{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "# Chapter 08:Principles of free convection"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.1:pg-355"
   ]
  },
  {
   "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 8, Example 1\n",
      "Grashoff number is\n",
      "GrL= 2175146201.53\n",
      "Rayleigh number is\n",
      "RaL= 9440134514.65\n",
      "Therefore the flow is turbulent\n",
      "Now we use [(hbarL*L)/k]=0.10*(GrL*Pr)**(1/3)\n",
      "The average heat transfer coefficient in W/(m**2*K) is\n",
      "hbarL= 0.314\n",
      "The rate of heat transfer in W is\n",
      "q= 0.5024\n"
     ]
    }
   ],
   "source": [
    "import math \n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 1\"\n",
    "#Water is heated by a vertical flat plate  length(L=200mm or .2m )by breadth(B=200mm) which is maintained at temprature,Tw=60°C\n",
    "Tw=60;\n",
    "L=.2;\n",
    "B=.2;# in metre\n",
    "#Area(A) is L*B \n",
    "A=L*B;\n",
    "#Water is at temprature,Tinf=20°C\n",
    "Tinf=20;\n",
    "#At mean film temprature 40°C The physical properties parameters can be taken as \n",
    "#conductivity(k=0.0628W/(m*K)),Prandtl number(Pr=4.34),density(rho=994.59kg/m**3),kinematic viscosity(nu=0.658*10**-6m**2/s),volume expnasion coefficient(Beta=3*10**-4K**-1))\n",
    "k=0.628;\n",
    "Pr=4.34;\n",
    "rho=994.59;\n",
    "nu=0.658*10**-6;\n",
    "Beta=3*10**-4;\n",
    "#g is acceleration due to gravity =9.81m/s**2\n",
    "g=9.81;\n",
    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
    "print\"Grashoff number is\"\n",
    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2\n",
    "print\"GrL=\",GrL\n",
    "#Rayleigh number is defined as RaL=GrL*Pr\n",
    "print\"Rayleigh number is\"\n",
    "RaL=GrL*Pr\n",
    "print\"RaL=\",RaL\n",
    "print\"Therefore the flow is turbulent\"\n",
    "print\"Now we use [(hbarL*L)/k]=0.10*(GrL*Pr)**(1/3)\"\n",
    "#hbarL is the average heat transfer coefficient\n",
    "print\"The average heat transfer coefficient in W/(m**2*K) is\"\n",
    "hbarL=(0.10*(GrL*Pr)**(1/3)*k)/L\n",
    "print\"hbarL=\",hbarL\n",
    "#The rate of heat transfer is given by q=hbarL*A*(Tw-Tinf)\n",
    "print\"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"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.2:pg-357"
   ]
  },
  {
   "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 8, Example 2\n",
      "The minimum spacing between the plates will be twice the thickness of the boundary layer at the trailing edge where x=0.09\n",
      "Grashoff number is\n",
      "GrL= 198210197.615\n",
      "Rayleigh number is\n",
      "RaL= 860232257.647\n",
      "Since Ra<10**9,Therefore the flow is laminar\n",
      "The thickness of the boundary layer in metre is\n",
      "delta= 4.11168026839e-10\n",
      "The minimum spacing in metre is\n",
      "spac= 8.22336053678e-10\n"
     ]
    }
   ],
   "source": [
    " \n",
    "import math \n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 2\"\n",
    "#The thin plates are kept at temprature(Tw)=60°C while the temprature of water bath(Tinf)=20°C\n",
    "Tw=60;\n",
    "Tinf=20;\n",
    "#The plates have length(L)=90mm or .09m\n",
    "L=.09;\n",
    "#The minimum spacing between the plates will be twice the thickness of the boundary layer at the trailing edge where x=0.09.\n",
    "print\"The minimum spacing between the plates will be twice the thickness of the boundary layer at the trailing edge where x=0.09\"\n",
