{ "cells": [ { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "# Chapter 5:Convection" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex5.3:pg-206" ] }, { "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 5, Example 3\n", "Reynolds number is\n", "ReL= 55263.1578947\n", "The average heat transfer coefficient over a length(L)= 1m ,in W/m**2 is\n", "hbarL= 4.16\n", "The rate of heat transfer in W/m of width is\n", "Q= 332.8\n" ] } ], "source": [ " \n", "import math \n", "from scipy.integrate import quad\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 3\"\n", "#Air at temprature (T1=20°C) and 1 atmospheric pressure flows over a flat plate with a free stream velocity(Uinf) of 1m/s.\n", "Uinf=1;\n", "T1=20;\n", "#The length of plate is 1m and is heated over its entire length to a constant temprature of T2=100°C.\n", "T2=100;\n", "#For air at 20°C(The mean temprature of 100°C and 20°C),viscosity(mu=1.9*10**-5kg/(m*s)),density(rho=1.05kg/m**3),conductivity(k=0.03W/(m*K)),specific heat(cp=1.007kJ/(kg*K))\n", "#Prandtl number is Pr=0.7\n", "mu=1.9*10**-5;\n", "rho=1.05;\n", "k=0.03;\n", "cp=1.007;\n", "Pr=0.7;\n", "#For laminar flow over a plate Nusselt number is Nux=0.332*Rex**0.5*Pr**(1/3)\n", "#The boundary layer flow over a flat plate will be laminar if Reynolds number is Rex=(rho*Uinf*x)/mu<5*10**5\n", "#First of all we have to check whether the flow is laminar or not.\n", "#Let us check at x=1m\n", "x=1.0;\n", "print\"Reynolds number is\"\n", "ReL=(rho*Uinf*x)/mu\n", "print\"ReL=\",ReL\n", "#There fore the flow is laminar and we can use the relationships of Nux,\n", "#Thus Rex=(1.05*1*x)/(1.9*10**-5)=0.5526*10**5*x\n", "#Therefore we can write Nux=(hx*x/k)=0.332*(0.5526*10**5*x)**0.5*Pr**(1/3)....or hx=2.08*x**(-1/2) W/(m**2*°C)\n", "#hbarL is the average heat transfer coefficient over a length(L)\n", "print\"The average heat transfer coefficient over a length(L)= 1m ,in W/m**2 is\"\n", "L=1;\n", "hbarL=(1.0/L)*quad(lambda x:2.08*x**(-1/2.0),0,L)[0]\n", "print\"hbarL=\",hbarL\n", "#Q is the rate of heat transfer\n", "print\"The rate of heat transfer in W/m of width is\"\n", "Q=hbarL*L*(T2-T1)\n", "print\"Q=\",Q\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex5.4:pg-207" ] }, { "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 5, Example 4\n", "The local heat transfer coefficient hx is hx=27.063*U**0.85\n", "The minimum flow velocity in m/s is\n", "U= 15.4806813943\n" ] } ], "source": [ " \n", "import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 4\"\n", "#Air at atmospheric pressure is required to flow over a circuit board to cool the electronics element mounted on it.\n", "#Chip has length (L)=3mm and width(B)=3mm located x=0.1m from the leading edge\n", "L=0.003;#in metre\n", "B=0.003;#in metre\n", "x=0.1;\n", "#The Nusselt no. is given by Nux=0.06*Rex**0.85*Pr**0.33\n", "#The chip has to dissipate E=50mW of energy while its surface temprature has to be kept below temprature,Ts=45°C and free strem Temptrature of air is Tinf=25°C\n", "Ts=45;\n", "Tinf=25;\n", "E=50*10**-3;#in watt\n", "#For air ,density(rho=1.2kg/m**3),viscosity(mu=1.8*10**5kg/(m*s)),conductivity(k=0.03W/(m*K)) and specific heat(cp=1000J/(kg*K))\n", "rho=1.2;\n", "mu=1.8*10**5;\n", "k=0.03;\n", "cp=1000;\n", "#Let the minimum flow velocity be U.