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diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb new file mode 100644 index 00000000..92b1940b --- /dev/null +++ b/Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb @@ -0,0 +1,734 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 Empirical and Practical Relations for Forced Convection Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.1" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reynolds number is 14749.0\n", + "so that the flow is turbulent\n", + "Heat transfer per unit length is 103.5 W/m\n", + "Bulk temperature increase over the length of 3 m on tube is 40.04 degree C\n" + ] + } + ], + "source": [ + "#Example Number 6.1\n", + "# turbulent heat transfer in a tube \n", + "# Variable declaration\n", + "\n", + "p = 2*101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 200+273.15\t\t\t# [K] temperature of air \n", + "d = 0.0254 \t\t\t# [m] diameter of tube \n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 10 \t\t\t\t# [m/s] velocity of air\n", + "dT = 20 \t\t\t# [deg C] temperature difference between wall and air \n", + "\n", + "\t# we first calculate the reynolds number to determine if the flow is laminar \t\tor turbulent, and then select the appropriate empirical correlation to \t\t\tcalculate the heat transfer \n", + "\n", + "\t# the properties of air at a bulk temperature of 473 K are\n", + "\n", + "#Calculaiton\n", + "\n", + "rho = p/(R*Ta) \t\t\t# [kg/cubic meter] density of gas\n", + "mu = 2.57*10**(-5)\t\t# [kg/m s] viscosity \n", + "k = 0.0386 \t\t\t# [W/m degree celsius]\n", + "Cp = 1025 \t\t\t# [J/kg K]\n", + "Pr = 0.681 \t\t\t# prandtl no.\n", + "Re_d = rho*u*d/mu \t\t# reynolds number\n", + "\n", + "print \"Reynolds number is\",round(Re_d) \n", + "print\"so that the flow is turbulent\"\n", + " \n", + "\t# we therefore use equation (6-4a) to calculate the heat transfer coefficient\n", + "\n", + "Nu_d = 0.023*Re_d**(0.8)*Pr**(0.4) \t# nusselt no.\n", + "h = Nu_d*k/d \t\t\t\t# [W/m**2 deg C] heat transfer coefficient\n", + "\n", + "\t# the heat transfer per unit length is then\n", + "\n", + "import math\n", + "q_by_L = h*math.pi*d*(dT) \t\t# [W/m]\n", + "L = 3 \t\t\t\t\t# [m] \n", + "\t# we can now make an energy balance to calculate the increase in bulk \t\ttemperature in a 3 m length of tube :\n", + "\t# q = m_dot*Cp*dT_b = L*(q_byL)\n", + "m_dot = rho*u*math.pi*d**(2)/4 \t\t# [kg/s] gas flow rate\n", + "\t# so that we insert the numerical values in the energy balance to obtain \n", + "dT_b = L*q_by_L/(m_dot*Cp) \t\t# [degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print\"Heat transfer per unit length is\",round(q_by_L,1),\" W/m\"\n", + "print\"Bulk temperature increase over the length of 3 m on tube is\",round(dT_b,2),\" degree C\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.3" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reynolds number is 823.0\n", + "so that the flow is laminar\n", + "\n", + "Total heat transfer is 3.49 W\n", + "\n", + "Exit wall temperature is 161.0 degree celsius\n", + "\n", + "Heat transfer coefficient is 26.45 W/sq meter degree C\n" + ] + } + ], + "source": [ + "#Example Number 6.3\n", + "# heating of air in laminar tube flow for constant heat flux\n", + "\n", + "#Variable declaration\n", + "\n", + "import math\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 27 \t\t\t# [degree celsius] temperature of air \n", + "d = 0.005 \t\t\t# [m] diameter of tube \n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 3 \t\t\t\t# [m/s] velocity of air\n", + "L = 0.1 \t\t\t# [m] length of tube\n", + "Tb = 77 \t\t\t# [degree celsius] exit bulk temperature \n", + "\t# we first must evaluate the flow regime and do so by taking properties at the \taverage bulk temperature \n", + "Tb_bar = (Ta+Tb)/2 \t\t# [degree celsius]\n", + "v = 18.22*10**(-6) \t\t# [square meter/s] kinematic viscosity\n", + "k = 0.02814 \t\t\t# [W/m degree celsius]\n", + "Cp = 1006 \t\t\t# [J/kg K]\n", + "Pr = 0.703 \t\t\t# prandtl no.