    "x=.09;\n",
    "#At mean film temprature 40°C The physical properties parameters can be taken as\n",
    "# conducivity(k=0.0628W/(m*K)),Prandtl number(Pr=4.34),Density(rho=994.59kg/m**3),kinematic viscosity(nu=0.658*10**-6m**2/s),Volume expansion coefficient(Beta=3*10**-4K**-1)\n",
    "k=0.628;\n",
    "Pr=4.34;\n",
    "rho=994.59;\n",
    "nu=0.658*10**-6;\n",
    "Beta=3*10**-4;\n",
    "#g is acceleration due to gravity =9.81m/s**2\n",
    "g=9.81;\n",
    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
    "print\"Grashoff number is\"\n",
    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2\n",
    "print\"GrL=\",GrL\n",
    "#Rayleigh number is defined as RaL=GrL*Pr\n",
    "print\"Rayleigh number is\"\n",
    "RaL=GrL*Pr\n",
    "print\"RaL=\",RaL\n",
    "print\"Since Ra<10**9,Therefore the flow is laminar\"\n",
    "#delta is the thickness of the boundary layer\n",
    "print\"The thickness of the boundary layer in metre is\"\n",
    "delta=x*3.93*Pr**(-1/2)*(0.952+Pr)**(1/4)*GrL**(-1/4)\n",
    "print\"delta=\",delta\n",
    "#spac is the minimum spacing \n",
    "print\"The minimum spacing in metre is\"\n",
    "spac=2*delta\n",
    "print\"spac=\",spac\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.3:pg-366"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 3\n",
      "Grashoff number is\n",
      "Grx= 517025.52213\n",
      "The boundary layer thickness in metre is\n",
      "delta= 60304.3038858\n",
      "The velocity at point x is ux in m/s is\n",
      "ux= 3247798354.51\n",
      "For maximum value of velocity,u\n",
      "Maximum velocity in m/s is\n",
      "Umax= 3.57089835848\n",
      "Mass flow rate at x=0.8m,in kG is\n",
      "mdot= 2.36830073295e+16\n"
     ]
    }
   ],
   "source": [
    "\n",
    "from scipy.integrate import quad\n",
    "print \"Introduction to heat transfer by S.K.Som, Chapter 8, Example 3\"\n",
    "#Considering question 5.7\n",
    "#A wall is exposed to nitrogen at one atmospheric pressure and temprature,Tinf=4°C.\n",
    "Tinf=4.0;\n",
    "#The wall is H=2.0m high and 2.5m wide and is maintained at temprature,Tw=56°C\n",
    "Tw=56.0;\n",
    "H=2.0;\n",
    "B=2.5;\n",
    "A=H*B;#Area of wall in m**2\n",
    "#The properties of nitrogen at mean film temprature (56+4)/2=30°C are given as \n",
    "#density(rho=1.142kg/m*3) ,conductivity(k=0.026W/(m*K)),kinematic viscosity(nu=15.630*10-6 m*2/s) ,prandtl number(Pr=0.713)\n",
    "rho=1.142;\n",
    "k=0.026;\n",
    "nu=15.630*10**-6;\n",
    "Pr=0.713;\n",
    "Tf=30.0;#mean film temprature\n",
    "Beta=1/(273.0+Tf);#volume expansion coefficient:unit K**-1\n",
    "#Now Grashoff number is Grx=(g*Beta*(Tw-Tinf)*x*3)/nu*2\n",
    "g=9.81;#acceleration due to gravity\n",
    "print \"Grashoff number is\"\n",
    "x=0.8;#distance from the bottom of wall\n",
    "Grx=(g*Beta*(Tw-Tinf)*x*3)/nu*2\n",
    "print\"Grx=\",Grx\n",
    "#Using equation delta=x*Pr*(-0.5)(0.952+Pr)*(0.25)*Grx*(-0.25)\n",
    "#delta is the boundary layer thickness\n",
    "print \"The boundary layer thickness in metre is\"\n",
    "delta=x*3.93*Pr*(-0.5)*(0.952+Pr)*(0.25)*Grx*(-0.25)\n",
    "print\"delta=\",delta\n",
    "#Now using equation ux=(g*Beta*delta*2(Tw-Tinf))/(4*nu)\n",
    "#ux is the velocity at point x\n",
    "print \"The velocity at point x is ux in m/s is\"\n",
    "ux=(g*Beta*delta*2*(Tw-Tinf))/(4*nu)\n",
    "print\"ux=\",ux\n",
    "# (u/ux)=(y/delta)*(1-y/delta)**2\n",