\n", "#The local heat transfer coefficient hx where the chip is mounted is determined as hx=(k/x)*0.06*(rho*U*x/mu)**0.85*(mu*cp/k)**0.33\n", "print\"The local heat transfer coefficient hx is hx=27.063*U**0.85\"\n", "#from an enrgy balance we can write as E=27.063*U**0.85*L*B*(Ts-Tinf)\n", "print\"The minimum flow velocity in m/s is\"\n", "U=(E/(27.063*L*B*(Ts-Tinf)))**(1/0.85)\n", "print\"U=\",U\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex5.6:pg-208" ] }, { "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 5, Example 6\n", "mdot= 0.0235619449019\n", "(dTb/dx)in °C/m is\n", "Y= 26.6666666667\n", "Therefore Exit bulk mean temprature Tb2 in °C is\n", "Tb2= 83.3333333333\n", "Heat flux(hx) in W/(m**2*°C) is \n", "hx= 100\n", "Overall Nusselt number is \n", "NuL= 87.7192982456\n" ] } ], "source": [ " \n", "import math\n", " \n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 6\"\n", "#Air at 1atm pressure and temprature(Tin)=30°C enters a tube of 25mm diameter(D) with a velocity(U) of 10m/s\n", "D=0.025;#in metre\n", "U=10;\n", "Tin=30;\n", "#Tube is heated so that a constant heat flux(q) of 2kW/m**2 is maintained at the wall whose temprature is deltaT=20°C above the bulk mean air temprature through the length of tube \n", "#Let Tw-Tb=T\n", "q=2000;\n", "deltaT=20;\n", "#The length(L)= 2m\n", "L=2;\n", "#For air density(rho=1.2kg/m**3),specific heat(cp=1000J/(kg*K))\n", "rho=1.2;\n", "cp=1000;\n", "#From an energy balance of a control volume of air we get mdot*cp*(Tb+(dTb/dx)*deltax-Tb)=q*pi*D*deltax or (dTb/dx)=(q*pi*D)/(mdot*cp)\n", "#mass flow rate=mdot\n", "mdot=rho*math.pi*D**2*U;\n", "print\"mdot=\",mdot\n", "#let (dTb/dx)=Y\n", "print\"(dTb/dx)in °C/m is\"\n", "Y=(4*q*math.pi*D)/(mdot*cp)\n", "print\"Y=\",Y\n", "#Tb2 is Exit bulk mean temprature\n", "print\"Therefore Exit bulk mean temprature Tb2 in °C is\"\n", "Tb2=Tin+Y*2\n", "print\"Tb2=\",Tb2\n", "#Again we can write at any section of the tube hx*(Tw-Tb)=q or hx=q/(Tw-Tb)\n", "#hx is heat flux\n", "print\"Heat flux(hx) in W/(m**2*°C) is \"\n", "hx=q/(deltaT)\n", "print\"hx=\",hx\n", "#Since Tw-Tb remains the same,The heat transfer coefficient at all sections are the same\n", "#Now Overall Nusselt number,NuL=hx*D/k\n", "#The thermal conductivity of air at mean temprature of (30+83.4)/2=56.7°C is k=0.0285 W/(m*K)\n", "k=0.0285;\n", "print\"Overall Nusselt number is \"\n", "NuL=hx*D/k\n", "print\"NuL=\",NuL\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex5.7:pg-210" ] }, { "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 5, Example 7\n", "Beta(The volumetric coefficient of expansion in K**-1 is\n", "Beta= 0\n", "Grashoff number is\n", "Gr= 0.0\n", "The average nusselt number is\n", "NuHbar= 0.13\n", "Heat flux hbar in W/(m**2*°C)\n", "hbar= 0.00169\n", "The heat loss from the plate in W is\n", "Q= 0.4394\n" ] } ], "source": [ " \n", "import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 7\"\n", "#A wall is exposed to nitrogen at one atmospheric pressure and temprature,Tinf=4°C.