\n", + "Re_d = u*d/v \t\t\t# reynolds number\n", + "print \"Reynolds number is\",round(Re_d) \n", + "print\"so that the flow is laminar\\n\" \n", + "\t\n", + "\t#the tube length is short, so we expect a thermal entrance effect and shall \t\tconsult figure(6-5)\n", + "\t# the inverse Graetz number is computed as \n", + "Gz_inverse = L/(Re_d*Pr*d) \n", + "\t# therefore, for qw = constant, we obtain the nusselt number at exit from \t\tfigure (6-5) as\n", + "Nu = 4.7 \n", + "\t# the total heat transfer is obtained in terms of the overall energy balance \n", + "\t# at entrance \n", + "rho = 1.1774 \t\t\t # [kg/cubic meter] density\n", + "\t# mass flow is\n", + "m_dot = rho*math.pi*d**(2)*u/4 \t # [kg/s]\n", + "q = m_dot*Cp*(Tb-Ta) \t\t # [W]\n", + "\t# thus we may find the heat transfer without the actually determining wall \t\ttemperatures or values of h. However, to determine Tw we must compute qw for \t\tinsertion in equation(b). we have\n", + "qw = q/(math.pi*d*L) \t\t # [W]\n", + "\t# now\n", + "Tw = Tb+(qw*d/(Nu*k)) \t\t # [degree celsius]\n", + "\t# and the heat transfer coefficient is\n", + "h = qw/(Tw-Tb) \t\t\t # [W/square meter degree celsius]\n", + "print \"Total heat transfer is\",round(q,2),\"W\"\n", + "print \"\\nExit wall temperature is\",round(Tw),\" degree celsius\" \n", + "print \"\\nHeat transfer coefficient is\",round(h,2),\" W/sq meter degree C\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reynolds number is 823.0\n", + "so that the flow is laminar\n", + "Exit wall temperature is 128.66 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 6.4\n", + "# heating of air with isothermal tube wall\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 27\t\t\t\t# [degree celsius] temperature of air \n", + "d = 0.005 \t\t\t# [m] diameter of tube \n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 3 \t\t\t\t# [m/s] velocity of air\n", + "L = 0.1 \t\t\t# [m] length of tube\n", + "Tb = 77 \t\t\t# [degree celsius] exit bulk temperature \n", + "\n", + "\t# we first must evaluate the flow regime and do so by taking properties at the \taverage bulk temperature \n", + "\n", + "Tb_bar = (Ta+Tb)/2 \t\t# [degree celsius]\n", + "v = 18.22*10**(-6) \t\t# [square meter/s] kinematic viscosity\n", + "k = 0.02814 \t\t\t# [W/m degree celsius]\n", + "Cp = 1006 \t\t\t# [J/kg K]\n", + "Pr = 0.703 \t\t\t# prandtl no.\n", + "Re_d = u*d/v \t\t\t# reynolds number\n", + "print \"Reynolds number is\",round(Re_d) \n", + "print \"so that the flow is laminar\" \n", + "\t# so that the flow is laminar\n", + "\t# now we determine Nu_d_bar for Tw = constant. for Gz_inverse = 0.0346 we read \n", + "Nu_d = 5.15 \n", + "\t# we thus calculate the average heat transfer coefficient as \n", + "\n", + "h_bar = Nu_d*k/d \t\t# [W/square meter degree celsius]\n", + "\t# we base the heat transfer on a mean bulk temperature of Tb_bar, so that\n", + "import math\n", + "Tw = 3.49/(h_bar*math.pi*d*L)+Tb_bar \t# [degree celsius]\n", + "\n", + "\n", + "print \"Exit wall temperature