    "#Putting value of ux we get velovity function,u=465.9*(y-116*y*2+3341*y*3)\n",
    "#For maximum value of u,du/dy=465.9*(1-232*y+10023*y**2)=0...this is a quadratic equation in which coefficients a=10023,b=232,c=1\n",
    "a=10023;\n",
    "b=232;\n",
    "c=1;\n",
    "#Solution for quadratic equation is given by y=(-b+-(b*2-4ac)*0.5)/2*a\n",
    "print \"For maximum value of velocity,u\"\n",
    "y=(b+(b*2-4*a*c)*0.5)/(2*a)#root of the quadratic equation\n",
    "y=(b-(b*2-4*a*c)*0.5)/(2*a)#root of the quadratic equation\n",
    "#The value of 0.0173 is at the edge of boundary layer,where u=0\n",
    "#Therefore the maximum value occurs at y=0.00573m i.e Umax=465.9*y*(1-57.8*y)**2\n",
    "y=0.00573;\n",
    "#Umax is maximum velocity\n",
    "print \"Maximum velocity in m/s is\"\n",
    "Umax=465.9*y*(1-57.8*y)*2#NOTE:The answer given in the book is incorrect,in this expresssion they considered square on y only,however it is on whole expression (1-57.8*y)*2\n",
    "#mdot is mass flow rate\n",
    "print\"Umax=\",Umax\n",
    "print \"Mass flow rate at x=0.8m,in kG is\"\n",
    "I=quad(lambda y:465.9*(y-116*y*2+3341*y*3),0,delta)\n",
    "mdot=rho*B*I[0]\n",
    "print\"mdot=\",mdot\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.4:pg-369"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Introduction to heat transfer by S.K.Som, Chapter 8, Example 4\n",
      "Grashoff number is\n",
      "Rayleigh number is\n",
      "Hence,the flow is laminar\n",
      "The thickness of the boundary layer in metre is\n",
      "The average heat transfer coeficient in W/(m**2*K) is\n",
      "0.101781170483\n"
     ]
    }
   ],
   "source": [
    " \n",
    "import math \n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 4\"\n",
    "#A square plate length,L=0.2m by breadth,B=0.2m is suspended vertically in a quiescent atmospheric air at a temprature(Tinf)=300K\n",
    "L=0.2;\n",
    "B=0.2;\n",
    "Tinf=300;\n",
    "#The Temprature of plate(Tw) is maintained at 400K\n",
    "Tw=400;\n",
    "#The required property value of air at a film temprature(Tf)=350K,kinematic viscosity (nu=20.75*10**-6),Prandtl number(Pr=0.69),conductivity(k=0.03W/(m*K))\n",
    "Tf=350;\n",
    "nu=20.75*10**-6;\n",
    "Pr=0.69;\n",
    "k=0.03;\n",
    "#volume expansion coefficient is Beta\n",
    "Beta=(1/Tf);\n",
    "#g is acceleration due to gravity =9.81m/s**2\n",
    "g=9.81;\n",
    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
    "print\"Grashoff number is\"\n",
    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2 \n",
    "print\"GrL=\",GrL\n",
    "#Rayleigh number is defined as RaL=GrL*Pr\n",
    "print\"Rayleigh number is\"\n",
    "RaL=GrL*Pr\n",
    "print\"Hence,the flow is laminar\"\n",
    "print\"RaL=\",RaL\n",
    "#delta is the thickness of the boundary layer\n",
    "print\"The thickness of the boundary layer in metre is\"\n",
    "x=0.2;#location of trailing edge of plate\n",
    "delta=(x*3.93*(0.952+Pr)**(1/4))/(Pr**(1/2)*(GrL)**(1/4))#NOTE:The answer in the book is incorrect(calculation mistake)\n",
    "print\"delta=\",delta\n",
    "#hL and hbarL are local and average heat transfer coefficient respectively\n",
    "print\"The average heat transfer coeficient in W/(m**2*K) is\"\n",
    "hL=(2*k)/delta;\n",
    "hbarL=(4.0/3)*(hL)#NOTE:The answer in the book is incorrect(calculation mistake)\n",
    "print\"hL=\",hL\n",
    "print\"hbarL=\",hbarL\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": [
    "## Ex8.5:pg-373"
   ]
  },
  {