\n", "Tinf=4;\n", "#The wall is H=2.0m high and B=2.5m wide and is maintained at temprature,Ts=56°C\n", "Ts=56;\n", "H=2;\n", "B=2.5;\n", "A=H*B;#area is(A)\n", "#The average nusselt number NuHbar over the height of the plate is given by NuHbar=0.13*(Gr*Pr)**(1/3)\n", "#The properties of nitrogen at mean film temprature(Tf) is (56+4)/2=30°C are given as density(rho=1.142kg/m**3) ,conductivity(k=0.026W/(m*K)),\n", "#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;\n", "#We first have to detrmine the value of Grashoff number,Gr.In consideration of nitrogen as an ideal gas,we can write\n", "#Beta(The volumetric coefficient of expansion)=1/T\n", "print\"Beta(The volumetric coefficient of expansion in K**-1 is\"\n", "Beta=1/(273+Tf)\n", "print\"Beta=\",Beta\n", "#Now Gr=(g*Beta*(Ts-Tinf)*H**3)/nu**2\n", "g=9.81;#acceleration due to gravity\n", "print\"Grashoff number is\"\n", "Gr=(g*Beta*(Ts-Tinf)*H**3)/nu**2\n", "print\"Gr=\",Gr\n", "print\"The average nusselt number is\"\n", "NuHbar=0.13*(Gr*Pr)**(1/3)\n", "print\"NuHbar=\",NuHbar\n", "#hbar is the heat flux\n", "print\"Heat flux hbar in W/(m**2*°C)\"\n", "hbar=NuHbar*k/H\n", "print\"hbar=\",hbar\n", "#Q is the heat loss from the plate\n", "print\"The heat loss from the plate in W is\"\n", "Q=hbar*A*(Ts-Tinf)\n", "print\"Q=\",Q\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex5.8:pg-211" ] }, { "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 5, Example 8\n", "The heat flux in W/(m**2*K) is\n", "hbar 120.0\n", "The heat loss from the wire is Q=hbar*pi*D*L*(Twire-Tair) in Watt\n", "Q= 1.88495559215\n", "The current in Ampere is\n", "I= 17.7245385091\n" ] } ], "source": [ " \n", " \n", " \n", " \n", " import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 5, Example 8\"\n", "#Eletric current passes through a L=0.5m long horizontal wire of D=0.1mm diameter.\n", "L=0.5;\n", "D=0.1*10**-3;\n", "#The wire is to be maintained at temprature,Twire=400K and the air is at temprature,Tair=300K.\n", "Twire=400;\n", "Tair=300;\n", "#The resistance of the wire(R) is 0.012ohm per meter.Nusselt number(NuL) over the length of wire to be 0.4.\n", "NuL=0.4;\n", "R=0.012;\n", "#At mean temprature of Tf=350K, The thermal conductivity of air is k=0.03W/(m*K)\n", "k=0.03;\n", "#Nusselt number is NuL=hbar*D/k\n", "#hbar is the heat flux\n", "print\"The heat flux in W/(m**2*K) is\"\n", "hbar=NuL*k/D\n", "print\"hbar\",hbar\n", "#Q is the heat loss from the wire\n", "print\"The heat loss from the wire is Q=hbar*pi*D*L*(Twire-Tair) in Watt\"\n", "Q=hbar*math.pi*D*L*(Twire-Tair)\n", "print\"Q=\",Q\n", "#At steady state the ohmic loss in the wire equals the heat loss from its surface Therfore I**2*R=Q\n", "#I is the current flow.\n", "print\"The current in Ampere is\"\n", "I=(Q/(R*L))**0.5\n", "print\"I=\",I\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n", "\n" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", 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