is\",round(Tw,2),\"degree celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The length of tube necessary to accomplish the heating is 1.4 m\n" + ] + } + ], + "source": [ + "#Example Number 6.5\n", + "# heat transfer in a rough tube \n", + "\n", + "# Variable declaration\n", + "\n", + "Tw = 90 \t\t\t# [degree celsius] temperature of tube wall \n", + "d = 0.02 \t\t\t# [m] diameter of tube \n", + "u = 3 \t\t\t\t# [m/s] velocity of air\n", + "Tw_i = 40 \t\t\t# [degree celsius] entering water temperature \n", + "Tw_f = 60 \t\t\t# [degree celsius] leaving water temperature\n", + "Cp = 4.174*10**3 \t\t# [J/kg K]\n", + "\n", + "#Calculation\n", + "\n", + "Tb_bar = (Tw_i+Tw_f)/2 \t\t# [degree celsius]\n", + "\n", + "\t#we first calculate the heat transfer from q = m_dot*Cp*dTb\n", + "\t\n", + "import math\n", + "q = 989*3*math.pi*0.01**(2)*4174*(Tw_f-Tw_i) \t# [W]\n", + "\n", + "\t# for the rough tube condition, we may employ the Petukhov relation, equation \t\t (6-7) The mean film temperaturee is \n", + "\n", + "Tf = (Tw+Tb_bar)/2 \t\t# [degree celsius]\n", + "\n", + "\t# and the fluid properties are \n", + "\n", + "rho = 978 \t\t\t# [kg/cubic meter] density of gas\n", + "mu = 4.0*10**(-4) \t\t# [kg/m s] viscosity \n", + "k = 0.664 \t\t\t# [W/m degree celsius]\n", + "Pr = 2.54 \t\t\t# prandtl no.\n", + "\n", + "\t# also\n", + "\n", + "mu_b = 5.55*10**(-4) \t\t# [kg/m s] viscosity \n", + "mu_w = 2.81*10**(-4) \t\t# [kg/m s] viscosity \n", + "\n", + "\t# the reynolds number is thus \n", + "\n", + "Re_d = rho*u*d/mu \n", + "\n", + "\t# consulting figure(6-14), we find the friction factor as \n", + "\n", + "f_f = 0.0218 \n", + "\n", + "\t# because Tw>Tb, we take \n", + "\n", + "n = 0.11 \n", + "\n", + "\t# and obtain\n", + "\n", + "Nu_d=((f_f*Re_d*2.54)/((1.07+12.7*(f_f/8)**(0.5)*(2.54**(2.0/3.0)-1))*8))*(mu_b/mu_w)**(n) \n", + "h = Nu_d*k/d \t\t\t# [W/square meter degree celsius]\n", + "\n", + "\t# the tube length is obtained from energy balance \n", + "\n", + "L = q/(h*math.pi*d*(Tw-Tb_bar)) # [m]\n", + "\n", + "print \"The length of tube necessary to accomplish the heating is\",round(L,2),\"m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reynolds number is 50988.0\n", + "so the flow is turbulent\n", + "The constant heat flux that must be applied at the tube surface to result in an air temperature rise of 5 degree celsius is 11870.8 W/square meter\n", + "average wall temperature is 370.0 K\n" + ] + } + ], + "source": [ + "#Example Number 6.6\n", + "# turbulent heat transfer in a short tube \n", + "\n", + "#Variable declaration\n", + "\n", + "p = 101325\t\t\t # [Pa] pressure of air\n", + "Ta = 300 \t\t\t # [K] temperature of air \n", + "d = 0.02 \t\t\t # [m] diameter of tube \n", + "u = 40 \t\t\t\t # [m/s] velocity of air\n", + "L = 0.1\t\t\t\t # [m] length of tube\n", + "dT = 5.0 \t\t\t\t # [degree celsius] rise in temperature \n", + "#\t we first must evaluate the air properties at 300 K \n", + "v = 15.69*10**(-6) \t\t # [square meter/s] kinematic viscosity\n", + "k = 0.02624 \t\t\t # [W/m degree celsius]\n", + "Cp = 1006.0\t\t\t # [J/kg K]\n", + "Pr = 0.70 \t\t\t# prandtl no.\n", + "rho = 1.18 \t\t\t# [kg/cubic meter]\n", + "Re_d = u*d/v \t\t\t# reynolds number\n", + "\n", + "print \"Reynolds number is\",round(Re_d)\n", + "print\"so the flow is turbulent\"\n", + "\n", + "\t# consulting figure (6-6) for this value of Re_d and L/d = 5 we find\n", + "Nu_x_by_Nu_inf = 1.15 \n", + "\t# or the heat transfer coefficient is about 15 percent higher that it would be \tfor thermally developed flow.