   "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 8, Example 5\n",
      "Grashoff number is\n",
      "503958851.066\n",
      "Rayleigh number is\n",
      "357810784.257\n",
      "Therefore the flow is laminar\n",
      "Nusselt number is\n",
      "75.3134665126\n",
      "Average heat transfer coefficient(hbarL)in W/(m**2*°C)\n",
      "The rate of heat transfer in W is \n",
      "Now if we use NuL2=0.59*RaL**(1/4) with the value of C=0.59,n=(1/4)\n",
      "Nusselt number is\n",
      "Average heat transfer coefficient(hbarL)in W/(m**2*°C)\n",
      "The rate of heat transfer in W is \n",
      "49.9857347505\n",
      "(b)For the horizontal plate facing up\n",
      "Now RaL2=Gr*Pr*(Lc/L)**3\n",
      "Rayleigh number is\n",
      "Nusselt number is given by NuL3=C*(GrL*Pr)**n\n",
      "Average heat transfer coefficient(hbarL)in W/(m**2*°C)\n",
      "The rate of heat transfer in W is \n",
      "64.6997833306\n",
      "(c)When the hot surface faces is down\n",
      "Nusselt number is given by NuL4=0.27*RaL2**(1/4)\n",
      "13.1290144745\n",
      "Average heat transfer coefficient(hbarL) in W/(m**2)\n",
      "2.9408992423\n",
      "The rate of heat transfer in W is \n",
      "32.3498916653\n"
     ]
    }
   ],
   "source": [
    " \n",
    "import math \n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 5\"\n",
    "#A square plate of length(L)=0.5m by breadth,B=0.5m in a room at temprature,Tinf=30°C\n",
    "#One side of plate is kept a uniform temprature(Tw)=74°C\n",
    "Tw=74;\n",
    "L=0.5;\n",
    "B=0.5;\n",
    "Tinf=30.0;\n",
    "#The required properties at the film temprature(Tf)=52°C are kinematic viscosity(nu=1.815*10**-5),Prandtl number(Pr=0.71),conductivity(k=0.028W/(m*°C))\n",
    "Tf=52.0;\n",
    "Pr=0.71;\n",
    "nu=1.815*10**-5;\n",
    "k=0.028;\n",
    "#Area(A)=L*B m**2\n",
    "A=L*B;\n",
    "#Volume expansion coefficient is Beta\n",
    "Beta=1/(273+Tf);\n",
    "#g is acceleration due to gravity =9.81m/s**2\n",
    "g=9.81;\n",
    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
    "print\"Grashoff number is\"\n",
    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2 \n",
    "print GrL\n",
    "#Rayleigh number is defined as RaL1=GrL*Pr\n",
    "print\"Rayleigh number is\"\n",
    "RaL1=GrL*Pr\n",
    "print RaL1\n",
    "print\"Therefore the flow is laminar\"\n",
    "#We make use of following equation to find Nusselt number,NuL1=(4/3)*(0.508*Pr**(-1/2)*(0.952+Pr)**(-1/4)*Gr**(1/4))\n",
    "print\"Nusselt number is\"\n",
    "NuL1=(4.0/3)*(0.508*Pr**(1.0/2)*(0.952+Pr)**(-1.0/4)*GrL**(1.0/4))\n",
    "#Average heat transfer coefficient(hbarL) is given by (NuL*k)/L\n",
    "print NuL1\n",
    "print\"Average heat transfer coefficient(hbarL)in W/(m**2*°C)\"\n",
    "hbarL=(NuL1*k)/L\n",
    "#The rate of heat transfer(Q) from the plate by free convection is given by Q=hbarL*A*(Tw-Tinf)\n",
    "print\"The rate of heat transfer in W is \"\n",
    "Q=hbarL*A*(Tw-Tinf)\n",
    "print\"Now if we use NuL2=0.59*RaL**(1/4) with the value of C=0.59,n=(1/4)\"\n",
    "print\"Nusselt number is\"\n",
    "NuL2=0.59*RaL1**(1.0/4)\n",
    "#Average heat transfer coefficient(hbarL) is given by (NuL*k)/L\n",
    "print\"Average heat transfer coefficient(hbarL)in W/(m**2*°C)\"\n",
    "hbarL=(NuL2*k)/L\n",
    "#The rate of heat transfer(Q) from the plate by free convection is given by Q=hbarL*A*(Tw-Tinf)\n",
    "print\"The rate of heat transfer in W is \"\n",
    "Q=hbarL*A*(Tw-Tinf)\n",
    "print Q\n",
    "print\"(b)For the horizontal plate facing up\"\n",
    "#Perimeter(P) for a square plate is P=4*L\n",
    "P=4*L;\n",
    "#Characterstic length(Lc)=A/P\n",
    "Lc=A/P\n",
    "print\"Now RaL2=Gr*Pr*(Lc/L)**3\"\n",
    "print\"Rayleigh number is\"\n",
    "RaL2=GrL*Pr*(Lc/L)**3\n",
    "#The values of constants,C=0.54 and n=(1/4)\n",
    "C=0.54;\n",
    "n=(1.0/4);\n",
    "print\"Nusselt number is given by NuL3=C*(GrL*Pr)**n\"\n",
    "NuL3=C*(RaL2)**n\n",
    "print\"Average heat transfer coefficient(hbarL)in W/(m**2*°C)\"\n",
    "hbarL=(NuL3*k)/Lc\n",
    "print\"The rate of heat transfer in W is \"\n",
    "Q=hbarL*A*(Tw-Tinf)\n",
    "print Q\n",
    "print\"(c)When the hot surface faces is down\"\n",
    "print\"Nusselt number is given by NuL4=0.27*RaL2**(1/4)\"\n",
    "NuL4=0.27*RaL2**(1.0/4)\n",
    "print NuL4\n",
    "print\"Average heat transfer coefficient(hbarL) in W/(m**2)\"\n",
    "hbarL=(NuL4*k)/Lc\n",
    "print hbarL\n",
    "print\"The rate of heat transfer in W is \"\n",
    "Q=hbarL*A*(Tw-Tinf)\n",
    "print 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",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.6:pg-375"
   ]
  },
  {
   "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 8, Example 6\n",
      "Grashoff number is\n",
      "GrL= 813719594.384\n",
      "Rayleigh number is\n",
      "RaL= 569603716.069\n",
      "Therefore the flow is laminar\n",
      "Now we use NuL=0.59*RaL**(1/4.0) with the value of constants C=0.59,n=(1/4.0)\n",
      "Nusselt number is\n",
      "NuL= 91.1475952489\n",
      "Average heat transfer coefficient in W/(m**2*K)\n",
      "hbarL1= 5.46885571493\n",
      "Grashoff number GrD=GrL*(D/L)**3\n",
      "GrD= 0.00650975675508\n",
      "The correction factor is\n",
      "F= 39.485281111\n",
      "The correct value of Average heat transfer coefficient(hbarL2)=hbarL1*F in W/(m**2*K) is\n",
      "hbarL2= 215.939305259\n",
      "The ohmic loss in W is \n",
      "q= 3.39196667512\n",
      "The current flowing in the wire in Ampere is\n",
      "I= 7.51882822777\n"
     ]
    }
   ],
   "source": [
    " \n",
    " \n",
    " \n",
    " \n",
    "import math \n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 6\"\n",
    "#A vertical wire of length(L)=0.5m and  Dimeter(D)=0.1mm is maintained at temprature, Tw=400K\n",
    "#The temprature of quicsent air is Tinf=300K\n",
    "#Resistance(R) per  meter length is 0.12ohm\n",
    "R=0.12;\n",
    "Tw=400.0;\n",
    "L=0.5;\n",
    "D=0.1*10**-3;#in metre\n",
    "Tinf=300;\n",
    "#The required properties at the film temprature(Tf)=350K are kinematic viscosity(nu=20.75*10**-6m**2/s),Prandtl number(Pr=0.70),conductivity(k=0.03W/(m*°C))\n",
    "Tf=350.0;\n",
    "Pr=0.70;\n",
    "nu=20.75*10**-6;\n",
    "k=0.03;\n",
    "#Area(A)=L*B m**2\n",
    "A=math.pi*D*L;\n",
    "#Volume expansion Coefficient is Beta\n",
    "Beta=1/(Tf);\n",
    "#g is acceleration due to gravity =9.81m/s**2\n",
    "g=9.81;\n",
    "#Grashoff number is given by GrL=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
    "print\"Grashoff number is\"\n",
    "GrL=(g*Beta*(Tw-Tinf)*L**3)/(nu)**2 \n",
    "print\"GrL=\",GrL\n",
    "#Rayleigh number is defined as RaL=GrL*Pr\n",
    "print\"Rayleigh number is\"\n",
    "RaL=GrL*Pr\n",
    "print\"RaL=\",RaL\n",
    "print\"Therefore the flow is laminar\"\n",
    "#NuL is nusselt number\n",
    "#C and n are constants\n",
    "print\"Now we use NuL=0.59*RaL**(1/4.0) with the value of constants C=0.59,n=(1/4.0)\"\n",
    "print\"Nusselt number is\"\n",
    "NuL=0.59*RaL**(1/4.0)\n",
    "print\"NuL=\",NuL\n",
    "#hbarL1 is the Average heat transfer coefficient\n",
    "print\"Average heat transfer coefficient in W/(m**2*K)\"\n",
    "hbarL1=(NuL*k)/L\n",
    "print\"hbarL1=\",hbarL1\n",
    "#Grashoff number GrD=GrL*(D/L)**3\n",
    "print\"Grashoff number GrD=GrL*(D/L)**3\"\n",
    "GrD=GrL*(D/L)**3\n",
    "print\"GrD=\",GrD\n",
    "#The correction factor is given By F=1.3*((L/D)/GrD)**(1/4.0)+1.0\n",
    "print\"The correction factor is\"\n",
    "F=1.3*((L/D)/GrD)**(1/4.0)+1.0\n",
    "print\"F=\",F\n",
    "print\"The correct value of Average heat transfer coefficient(hbarL2)=hbarL1*F in W/(m**2*K) is\"\n",
    "hbarL2=hbarL1*F\n",
    "print\"hbarL2=\",hbarL2\n",
    "#The ohmic power loss is given by energy balance I**2*R=q=hbar2*A*(Tw-Tinf)\n",
    "#q is the ohmic power loss\n",
    "print\"The ohmic loss in W is \"\n",
    "q=hbarL2*A*(Tw-Tinf)\n",
    "print\"q=\",q\n",
    "#The current flowing in the wire I=(q/(R*L)**(1/2.0)\n",
    "print\"The current flowing in the wire in Ampere is\"\n",
    "I=(q/(R*L))**(1/2.0)\n",
    "print\"I=\",I\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.7:pg-378"
   ]
  },
  {
   "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 8, Example 7\n",
      "Grashoff number is\n",
      "GrD= 53311595.6796\n",
      "Rayleigh number is\n",
      "RaD= 37318116.9757\n",
      "The flow is laminar over the entire cylinder\n",
      "we use following equation to find Nusselt number NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\n",
      "NuD= 0.974169\n",
      "Average heat transfer coefficient in W/(m**2*K)\n",
      "hbar= 0.14612535\n",
      "The heat loss per meter length in W is\n",
      "q= 9.64039284733\n"
     ]
    }
   ],
   "source": [
    " \n",
    "import math \n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 7\"\n",
    "#A long horizontal pressurized hot water of diameter(D)=200mm passes through a room where the air temprature is Tinf=25°C\n",
    "D=.2;\n",
    "Tinf=25;\n",
    "#Length(L)=1m ,since the unit length is considered\n",
    "L=1;\n",
    "#Area(A)=pi*L*D\n",
    "A=math.pi*L*D;\n",
    "#The pipe surface temprature is Tw=130°C\n",
    "Tw=130;\n",
    "#The properties of air at the film temprature Tf=77.5°C are kinematic viscosity(nu=21*10**-6m**2/s),Prandtl number(Pr=0.70),Conductivity(k=0.03W/(m*K))\n",
    "Tf=77.5;\n",
    "nu=21*10**-6;\n",
    "k=0.03;\n",
    "Beta=(1/(273+Tf));#Volume expansion coefficient in k**-1)\n",
    "Pr=0.70;\n",
    "#g is acceleration due to gravity =9.81m/s**2\n",
    "g=9.81;\n",
    "#Grashoff number is given by GrD=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
    "print\"Grashoff number is\"\n",
    "GrD=(g*Beta*(Tw-Tinf)*D**3)/(nu)**2 \n",
    "print\"GrD=\",GrD\n",
    "#Rayleigh number is defined as RaD=GrD*Pr\n",
    "print\"Rayleigh number is\"\n",
    "RaD=GrD*Pr\n",
    "print\"RaD=\",RaD\n",
    "print\"The flow is laminar over the entire cylinder\"\n",
    "#NuD is the nusselt number\n",
    "print\"we use following equation to find Nusselt number NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\"\n",
    "NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\n",
    "print\"NuD=\",NuD\n",
    "#hbar is the avearge heat transfer coefficient\n",
    "print\"Average heat transfer coefficient in W/(m**2*K)\"\n",
    "hbar=(NuD*k)/D\n",
    "print\"hbar=\",hbar\n",
    "#The heat loss per meter length is given by q=hbar*A*(Tw-Tinf)\n",
    "print\"The heat loss per meter length in W is\"\n",
    "q=hbar*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"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex8.8:pg-381"
   ]
  },
  {
   "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 8, Example 8\n",
      "Let us take first trial Tw=64°C\n",
      "Grashoff number is\n",
      "GrD= 226303.67232\n",
      "Rayleigh number is\n",
      "The flow is laminar \n",
      "RaD= 941423.276851\n",
      "we use following equation to find Nusselt number NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\n",
      "NuD= 0.974169\n",
      "Average heat transfer coefficient in W/(m**2*K)\n",
      "hbarD= 77.20289325\n",
      "Hence,steady state Surface temprature in °C is\n",
      "Hence we see that our guess is in excellent agreement with the calculated value\n",
      "Tw= 793.068225127\n"
     ]
    }
   ],
   "source": [
    " \n",
    "import math \n",
    " \n",
    "print\"Introduction to heat transfer by S.K.Som, Chapter 8, Example 8\"\n",
    "#An electric immersion heater diameter(D)=8mm and length(L)=300mm is rated at power input,P=450W\n",
    "P=450;\n",
    "L=0.3;#in metre\n",
    "D=0.008;#in metre\n",
    "#If the heater is horizontally positioned in a large tank of stationery water at temprature,Tinf=20°C\n",
    "Tinf=20;\n",
    "#At steady state ,The electrical power input(P)=(Q)Heat loss from the heater\n",
    "#P=Q\n",
    "#Q=hbarD*(pi*D)*L*(Tw-Tinf)\n",
    "#This gives Tw(surface temprature)=Tinf+(P/(hbarD*pi*D*L))\n",
    "#So we need to find Average heat transfer coefficient,hbarD.\n",
    "#In this problem we need to take guess of steady state surface temprature(Tw) and iterate the solution for Tw till a desired convergence is achieved.\n",
    "print\"Let us take first trial Tw=64°C\"\n",
    "Tw=64;\n",
    "Tf=(Tw+Tinf)/2;#mean film temprature\n",
    "#At this temprature of 42°C,The required properties of water kinematic viscosity(nu=6.25*10**-7m**2/s),Prandtl number(Pr=4.16),Conductivity(k=0.634W/(m*K)),Beta=4*10**-4K**-1\n",
    "Beta=4*10**-4;#Volume expansion coefficient\n",
    "nu=6.25*10**-7;\n",
    "Pr=4.16;\n",
    "k=0.634;\n",
    "#g is acceleration due to gravity =9.81m/s**2\n",
    "g=9.81;\n",
    "#Grashoff number is given by GrD=(g*beta*(Tw-Tinf)*L**3)/(nu)**2\n",
    "print\"Grashoff number is\"\n",
    "GrD=(g*Beta*(Tw-Tinf)*D**3)/(nu)**2 \n",
    "print\"GrD=\",GrD\n",
    "#Rayleigh number is defined as RaD=GrD*Pr\n",
    "print\"Rayleigh number is\"\n",
    "RaD=GrD*Pr\n",
    "print\"The flow is laminar \"\n",
    "print\"RaD=\",RaD\n",
    "#/NuD is nusselt number\n",
    "#hbarD is Average heat transfer coefficient\n",
    "print\"we use following equation to find Nusselt number NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\"\n",
    "NuD=(0.60+((0.387*RaD**(1/6))/(1+(0.559/Pr**(9/16)))**(8/27)))**2\n",
    "print\"NuD=\",NuD\n",
    "print\"Average heat transfer coefficient in W/(m**2*K)\"\n",
    "hbarD=(NuD*k)/D\n",
    "print\"hbarD=\",hbarD\n",
    "print\"Hence,steady state Surface temprature in °C is\"\n",
    "Tw=Tinf+(P/(hbarD*math.pi*D*L))\n",
    "print\"Hence we see that our guess is in excellent agreement with the calculated value\"\n",
    "print\"Tw=\",Tw\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  }
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