    "#Reynolds number is given by Re=(Uinf*L)/nu\n",
    "print\"Reynold number is\"\n",
    "Re=(Uinf*L)/nu\n",
    "print\"Re=\",Re\n",
    "print\"which is less than critical reynolds number,So the flow is laminar.\"\n",
    "#NuL is the average nusselt number\n",
    "print\"The average nusselt number over the entire length under the situation is given by NuL=0.664*Re**0.5*Pr**(1/3)\"\n",
    "NuL=0.664*Re**0.5*Pr**(1/3)\n",
    "print\"NuL=\",NuL\n",
    "#Heat flux is given by h=(k/L)*NuL\n",
    "print\"Heat flux in W/(m**2*K) is\"\n",
    "h=(k/L)*NuL\n",
    "print\"h=\",h\n",
    "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n",
    "#Since unit width is considerd so B=1\n",
    "#Area(A)=L*B\n",
    "B=1;\n",
    "A=L*B;\n",
    "print\"The rate of heat transfer per unit width in W is\"\n",
    "Q=h*A*(Tinf-Tw)\n",
    "print\"Q=\",Q"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.2:pg-298"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 2\n",
      "The location x in m where the transition occurs\n",
      "x= 0.275\n",
      "The average Nusselt number for the laminar zone is\n",
      "Nux= 469.518902708\n",
      "Heat flux in W/(m**2*K) is\n",
      "h= 44.3908780742\n",
      "The reynolds number at L=2m is\n",
      "ReL= 3636363.63636\n",
      "The average heat transfer coefficient over L=2m in W/(m**2*K)\n",
      "hbarL= -11.322519\n",
      "The rate of heat transfer per unit width in W is\n",
      "Q= -2264.5038\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 2\"\n",
    "#Atmospheric air at temprature,Tinf=300K and with a free stream Velocity Uinf=30m/s flows over a flat plate parallel to a side of length(L)=2m.\n",
    "Tinf=300;\n",
    "Uinf=30;\n",
    "L=2;\n",
    "#It is maintained at a uniform temprature of Tw=400K.\n",
    "Tw=400;\n",
    "#The properties of air at the film temprature of 350K are Prandtl number(Pr=0.705),conductivity(k=0.026W/(m*°C)),kinematic viscosity(nu=16.5*10**-6m**2/s)\n",
    "Pr=0.705; \n",
    "k=0.026;\n",
    "nu=16.5*10**-6;\n",
    "#We first find the location x(for reynolds number,Re=5*10**5) where the transition occurs\n",
    "#Rex is reynolds number\n",
    "print\"The location x in m where the transition occurs\"\n",
    "Rex=5*10**5;\n",
    "x=(nu*Rex)/Uinf\n",
    "print\"x=\",x\n",
    "#The average Nusselt number for the laminar zone is given by Nux=0.664*Re**0.5*Pr**(1/3)\n",
    "print\"The average Nusselt number for the laminar zone is\"\n",
    "Nux=0.664*Rex**0.5*Pr**(1/3)\n",
    "print\"Nux=\",Nux\n",
    "#Heat flux is given by h=(k/x)*Nux\n",
    "print\"Heat flux in W/(m**2*K) is\"\n",
    "h=(k/x)*Nux\n",
    "print\"h=\",h\n",
    "#Reynolds number is given by ReL=(Uinf*L)/nu\n",
    "print\"The reynolds number at L=2m is\"\n",
    "ReL=(Uinf*L)/nu\n",
    "print\"ReL=\",ReL\n",
    "#The average heat transfer coefficient over L=2m is determined from hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n",
    "print\"The average heat transfer coefficient over L=2m in W/(m**2*K)\"\n",
    "hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n",
    "print\"hbarL=\",hbarL\n",
    "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n",
    "#Since unit width is considerd so B=1\n",
    "#Area(A)=L*B\n",
    "B=1;\n",
    "A=L*B;\n",
    "print\"The rate of heat transfer per unit width in W is\"\n",
    "Q=hbarL*A*(Tw-Tinf)\n",
    "print\"Q=\",Q\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.3:pg-314"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 3\n",
      "(a)When the air flows parallel to the long side we have L=5 and the Reynolds no. becomes\n",
      "ReL= 1250000.0\n",
      "which is greater than critical Reynolds number.\n",
      "The average heat transfer coefficient over L=5m in W/(m**2*K)\n",
      "hbarL= -5.225778\n",
      "The rate of heat transfer per unit width in W is\n",
      "Q= -3135.4668\n",
      "(b)When the air flow is parallel to the 1m side we have L=1 an the Reynolds no. becomes \n",
      "which is less than critical Reynolds number.\n",
      "ReL= 250000.0\n",
      "Heat flux in W/(m**2*K) is\n",
      "h= 9.96\n",
      "The rate of heat transfer per unit width in W is\n",
      "Q= 5976.0\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 3\"\n",
    "#Air at a pressure of 101kPa and temprature,Tinf=20°C flows with a velocity(Uinf) of 5m/s over a flat plate whose temprature is kept constant at Tw=140°C.\n",
    "Tw=140;\n",
    "Tinf=20;\n",
    "Uinf=5;\n",
    "#The properties at the film temprature of 80°C are Prandtl number(Pr=0.706),Conductivity(k=0.03W/(m*°C)),kinematic viscosity(nu=2*10**-5m**2/s)\n",
    "Pr=0.706;\n",
    "k=0.03;\n",
    "nu=2*10**-5;\n",
    "#ReL is reynolds number and L is length of flat plate\n",
    "print\"(a)When the air flows parallel to the long side we have L=5 and the Reynolds no. becomes\"\n",
    "L=5;\n",
    "ReL=(Uinf*L)/nu\n",
    "print\"ReL=\",ReL\n",
    "print\"which is greater than critical Reynolds number.\"\n",
    "#Thus we have combined laminar and tubulent flow.\n",
    "# So The average heat transfer coefficient over L=5m is determined from hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n",
    "print\"The average heat transfer coefficient over L=5m in W/(m**2*K)\"\n",
    "hbarL=(k/L)*(0.037*(ReL)**(4/5)-871)*Pr**(1/3)\n",
    "print\"hbarL=\",hbarL\n",
    "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n",
    "#Since width is 1m so B=1\n",
    "#Area(A)=L*B\n",
    "B=1;\n",
    "A=L*B;\n",
    "#Q is the rate of heat transfer\n",
    "print\"The rate of heat transfer per unit width in W is\"\n",
    "Q=hbarL*A*(Tw-Tinf)\n",
    "print\"Q=\",Q\n",
    "#When the air flow is parallel to the 1m side we have L=1\n",
    "print\"(b)When the air flow is parallel to the 1m side we have L=1 an the Reynolds no. becomes \"\n",
    "L=1;\n",
    "ReL=(Uinf*L)/nu\n",
    "print\"which is less than critical Reynolds number.\"\n",
    "print\"ReL=\",ReL\n",
    "#Thus we have laminar flow\n",
    "#Heat flux is given by h=(k/L)*0.664*ReL**0.5*Pr**(1/3)\n",
    "print\"Heat flux in W/(m**2*K) is\"\n",
    "h=(k/L)*0.664*ReL**0.5*Pr**(1/3)\n",
    "print\"h=\",h\n",
    "#The rate of heat transfer per unit width is Q=h*A*(Tinf-Tw)\n",
    "#Now width is 5m so B=5\n",
    "#Area(A)=L*B\n",
    "B=5;\n",
    "A=L*B;\n",
    "#Q is the rate of heat transfer\n",
    "print\"The rate of heat transfer per unit width in W is\"\n",
    "Q=h*A*(Tw-Tinf)\n",
    "print\"Q=\",Q\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.4:pg-322"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 4\n",
      "(a)Reynolds number is\n",
      "ReL= 6000.0\n",
      "The boundary layer thickness in m is\n",
      "delta= 0.387298334621\n",
      "Prandtl no. is\n",
      "Pr= 831.024930748\n",
      "The thermal boundary layer thickness in m is\n",
      "deltaT= 0.387298334621\n",
      "(b)Since the prandtl number is high So Nusselt no. is\n",
      "NuL= 26.2588270873\n",
      "Heat flux in W/(m**2*K) is\n",
      "hL= 0.919058948055\n",
      "hbarL in W/(m**2*K) is\n",
      "hbarL= 1.83811789611\n",
      "(c)The rate of heat transfer in W is\n",
      "Q= 661.7224426\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 4\"\n",
    "#Castor oil at temprature,Tinf=36°C flows over a heated plate of length,L=6m and breadth,B=1m at velocity,Uinf=0.06m/s\n",
    "Tinf=36;\n",
    "L=6;\n",
    "B=1;\n",
    "Uinf=0.06;\n",
    "#For a surface temprature at Tw=96°C\n",
    "Tw=96;\n",
    "#The properties at film temprature 66°C conductivity(k=0.21W/(m*K)),kinematic viscosity(nu=6*10**-5m**2/s),Thermal diffusivity(alpha=7.22*10**-8 m**2/s)\n",
    "nu=6*10**-5;\n",
    "k=0.21;\n",
    "alpha=7.22*10**-8;\n",
    "#ReL is reynolds number\n",
    "print\"(a)Reynolds number is\"\n",
    "ReL=(Uinf*L)/nu\n",
    "print\"ReL=\",ReL\n",
    "#Therefore the boundary layer is laminar over the entire plate.\n",
    "#delta is the boundary layer thickness\n",
    "print\"The boundary layer thickness in m is\"\n",
    "delta=(5*L)/(ReL)**0.5\n",
    "print\"delta=\",delta\n",
    "#Pr is prandtl number.\n",
    "print\"Prandtl no. is\"\n",
    "Pr=nu/alpha\n",
    "print\"Pr=\",Pr\n",
    "#deltaT is thermal boundary layer thickness\n",
    "print\"The thermal boundary layer thickness in m is\"\n",
    "deltaT=delta/(Pr**(1/3))#NOTE:Answer in the book is incorrect(calculation mistake)\n",
    "print\"deltaT=\",deltaT\n",
    "#NuL is the nusselt number\n",
    "print\"(b)Since the prandtl number is high So Nusselt no. is\"\n",
    "NuL=0.339*(ReL)**0.5*Pr**(1/3)\n",
    "print\"NuL=\",NuL\n",
    "#Heat flux is given by hL=(k/L)*NuL\n",
    "print\"Heat flux in W/(m**2*K) is\"\n",
    "hL=(k/L)*NuL\n",
    "print\"hL=\",hL\n",
    "#hbarL is the average heat flux over length L\n",
    "print\"hbarL in W/(m**2*K) is\"\n",
    "hbarL=2*hL\n",
    "print\"hbarL=\",hbarL\n",
    "#The rate of heat transfer is Q=h*A*(Tinf-Tw)\n",
    "#Area(A)=L*B\n",
    "A=L*B;\n",
    "print\"(c)The rate of heat transfer in W is\"\n",
    "Q=hbarL*A*(Tw-Tinf)\n",
    "print\"Q=\",Q\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.5:pg-322"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 5\n",
      "Reynolds number is\n",
      "ReL= 11181.8181818\n",
      "Therefore the flow is turbulent over the module \n",
      "The local heat transfer coefficient at L in W/(m**2*K)is\n",
      "hL= 8.32911955901e+30\n",
      "The required power generation in W/m**3 is\n",
      "qm= 1.6658239118e+31\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 5\"\n",
    "#A flat plate of width B=1m is maintained at a uniform surface temprtaure(Tw)=225°C\n",
    "Tw=225;\n",
    "B=1;\n",
    "#Heating is done by rectangular modules of thickness t=10mm and length l=40mm.\n",
    "t=10;\n",
    "l=40;\n",
    "#atmospheric air at temprature,Tinf=25°C flows over the plate at velocity(Uinf)=30m/s.\n",
    "Tinf=25;\n",
    "Uinf=30;\n",
    "#The thermophysical properties of module are conductivity(km=5.2W/(m*K)),specific heat(cp=320J/(kg/K)),density(rho=2300kg/m**3).\n",
    "km=5.2;\n",
    "cp=320;\n",
    "rho=2300;\n",
    "#Assume the air properties at the film temprature of 125°C conductivity(ka=0.031W/(m*K)),kinematic viscosity(nu=22*10**-6m**2/s),Prandtl number(Pr=0.7)\n",
    "ka=0.031;\n",
    "nu=22*10**-6;\n",
    "Pr=0.7;\n",
    "#Module is placed at a distance of 800mm from the leading edge\n",
    "#The distance from leading edge to the centre-line of the module,L=800+20=820mm.\n",
    "L=0.0082;#in metre\n",
    "#ReL is the reynolds number \n",
    "print\"Reynolds number is\"\n",
    "ReL=(Uinf*L)/nu\n",
    "print\"ReL=\",ReL\n",
    "print\"Therefore the flow is turbulent over the module \"\n",
    "#The local heat transfer coefficient at L is calculated using hL=(k/L)*0.0296*(ReL)**(4/5)*(Pr)**(1/3)\n",
    "print\"The local heat transfer coefficient at L in W/(m**2*K)is\"\n",
    "hL=(ka/L)*0.0296*(ReL)**(4/0.5)*(Pr)**(1/0.3)\n",
    "print\"hL=\",hL\n",
    "#We consider that the local heat transfer coefficient at L=0.82m remains the same over the module which extends from L=0.80m to 0.84m \n",
    "#If qm be the power generation in W/m**2 within the module ,we can write from energy balance qm*(t/0.1000)*(l/0.1000)*(B)=hbarL*(t/0.1000)*(B)*(Tw-Tinf)\n",
    "print\"The required power generation in W/m**3 is\"\n",
    "qm=(hL*(l/0.1000)*(B)*(Tw-Tinf))/((t/0.1000)*(l/0.1000)*(B))\n",
    "print\"qm=\",qm\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.6:pg-327"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 6\n",
      "Reynolds number is\n",
      "ReL= 15000000.0\n",
      "Since ReL>Rec(=5*10**5) the flow is approximated as turbulent over the entire surface of the wing \n",
      "Nux= 0.0308\n",
      "Nusselt number is \n",
      "NubarL= 0.0308\n",
      "Average heat transfer coefficient in W/(m**2*K) is\n",
      "hbarL= 0.0003696\n",
      "Surface temprature of wing in kelvin is\n",
      "Tw= 1217800.46753\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    "\n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 6\"\n",
    "#An aircraft is moving at a velocity of Uinf=150m/s in air at an altitude where the pressure is 0.7bar and the temprature is Tinf=-5°C.\n",
    "Tinf=-5;\n",
    "Uinf=150;\n",
    "#The top surface of the wing absorbs solar radiation at a rate of Qr=900W/m**2.\n",
    "Qr=900;\n",
    "#Considering the wing as a flat plate of length(L)=2m and to be of solid construction with a single uniform surface temprature .\n",
    "L=2;\n",
    "#The properties of air at 268K and 0.7 bar are conductivity(k=0.024W/(m*K)),kinematic viscosity(nu=2*10**-5m**2/s),Prandtl number(Pr=0.72)\n",
    "k=0.024;\n",
    "nu=2*10**-5;\n",
    "Pr=0.72;\n",
    "#ReL is reynolds number\n",
    "print\"Reynolds number is\"\n",
    "ReL=Uinf*L/nu\n",
    "print\"ReL=\",ReL\n",
    "#Rec is critical reynolds number\n",
    "print\"Since ReL>Rec(=5*10**5) the flow is approximated as turbulent over the entire surface of the wing \"\n",
    "#Nusselt number is given by Nux=0.0308*ReL**(4/5)*Pr**(1/3)\n",
    "Nux=0.0308*ReL**(4/5)*Pr**(1/3);\n",
    "print\"Nux=\",Nux\n",
    "#NubarL is average nusselt number over length L\n",
    "print\"Nusselt number is \"\n",
    "NubarL=(5/4)*Nux\n",
    "print\"NubarL=\",NubarL\n",
    "#Average heat transfer coefficient is given by hbarL=(k/L)*NubarL\n",
    "print\"Average heat transfer coefficient in W/(m**2*K) is\"\n",
    "hbarL=(k/L)*NubarL\n",
    "print\"hbarL=\",hbarL\n",
    "#From an energy balance the airfoil at steady state,Qr*As=2*hbarL*As*(Tw-Tinf) where Qr=radiation flux,As=upper or lower surface area.\n",
    "#Therefore we can write Surface temprature of wing, Tw=Tinf+(Qr/(2*hbarL))\n",
    "print\"Surface temprature of wing in kelvin is\"\n",
    "Tw=(273+Tinf)+(Qr/(2*hbarL))\n",
    "print\"Tw=\",Tw\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.7:pg-331"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 7\n",
      "Reynolds number is\n",
      "Re= 141.176470588\n",
      "Nusselt number is\n",
      "NuD= 6.85819682626\n",
      "The average Heat transfer coefficient in W/(m**2*K) is\n",
      "hbar= 4629.28285773\n",
      "Heat transfer per unit length in W/m is\n",
      "qL= 14.5433210172\n",
      "If we use eq NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)\n",
      "NuD= 7.66669771975\n",
      "The average Heat transfer coefficient in W/(m**2*K) is\n",
      "hbar= 5175.02096083\n",
      "Heat transfer per unit length in W/m is\n",
      "qL= 16.2578078327\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 7\"\n",
    "#A fine wire having a diameter(D)=0.04mm is placed in an air stream at temprature,Tinf=25°C having a flow velocity of Uinf=60m/s perpendicular to wire.\n",
    "D=0.04;\n",
    "Tinf=25;\n",
    "Uinf=60;\n",
    "#An electric current is passed through the wire ,raising its surface temprature to Tw=50°C\n",
    "Tw=50;\n",
    "#For air at the film temprature of 37.5°C,conductivity(k=0.027 W/(m*K)),kinematic viscosity(nu=17*10**-6m**2/s) and Prandtl number(Pr=0.71)\n",
    "k=0.027;\n",
    "nu=17*10**-6;\n",
    "Pr=0.71;\n",
    "#Re is reynolds number\n",
    "print\"Reynolds number is\"\n",
    "Re=Uinf*(D*10**-3)/nu\n",
    "print\"Re=\",Re\n",
    "#C and n are constants\n",
    "#The values of C and n are found  for Re=141 are C=0.683 and n=0.466\n",
    "#NuD is nusselt number\n",
    "print\"Nusselt number is\"\n",
    "NuD=(0.683)*Re**0.466*Pr**(1/3)\n",
    "print\"NuD=\",NuD\n",
    "#hbar is the average Heat transfer coefficient\n",
    "print\"The average Heat transfer coefficient in W/(m**2*K) is\"\n",
    "hbar=(k/(D*10**-3))*NuD\n",
    "print\"hbar=\",hbar\n",
    "#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)\n",
    "print\"Heat transfer per unit length in W/m is\"\n",
    "qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)\n",
    "print\"qL=\",qL\n",
    "#NuD is nusselt number\n",
    "print\"If we use eq NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)\"\n",
    "NuD=0.3+((0.62*Re**0.5*Pr**(1/3))/(1+(0.4/Pr)**(2/3))**(1/4))*(1+(Re/282000)**(5/8))**(4/5)\n",
    "print\"NuD=\",NuD\n",
    "#hbar is the average Heat transfer coefficient\n",
    "print\"The average Heat transfer coefficient in W/(m**2*K) is\"\n",
    "hbar=(k/(D*10**-3))*NuD\n",
    "print\"hbar=\",hbar\n",
    "#Heat transfer per unit length(qL) is given by pi*D*hbar*(Tw-Tinf)\n",
    "print\"Heat transfer per unit length in W/m is\"\n",
    "qL=math.pi*(D*10**-3)*hbar*(Tw-Tinf)\n",
    "print\"qL=\",qL\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.8:pg-334"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 8\n",
      "Reynolds number for mercury is\n",
      "ReHg= 1000.0\n",
      "Reynolds number for oil is\n",
      "Reoil= 15.3846153846\n",
      "The hydrodynamic entry length for mercury in m is\n",
      "LeHg= 1.25\n",
      "The hydrodynamic entry length for oil in m is\n",
      "Leoil= 0.0192307692308\n",
      "The thermal entry length for mercury in m is \n",
      "LtHg= 0.02375\n",
      "The thermal entry length for oil in m is\n",
      "Ltoil= 1.63461538462\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    "\n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 8\"\n",
    "#Mercury and a light oil flowing at Uinf=4mm/s in a smooth tube having diameter(D)=25mm at a bulk temprature of 80°C.\n",
    "Uinf=4*10**-3;#in metre\n",
    "D=25*10**-3;#in metre\n",
    "#The pertinent properties of the fluid at that temprature are kinematic viscosity of mercury(nuHg=1*10**-7m**2/s),kinematic viscosity of oil(nuoil=6.5*10**-6m**2/s)\n",
    "#Prandtl number of mercury(PrHg=0.019),Prandtl number of oil(Proil=85).\n",
    "nuHg=1*10**-7;\n",
    "nuoil=6.5*10**-6;\n",
    "PrHg=0.019;\n",
    "Proil=85;\n",
    "#ReHg is Reynolds number for mercury\n",
    "print\"Reynolds number for mercury is\"\n",
    "ReHg=Uinf*D/nuHg\n",
    "print\"ReHg=\",ReHg\n",
    "#Reoil is Reynolds number for oil\n",
    "print\"Reynolds number for oil is\"\n",
    "Reoil=Uinf*D/nuoil\n",
    "print\"Reoil=\",Reoil\n",
    "#The hydrodynamic length are given by L=0.05*Re*D\n",
    "#LeHg is the hydrodynamic entry length for mercury\n",
    "print\"The hydrodynamic entry length for mercury in m is\"\n",
    "LeHg=0.05*ReHg*D\n",
    "print\"LeHg=\",LeHg\n",
    "#Leoil the hydrodynamic entry length for oil\n",
    "print\"The hydrodynamic entry length for oil in m is\"\n",
    "Leoil=0.05*Reoil*D\n",
    "print\"Leoil=\",Leoil\n",
    "#The thermal entry length are given by L=0.05*Re*Pr*D\n",
    "#LtHg is the thermal entry length for mercury\n",
    "print\"The thermal entry length for mercury in m is \"\n",
    "LtHg=0.05*ReHg*PrHg*D\n",
    "print\"LtHg=\",LtHg\n",
    "#Ltoil is the thermal entry length for oil\n",
    "print\"The thermal entry length for oil in m is\"\n",
    "Ltoil=0.05*Reoil*Proil*D\n",
    "print\"Ltoil=\",Ltoil\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.9:pg-336"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 9\n",
      "Reynold number is\n",
      "Re= 348.623853211\n",
      "Therefore the flow is laminar.The hydrodynamic entrance length in m is\n",
      "Leh= 0.0697247706422\n",
      "The thermal entrance length in m is\n",
      "Let= 0.0488073394495\n",
      "The heat transfer coefficient in W/(m**2*K) is \n",
      "h= 32.7\n",
      "The mass flow rate of air in kg/s is\n",
      "mdot= 2.38761041673e-05\n",
      "Therefore the constant surface heat flux qw in W/m**2 is\n",
      "qw= 95.95\n",
      "The tube surface temprature at the exit plane in °C is \n",
      "Twe= 127.934250765\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 9\"\n",
    "#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n",
    "Tbi=75;\n",
    "D=4*10**-3;#in metre\n",
    "U=2;\n",
    "#The tube length is L=1.0m and a constant heat flux is imposed by the tube surface on the air over the entire length.\n",
    "L=1;\n",
    "#An exit bulk mean temprature(Tbo)=125°C is required.\n",
    "Tbo=125;\n",
    "#The properties of air 100°C are density(rho=0.95kg/m**3),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*K)),viscosity(mu=2.18*10**-5kg/(m*s)),specific heat(cp=1.01kJ/(kg/K))\n",
    "rho=0.95;\n",
    "Pr=0.70;\n",
    "k=0.03;\n",
    "mu=2.18*10**-5;\n",
    "cp=1.01*10**3;\n",
    "#Re is reynolds number\n",
    "print\"Reynold number is\"\n",
    "Re=rho*U*D/mu\n",
    "print\"Re=\",Re\n",
    "#Leh is the hydrodynamic entrance length\n",
    "print\"Therefore the flow is laminar.The hydrodynamic entrance length in m is\"\n",
    "Leh=0.05*Re*D\n",
    "print\"Leh=\",Leh\n",
    "#Let is the thermal entrance length\n",
    "print\"The thermal entrance length in m is\"\n",
    "Let=0.05*Re*Pr*D\n",
    "print\"Let=\",Let\n",
    "#The length of tube is given as 1m.A reasonable approach is to consider the flow to be fully developed for both velocity and tempratures over the entire profile lengths.\n",
    "#For a fully developed flow with constant surface heat flux,Nusselt number is Nu=4.36\n",
    "Nu=4.36;\n",
    "#h is the heat transfer coefficient\n",
    "print\"The heat transfer coefficient in W/(m**2*K) is \"\n",
    "h=Nu*(k/D)\n",
    "print\"h=\",h\n",
    "#Here h=hL Since the heat transfer coefficient is constant over the entire length of tube.\n",
    "#hL is the local heat transfer coefficient\n",
    "hL=h;\n",
    "#from an energy balance qw*pi*D*L=mdot*cp*(Tbo-Tbi)\n",
    "#mdot is mass flow rate\n",
    "print\"The mass flow rate of air in kg/s is\"\n",
    "mdot=rho*(math.pi/4)*D**2*U\n",
    "print\"mdot=\",mdot\n",
    "#qw is the constant surface heat flux\n",
    "print\"Therefore the constant surface heat flux qw in W/m**2 is\"\n",
    "qw=(mdot*cp*(Tbo-Tbi))/(math.pi*D*L)\n",
    "print\"qw=\",qw\n",
    "#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n",
    "print\"The tube surface temprature at the exit plane in °C is \"\n",
    "Twe=Tbo+(qw/hL)\n",
    "print\"Twe=\",Twe\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.10:pg-338"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 7, Example 10\n",
      "Reynold number is\n",
      "Re= 348.623853211\n",
      "Therefore the flow is laminar.The hydrodynamic entrance length in m is\n",
      "Leh= 0.0697247706422\n",
      "The thermal entrance length in m is\n",
      "Let= 0.0488073394495\n",
      "The thermal entrance length is greater than the tube length Therefore the flow is hydrodynamically developed but not thermally developed\n",
      "The inverse of graetz number Gr_1 is\n",
      "Gr_1= 0.040977443609\n",
      "Therefore the local heat transfer coefficient in W/(m**2*K) is\n",
      "hL= 35.25\n",
      "The mass flow rate of air in kg/s is\n",
      "mdot= 2.38761041673e-05\n",
      "Therefore surafce heat flux qw in W/m**2 is\n",
      "qw= 2398.75\n",
      "The tube surface temprature at the exit plane in °C is \n",
      "Twe= 193.04964539\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 10\"\n",
    "#Air at one atmospheric pressure and temprature(Tbi=75°C) enters a tube of internal diameter(D)=4.0mm with average velocity(U)=2m/s\n",
    "Tbi=75;\n",
    "D=4*10**-3;\n",
    "U=2;\n",
    "#The heated tube length is L=0.04m and a constant heat flux is imposed by the tube surface on the air over the entire length.\n",
    "L=0.04;\n",
    "#An exit bulk mean temprature(Tbo)=125°C is required.\n",
    "Tbo=125;\n",
    "#The properties of air 100°C are density(rho=0.95kg/m**3),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*K)),viscosity(mu=2.18*10**-5kg/(m*s)),specific heat(cp=1.01kJ/(kg/K))\n",
    "rho=0.95;\n",
    "Pr=0.70;\n",
    "k=0.03;\n",
    "mu=2.18*10**-5;\n",
    "cp=1.01*10**3;\n",
    "#Re is the reynolds number \n",
    "print\"Reynold number is\"\n",
    "Re=rho*U*D/mu\n",
    "print\"Re=\",Re\n",
    "#Leh is the hydrodynamic entrance length\n",
    "print\"Therefore the flow is laminar.The hydrodynamic entrance length in m is\"\n",
    "Leh=0.05*Re*D\n",
    "print\"Leh=\",Leh\n",
    "#Let is thermal entrance length\n",
    "print\"The thermal entrance length in m is\"\n",
    "Let=0.05*Re*Pr*D\n",
    "print\"Let=\",Let\n",
    "print\"The thermal entrance length is greater than the tube length Therefore the flow is hydrodynamically developed but not thermally developed\" \n",
    "#We calculate the inverse graetz number at x=L=0.04m\n",
    "x=0.04;\n",
    "#Gr_1 is inverse of graetz number\n",
    "print\"The inverse of graetz number Gr_1 is\"\n",
    "Gr_1=(x/D)*(1/(Re*Pr))\n",
    "print\"Gr_1=\",Gr_1\n",
    "#For constant surface heat flux nusselt number is Nu=4.7 and Graetz number is Gr=4.1*10**-2\n",
    "Nu=4.7;\n",
    "Gr=4.1*10**-2;\n",
    "#hL is the local heat transfer coefficient\n",
    "print\"Therefore the local heat transfer coefficient in W/(m**2*K) is\"\n",
    "hL=Nu*(k/D)\n",
    "print\"hL=\",hL\n",
    "#from an energy balance qw*pi*D*L=mdot*cp*(Tbo-Tbi)\n",
    "#mdot is the mass flow rate\n",
    "print\"The mass flow rate of air in kg/s is\"\n",
    "mdot=rho*(math.pi/4)*D**2*U\n",
    "print\"mdot=\",mdot\n",
    "#qw is the surface heat flux\n",
    "print\"Therefore surafce heat flux qw in W/m**2 is\"\n",
    "qw=(mdot*cp*(Tbo-Tbi))/(math.pi*D*L)\n",
    "print\"qw=\",qw\n",
    "#Let Twe be the surface temprature at the exit plane.Then we can write hL*(Twe-Tbo)=qw\n",
    "print\"The tube surface temprature at the exit plane in °C is \"\n",
    "Twe=Tbo+(qw/hL)\n",
    "print\"Twe=\",Twe\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex7.11:pg-339"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " Introduction to heat transfer by S.K.Som, Chapter 7, Example 11\n",
      "Reynold number is\n",
      "Re1= 13541.3214942\n",
      "Nusselt number is\n",
      "Nubar1= 56.808608087\n",
      "The heat transfer transfer coefficient in W/(m**2*°C) \n",
      "hbar1= 522.6391944\n",
      "Outlet fluid temprature in first iteration is Tbo2 in °C is\n",
      "Tb2 in °C is\n",
      "Tb2= -30.4912413164\n",
      "Reynold number is\n",
      "Re2= 13938.8493187\n",
      "Nusselt number is\n",
      "The heat transfer transfer coefficient in W/(m**2*°C) \n",
      "hbar2= 784.03829067\n",
      "Outlet fluid temprature in second iteration is Tbo3 in °C is\n",
      "Tbo3= -16.646852652\n",
      "Tb3 in °C is\n",
      "The Exit fluid temprature after second iteration is obtained as Tbo=-16.67°C\n",
      "Tb3= -28.323426326\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    " import math\n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 7, Example 11\"\n",
    "#Liquid sulphur di oxide in a saturated state flows inside a L=5m long tube and D=25mm internal diameter with a mass flow rate(mdot) of 0.15 kg/s.\n",
    "#The tube is heated at a constant surface temprature(Tw) of -10°C and the inlet fluid temprature is Tbi=-40°C\n",
    "Tw=-10;\n",
    "Tbi=-40;\n",
    "mdot=0.15;\n",
    "D=0.025;#in metre\n",
    "L=5;\n",
    "#The properties to be used shoud be estimated at a temprature which is arithmetic mean of Tbi and Tbo.\n",
    "#Since (outlet fluid temprature Tbo) is not known a priori,the solution has to be based on an iterative method starting with a guess value of Tb1=(Tbi+Tbo)/2\n",
    "#Here we denote bulk mean temprature as Tb.The superscript refers to the no. of trials\n",
    "#For first trial,guess Tbo1=-20°C;so Tb1=-30°C\n",
    "#We have the property values as follows at a temprature of -30°C.\n",
    "rhob1=1520.64;#density in kg/m**3\n",
    "nub1=0.371*10**-6;#kinematic viscosity in m**2/s\n",
    "kb1=0.23;#conductivity in W/(m*°C)\n",
    "Prb1=3.31;#Prandtl number\n",
    "mub1=nub1*rhob1;#viscosity in kg/(m*s)\n",
    "cpb1=1361.6;#specific heat in J/(kg*K)\n",
    "#muw=nuw*rhow at Tw=10°C\n",
    "nuw=0.288*10**-6;#kinematic viscosity at Tw in m**2/s\n",
    "rhow=1463.61;#density at Tw in kg/m**3\n",
    "muw=nuw*rhow;#viscosity at Tw in kg/(m*s)\n",
    "#The reynolds number is found as Re1=(4*mdot)/(math.pi*D*mub1)\n",
    "print\"Reynold number is\"\n",
    "Re1=(4*mdot)/(math.pi*D*mub1)\n",
    "print\"Re1=\",Re1\n",
    "#Hence the flow is turbulent\n",
    "#Now using equation, nusselt number is,Nubar1=0.027*(Re1)**0.8*Prb1**(1/3)*(mub1/muw)**0.14\n",
    "print\"Nusselt number is\"\n",
    "Nubar1=0.027*(Re1)**0.8*Prb1**(1/3)*(mub1/muw)**0.14\n",
    "print\"Nubar1=\",Nubar1\n",
    "#The heat transfer transfer coefficient hbar1=(kb1/D)*Nubar1\n",
    "print\"The heat transfer transfer coefficient in W/(m**2*°C) \"\n",
    "hbar1=(kb1/D)*Nubar1\n",
    "print\"hbar1=\",hbar1\n",
    "#The outlet fluid temprature can be found by making use of eqn Tbo2=Tw-(Tw-Tbi)*math.e((-math.pi*D*L*hbar1)/(mdot*cpb1))\n",
    "print\"Outlet fluid temprature in first iteration is Tbo2 in °C is\"\n",
    "Tbo2=Tw-(Tw-Tbi)*math.e**((-math.pi*D*L*hbar1)/(mdot*cpb1))\n",
    "#Tb2 is the bulk mean temprature.\n",
    "print\"Tb2 in °C is\"\n",
    "Tb2=(Tbi+Tbo2)/2\n",
    "print\"Tb2=\",Tb2\n",
    "#Since the value differs from the assumed value of Tb1=-30°C,WE require furtheriteration,Therfore we start second trial with Tb2=-28.36°C\n",
    "#We have the property value at a temprature of -28.36°C as follows\n",
    "rhob2=1514;#density in kg/m**3\n",
    "nub2=0.362*10**-6;#kinematic viscosity in m**2/s\n",
    "kb2=0.229;#conductivity in W/(m*°C)\n",
    "Prb2=3.23;#Prandtl number\n",
    "mub2=nub2*rhob2;#viscosity in kg/(m*s)\n",
    "cpb2=1362;#specific heat in J/(kg*K)\n",
    "#muw=nuw*rhow at Tw=10°C\n",
    "nuw=0.288*10**-6;#viscosity at Tw in m**2/s\n",
    "rhow=1463.61;#density at Tw in kg/m**3\n",
    "muw=nuw*rhow;#kinematic viscosity at Tw in kg/(m*s)\n",
    "#The reynolds number is found as Re2=(4*mdot)/(math.pi*D*mub2)\n",
    "print\"Reynold number is\"\n",
    "Re2=(4*mdot)/(math.pi*D*mub2)\n",
    "print\"Re2=\",Re2\n",
    "#Now using equation, nusselt number is,Nubar2=0.027*(Re2)**0.8*Prb2**(1/3.0)*(mub2/muw)**0.14\n",
    "print\"Nusselt number is\"\n",
    "Nubar2=0.027*(Re2)**0.8*Prb2**(1/3.0)*(mub2/muw)**0.14\n",
    "#The heat transfer transfer coefficient hbar2=(kb2/D)*Nubar2\n",
    "print\"The heat transfer transfer coefficient in W/(m**2*°C) \"\n",
    "hbar2=(kb2/D)*Nubar2\n",
    "print\"hbar2=\",hbar2\n",
    "#The outlet fluid temprature can be found by making use of eqn Tbo3=Tw-(Tw-Tbi)*math.e((-math.pi*D*L*hbar2)/(mdot*cpb2))\n",
    "print\"Outlet fluid temprature in second iteration is Tbo3 in °C is\"\n",
    "Tbo3=Tw-(Tw-Tbi)*math.e**((-math.pi*D*L*hbar2)/(mdot*cpb2))\n",
    "print\"Tbo3=\",Tbo3\n",
    "#Tb3 is the bulk mean temprature.\n",
    "print\"Tb3 in °C is\"\n",
    "Tb3=(Tbi+Tbo3)/2\n",
    "#We see that difference between Tbo2 and Tbo3 and that between Tb2 and Tb3 is marginal.Therfore we can stop iteration and present the result as Tbo=-16.67°C\n",
    "print\"The Exit fluid temprature after second iteration is obtained as Tbo=-16.67°C\"\n",
    "print\"Tb3=\",Tb3"
   ]
  }
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