\n", + "\t# we calculate heat-transfer for developed flow using \n", + "Nu_d = 0.023*Re_d**(0.8)*Pr**(0.4) \n", + "\t# and \n", + "h = k*Nu_d/d \t\t\t # [W/square meter degree celsius]\n", + "\t# increasing this value by 15 percent\n", + "h = 1.15*h \t\t\t # [W/square meter degree celsius]\n", + "\t# the mass flow is\n", + "import math\n", + "Ac = math.pi*d**(2)/4 \t\t # [square meter] \n", + "m_dot = rho*u*Ac \t\t # [kg/s]\n", + "\t# so the total heat transfer is\n", + "A = math.pi*d*L\t\t\t # [square meter] \n", + "q_by_A = m_dot*Cp*dT/A\t\t # [W/square meter]\n", + "print \"The constant heat flux that must be applied at the tube surface to result in an air temperature rise of 5 degree celsius is\",q_by_A,\" W/square meter\"\n", + "Tb_bar = (Ta+(Ta+dT))/2 \t # [K]\n", + "Tw_bar = Tb_bar+q_by_A/h \t # [K] \n", + "print \"average wall temperature is\",round(Tw_bar),\"K\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss per unit length of cylinder is 3100.0 W/m\n" + ] + } + ], + "source": [ + "#Example Number 6.7\n", + "# airflow across isothermal cylinder\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 35+273.15 \t\t\t# [K] temperature of air \n", + "d = 0.05 \t\t\t# [m] diameter of tube \n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 50 \t\t\t\t# [m/s] velocity of air\n", + "Tc = 150+273.15 \t\t# [degree celsius] cylinder temperature\n", + "\n", + "\t# we first find the reynolds number and then find the applicable constants \tfrom table(6-2) for use with equation (6-17) \n", + "\t# the properties of air are evaluated at the film temperature:\n", + "\n", + "#Calculation\n", + "\n", + "Tf = (Ta+Tc)/2 \t\t\t# [K]\n", + "rho_f = p/(R*Tf) \t\t# [kg/cubic meter]\n", + "mu_f = 2.14*10**(-5) \t\t# [kg/m s]\n", + "k_f = 0.0312 \t\t\t# [W/m degree celsius]\n", + "Pr_f = 0.695 \t\t\t# prandtl number\n", + "Re_f = rho_f*u*d/mu_f \t\t# reynolds number\n", + "\n", + "\t# from table (6-2) table\n", + "\n", + "C = 0.0266 \n", + "n = 0.805 \n", + "\n", + "\t# so from equation (6-17)\n", + "\n", + "h = C*(Re_f)**(n)*(Pr_f)**(1.0/3.0)*k_f/d # [W/sq m deg C] heat transfer coefficient\n", + "\n", + "\t# the heat transfer per unit length is \n", + "import math\n", + "\n", + "q_by_L = h*math.pi*d*(Tc-Ta) \t\t# [W/m]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat loss per unit length of cylinder is\",round(q_by_L),\"W/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.8" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost per unit length by the wire is 11.88 W/m\n" + ] + } + ], + "source": [ + "#Example Number 6.8\n", + "# heat transfer from electrically heated\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Tw = 25+273.15 \t\t\t# [K] temperature of air \n", + "d = 3.94*10**(-5) \t\t# [m] diameter of wire\n", + "R = 287 \t\t\t# [] gas constant\n", + "u = 50 \t\t\t\t# [m/s] velocity of air perpendicular to the air\n", + "Tr = 50+273.15 \t\t\t# [degree celsius] rise in surface temperature\n", + "\t# we first obtain the properties at the film temperature :\n", + "\n", + "#Calculation\n", + "\n", + "Tf = (Tw+Tr)/2 \t\t\t# [K]\n", + "v_f = 16.7*10**(-6) \t\t# [square meter/s]\n", + "k = 0.02704 \t\t\t# [W/m degree celsius]\n", + "Pr_f = 0.706 \t\t\t# prandtl number\n", + "Re_d = u*d/v_f \t\t\t# reynolds number\n", + "\t# the Peclet number is \n", + "Pe = Re_d*Pr_f \n", + "\t# and we find that equations (6-17),(6-21), or (6-19) apply.\n", + "\t# let us make the calculation with both the simplest expression, (6-17),and \t\tthe most complex,(6-21), and compare results.\n", + "\t# using equation (6-17) with \n", + "C = 0.683 \n", + "n = 0.466 \n", + "\t# we have\n", + "Nu_d = 0.683*Re_d**(n)*Pr_f**(1/3) \n", + "\t# the value of heat transfer coefficient is\n", + "h = Nu_d*k/d \t\t\t # [W/square meter degree celsius]\n", + "\t# the heat transfer per unit length is then \n", + "import math\n", + "q_by_L = math.pi*d*h*(Tr-Tw)\t # [W/m]\n", + "\t# using equation (6-21), we calculate the nusselt no as \n", + "Nu_d1=0.3+((0.62*Re_d**(1.0/2.0)*Pr_f**(1.0/3.0))/((1+(0.4/Pr_f)**(2.0/3.0))**(1.0/4.0))*((1+(Re_d/282000)**(5.0/8.0))**(4.0/5.0))) \n", + "h1 = Nu_d1*k/d \t\t\t # [W/square meter degree celsius]\n", + "\t# and\n", + "q_by_L1 = h1*math.pi*d*(Tr-Tw) \t # [W/m]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat lost per unit length by the wire is\",round(q_by_L1,2),\"W/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.9" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost by the sphere is 1.554 W\n" + ] + } + ], + "source": [ + "#Example Number 6.9\n", + "# heat transfer from sphere \n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "Ta = 27+273.15 \t\t\t# [K] temperature of air \n", + "d = 0.012 \t\t\t# [m] diameter of sphere\n", + "u = 4 \t\t\t\t# [m/s] velocity of air \n", + "Ts = 77+273.15 \t\t\t# [degree celsius] surface temperature of sphere\n", + "\t# consulting equation (6-30) we find that the reynolds number is evaluated at \t\tthe free-stream temperature.\n", + "\t# we therefore need the following properties at Ta = 300.15K\n", + "v = 15.69*10**(-6) \t\t# [square meter/s]\n", + "k = 0.02624 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.708 \t\t\t# prandtl number\n", + "mu_inf = 1.8462*10**(-5) \t# [kg/m s]\n", + "\t# at Ts = 350K\n", + "mu_w = 2.075*10**(-5) \t\t# [kg/m s]\n", + "Re_d = u*d/v \t\t\t# reynolds number\n", + "\n", + "#Calculation\n", + "\t# from equation (6-30),\n", + "Nu_bar=2+((0.4)*(Re_d)**(1.0/2.0)+0.06*(Re_d)**(2.0/3.0))*(Pr**(0.4))*((mu_inf/mu_w)**(1.0/4.0)) \n", + "\n", + "\n", + "\t# and\n", + "\n", + "h_bar = Nu_bar*k/d \t\t# [W/sq m degree celsius] heat transfer coefficient\n", + "\n", + "\t# the heat transfer is then \n", + "\n", + "import math\n", + "\n", + "A = (4*math.pi*d**(2))/4 \t# [square meter] area of sphere\n", + "\n", + "q = h_bar*A*(Ts-Ta) \t\t# [W]\n", + "\n", + "\n", + "\t# for comparison purposes let us also calculate the heat-transfer coefficient \t\t using equation(6-25). the film temperature is \n", + "Tf = (Ta+Ts)/2 \t\t\t# [K]\n", + "v_f = 18.23*10**(-6) \t\t# [square meter/s]\n", + "k_f = 0.02814 \t\t\t# [W/m degree celsius] \n", + "\t# reynolds number is \n", + "Re_d1 = u*d/v_f \n", + "\t# from equation (6-25)\n", + "Nu_f = 0.37*(u*d/v_f)**(0.6) \n", + "\t# and h_bar is calculated as\n", + "\n", + "h_bar = Nu_f*k_f/d \t\t# [W/sq m degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat lost by the sphere is\",round(q,3),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.11" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer coefficient is 185.6 W/square meter degree celsius\n", + "Heat transfer coefficient for previous problem is 163.5 W/sq meter degree C\n", + "Percentage increase in value of h is 14.0\n" + ] + } + ], + "source": [ + "#Example Number 6.11\n", + "# alternate calculation method \n", + "\n", + "# Variable declaration\n", + "\n", + "\t# data for this example is taken from previous example (6-10)\n", + "\t# properties for use in equation (6-34) are evaluated at free-atream \tconditions of 10 degree celsius\n", + "v = 14.2*10**(-6)\t\t\t # [square meter/s]\n", + "k = 0.0249 \t\t\t\t # [W/m degree celsius]\n", + "Pr = 0.712 \t\t\t\t # prandtl number\n", + "Pr_w = 0.70 \t\t\t\t# prandtl number\n", + "u = 7 \t\t\t\t\t# [m/s] velocity of air \n", + "Sp = 0.0381 \t\t\t\t# [m] spacing between normal and parallel \t\t\t\t\t\tdirection to the flow\n", + "Sn = 0.0381 \t\t\t\t# spacing between normal and parallel \t\t\t\t\t direction to the flow\n", + "d = 0.0254 \t\t\t\t# [m] diameter of tube\n", + "\t#maximum velocity is \n", + "u_max = u*(Sn/(Sn-d)) \t\t\t# [m/s]\n", + "\t# the reynolds number is \n", + "Re_d_max = u_max*d/v \n", + "\t# so that the constants for equation (6-34) are\n", + "C = 0.27 \n", + "n = 0.63 \n", + "\t# inserting values we obtain\n", + "h = C*Re_d_max**(n)*(Pr/Pr_w)**(1/4)*k/d \t# [W/sq m degree C] heat transfer \t\t\t\t\t\t\tcoefficient\n", + "\t# multiplying by 0.92 from table 6-7 (page no.-300) to correct for only five \ttube rows gives\n", + "h = 0.92*h \t\t\t\t\t# [W/square meter degree celsius]\n", + "print \"Heat transfer coefficient is\",round(h,1),\"W/square meter degree celsius\"\n", + "\n", + "h_in = 163.46432 \t\t\t# [W/sq m deg C] from previous example\n", + "\n", + "print \"Heat transfer coefficient for previous problem is\",round(h_in,1),\"W/sq meter degree C\" \n", + "P = (h-h_in)*100/h_in \n", + "print \"Percentage increase in value of h is\",round(P) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 6.12" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of tube required to effect the heat transfer is 1.56 m\n" + ] + } + ], + "source": [ + "#Example Number 6.12\n", + "# heating of liquid bismuth in tube \n", + "\n", + "# variable declaration\n", + "\n", + "m_dot = 4.5 \t\t\t# [Kg/s] flow rate of bismuth\n", + "d = 0.05 \t\t\t# [m] diameter of steel tube\n", + "\n", + "Ti = 415 \t\t\t# [degree celsius] initial temperature of bismuth\n", + "Tf = 440 \t\t\t# [degree celsius] final temperature of bismuth\n", + "\t# because a constant heat flux is maintained, we may use equation 6-47 to \t\tcalculate the heat transfer coefficient.\n", + "\t# the properties of bismuth are evaluated at the average bulk temperature of \n", + "#Calculation\n", + "\n", + "Ta = (Ti+Tf)/2 \t\t\t# [degree celsius]\n", + "mu = 1.34*10**(-3) \t\t# [Kg/m s] viscosity\n", + "Cp = 149 \t\t\t# [J/Kg degree celsius] heat \n", + "k = 15.6 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.013 \t\t\t# prandtl number\n", + "\t# the total transfer is calculated from\n", + "q = m_dot*Cp*(Tf-Ti) \t\t# [W]\n", + "\t# we calculate reynolds and peclet number as \n", + "\n", + "import math\n", + "G = m_dot/(math.pi*d**(2)/4) \n", + "Re_d = d*G/mu \n", + "Pe = Re_d*Pr \n", + "\t# the heat transfer coefficient is calculated from equation 6-47\n", + "Nu_d = 4.82+0.0185*Pe**(0.827) \n", + "h = Nu_d*k/d \t\t\t\t# [W/square meter degree celsius]\n", + "\t# the total required surface area of the tube may now be computed from q=h*A*DT\n", + "\t# where we use the temperature difference of\n", + "DT = 20 \t\t\t# [degree celsius] \n", + "A = q/(h*DT) \t\t\t# [square meter] \n", + "\t# the area in turn can be expressed in terms of tube length \n", + "L = A/(math.pi*d) \t\t# [m]\n", + "\n", + "#Result\n", + "print\"Length of tube required to effect the heat transfer is\",round(L,2),\"m\" " + ] + } + ], + "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", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |