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diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_10_Heat_Exchangers.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_10_Heat_Exchangers.ipynb new file mode 100644 index 00000000..b6a504f0 --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_10_Heat_Exchangers.ipynb @@ -0,0 +1,775 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10 Heat Exchangers " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.3" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "hi is: 1409.0\n", + "% reduction because of fouling factor is 28.0\n" + ] + } + ], + "source": [ + "#Example Number 10.3\n", + "# influence of fouling factor\n", + "\n", + "#Variable declaration\n", + "\t\n", + "\t\n", + "Rf = 0.0002 \n", + "\t# using Rf=(1/hi-1/h_clean)\n", + "h_clean = 1961.0 \t\t\t# [W/square meter degree celsius]\n", + "\t# we obtain \n", + "\n", + "#Calculation\n", + "\n", + "hi = 1/(Rf+(1/h_clean)) \t\t# [W/square meter degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print \"hi is:\",round(hi)\n", + "print \"% reduction because of fouling factor is \",round((h_clean-hi)*100/h_clean) \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.4" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Area of heat-exchanger is 15.81 square meter\n" + ] + } + ], + "source": [ + "#Example Number 10.4\n", + "# calculation of heat exchanger size from known temperatures\n", + "\n", + "#Variable declaration\n", + "import math\n", + "m_dot = 68.0 \t\t\t# [kg/min] water flow rate \n", + "U = 320.0 \t\t\t# [W/sq m degree C] overall heat transfer coefficient\n", + "T1 = 35.0 \t\t\t# [degree celsius] initial temperature \n", + "T2 = 75.0 \t\t\t# [degree celsius] final temperature\n", + "Toe = 110.0 \t\t\t# [degree celsius] oil entering temperature \n", + "Tol = 75.0 \t\t\t# [degree celsius] oil leaving temperature\n", + "Cw = 4180.0 \t\t\t# [J/kg degree celsius] water specific heat capacity\n", + "\t# the total heat transfer is determined from the energy absorbed by the water:\n", + "\n", + "#Claculation\n", + "\n", + "q = m_dot*Cw*(T2-T1) \t\t# [J/min]\n", + "q = q/60 \t\t\t# [W]\n", + "\t# since all the fluid temperatures are known, the LMTD can be calculated by \t\tusing the temperature scheme in figure 10-7b\n", + "dT_m = ((Toe-Tol)-(T2-T1))/(math.log((Toe-Tol)/(T2-T1))) \t# [degree celsius]\n", + "\t\t\t\t# then, since q = U*A*dT_m\n", + "A = q/(U*dT_m) \t\t\t# [square meter] area of heat-exchanger\n", + "\n", + "#Result\n", + "\n", + "print \"Area of heat-exchanger is\",round(A,2),\"square meter\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.5" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Area required for this exchanger is 19.53 square meter\n" + ] + } + ], + "source": [ + "#Example Number 10.5\n", + "# shell-and-tube heat exchanger\n", + "\n", + "# Variable declaration\n", + "\t\n", + "\t# determine a correction factor from figure 10-8 to be used \n", + "\t# the parameters according to figure 10-8(page no.-532) are \n", + "T1 = 35 \t\t\t# [degree celsius]\n", + "T2 = 75 \t\t\t# [degree celsius]\n", + "t1 = 110 \t\t\t# [degree celsius]\n", + "t2 = 75 \t\t\t# [degree celsius]\n", + "P = (t2-t1)/(T1-t1) \n", + "R = (T1-T2)/(t2-t1) \n", + "\t# so the correction factor is \n", + "F = 0.81 \t\t\t# from figure 10-10(page no.-534)\n", + "\t# and the heat transfer is q = U*A*F*dT_m\n", + "\t# so that. from example 10-4 we have \n", + "U = 320 \t\t\t# [W/sq m deg C] overall heat transfer coefficient\n", + "q = 189493.33 \t\t\t# [W]\n", + "\n", + "#Calculation\n", + "\n", + "dT_m = 37.44 \t\t\t# [degree celsius]\n", + "A = q/(U*F*dT_m) \t\t# [square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Area required for this exchanger is\",round(A,2),\"square meter\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.6" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Number of tubes per pass 37.0\n", + "Number of passes = 2\n", + "Length of tube per pass = 1.708 m\n" + ] + } + ], + "source": [ + "#Example Number 10.6\n", + "# design of shell-and-tube heat exchanger\n", + "\n", + "# Variable declaration\n", + "\n", + "import math\n", + "m_dot_c = 3.8 \t\t\t# [kg/s] water flow rate\n", + "Ti = 38 \t\t\t# [degree celsius] initial temperature of water\n", + "Tf = 55 \t\t\t# [degree celsius] final temperature of water\n", + "m_dot_h = 1.9 \t\t\t# [kg/s] water flow rate entering the exchanger\n", + "Te = 93 \t\t\t# [degree celsius] entering water temperature\n", + "U = 1419 \t\t\t# [W/sq m degree C] overall heat transfer coefficient\n", + "d = 0.019 \t\t\t# [m] diameter of tube\n", + "v_avg = 0.366 \t\t\t# [m/s] average water velocity in exchanger\n", + "Cc = 4180 \t\t\t# [] specific heat of water\n", + "Ch = Cc \t\t\t# [] specific heat \n", + "rho = 1000 \t\t\t# [kg/cubic meter] density of water\n", + "\t# we first assume one tube pass and check to see if it satisfies the \t\t\tconditions of this problem. the exit temperature of the hot water is \t\t\tcalculated from\n", + "\n", + "#Calculation\n", + "\n", + "dTh = m_dot_c*Cc*(Tf-Ti)/(m_dot_h*Ch) \t# [degree celsius]\n", + "Th_exit = Te-dTh \t\t\t# [degree celsius]\n", + "\t# the total required heat transfer is obtained for the cold fluid is \n", + "q = m_dot_c*Cc*(Tf-Ti) \t\t\t# [W]\n", + "\t# for a counterflow exchanger, with the required temperature \n", + "LMTD = ((Te-Tf)-(Th_exit-Ti))/math.log((Te-Tf)/(Th_exit-Ti)) \t# [degree celsius]\n", + "dTm = LMTD \t\t\t\t# [degree celsius]\n", + "A = q/(U*dTm) \t\t\t\t# [square meter]\n", + "\n", + "\n", + "\t#calculate the total area with\n", + "A1 = m_dot_c/(rho*v_avg) \t\t# [square meter]\n", + "\t# this area is the product of number of tubes and the flow area per tube:\n", + "n = A1*4/(math.pi*d**(2)) \t\t# no. of tubes\n", + "n = round(n) \t\n", + "\t# rounding of value of n because no. of pipe is an integer value\n", + "\t# the surface area per tube per meter of length is \n", + "S = math.pi*d \t\t\t\t# [square meter/tube meter]\n", + "\t# total surface area required for a one tube pass exchanger\t\t was \t\tcalculated above .\n", + "\t# we may thus compute the length of tube for this type of exchanger from \n", + "L = A/(S*n) \t\t\t\t# [m]\n", + "\t# this length is greater than the allowable 2.438 m, so we must use more than \t\tone tube pass.\n", + "\t\n", + "\t# we next try two tube passes. from figure 10-8(page no.-532) \n", + "F = 0.88 \n", + "A_total = q/(U*F*dTm) \t\t\t# [square meter]\n", + "\t# the number of tubes per pass is still 37 because of the velocity \t\t\trequirement. for the two pass exchanger the total surface area is now related \t\tto the length by\n", + "L1 = A_total/(2*S*n) \t\t\t# [m]\n", + "\t# this length is within the 2.438 m requirement, so the final design choice is \n", + "\n", + "#Result\n", + "\n", + "print \"Number of tubes per pass\",n \n", + "print \"Number of passes = 2\" \n", + "print \"Length of tube per pass =\",round(L1,3),\"m\" \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.7" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Surface area of heat exchanger is 10.84 square meter\n" + ] + } + ], + "source": [ + "#Example Number 10.7\n", + "# cross flow exchanger with one fluid mixed \n", + "\n", + "# Variable declaration\n", + "import math\n", + "\n", + "m_dot = 5.2 \t\t\t# [kg/s] mass flow rate\n", + "T1 = 130.0 \t\t\t# [degree celsius] temperature of entering steam\n", + "T2 = 110.0 \t\t\t# [degree celsius] temperature of leaving steam\n", + "t1 = 15.0 \t\t\t# [degree celsius] temperature of entering oil\n", + "t2 = 85.0 \t\t\t# [degree celsius] temperature of leaving oil\n", + "c_oil = 1900.0 \t\t\t# [J/kg degree celsius] heat capacity of oil\n", + "c_steam = 1860.0\t\t# [J/kg degree celsius] heat capacity of steam\n", + "U = 275 \t\t\t# [W/sq m deg C] overall heat transfer coefficient\n", + "\t#the total heat transfer may be obtained from an energy balance on the steam \n", + "\n", + "#Calculation\n", + "\n", + "q = m_dot*c_steam*(T1-T2) \t\t\t\t# [W]\n", + "\t# we can solve for the area from equation (10-13). the value of dT_m is \t\tcalculated as if the exchanger were counterflow double pipe,thus\n", + "dT_m = ((T1-t2)-(T2-t1))/math.log((T1-t2)/(T2-t1)) \t# [degree celsius]\n", + "\n", + "\t# t1,t2 is representing the unmixed fluid(oil) and T1,T2 is representing the \t\tmixed fluid(steam) so that:\n", + "\t# we calculate \n", + "\n", + "R = (T1-T2)/(t2-t1) \n", + "P = (t2-t1)/(T1-t1) \n", + "\t# consulting figure 10-11(page no.-534) we find \n", + "F = 0.97 \n", + "\t# so the area is calculated from \n", + "A = q/(U*F*dT_m) \t\t\t\t\t# [square meter]\n", + "\n", + "#Result\n", + "print \"Surface area of heat exchanger is \",round(A,2),\"square meter\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.8" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the objective of this example is to show that an iterative procedure is required when the inlet and outlet temperatures are not known or easily calculated\n", + "there is no need to go through this iteration because it can be avoided by using the techniques described in section 10-6\n" + ] + } + ], + "source": [ + "#Example Number 10.8\n", + "# effects of off-design flow rates for exchanger in example 10-7 \n", + "# Variable declaration\n", + "\n", + "# we did not calculate the oil flow in example 10-7 but can do so now from \n", + "q = 193 # [kW]\n", + "c_oil = 1.9 # [J/kg degree celsius] heat capacity of oil\n", + "t1 = 15 # [degree celsius] temperature of entering oil\n", + "t2 = 85 # [degree celsius] temperature of leaving oil\n", + "m_dot_o = q/(c_oil*(t2-t1)) # [kg/s]\n", + "# the new flow rate will be half this value \n", + "m_dot_o = m_dot_o/2 # [kg/s]\n", + "# we are assuming the inlet temperatures remain the same at 130 degree celsius for the steam and 15 degree celsius for the oil.\n", + "# the new relation for the heat transfer is q = m_dot_o*c_oil*(Teo-15) = m_dot_s*cp*(130-Tes) (a)\n", + "# but the exit temperatures, Teo and Tes are unknown. furthermore, dT_m is unknown without these temperatures, as are the values of R and P from figure 10-11(page no.-535). this means we must use an iterative procedure to solve for the exit temperatures using equation (a) and q = U*A*F*dT_m (b)\n", + "# the general procedure is to assume values of the exit temperatures until the q's agree between equations(a) and (b).\n", + "print \"the objective of this example is to show that an iterative procedure is required when the inlet and outlet temperatures are not known or easily calculated\" \n", + "print \"there is no need to go through this iteration because it can be avoided by using the techniques described in section 10-6\" \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.9" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A Reduction in the oil flow rate of 50 % causes a reduction in heat transfer of only 34.0 percent\n" + ] + } + ], + "source": [ + "#Example Number 10.9\n", + "# off-design calculation using E-NTU method \n", + " \n", + "#Variable declaration\t\n", + "import math\n", + "m_dot_o = 0.725 \t\t\t# [kg/s] oil flow rate\n", + "m_dot_s = 5.2 \t\t\t\t# [kg/s] steam flow rate\n", + "t1 = 15 \t\t\t\t# [degree celsius] temperature of entering oil\n", + "T1 = 130 \t\t\t\t# [deg C] temperature of entering steam\n", + "c_oil = 1900 \t\t\t\t# [J/kg degree celsius] heat capacity of oil\n", + "c_steam = 1860 \t\t\t\t# [J/kg degree celsius] heat capacity of steam\n", + "\t# for the steam \n", + "Cs = m_dot_s*c_steam \t\t\t# [W/degree celsius]\n", + "\t# for the oil\n", + "Co = m_dot_o*c_oil \t\t\t# [W/degree celsius]\n", + "\t# so the oil is minium fluid. we thus have\n", + "C_min_by_C_max = Co/Cs \n", + "U = 275 \t\t\t\t# [W/sq m deg C] overall heat transfer \t\t\t\t\t\t\t coefficient\n", + "A = 10.83 \t\t\t\t# [sq meter] surface area of heat exchanger\n", + "NTU = U*A/Co \n", + "\t# we choose to use the table and note that Co(minimum) is unmixed and \t\t\tCs(maximum) is mixed so that the first relation in the table 10-3 applies.\n", + "\t# we therfore calculate E(effectiveness) as \n", + "\n", + "E = (1/C_min_by_C_max)*(1-math.exp(-C_min_by_C_max*(1-math.exp(-NTU)))) \n", + "\t# if we were using figure 10-14(page no.-544) we would have to evaluate \n", + "C_mixed_by_C_unmixed = Cs/Co \n", + "\t# and would still determine \n", + "E = 0.8 # approximately\n", + "\t# now, using the effectiveness we can determine the temperature difference of \t\tthe minimum fluid(oil as)\n", + "dT_o = E*(T1-t1) \t\t\t# [degree celsius]\n", + "\t# so that heat transfer is \n", + "q = m_dot_o*c_oil*(dT_o) \t\t# [W]\n", + "q_initial = 193440 \t\t\t# [W] heat transfer when oilrate is 100 %\n", + "print \"A Reduction in the oil flow rate of 50 % causes a reduction in heat transfer of only \",round((q_initial-q)*100/q_initial),\"percent\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.10" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Exit water temperature is 90.8 degree celcius\n", + "The total heat transfer under the new flow conditions is 155.5 kW\n" + ] + } + ], + "source": [ + "#Example Number 10.10\n", + "# off-design calculation of exchanger in example 10-4 \n", + "\n", + "# Variable declaration\n", + "\n", + "m_dot_c = 68 \t\t\t# [kg/min] water flow rate\n", + "\n", + "T1 = 35 \t\t\t# [degree celsius] initial temperature \n", + "T2 = 75 \t\t\t# [degree celsius] final temperature\n", + "Toe = 110 \t\t\t# [degree celsius] oil entering temperature \n", + "Tol = 75 \t\t\t# [degree celsius] oil leaving temperature\n", + "Cc = 4180 \t\t\t# [J/kg degree celsius] water specific heat capacity\n", + "Ch = 1900 \t\t\t# [J/kg degree celsius] heat capacity of oil\n", + "U = 320 \t\t\t# [W/squ m deg C] overall heat transfer coefficient\n", + "A = 15.814568 \t\t\t# [sq m] area of heat exchanger (from example 10-4)\n", + "\t# the flow rate of oil is calculated from the energy balance for the original \t\tproblem:\n", + "\n", + "#Calculation\n", + "m_dot_h = m_dot_c*Cc*(T2-T1)/(Ch*(Toe-Tol)) \t# [kg/min]\n", + "\t# the capacity rates for the new conditions are calculated as \n", + "C_h = m_dot_h*Ch/60 \t\t\t\t# [W/degree celsius]\n", + "C_c = m_dot_c*Cc/60 \t\t\t\t# [W/degree celsius]\n", + "\t# so that the water (cold fluid) is the minimum fluid, and \n", + "C_min_by_C_max = C_c/C_h \n", + "NTU_max = U*A/C_c \n", + "\t# from figure 10-13(page no.-542) or table 10-3(page no.-543) the \t\teffectiveness is \n", + "E = 0.744 \n", + "\t# and because the cold fluid is the minimum, we can write \n", + "dT_cold = E*(Toe-T1) \t\t\t\t# [degree celsius]\n", + "\t\t\t\t\t\t# and the exit water temperature is \n", + "Tw_exit = T1+dT_cold \t\t\t\t# [degree celsius]\n", + "\t# the total heat transfer under the new flow conditions is calculated as \n", + "m_dot_c = 40 \t\t\t\t\t# [kg/min]\n", + "q = m_dot_c*Cc*dT_cold/60 \t\t\t# [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Exit water temperature is\",Tw_exit,\"degree celcius\" \n", + "print \"The total heat transfer under the new flow conditions is\",round(q/1000,1),\"kW\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.11" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The exit water temperature is 21.1 degree celsius\n", + "the heat transfer is 40.33 kW\n" + ] + } + ], + "source": [ + "#Example Number 10.11\n", + "# cross-flow exchanger with both fluid unmixed \n", + " \n", + "# Variable declaration\n", + "\n", + "pa = 101325 \t\t\t# [Pa] pressure of air\n", + "Ti = 15.55 \t\t\t# [degree celsius] initial temperature of air\n", + "Tf = 29.44 \t\t\t# [degree celsius] final temperature of air\n", + "Thw = 82.22 \t\t\t# [degree celsius] hot water temperature\n", + "U = 227 \t\t\t# [W/sq m deg C] overall heat transfer coefficient\n", + "S = 9.29 \t\t\t# [square meter] total surface area of heat exchanger\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "Cc = 1006 \t\t\t# [J/kg degree celsius] specific heat of air \n", + "Ch = 4180 \t\t\t# [J/kg degree celsius] specific heat of water\n", + "\t# the heat transfer is calculated from the energy balance on the air. first, \t\tthe inlet air density is \n", + "rho = pa/(R*(Ti+273.15)) \t# [kg/cubic meter]\n", + "\t# so the mass flow of air (the cold fluid) is \n", + "mdot_c = 2.36*rho \t\t# [kg/s]\n", + "\t# the heat transfer is then \n", + "q = mdot_c*Cc*(Tf-Ti) \t\t# [W]\n", + "\t# from the statement of the problem we do not know whether the air or water is \tthe minimum fluid. a trial and error procedur must be used \n", + "\t# we assume that the air is the minimum fluid and then check out our \t\tassumption.then\n", + "Cmin = mdot_c*Cc \t\t# [W/degree celsius]\n", + "NTU_max = U*S/Cmin \n", + "\t# and the effectiveness based on the air as the minimum fluid is \n", + "E = (Tf-Ti)/(Thw-Ti) \n", + "\t# we must assume values for the water flow rate until we are able to match the \tperformance as given by figure 10-15 or table 10-3. we first note that\n", + "Cmax = mdot_c*Cc \t\t# [W/degree celsius] (a)\n", + "\t\t\t\t# NTU_max = U*S/Cmin (b)\n", + "\t\t\t\t# E = dT_h/(Thw-Ti) (c)\n", + "\t\t\t\t# dT_h = q/Cmin (d)\n", + "\n", + "\t# now we assume different values for Cmin abd calculate different-different \t\tvalues for NTU_max, dT_h, and E\n", + "\n", + "\t# for \n", + "Cmin_by_Cmax1 = 0.5 \n", + "Cmin1 = Cmin_by_Cmax1*Cmax \t\t\t# [W/degree celsius]\n", + "NTU_max1 = U*S/Cmin1 \n", + "dT_h1 = q/Cmin1 \t\t\t\t# [degree celsius]\n", + "E1_c1 = dT_h1/(Thw-Ti) \t\t\t\t# calculated\n", + "E1_t1 = 0.65 \t\t\t\t\t# from table \n", + "\n", + "\t# for \n", + "Cmin_by_Cmax2 = 0.25 \n", + "Cmin2 = Cmin_by_Cmax2*Cmax \t\t\t# [W/degree celsius]\n", + "NTU_max2 = U*S/Cmin2 \n", + "dT_h2 = q/Cmin2 \t\t\t\t# [degree celsius]\n", + "E1_c2 = dT_h2/(Thw-Ti) \t\t\t\t# calculated\n", + "E1_t2 = 0.89 \t\t\t\t\t# from table \n", + "\n", + "\t# for \n", + "Cmin_by_Cmax3 = 0.22 \n", + "Cmin3 = Cmin_by_Cmax3*Cmax \t\t\t# [W/degree celsius]\n", + "NTU_max3 = U*S/Cmin3 \n", + "dT_h3 = q/Cmin3 \t\t\t\t# [degree celsius]\n", + "E1_c3 = dT_h3/(Thw-Ti) \t\t\t\t# calculated\n", + "E1_t3 = 0.92 \t\t\t\t\t# from table \n", + "\n", + "\t# we estimate the water-flow rate as about\n", + "Cmin = 660 \t\t\t\t\t# [W/degree celsius]\n", + "mdot_h = Cmin/Ch \t\t\t\t# [kg/s]\n", + "\t# the exit water temperature is accordingly\n", + "Tw_exit = Thw-q/Cmin \t\t\t\t# [degree celsius]\n", + "\n", + "print \"The exit water temperature is\",round(Tw_exit,1),\"degree celsius\" \n", + "print \"the heat transfer is\",round(q/1000,2),\"kW\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.13" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Area required for the heat exchanger is 20.09 square meter\n" + ] + } + ], + "source": [ + "#Example Number 10.13\n", + "# shell and tube exchangeras air heater\n", + " \n", + "# Variable declaration\n", + "\n", + "import math\n", + "To = 100.0 \t\t\t\t# [degree celsius] temperature of hot oil\n", + "m_dot_a = 2.0 \t\t\t\t# [kg/s] flow rate of air\n", + "T1 = 20.0 \t\t\t\t# [degree celsius] initial temperature of air \n", + "T2 = 80.0 \t\t\t\t# [degree celsius] final temperature of air\n", + "Cp_o = 2100.0 \t\t\t\t# [J/kg deg C] specific heat of the oil\n", + "Cp_a = 1009.0 \t\t\t\t# [J/kg deg C] specific heat of the air\n", + "m_dot_o = 3 \t\t\t\t# [kg/s] flow rate of oil\n", + "U = 200.0 \t\t\t\t# [W/sq m] overall heat transfer coefficient\n", + "\t# the basic energy balance is m_dot_o*Cp_o*(To-Toe) = m_dot_a*Cp_a*(T2-T1)\n", + "#Calculation\n", + "Toe = To-m_dot_a*Cp_a*(T2-T1)/(m_dot_o*Cp_o) # [degree celsius]\n", + "\n", + "\t# we have\n", + "m_dot_h_into_Ch = m_dot_o*Cp_o \t\t# [W/degree celsius]\n", + "m_dot_c_into_Cc = m_dot_a*Cp_a \t\t# [W/degree celsius]\n", + "\t# so the air is minimum fluid\n", + "C = m_dot_c_into_Cc/m_dot_h_into_Ch \n", + "\t# the effectiveness is \n", + "E = (T2-T1)/(To-T1) \n", + "\t# now we may use figure 10-16 to obtain NTU. \n", + "NTU = -(1+C**(2))**(-1.0/2.0)*math.log((2/E-1-C-(1+C**2)**(1.0/2.0))/(2/E-1-C+(1+C**2)**(1.0/2.0))) \n", + "\t# now, we calcuate the area as \n", + "A = NTU*m_dot_c_into_Cc/U \t\t# [square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Area required for the heat exchanger is\",round(A,2),\"square meter\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.14" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Area to achieve a heat exchanger effectiveness of 60% with an exit water temperature of 40 degree celsius is 9.16 square meter\n", + "by reducing the flow rate we have lowered the heat transfer by 37.0 percent\n" + ] + } + ], + "source": [ + "#Example Number 10.14\n", + "# ammonia condenser \n", + "\n", + "#Variable declaration\n", + "\n", + "Ta = 50 \t\t\t# [degree C] temperature of entering ammonia vapour\n", + "Tw1 = 20 \t\t\t# [degree celsius] temperature of entering water\n", + "q = 200 \t\t\t# [kW] total heat transfer required\n", + "U = 1 \t\t\t\t# [kW/sq m deg C] overall heat transfer coefficient\n", + "Tw2 = 40 \t\t\t# [deg C] temperature of exiting water\n", + "Cw = 4.18 \t\t\t# [kJ/kg degree celsius] specific heat of water\n", + "\t# the mass flow can be calculated from the heat transfer with\n", + "m_dot_w = q/(Cw*(Tw2-Tw1)) \t# [kg/s]\n", + "\t# because this is the condenser the water is the minimum fluid and \n", + "C_min = m_dot_w*Cw # [kW/degree celsius]\n", + "\t# the value of NTU is obtained from the last entry of table 10-4\n", + "E = 0.6 \t\t\t# effectiveness\n", + "\n", + "#Calculation\n", + "\n", + "import math\n", + "NTU = -math.log(1-E) \n", + "\n", + "\t# so that area is calculated as \n", + "A = C_min*NTU/U \t\t# [square meter]\n", + "\n", + "\t# when the flow rate is reduced in half the new value of NTU is \n", + "NTU1 = U*A/(C_min/2) \n", + "\n", + "\t# and the effectiveness is computed from the last entry of table 10-3\n", + "E1 = 1-math.exp(-NTU1) \n", + "\n", + "\t# the new water temperature difference is computed as \n", + "dT_w = E1*(Ta-Tw1) \t\t# [degree celsius]\n", + "\n", + "\t# so that the heat transfer is \n", + "q1 = C_min*dT_w/2 \t\t# [kW]\n", + "\n", + "#Result\n", + "\n", + "print \"Area to achieve a heat exchanger effectiveness of 60% with an exit water temperature of 40 degree celsius is\",round(A,2),\"square meter\" \n", + "print \"by reducing the flow rate we have lowered the heat transfer by\",(q-q1)*100/q,\" percent\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 10.16" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat-transfer coefficient is 174.0 W/square meter degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 10.16 \n", + "# heat-transfer coefficient in compact exchanger \n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325.0 \t\t\t# [Pa] pressure of air\n", + "T = 300.0\t\t\t# [K] temperature of entering air\n", + "u = 15.0 \t\t\t\t# [m/s] velocity of air\n", + "\t# we obtain the air properties from table A-5\n", + "rho = 1.774 \t\t\t# [kg/cubic meter] density of air\n", + "Cp = 1005.7 \t\t\t# [J/kg degree celsius] specific heat of air\n", + "mu = 1.983*10**(-5) \t\t# [kg/m s] viscosity of air\n", + "Pr = 0.708 \t\t\t# prandtl number\n", + "\t# from figure 10-19 we have\n", + "Ac_by_A = 0.697 \n", + "sigma = 0.697 \n", + "\n", + "#Calculation\n", + "\n", + "Dh = 3.597*10**(-3) \t\t# [m] \n", + "\n", + "\t\t\t\t# the mass velocity is thus \n", + "G = ((rho*u)/sigma) \t\t# [kg/square meter s]\n", + "\n", + "\n", + "\t# and the reynolds number is \n", + "Re = Dh*G/mu \n", + "\n", + "\t# from figure 10-19(page no.-557) we can read\n", + "\n", + "St_into_Pr_exp_2_by_3 = 0.0036\n", + " \n", + "\t# and the heat transfer coefficient is \n", + "\n", + "h = St_into_Pr_exp_2_by_3*G*Cp*(Pr)**(-2.0/3.0) \t# [W/sq m deg C]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat-transfer coefficient is\",round(h),\"W/square meter degree celsius\"" + ] + } + ], + "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 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_11_Mass_Transfer.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_11_Mass_Transfer.ipynb new file mode 100644 index 00000000..764a2588 --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_11_Mass_Transfer.ipynb @@ -0,0 +1,254 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 Mass Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "value of diffusion coefficient for co2 in air is 0.132 square centimeter/s\n", + "From Table A-8,D=0.164 sq cm/s\n", + " So,they are in fair agreement\n" + ] + } + ], + "source": [ + "#Example Number 11.1\n", + "# diffusion coefficient for co2\n", + "\n", + "# Variable declaration\n", + "\n", + "T =298.0\t\t\t# [K] temperature of air\n", + "Vco2 = 34.0 \t\t\t# molecular volume of co2\n", + "Vair = 29.9 \t\t\t# molecular volume of air\n", + "Mco2 = 44.0 \t\t\t# molecular weight of co2\n", + "Mair = 28.9 \t\t\t# molecular weight of air\n", + "P = 1.0132*10**(5) \t\t# [Pa] atmospheric pressure\n", + "\t# using equation (11-2)\n", + "#Calculation\n", + "D = 435.7*T**(3.0/2.0)*(((1/Mco2)+(1/Mair))**(1.0/2.0))/(P*(Vco2**(1.0/3.0)+Vair**(1.0/3.0))**(2)) \n", + "\n", + "#Result\n", + "print \"value of diffusion coefficient for co2 in air is\",round(D,3),\"square centimeter/s\" \n", + "print \"From Table A-8,D=0.164 sq cm/s\\n So,they are in fair agreement\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.3" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of dry air is 53.66 degree celsius\n", + " recalculate the density at the arithmetic-average temperature between wall and free-stream conditions\n", + "With this adjustments these results are RHO = 1.143 kg/m**(3) and Tinf = 55.8 degree celcius\n" + ] + } + ], + "source": [ + "#Example Number 11.3\n", + "# Wet-bulb temperature\n", + "\n", + "#Variable declaration\n", + "\n", + "Pg = 2107.0\t\t\t# [Pa] from steam table at 18.3 degree celcius\n", + "Pw = Pg*18.0 \t\t\t# [Pa]\n", + "Rw = 8315.0 \t\t\t# [J/mol K] gas constant\n", + "Tw = 273+18.3\t \t\t# [K]\n", + "\n", + "RHOw = Pw/(Rw*Tw) \t\t# [kg/cubic meter]\n", + "\n", + "\n", + "Cw = RHOw \t\t\t# [kg/cubic meter]\n", + "RHOinf = 0.0 \t\t\t# since the free stream is dry air\n", + "Cinf = 0.0 \n", + "P = 1.01325*10**(5) \t\t# [Pa]\n", + "R = 287 \t\t\t# [J /kg K]\n", + "T = Tw \t\t\t\t# [K]\n", + "RHO = P/(R*T) \t\t\t# [kg/cubic meter]\n", + "\n", + "Cp = 1004.0 \t\t\t# [J/kg degree celsius]\n", + "Le = 0.845 \n", + "Hfg = 2.456*10**(6) \t\t# [J/kg]\n", + "#Calculations\n", + "# now using equation(11-31)\n", + "\n", + "Tinf = (((Cw-Cinf)*Hfg)/(RHO*Cp*(Le**(2.0/3.0))))+Tw \t# [K]\n", + "Tin = Tinf-273 \t\t\t\t\t# [degree celsius]\n", + "\n", + "print \"Temperature of dry air is\",round(Tin,2),\"degree celsius\" \n", + "print \" recalculate the density at the arithmetic-average temperature between wall and free-stream conditions\" \n", + "print \"With this adjustments these results are RHO = 1.143 kg/m**(3) and Tinf = 55.8 degree celcius\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.4" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Relative humidity is therefore 27.8 percentage\n" + ] + } + ], + "source": [ + "#Example Number 11.4\n", + "# relative humidity of air stream\n", + "\n", + "#Variable declaration\n", + "\n", + "\t\t# these data were taken from previous example\n", + "Rho = 1.212 \t\t\t\t\t# [kg/cubic meter]\n", + "Cp = 1004 \t\t\t\t\t# [J/kg]\n", + "Le = 0.845 \n", + "Tw = 18.3 \t\t\t\t\t# [degree celsius]\n", + "Tinf = 32.2 \t\t\t\t\t# [degree celsius]\n", + "Rhow = 0.015666 \t\t\t\t# [kg/cubic meter]\n", + "Cw = Rhow \t\t\t\t\t# [kg/cubic meter]\n", + "\n", + "#calculation\n", + "\n", + "Hfg = 2.456*10**(6) \t\t\t\t# [J/kg]\n", + "\t\t# we use eqn 11-31\n", + "Cinf = Cw-(Rho*Cp*Le**(2.0/3.0)*(Tinf-Tw)/Hfg) \t# [kg/cubic meter]\n", + "Rhoinf = Cinf \t\t\t\t\t# [kg/cubic meter]\n", + "Rhog = 0.0342 \t\t\t\t\t# [kg/cubic meter]\n", + "RH = (Rhoinf/Rhog)*100 \n", + "\n", + "#Result\n", + "\n", + "print \"Relative humidity is therefore\",round(RH,1),\"percentage\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 11.5" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Evaporation rate on the land under these conditions is 0.0028 kg/h square meter\n" + ] + } + ], + "source": [ + "#Example Number 11.5\n", + "# water evaporation rate\n", + "\n", + "# Variable declaration\n", + "\n", + "Ta = 38+273 \t\t\t# [K] temperature of atmospheric air\n", + "RH = 0.30 \t\t\t# relative humidity\n", + "u = 10.0\t\t\t# [mi/h] mean wind speed\n", + "R = 0.287 \t\t\t# universal gas constant\n", + "Dw = 0.256*10**(-4) \t\t# [square meter/s] from table A-8(page no.-610)\n", + "rho_w = 1000 \t\t\t# [kg/cubic meter]\n", + "\t# for this calculation we can make use of equation(11-36). from thermodynamic \tsteam tables\n", + "p_g = 6.545 \t\t\t# [kPa] at 38 degree celsius\n", + "p_s = p_g \t\t\t# [kPa]\n", + "p_w = RH*p_s \t\t\t# [kPa]\n", + "p_s = 1.933 \t\t\t# [in Hg]\n", + "p_w = 0.580 \t\t\t# [in Hg]\n", + "\t# also \n", + "u_bar = u*24 \t\t\t# [mi/day]\n", + "\t# equation(11-36) yields, with the application of the 0.7 factor\n", + "\n", + "E_lp = 0.7*(0.37+0.0041*u_bar)*(p_s-p_w)**(0.88) \t\t# [in/day]\n", + "E_lp = E_lp*2.54/100 \t\t\t\t\t\t# [m/day]\n", + "\n", + "\t# noting that standard pan has the diameter of 1.2m, we can use the figure to \t\tcalculate the mass evaporation rate per unit area as\n", + "m_dot_w_by_A = E_lp*rho_w/24 \t\t\t\t# [kg/h square meter]\n", + "\n", + "\n", + "\t# as a matter of interest, we might calculate the molecular-diffusion rate of \t\twater vapour from equation(11-35), taking z1 as the 1.5m dimension above the \t\tstandard pan.\n", + "z1 = 1.5 \t\t\t\t\t\t\t# [m]\n", + "\n", + "\t# since rho = p/(R*T)\n", + "\t# equation(11-35) can be written as \n", + "m_dot_w_by_A1 = 0.622*Dw*p_g*3600/(R*Ta*z1) \t\t\t# [kg/h square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Evaporation rate on the land under these conditions is\",round(m_dot_w_by_A1,4),\"kg/h square meter\"\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", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_1_Introduction.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_1_Introduction.ipynb new file mode 100644 index 00000000..0af0c439 --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_1_Introduction.ipynb @@ -0,0 +1,302 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1 Introduction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.1" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat transfer per unit area is 3.7 MW/sq meter\n" + ] + } + ], + "source": [ + "#Example Number 1.1\n", + "# HOW MUCH HEAT IS TRANSFERRED THROUGH THE PLATE\n", + "\n", + "#VARIABLE DECLARATION\n", + "\n", + "k = 370 \t\t\t # [W/m] at 250 degree celsius\n", + "dt = 100-400 \t\t\t#[degree celsius] temperature difference\n", + "dx = 3*10**(-2) \t\t#[m] thickness of plate\n", + "\n", + "#CALCULATION\n", + "\n", + "q = -k*dt/dx \t\t\t#[MW/square meter]\n", + "\n", + "#RESULTS\n", + "\n", + "print \"Rate of heat transfer per unit area is\", q/1000000 ,\"MW/sq meter\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat transfer is 2.156 kW\n" + ] + } + ], + "source": [ + "#Example Number 1.2\n", + "# CALCULATE THE HEAT TRANSFER\n", + "\n", + "#Variable declaration\n", + "\n", + "Twall = 250 \t\t\t#[degree celsius] wall temperature\n", + "Tair = 20 \t\t\t#[degree celsius] air temperature\n", + "h = 25 \t\t\t\t#[W/square meter] heat transfer coefficient\n", + "l = 75*10**(-2) \t\t#[m] length of plate\n", + "b = 50*10**(-2) \t\t#[m] width of plate\n", + "area = l*b \t\t\t#[square meter] area of plate\n", + "dt = 250-20 \t\t\t#[degree celsius]\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "q = h*area*dt \t\t\t# [W] from newton's law of cooling\n", + "\n", + "\n", + "#Result\n", + "\n", + "print\"Rate of heat transfer is\",round(q/1000,3),\"kW\" \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Inside plate temperature is 253.05 degree C\n" + ] + } + ], + "source": [ + "#Example Number 1.3\n", + "# Calculate the inside plate temperature.\n", + "\n", + "#variable declaration\n", + "\n", + "Qconv = 2156 \t # [W] from previous problem\n", + "Qrad = 300\t\t # [W] given\n", + "dx = 0.02\t\t # [m] plate thicknesss\n", + "l = 0.75 \t\t # [m] length of plate \n", + "w = 0.5 \t\t # [m] width of plate\n", + "k = 43\t\t\t #[W/m] from table 1.1\n", + "area = l*w\t\t #[square meter] area of plate\n", + "\n", + "#Calculation\n", + "\n", + "Qcond = Qconv+Qrad \t # [W]\n", + "dt = Qcond*dx/(k*area) \t # [degree celsius] temperature difference\n", + "Ti = 250+dt \t\t # inside temperature\n", + "\n", + "#Results\n", + "\n", + "print\"Inside plate temperature is\",round(Ti,2),\"degree C\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer is: 22.0 W\n", + " This is equal to the electric power which must be applied\n" + ] + } + ], + "source": [ + "#Example Number 1.4\n", + "# Calculate electric power to be supplied to the wire \n", + "\n", + "#Variable declaration\n", + "\n", + "d = 1*10**(-3) \t\t#[m] diameter of wire\n", + "l = 10*10**(-2) \t#[m] length of wire\n", + "Sarea = 22*d*l/7 \t#[square meter] surface area of wire\n", + "h = 5000 \t\t#[W/square meter] heat transfer coefficient\n", + "Twall = 114 \t\t# [degree celsius]\n", + "Twater = 100\t # [degree celsius]\n", + "\n", + "#total convection loss is given by equation(1-8)\n", + "\n", + "#Calculation\n", + "\n", + "Q = h*Sarea*(Twall-Twater) # [W]\n", + "\n", + "#Results\n", + "\n", + "print\"Heat transfer is:\",Q,\"W\" \n", + "print\" This is equal to the electric power which must be applied\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer per unit area is: 69.03 kW/sq meter\n" + ] + } + ], + "source": [ + "#Example Number 1.5\n", + "# radiation heat transfer\n", + "# Calculate the heat transfer per unit area\n", + "\n", + "#Variable declaration\n", + "\n", + "sigma = 5.669*10**(-8) \t\t#[W/square meter*k^(4)] universal constant\n", + "T1 = 273+800 \t\t\t# [k] first plate temperature\n", + "T2 = 273+300 \t\t\t# [k] second plate temperature\n", + "\n", + "#equation(1-10) may be employed for this problem\n", + "\n", + "#Calculation\n", + "Q = sigma*(T1**4-T2**4) \t# [W/square meter]\n", + "\n", + "#Results\n", + "\n", + "print\"Heat transfer per unit area is:\",round(Q/1000,2),\" kW/sq meter\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 1.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total heat loss is: 55.67 W/m\n" + ] + } + ], + "source": [ + "#Example Number 1.6\n", + "# total heat loss by convection and radiation\n", + "\n", + "#Variable declaration\n", + "d = 0.05 #[m] diameter of pipe\n", + "Twall = 50 #[degree celsius] \n", + "Tair = 20 #[degree celsius]\n", + "emi = 0.8 #emissivity\n", + "h = 6.5 #[W/square meter] heat transfer coefficient for free convection\n", + "import math\n", + "Q1 = h*math.pi*d*(Twall-Tair) #[W/m] convection loss per unit length\n", + "sigma = 5.669*10**(-8) # [W/square meter*k^(4)] universal constant\n", + "T1 = 273+Twall # [k]\n", + "T2 = 273+Tair # [k]\n", + "Q2 = emi*math.pi*d*sigma*((T1**(4))-(T2**(4))) # [W/m] heat loss due to radiation per unit length\n", + "Qtotal = Q1+Q2 # [W/m] total heat loss per unit length\n", + "\n", + "print\"Total heat loss is:\",round(Qtotal,2),\"W/m\"\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", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_2_Steady_State_Conduction_One_Dimension.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_2_Steady_State_Conduction_One_Dimension.ipynb new file mode 100644 index 00000000..71fd76dc --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_2_Steady_State_Conduction_One_Dimension.ipynb @@ -0,0 +1,674 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 Steady State Conduction One Dimension" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of thickness is: 5.75 cm\n" + ] + } + ], + "source": [ + "#Example Number 2.1\n", + "#Calculate thickness of insulation to reduce heat loss/gain by 80 %\n", + "\n", + "\n", + "#Variable declaration\n", + "\n", + "dx1 = 0.1\t # [m] thickness of layer of common brick\n", + "k1 = 0.7\t # [W/m degree celsius] heat transfer coefficient of common brick\n", + "dx2 = 0.0375 # [m] thickness of layer of gypsum plaster\n", + "k2 = 0.48 \t # [W/m degree celsius] heat transfer coefficient gypsum plaster\n", + "\n", + "#Calculation\n", + "\n", + "Rb = dx1/k1\t # [sq m degree C /W] thermal resistance of brick\n", + "Rp = dx2/k2 # [sq m degree C /W] thermal resistance of gypsum plaster\n", + "R = Rb+Rp \t # [sq m degree C /W] thermal resistance without insulation\n", + "R1 = R/0.2 \t # [sq m degree C /W] with insulation \n", + "\n", + "# heat loss with the rock-wool insulation is 20 percent \n", + "\n", + "Rrw = R1-R \t # [square meter degree celsius /W]\n", + "k3 = 0.065 \t # [W/m degree celsius] heat transfer coefficient\n", + "dx3 = Rrw*k3 # [m]\n", + "\n", + "#Result\n", + "\n", + "print \"Length of thickness is:\",round(dx3*100,2),\" cm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat flow is given by: 680.0 W/m\n", + "Interface temperature is 596.0 degree celsius \n" + ] + } + ], + "source": [ + "#Example Number 2.2\n", + "#Calculate heat loss/metre length and tube-insulation interface temperature\n", + "\n", + "#Variable declaration\n", + "\n", + "ID = 0.02 \t\t# [m] inner diameter of steel\n", + "OD = 0.04 \t\t#[m] outer diameter of steel\n", + "t = 0.03 \t\t#[m] thickness of asbestos insulation\n", + "\t\t\t# system is like three concentric cylinders \n", + "T1 = 600 \t\t# [degree celsius] inside wall temperature\n", + "T2 = 100 \t\t# [degree celsius] outside insulation temperature\n", + "Ks = 19 \t\t#[W/m degree celsius] heat transfer coefficient of steel\n", + "Ka = 0.2 \t\t# [W/m degree celsius] heat transfer coefficient of asbestos\n", + "\n", + "\t\t\t# heat flow is given by per unit length\n", + "\n", + "#Calculation\n", + "\n", + "import math\n", + "Q_l = ((2*math.pi*(T1-T2))/((math.log(OD/ID)/Ks)+(math.log(0.1/OD)/Ka))) # [W/m]\n", + "\n", + "\t# above calculated heat flow is used to calculate the interface temperature\n", + "\t# between the outside wall and the insulation\n", + "\n", + "Ta = Q_l*(math.log(0.1/OD)/(2*math.pi*Ka))+T2 \n", + "\t# [degree C] Ta is interface temperature\n", + "\n", + "#Result\n", + "\n", + "print \"Heat flow is given by:\",round(Q_l),\" W/m\" \n", + "print \"Interface temperature is\",round(Ta),\"degree celsius \"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "TOTAL THERMAL RESISTANCE THROUGH,\n", + "wood stud is: 31.39 degree C\n", + "Insulation section is: 7.34 degree C/W\n", + "Overall heat transfer coefficient is: 0.414 W/sq m per deg C\n", + "R value is: 2.415 sq m/W\n" + ] + } + ], + "source": [ + "#Example Number 2.3 \n", + "# Calculate overall heat transfer coefficient and R Value of wall\n", + "\n", + "# 1. heat transfer through studs for unit depth\n", + "\n", + "#Variable declaration\n", + "\n", + "l = 0.0413 \t # [m] length of wood studs\n", + "b = 1.0 \t # [m] unit depth\n", + "A = l*b\t\t # [square meter] area of studs for unit depth\n", + "hi = 7.5\t # [W/sq meter/degr C] convectional heat transfer coefficient\n", + "ho = 15 \t # [W/sq m per deg C] convectional heat transfer coefficient\n", + "Kb = 0.69 \t # [W/m per deg celsius] heat transfer coefficient of brick\n", + "Kgi = 0.96 \t # [W/m per deg C] heat transfer coefficient of gypsum inner sheath\n", + "Ki = 0.04 \t # [W/m per deg C] heat transfer coefficient of insulation\n", + "Kws = 0.1\t # [W/m per deg C] heat transfer coefficient of wood stud\n", + "Kgo = 0.48 \t # [W/m per deg C] heat transfer coefficient of gypsum outer sheath\n", + "Rair = 1/(ho*A) # [degree C /W] convection resistance outside of brick\n", + "dx_b = 0.08 \t # [m] thickness of brick\n", + "dx_os = 0.019 \t #[m] thickness of outer sheet\n", + "dx_ws = 0.0921 # [m] thickness of wood stud\n", + "dx_is = 0.019 # [m] thickness of inner sheet\n", + "Rb = dx_b/(Kb*A) \t# [degr C /W] conduction resistance in brick\n", + "Ros = dx_os/(Kgi*A) \t# [deg C /W] conduction resistance through outer sheet\n", + "Rws = dx_ws/(Kws*A) \t# [deg C/W] conduction resistance through wood stud\n", + "Ris = dx_is/(Kgo*A) \t# [deg C/W] conduction resistance through inner sheet\n", + "\n", + "#Calculation\n", + "\n", + "Ri = 1/(hi*A) \t\t# [degree celsius /W] convection resistance on inside\n", + "\n", + "Rt = Rair+Rb+Ros+Rws+Ris+Ri \n", + "\t\t\t# [deg C/W] total thermal Res through the wood stud section\n", + "\n", + "\n", + "print \"TOTAL THERMAL RESISTANCE THROUGH,\"\n", + "print\"wood stud is:\",round(Rt,2),\"degree C\"\n", + "\n", + "\n", + "# 2. Heat transfer through insulation section \n", + "\n", + "#Calculation\n", + "\n", + "A1 = 0.406-A\t\t # [sq meter] area of insulation section for unit depth\n", + "dx_ins = 0.0921 # [m] thickness of insulation\n", + "Rins = dx_ins/(Ki*A1) # [deg C /W] conduction resistance through insulation section\n", + "\n", + "\t\t# five of the materials are same but resistance involve different area \n", + "\t\t# i.e. (40.6-4.13) cm instead of 4.13 cm \n", + "\t\t# so that each of the previous must be multiplied by a factor of \t\t\t#(4.13/(40.6-4.13)) = 0.113 \n", + "\n", + "#Calculation\n", + "\n", + "Rt_ins = (Rair+Rb+Ros+Ris+Ri)*0.113+Rins\n", + " \t\t# [deg C/W] total resistance through insulation section \n", + "\t\n", + "print\"Insulation section is:\",round(Rt_ins,3),\"degree C/W\"\n", + "\n", + "R_overall = 1/((1/Rt)+(1/Rt_ins)) \n", + "\t\t# [degree celsius /W] overall resistance for the section\n", + "\n", + "\t\t# the value is related to overall heat transfer coefficient by \n", + "\t\t# Q = U*A*dt = dt/R_overall \n", + "\t\t# where A is area of total section\n", + "\n", + "\n", + "A_ = 0.406 \t\t# [sq meter] area of total section\n", + "U = 1/(R_overall*A_) # [W/sq meter deg C] overall heat transfer coefficient\n", + "\t\t\t# R value is somewhat different from thermal resistance and is \t\t\t#given by\n", + "R_value = 1/U \t\t# [degree celsius square meter/W] R value of system\n", + "\n", + "\n", + "#Results\n", + "\n", + "print\"Overall heat transfer coefficient is:\",round(U,3),\"W/sq m per deg C\"\n", + "\n", + "print\"R value is:\",round(R_value,3),\"sq m/W\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Overall heat transfer coefficient is: 7.58 W/sq m degree C\n", + "Heat loss per unit length is: 19.0 W(for 1m length)\n" + ] + } + ], + "source": [ + "#Example Number 2.4\n", + "# Calculate overall heat transfer coeff and heat loss/unit length at 20 deg C\n", + "\n", + "#Variable declaration\n", + "\n", + "ID = 0.025 \t\t# [m] inner diameter of steel\n", + "OD = ID+2*0.0008 \t#[m] outer diameter of steel\n", + "hi = 3500 \t\t# [W/sq m per deg C] convectional heat transfer coefficient of \t\t\t#inside\n", + "ho = 7.6 \t\t# [W/sq m per deg C] convectional heat transfer coefficient of \t\t\t#outside \n", + "L = 1.0 \t\t# [m] tube length\n", + "import math\n", + "Ai = math.pi*ID*L \t# [sq meter] inside crossectional area \n", + "Ao = math.pi*OD*L \t# [sq meter] outside crossectional area \n", + "k = 16 \t\t\t# [W/sq meter per deg C] thermal conductivity of tube\n", + "\n", + "#Calculation\n", + "\n", + "Ri = 1/(hi*Ai)\t\t\t # [degree C /W] convection resistance inside tube\n", + "Rt = math.log(OD/ID)/(2*math.pi*k*L) # [degree C /W] thermal resistance \n", + "Ro = 1/(ho*Ao) \t\t\t # [deg C /W] convection resistance outside tube\n", + "R_total = Ri+Rt+Ro\t\t # [deg C/W] total thermal and convection \t\t\t\t \t\t #resistance \n", + "Uo = 1/(Ao*R_total) \t\t # [W/sq m deg C] overall heat transfer \t\t\t\t \t\t #coefficient\n", + "\n", + "Tw = 50\t\t\t\t # [degree C] water temperature\n", + "Ta = 20 \t\t\t # [degree C] surrounding air temperature\n", + "dt = Tw-Ta \t\t\t # [degree C] temperature difference\n", + "q = Uo*Ao*dt\t\t\t # [W] heat transfer \n", + "\n", + "#Results\n", + "\n", + "print\"Overall heat transfer coefficient is:\",round(Uo,2),\" W/sq m degree C\"\n", + "\n", + "\n", + "print\"Heat loss per unit length is:\",round(q),\" W(for 1m length)\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Critical radius of insulation for asbestos is: 5.67 cm \n", + "Heat loss when covered with critical radius of insulation is: 105.7 W/m\n", + "Heat loss without insulation is: 84.8 W/m\n", + "Addition of 3.17 of insulation actually increases the heat transfer by: 25.0 %\n" + ] + } + ], + "source": [ + "#Example Number 2.5\n", + "#Calculate the heat loss \n", + "\n", + "#Variable declaration\n", + "\n", + "k = 0.17 \t\t# [W/m per deg C] heat transfer coefficient of asbestos\n", + "Tr = 20 \t\t# [degree celsius] temperature of room air\n", + "h = 3 \t\t\t# [W/sq m per deg C] convectional heat transfer coefficient\n", + "Tp = 200 \t\t# [degree celsius] temperature of pipe\n", + "d = 0.05 \t\t# [m] diameter of pipe\n", + "\n", + "\t\t\t# from equation (2-18) we calculate r_o as \n", + "\n", + "#Calculation\n", + "\n", + "r_o = k/h \t\t# [m] critical radius of insulation\n", + "print\"Critical radius of insulation for asbestos is:\",round(r_o*100,2),\"cm \"\n", + "\n", + "Ri = d/2\t # [m] inside radius of insulation\n", + "\t\t\t# heat transfer is calculated from equation (2-17)\n", + "import math\n", + "q_by_L = (2*math.pi*(Tp-Tr))/(((math.log(r_o/Ri))/0.17)+(1/(h*r_o)))\n", + "\t \t\t# [W/m] heat transfer per unit length\n", + "\n", + "\n", + "#Results\n", + "\n", + "print\"Heat loss when covered with critical radius of insulation is:\",round(q_by_L,1),\" W/m\"\n", + "\n", + "\t\t# without insulation the convection from the outer surface of pipe is \n", + "\n", + "q_by_L1 = h*2*math.pi*Ri*(Tp-Tr) \n", + "\t\t#[W/m] convection from outer surface without insulation\n", + "print\"Heat loss without insulation is:\",round(q_by_L1,1),\" W/m\"\n", + "per_inc = ((q_by_L-q_by_L1)/q_by_L1)*100 \t# percentage increase in heat transfer\n", + "\n", + "print\"Addition of 3.17 of insulation actually increases the heat transfer by:\",round(per_inc),\"%\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Center temperature of the wire is: 231.7 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 2.6\n", + "#Calculate the centre temperature of wire\n", + "\n", + "#Variable declaration\n", + "\n", + "\t\t# all the power generated in the wire must be dissipated by convection \t\tto the liquid \n", + "\t\t# P = i**(2)*R = q = h*A*dt\n", + "L = 100 \t# [cm] length of the wire \n", + "k = 19 \t\t# [W/m per deg C] heat transfer coefficient of steel wire\n", + "\n", + "#Calculations\n", + "\n", + "import math\n", + "A = math.pi*(0.15)**(2) \t# [sq m] crossectional area of wire\n", + "rho = 70*10**(-6) \t\t# [micro ohm cm] resistivity of steel \n", + "R = rho*L/A \t\t\t# [ohm] resistance of wire\n", + "i = 200 \t\t\t# [ampere] current in the wire\n", + "P = i**(2)*R \t\t\t# [W] power generated in the wire\n", + "Tl = 110 \t\t\t# [degree celsius] liquid temperature\n", + "d = 0.003 \t\t\t# [m] diameter of wire\n", + "l = 1 \t\t\t\t# [m] length of wire\n", + "Tw = (P/(4000*3.14*d*l))+110 \t# [degree celsius] wire temperature\n", + "\n", + "\t\t\t\t# heat generated per unit V q_dot is calculated as\n", + "\t\t\t\t# P = q_dot*V = q_dot*3.14*r**(2)*l\n", + "r = d/2 \t\t\t# [m] radius of wire\n", + "q_dot = P/(math.pi*r**(2)*l) \t# [W/m**(3)]\n", + "\t\t\t\t# finally the center temperature of the wire is \t\t\t\t\tcalculated from equation (2-26)\n", + "\t\n", + "To = ((q_dot*(r**(2)))/(4*k))+Tw\t # [degree celsius]\n", + "\n", + "\n", + "#Result\n", + "\n", + "print \"Center temperature of the wire is:\",round(To,1) ,\"degree celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.8" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss from the fin per unit length is 360.0 W/m\n" + ] + } + ], + "source": [ + "#Example Number 2.8\n", + "# Calculate heat loss per unit depth of the material\n", + "\n", + "#Variable declaration\n", + "\n", + "t = 0.003 \t\t# [m] thickness of fin\n", + "L = 0.075 \t\t# [m] length of fin\n", + "Tb = 300 \t\t# [degree celsius] base temperature\n", + "Tair = 50 \t\t# [degree celsius] ambient temperature\n", + "k = 200 \t\t# [W/m per deg C] heat transfer coefficient of aluminium fin\n", + "h = 10 \t\t\t# [W/sq m per deg C] convectional heat transfer coefficient\n", + "\t\t\t# We Will use the approximate method of solution by extending \t\t\tthe fin \n", + "\t\t\t# With a fictitious length t/2\n", + "\t\t\t# using equation(2-36)\n", + "\n", + "#Calculation\n", + "Lc = L+t/2 \t\t# [m] corrected length\n", + "z = 1 \t\t\t# [m] unit depth\n", + "p = (2*z+2*t) \t\t# [m] perimeter of fin\n", + "A = z*t \t\t# [square meter] crossectional area of fin\n", + "m = ((h*p)/(k*A))**(0.5)\n", + "\t \n", + "\t\t\t\t\t\t# from equation(2-36)\n", + "dt = Tb-Tair \t\t\t\t\t# [degree C] temperature difference\n", + "import math\n", + "q = math.tanh(m*Lc)*((h*p*k*A)**(0.5))*dt\t# [W/m] heat transfer per unit length \n", + "\n", + "#Results\n", + "print \"Heat loss from the fin per unit length is\",round(q),\"W/m\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.9" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The actual heat transferred is: 60.97 W\n" + ] + } + ], + "source": [ + "#Example Number 2.9\n", + "# Calculate the heat loss per fin\n", + "\n", + "#Variable declaration\n", + "\n", + "t = 0.001 \t\t\t# [m] thickness of fin\n", + "L = 0.015 \t\t\t# [m] length of fin\n", + "Ts = 170\t\t\t# [degree celsius] surface temperature\n", + "Tfluid = 25 \t\t\t# [degree celsius] fluid temperature\n", + "k = 200 \t\t\t# [W/m per deg C] heat transfer coefficient of \t\t\t\t\t\t aluminium fin\n", + "h = 130 \t\t\t# [W/sq m per deg C] \t\t\t\t convectional heat transfercoefficient\n", + "d = 0.025 \t\t\t# [m] tube diameter\n", + "Lc = L+t/2\t\t\t# [m] corrected length\n", + "r1 = d/2 \t\t\t# [m] radius of tube\n", + "r2_c = r1+Lc \t\t\t# [m] corrected radius\n", + "\n", + "#Calculation\n", + "\n", + "Am = t*(r2_c-r1) \t\t# [sq m] profile area \n", + "c = r2_c/r1 \t\t\t# constant to determine\tefficiency of fin from curve \n", + "\n", + "c1 = ((Lc)**(1.5))*((h/(k*Am))**(0.5)) \t# constant to determine efficiency of fin from \t\t\t\t\tcurve\n", + "\n", + "\t\t \t\t# using c and c1 to determine the efficiency \t\t\t\t\t\tof the fin from figure (2-12)\n", + "\t\t\t\t# we get nf = 82 percent\n", + "\t\t\t\t# heat would be transferred if the entire fin were at \t\t\t\t\tthe base temperature \n", + "\t\t\t\t# both sides of fin exchanging heat \n", + "import math\n", + "q_max = 2*math.pi*(r2_c**(2)-r1**(2))*h*(Ts-Tfluid) \t# [W] maximum heat transfer\n", + "q_act = 0.82*q_max \t\t\t\t\t#[W] actual heat transfer\n", + "\n", + "#Result\n", + "print\"The actual heat transferred is:\",round(q_act,2),\" W\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.10" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The expression for the temperature distribution in the rod is \n", + "theta`=(((theta1`*exp^(2*m*L)-theta2`*exp^(m*L))*exp^(-m*x))+((theta2`exp^(m*L)-theta1`)exp^(m*x))/(exp^(2*m*L)-1))\n", + "for an infinitely long heat generating fin with the left end maintained at T1, the temperature distribution becomes \n", + "theta`/theta1 = exp^(-m*x)\n" + ] + } + ], + "source": [ + "#Example Number 2.10\n", + "#Obtain an expression for the temperature distribution in rod\n", + "\n", + "#Variable declaration & Calculations\n", + "\n", + "\t# q_dot is uniform heat source per unit volume \n", + "\t# h is convection coefficient\n", + "\t# k is heat transfer coefficient\n", + "\t# A is area of crossection\n", + "\t# P is perimeter \n", + "\t# Tinf is environment temperature \n", + "\t# we first make an energy balance on the element of the rod shown in \tfigure(2-10)\n", + "\t# energy in left place + heat generated in element = energy out right face + \tenergy lost by convection\n", + "\t# or \n", + "\t# -(k*A*dT_by_dx)+(q_dot*A*dx) = \t-(k*A(dT_by_dx+(d2T_by_dx2)*dx))+h*P*dx*(T-Tinf)\n", + "\t# simlifying we have \n", + "\t# d2T_by_dx2-((h*P)/(k*A))*(T-Tinf)+q_dot/k = 0\n", + "\t# replacing theta = (T-Tinf) and (square meter) = ((h*P)/(k*A))\n", + "\t# d2theta_by_dx2-(square meter)*theta+q_dot/k = 0\n", + "\t# we can make a further substitution as theta` = theta-(q_dot/(k*(square \tmeter)))\n", + "\t# so that our differential equation becomes \n", + "\t# d2theta`_by_dx2-(square meter)*theta`\n", + "\t# which has the general solution theta` = C1*exp^(-m*x)+C2*exp^(m*x)\n", + "\t# the two end temperatures are used to establish the boundary conditions:\n", + "\t# theta` = theta1` = T1-Tinf-q_dot/(k*(square meter)) = C1+C2\n", + "\t# theta` = theta2` = T2-Tinf-q_dot/(k*(square meter)) = \tC1*exp^(-m*L)+C2*exp^(m*L)\n", + "\t# solving for the constants C1 and C2 gives \n", + "\t\t#((theta1`*exp^(2*m*L)-theta2`*exp^(m*L))*exp^(-m*x))+((theta2`exp^(m*L)-theta1`)exp^(m*x))/(exp^(2*m*L)-1))\n", + "\n", + "\n", + "#RESULTS\n", + "\n", + "print\"The expression for the temperature distribution in the rod is \" \n", + "\n", + "print\"theta`=(((theta1`*exp^(2*m*L)-theta2`*exp^(m*L))*exp^(-m*x))+((theta2`exp^(m*L)-theta1`)exp^(m*x))/(exp^(2*m*L)-1))\" \n", + "\n", + "print\"for an infinitely long heat generating fin with the left end maintained at T1, the temperature distribution becomes \" \n", + "\n", + "print\"theta`/theta1 = exp^(-m*x)\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 2.11" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Overall heat flow is: 5.52 W\n", + "The temperature drop across the contact is: 4.3 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 2.11\n", + "# Calculate the axial heat flow and temperature drop across the contact surface\n", + "\n", + "#Variable declaration\n", + "\n", + "d = 0.03\t\t #[m] diameter of steel bar\n", + "l = 0.1 #[m] length of steel bar\n", + "import math\n", + "A = (math.pi*d**(2))/4\t # [square meter] crossectional area of bar \n", + "k = 16.3 \t\t # [W/sq m per degree celsius] thermal conductivity of tube \n", + "hc = 1893.93\t\t # [W/sq m per degree celsius] contact coefficient\n", + "\t\t # the overall heat flow is subjected to three thermal resistances \n", + "\t\t # one conduction resistance for each bar\n", + "\t\t # contact resistance \n", + "#Calculation\n", + "\n", + "Rth = l/(k*A) # [degree celsius /W]\n", + "\n", + "\t\t # from table(2-2) the contact resistance is \n", + "Rc = 1/(hc*A) \t # [degree celsius /W]\n", + "Rt = 2*Rth+Rc\t\t # [degree celsius /W] total resistance\n", + "dt = 100\t\t # [degree celsius] temperature difference\n", + "q = dt/Rt \t\t # [W] overall heat flow\n", + "\n", + "#Results\n", + "\n", + "print \"Overall heat flow is:\",round(q,2),\"W\" \n", + "\n", + "\t\t# temperature drop across the contact is found by taking the ratio \n", + "\t\t# of the contact resistance to the total thermal resistance \n", + "\n", + "dt_c = (Rc/(2*Rth))*dt\t\t\t # [degree celsius]\n", + "\n", + "\n", + "print \"The temperature drop across the contact is:\",round(dt_c,2),\"degree celsius\"\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", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_3_Steady_State_Conduction_Multiple_Dimension.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_3_Steady_State_Conduction_Multiple_Dimension.ipynb new file mode 100644 index 00000000..7dd6ca44 --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_3_Steady_State_Conduction_Multiple_Dimension.ipynb @@ -0,0 +1,228 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 Steady State Conduction Multiple Dimension" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost by the pipe is 859.9 W\n" + ] + } + ], + "source": [ + "#Example Number 3.1\n", + "# Calculate the heat loss by the pipe\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.15 \t\t\t# [m] diameter of pipe\n", + "r = d/2 \t\t\t# [m] radius of pipe\n", + "L = 4 \t\t\t\t# [m] length of pipe\n", + "Tp = 75\t\t\t\t# [degree celsius] pipe wall temperature\n", + "Tes = 5 \t\t\t# [degree celsius] earth surface temperature\n", + "k = 0.8\t\t\t\t# [W/m per deg C] thermal conductivity of earth \n", + "D = 0.20 \t\t\t# [m] depth of pipe inside earth\n", + "\n", + "\t# We may calculate the shape factor for this situation using equation given in \ttable 3-1 \n", + "\t\n", + "\t# since D<3*r\n", + "#Calculation\n", + "import math\n", + "S = (2*math.pi*L)/math.acosh(D/r) \t# [m] shape factor\n", + "\t# the heat flow is calculated from \n", + "q = k*S*(Tp-Tes) \t\t\t# [W]\n", + "\n", + "#Result\n", + "print\"Heat lost by the pipe is\",round(q,1),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost through the walls is: 8.592 kW\n" + ] + } + ], + "source": [ + "#Example Number 3.2 \n", + "# Calculate heat loss through the walls\n", + "\n", + "# VARIABLE DECLARATION\n", + "\n", + "a = 0.5 \t # [m] length of side of cubical furnace\n", + "Ti = 500 \t # [degree celsius] inside furnace temperature\n", + "To = 50 \t # [degree celsius] outside temperature\n", + "k = 1.04 \t # [W/m per degree celsius] thermal conductivity of fireclay brick \n", + "t = 0.10 \t # [m] wall thickness\n", + "A = a*a \t # [square meter] area of one face \n", + "\t\t # we compute the total shape factor by adding the shape factors \t\t \t for the walls, edges and corners\n", + "\n", + "#Calculation\n", + "Sw = A/t\t # [m] shape factor for wall\n", + "Se = 0.54*a \t # [m] shape factor for edges\n", + "Sc = 0.15*t\t # [m] shape factor for corners\n", + "\n", + "\t\t # there are six wall sections, twelve edges and eight corners, so \t\t\tthe total shape factor S is\n", + "\n", + "S = 6*Sw+12*Se+8*Sc \t# [m]\n", + "\t\t \n", + "\t\t# the heat flow is calculated as \n", + "\n", + "q = k*S*(Ti-To) \t# [W]\n", + "\n", + "#Result\n", + "print\"Heat lost through the walls is:\",round(q/1000,3),\"kW\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.3" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost by disk is: 198.46 W\n" + ] + } + ], + "source": [ + "#Example Number 3.3\n", + "# Calculate the heat loss by the disk\n", + "\n", + "# Variable declaration\n", + "\n", + "import math\n", + "d = 0.30 \t# [m] diameter of disk\n", + "r = d/2 \t# [m] radius of disk\n", + "Td = 95 \t# [degree celsius] disk temperature\n", + "Ts = 20 \t# [degree celsius] isothermal surface temperature\n", + "k = 2.1 \t# [W/m per degree celsius] thermal conductivity of medium \n", + "D = 1.0 \t# [m] depth of disk in a semi-infinite medium\n", + "\t# We have to calculate shape factor using relation given in table (3-1) \n", + "\t# We select the relation for the shape factor is for the case D/(2*r)>1\n", + "\n", + "#Calculation\n", + "S = (4*math.pi*r)/((math.pi/2)-math.atan(r/(2*D)))\t # [m] shape factor\n", + "\t# heat lost by the disk is \n", + "q = k*S*(Td-Ts) \t\t\t\t\t # [W]\n", + "\n", + "#Result\n", + "print\"Heat lost by disk is:\",round(q,2),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 3.4" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer between the disks is: 308.4 W\n" + ] + } + ], + "source": [ + "#Example Number 3.4 \n", + "#Calculate the heat transfer betwwen the disks\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.50\t # [m] diameter of both disk\n", + "r = d/2 \t # [m] radius of disk\n", + "Td1 = 80\t # [degree celsius] first disk temperature\n", + "Td2 = 20 \t # [degree celsius] second disk temperature\n", + "k = 2.3 \t # [W/m per degree celsius] thermal conductivity of medium \n", + "D = 1.5\t\t # [m] seperation of disk in a infinite medium\n", + "\t# We have to calculate shape factor using relation given in table (3-1) \n", + "\t# We select the relation for the shape factor is for the case D>5*r\n", + "#Calculation\n", + "import math\n", + "\n", + "S = (4*math.pi*r)/((math.pi/2)-math.atan(r/D)) # [m] shape factor\n", + "q = k*S*(Td1-Td2) # [W]\n", + "\n", + "#Result\n", + "print\"Heat transfer between the disks is:\",round(q,1),\"W\" \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", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_4_Unsteady_State_Conduction.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_4_Unsteady_State_Conduction.ipynb new file mode 100644 index 00000000..3363aee9 --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_4_Unsteady_State_Conduction.ipynb @@ -0,0 +1,669 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 Unsteady State Conduction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.1" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "hV/A/k is: 0.0024 which is less than 0.1,So by eq 4.5\n", + "hA/pcV is 0.000334448160535 per second\n", + "Time is 5818.0 seconds\n" + ] + } + ], + "source": [ + "#Example 4.1\n", + "# Calculate the time required for ball to attain a temperature of 150 deg C.\n", + "\n", + "#variables:\n", + "\n", + "c=460.0 \t# kJ/kg\n", + "k=35.0 \t#W/m\n", + "d=0.05 \t# diameter in meter\n", + "r=d/2 \t\t# radius in meter\n", + "h=10.0 \t#convection heat transfer coeff in W/sq meter\n", + "T=150.0 \t#Temperature in deg C\n", + "p=7800 \t#Density in kg/cubic meter\n", + "Ti=100.0 \t# T,infinity ..in celsius\n", + "To=450.0 \t# in Celsius\n", + "\n", + "#CALCULATION\n", + "\n", + "import math \n", + "A=(4*math.pi) * (r**2) \t\t\t #Arear in sq meter\n", + "V=(A*r/3) \t\t\t\t # Volume in cubic meters\n", + "c1=(h*(V/A))/k \t \t\t #it is less than 0.1 So,\n", + "print \"hV/A/k is:\",round(c1,4),\"which is less than 0.1,So by eq 4.5\"\n", + "c2=((h*A)/(p*c*V)) \t\t # assumed variable for easiness\n", + "\n", + "#RESULTS\n", + "\n", + "print \"hA/pcV is\",c2,\"per second\"\n", + "t=(-1/c2)*(math.log((T-Ti)/(To-Ti))) \n", + "print \"Time is\",round(t),\"seconds\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.2" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at depth of 0 .025 m after 30 s for case 1 is 118.5 degree celsius \n", + "Temperature at depth of 0 . 0 2 5 m after 30 second for case 2 is 79.3 degree celsius\n", + "Surface temperature after 30 second is: 199.4 degree celsius \n" + ] + } + ], + "source": [ + "#Example 4.2\n", + "#Calculate the temperature at a depth of 2.5 cm after 0.5 min for both cases\n", + "import math\n", + "\n", + "#variable declaration\n", + "\n", + "k = 45 \t\t\t\t# [W/m per deg C ] thermal conductivity of steel block\n", + "alpha = 1.4*10**( -5) ; \t# [ square meter / s ] constant\n", + "Tb = 35 \t\t\t# [ degree celsius ] block temperature\n", + "x = 0.025 \t\t\t# [m] depth at which temperature is calculated\n", + "t = 30 \t\t\t\t# [s]time after which temperature is to be calculated\n", + "To = 250 \t\t\t# [ degree celsius]\n", + "\n", + "\n", + "#CALCULATION & rESULT\n", + "\n", + "er=(x/(2*math.sqrt(alpha*t)))\t \t # error function short form\n", + "T_x_t = (To +(Tb -To)*( math.erf(er))) \n", + "\t\t\t\t\t # for the constant heat flux case B we \t\t\t\t\t\t make use of eq4.13 a\n", + "\t\t\t # since qo/A is given\n", + "\n", + "print \"Temperature at depth of 0 .025 m after 30 s for case 1 is\",round(T_x_t,1),\" degree celsius \"\n", + "q_by_A = 3.2*10**(5) ; \t\t\t\t# [W/ s qua r e meter ]\n", + "\n", + "T_x_t1 =(Tb +(2* q_by_A*math.sqrt(alpha*t/math.pi)*math.exp(-(x**2)/(4*alpha *t))/k) -(q_by_A *x*(1 - math.erf(er))/k)) \t\t# [ degree celsius ]\n", + "\n", + "print \"Temperature at depth of 0 . 0 2 5 m after 30 second for case 2 is\",round(T_x_t1,1),\"degree celsius\" \n", + "\t\t\t\t# for the constant heat flux case the surface \t\t\t\t temperature after 30 s would be evaluated with x= 0 \t\t\t\t in equation(4.13 a )\n", + "x = 0 \t\t\t\t\t\t\t\t# [m] at the surface\n", + "\n", + "T_x_o = Tb +(2* q_by_A *math.sqrt(alpha*t/math.pi) * math.exp (-(x **2)/(4*alpha*t))/k) -(q_by_A*x*(1 -math.erf (er))/k) \t\t# [degree celsius]\n", + "\n", + "print \"Surface temperature after 30 second is:\",round(T_x_o,1) ,\"degree celsius \" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.3" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ASSUMPTION: Negligible radiation effects\n", + "At x=0\n", + "At x=2mm ie 0.002 m\n", + "0.002\n", + "T =: 513.0 Celsius\n" + ] + } + ], + "source": [ + "#Example 4.3\n", + "#Estimate the temperature at the surface and at a depth of 2 mm after 2 sec\n", + " \t\t\t#Qo/A=10^7 J/sq meter\n", + "import math \t \t# import math file\n", + "\n", + "#variable declaration\n", + "\n", + "c1=10 \t\t# =Qo/A laser pulse in MJ/sq meter\n", + "c1=(c1*10**6) \t\t# convert in joule from MJ\n", + "p=7800 \t\t\t# Density in kg/cubic meter\n", + "c=460\t\t\t#J/kg\n", + "a=(0.44*(10**(-5))) \t# sq m/sec\n", + "t=2 \t\t\t# time in seconds\n", + "Ti=40.0 \t\t# initial temp in deg C\n", + "\n", + "print \"ASSUMPTION: Negligible radiation effects\"\n", + "#From eqq 4.13,:\n", + "print \"At x=0\"\n", + "\n", + "#calculation\n", + "To=Ti+(c1/(p*c*math.sqrt((math.pi)*a*t)))\n", + "print \"At x=2mm ie 0.002 m\"\n", + "x=0.002\n", + "print x\n", + "T=(Ti+(c1/(p*c*math.sqrt((math.pi)*a*t)))*(math.exp(-(x**2)/(4*a*t))))\n", + "\n", + "#RESULTS\n", + "print \"T =:\", round(T),\"Celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.5" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The iteration are listed below\n", + " h*math.sqrt(alpha*t)/k x/(2*sqrt(alpha*t)) (T_x_t-Ts)/(Te-Ts)\n", + "1000 0.708 0.069 0.41\n", + "3000 1.226 0.04 0.61\n", + "4000 1.415 0.035 0.68\n", + "Consequently the time required is approximately 3000 second\n" + ] + } + ], + "source": [ + "#Example 4.5:Sudden exposure of semi-infinite solid slab to convection\n", + "# Calculate time req for the temp to reach 120 deg C at depth of 4 cm\n", + "\n", + "#VARIABLE DECLARATION\n", + "\n", + "alpha = 8.4*10**(-5) \t\t# [square meter/s] constant\n", + "Ts = 200\t\t # [degree celsius] initial temperature of of slab\n", + "Te = 70\t\t\t\t# [degree celsius] environment temperature \n", + "k = 215 \t\t\t# [W/m deg C] heat transfer coefficient of slab\n", + "h = 525 \t\t\t# [W/sq m degree celsius] heat transfer coefficient \n", + "x = 0.04\t\t\t# [m] depth at which temperature is calculated\n", + "T_x_t = 120 \t\t\t# [degree celsius] temperature at depth 0.04 m\n", + "\n", + "\t# using eq 4-15 or figure (4-5) for solution of this problem\n", + "\t# by using figure it is easier to calculate it involves iterative method to \tsolve because time appeares in both the variables \n", + "\n", + "\t# h*sqrt(alpha*t)/k and x/(2*sqrt(alpha*t))\n", + "K = (T_x_t-Ts)/(Te-Ts) \n", + "\t# Seek the values of t such that the above value of K is equal to the value of \tK which comes out from graph\n", + "\n", + "# values of t and obtain other readings\n", + "\n", + "#CALCULATION & RESULT\n", + "\n", + "print \"The iteration are listed below\"\n", + "\t# at t = 1000s\n", + "import math\n", + "t = 1000 # [s] time\n", + "A = h*math.sqrt(alpha*t)/k \n", + "B = x/(2*math.sqrt(alpha*t)) \n", + "\n", + "print \" h*math.sqrt(alpha*t)/k x/(2*sqrt(alpha*t)) (T_x_t-Ts)/(Te-Ts)\"\n", + "\n", + "print t,\" \",round(A,3),\" \",round(B,3),\" \",\"0.41\"\n", + "\n", + "t = 3000 \t\t\t# [s] time\n", + "A = h*math.sqrt(alpha*t)/k \n", + "B = x/(2*math.sqrt(alpha*t)) \n", + "\n", + "print t,\" \",round(A,3),\" \",round(B,3),\" \",\"0.61\"\n", + "\n", + "t = 4000 \t\t\t# [s] time\n", + "A = h*math.sqrt(alpha*t)/k \n", + "B = x/(2*math.sqrt(alpha*t)) \n", + " \n", + "print t,\" \",round(A,3),\" \",round(B,3),\" \",\"0.68\"\n", + "print \"Consequently the time required is approximately 3000 second\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.6" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at a depth of 1.25 cm from one of faces after 1 min of exposure of plate to the environment is 147.7 degree celsius\n", + "Energy removed per unit area from the plate in this time is 6475950.0 J/square meter\n" + ] + } + ], + "source": [ + "#Example 4.6-aluminium plate suddenly exposed to convection\n", + "#Calculate Energy removed /Area from the plate at this time\n", + "\n", + "#VARIABLE DECLARATION\n", + "alpha = 8.4*10**(-5) \t\t\t# [square meter/s] constant\n", + "Ts = 200 \t\t\t\t# [deg celsius] initial temperature of plate\n", + "Te = 70 \t\t\t\t# [degree celsius] environment temperature \n", + "k = 215 \t\t\t\t# [W/m deg C] heat transfer coeff of plate\n", + "h = 525 \t\t\t\t#[W/sq m deg C] heat transfer coefficient \n", + "x = 0.0125 \t\t\t\t#[m] depth at which temperature is calculated\n", + "t = 60\t\t\t\t\t#[s]time after which temperature is calculated\n", + "L = 0.025 \t\t\t\t# [m] thickness of plate\n", + "\n", + "#CALCULATION\n", + "\n", + "theta_i = Ts-Te\t\t\t\t # [degree celsius]\n", + "\n", + "\t\t\t# then \n", + "Z = alpha*t/L**2 \n", + "X = k/(h*L) \n", + "x_by_L = x/L \n", + "\t\n", + "\t\t\t# from figure 4-7(page no.-144-145)\n", + "\n", + "theta_o_by_theta_i = 0.61 \n", + "theta_o = theta_o_by_theta_i*theta_i # [degree celsius]\n", + "\t\t\t# from figure 4-10(page no.-149) at x/L = 0.5,\n", + "theta_by_theta_o = 0.98 \n", + "theta = theta_by_theta_o*theta_o \t# [degree celsius]\n", + "T = Te+theta \t\t\t\t# [degree celsius]\n", + "\n", + "\t\t\t# using Figure 4-14(page no.-152). For this calculation we \t\t\trequire the following properties of aluminium:\n", + "\t\n", + "rho = 2700 \t\t\t\t# [kg/cubic meter]\n", + "C = 900 \t\t\t\t# [J/kg degree celsius]\n", + "\n", + "\t\t\t# for figure 4-14(page no.-152) we need \n", + "\n", + "V = h**2*alpha*t/(k**2) \n", + "B = h*L/k \n", + "\n", + "\t\t\t# from figure 4-14(page no.-152)\n", + "\n", + "Q_by_Qo = 0.41 \n", + "\n", + "\t\t\t# for unit area \n", + "Qo_by_A = rho*C*2*L*theta_i \t\t# [J/square meter]\n", + "\n", + "# Now, heat removed per unit surface area is \n", + "Q_by_A = Qo_by_A*Q_by_Qo \t\t# [J/square meter]\n", + "\n", + "#RESULTS\n", + "\n", + "print\"Temperature at a depth of 1.25 cm from one of faces after 1 min of exposure of plate to the environment is\",round(T,1),\" degree celsius\" \n", + "\n", + "print\"Energy removed per unit area from the plate in this time is\",Q_by_A,\" J/square meter\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.7" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at a radius of 1.25 cm is 118.4 degree celsius\n", + "Heat lost per unit length 1 minute after the cylinder is exposed to the environment is 403174.0 J/m\n" + ] + } + ], + "source": [ + "#Example4.7-Long cylinder suddenly exposed to convection \n", + "\n", + "#VARIABLE DECLARATION\n", + "\n", + "d = 0.05 \t\t\t# [m] diameter of cylinder\n", + "Ti = 200 \t\t\t# [degree C] initial temperature of aluminium cylinder\n", + "Tinf = 70 \t\t\t# [degree celsius] temperature of environment\n", + "h = 525 \t\t\t# [W/sq m degree celsius] heat transfer coefficient\n", + "\n", + "#CALCULATION\n", + "\n", + "# we have\n", + "theta_i = Ti-Tinf \t\t# [degree celsius]\n", + "alpha = 8.4*10**(-5) \t\t# [square meter/s]\n", + "ro = d/2 \t\t\t# [m]\n", + "t = 60 \t\t\t\t# [s]\n", + "k = 215\t\t # [W/m degree celsius]\n", + "r = 0.0125 \t\t\t# [m]\n", + "rho = 2700 \t\t\t# [kg/cubic meter]\n", + "C = 900 \t\t\t# [J/kg degree celsius]\n", + "\n", + "\t# we compute\n", + "\n", + "Z = alpha*t/ro**2 \n", + "X = k/(h*ro) \n", + "r_by_ro = r/ro \n", + "\n", + "\n", + "\t# from figure 4-8(page no.-146)\n", + "theta_o_by_theta_i = 0.38 \n", + "\n", + "\t# from figure 4-11(page no.-150) at r/ro = 0.5\n", + "theta_by_theta_o = 0.98 \n", + "\n", + "\t# so that \n", + "theta_by_theta_i = theta_o_by_theta_i*theta_by_theta_o \n", + "theta = theta_i*theta_by_theta_i \t# [degree celsius]\n", + "T = Tinf+theta \t\t\t\t# [degree celsius]\n", + "\n", + "\t# to compute the heat lost, we determine\n", + "V = h**2*alpha*t/k**2 \n", + "B = h*ro/k \n", + "\n", + "\t# from figure 4-15(page no.-153)\n", + "Q_by_Qo = 0.65 \n", + "\n", + "\t# for unit length\n", + "import math\n", + "Qo_by_L = rho*C*math.pi*ro**2*theta_i \t# [J/m]\n", + "\n", + "\t#actual heat lost per unit length is \n", + "Q_by_L = Qo_by_L*Q_by_Qo \t\t# [J/m]\n", + "\n", + "\n", + "\n", + "#RESULTS\n", + "\n", + "print\"Temperature at a radius of 1.25 cm is\",round(T,1),\" degree celsius\" \n", + "print\"Heat lost per unit length 1 minute after the cylinder is exposed to the environment is\",round(Q_by_L),\"J/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.8" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The temperature at the axis is: 117.6 degree celsius\n", + "The temperature at the surface is 116.2 degree celsius\n" + ] + } + ], + "source": [ + "#Example 4.8-\n", + "#Calculate temperature at axis and surface of th cylinder 10 cm from end 1 min after exposure\n", + "\n", + "#VARIABLES DECLARATION\n", + "import math\n", + "\n", + "d = 0.05 \t\t# [m] diameter of aluminium cylinder \n", + "Ti = 200 \t\t# [degree celsius] initial temperature of of cylinder\n", + "Te = 70 \t\t# [degree celsius] environment temperature\n", + "k = 215 \t\t# [W/m degree celsius] heat transfer coefficient of plate\n", + "h = 525 \t\t# [W/sq m degree celsius] convection heat transfer coefficient \n", + "alpha = 8.4*10**(-5) # [square meter/s] constant\n", + "x = 0.10 \t\t# [m] distance at which temperature is calculated from end\n", + "t = 60 \t\t\t# [s] time after which temperature is measured\n", + "\t# so that the parameters for use with figure(4-5)\n", + "\n", + "#CALCULATION\n", + "A = h*math.sqrt(alpha*t)/k \n", + "B = x/(2*math.sqrt(alpha*t)) \n", + "# from figure (4-5)\n", + "z = 1-0.036 \n", + "S_of_X = z \n", + "\t# for the infinite cylinder we seek both the axis- and surface-temperature \tratios.\n", + "\t# the parameters for use with fig(4-8) are \n", + "r_o = d/2 \t\t\t# [m] radius of aluminium cylinder \n", + "r = d/2 \t\t\t# [m] for surface temperature ratio\n", + "C = k/(h*r_o) \n", + "D = (alpha*t/r_o**(2)) \n", + "y = 0.38 \n", + "\n", + "\t# this is the axis temperature ratio.\n", + "\t# to find the surface-temperature ratio,we enter figure (4-11),using \n", + "R = r/r_o \n", + "u = 0.97 \n", + "# thus \n", + "w = y \t\t\t# at r = 0\n", + "v = y*u \t\t# at r = r_o\n", + "C_of_O_axis = w \t# at r = 0\n", + "C_of_O_r_o = v \t\t# at r = r_o\n", + "\t# combining the solutions for the semi-infinite slab and infinite cylinder, we \thave \n", + "t = S_of_X*C_of_O_axis \n", + "s = S_of_X*C_of_O_r_o \n", + "\t# the corresponding temperatures are \n", + "T_axis = Te+t*(Ti-Te) \n", + "T_r_o = Te+s*(Ti-Te) \n", + "\n", + "\n", + "#RESULTS\n", + "\n", + "print\"The temperature at the axis is:\",round(T_axis,1),\" degree celsius\"\n", + "print\"The temperature at the surface is\",round(T_r_o,1),\" degree celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.9" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature at a radial position of 0.0125 m and a distance of 0.00625m from one end of cylinder 60 second after exposure to environment is 104.5 degree celsius\n" + ] + } + ], + "source": [ + "#Example 4.9\n", + "# finite length cylinder suddenly exposed to convection \n", + "\n", + "#Variable declaration\n", + "\n", + "d = 0.05 \t\t\t# [m] diameter of aluminium cylinder \n", + "Ti = 200 \t\t\t# [degree celsius] initial temperature of of cylinder\n", + "Te = 70\t\t\t # [degree celsius] environment temperature\n", + "k = 215 \t\t\t# [W/m deg celsius] heat transfer coefficient of plate\n", + "h = 525 \t\t\t# [W/sq m deg C] convection heat transfer coefficient \n", + "alpha = 8.4*10**(-5) \t\t#[square meter/s] constant\n", + "x1 = 0.00625 \t\t\t#[m]dist at which temperature is calculated from end\n", + "t = 60 \t\t\t\t# [s] time after which temperature is measured\n", + "r = 0.0125 \t\t\t# [m] radius at which temperature is calculated\n", + "\t# to solve this problem we combine the solutions from heisler charts for an \t\tinfinite cylinder and an infinite plate in accordance with the combination \t\tshown in fig (4-18f)\n", + "\t# for the infinite plate problem \n", + "L = 0.05\t\t\t # [m]\n", + "\n", + "#CALCULATION\n", + "\n", + "\t# the x position is measured fromthe center of the plate so that\n", + "x = L-x1 \t\t\t# [m]\n", + "A = k/(h*L) \n", + "B = (alpha*t/L**(2)) \n", + "\t# from figures (4-17) and (4-10) respectively\n", + "thetha_o_by_i = 0.75 \n", + "thetha_by_i = 0.95 \n", + "\t# so that\n", + "thetha_by_i_plate = thetha_o_by_i*thetha_by_i \n", + "\t# for the cylinder \n", + "r_o = d/2 \t\t\t# [m] radius of the cylinder\n", + "R = r/r_o \n", + "C = k/(h*r_o) \n", + "D = (alpha*t/r_o**(2)) \n", + "\t# and from figures (4-8) and (4-11), respectively\n", + "thetha_o_by_i_cyl = 0.38 \n", + "thetha_by_o = 0.98 \n", + "\t# so that\n", + "thetha_by_i_cyl = thetha_o_by_i_cyl*thetha_by_o \n", + "\t# combining the solutions for the plate and cylinder gives\n", + "thetha_by_i_short_cyl = thetha_by_i_plate*thetha_by_i_cyl \n", + "\t#thus\n", + "T = Te+thetha_by_i_short_cyl*(Ti-Te) \n", + "\n", + "#RESULTS\n", + "print\"Temperature at a radial position of 0.0125 m and a distance of 0.00625m from one end of cylinder 60 second after exposure to environment is\",round(T,1),\"degree celsius\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 4.10" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The actual heat loss in the 1-minute: 40.2 kJ\n" + ] + } + ], + "source": [ + "#Example 4.10\n", + "# heat loss for finite-length cylinder \n", + "\n", + "#Variable declaration\n", + "d = 0.05 \t\t# [m] diameter of aluminium cylinder\n", + "l = 0.1 \t\t# [m] length of aluminium cylinder \n", + "Ti = 200\t\t# [degree celsius] initial temperature of of cylinder\n", + "Te = 70 \t\t# [degree celsius] environment temperature\n", + "k = 215 \t\t# [W/m degree celsius] heat transfer coefficient of plate\n", + "h = 525 \t\t# [W/sq m deg celsius] convection heat transfer coefficient \n", + "alpha = 8.4*10**(-5) \t#[square meter/s] constant\n", + "x1 = 0.00625 \t\t#[m] distance at which temperature is calculated from end\n", + "t = 60 \t\t\t#[s]time after which temperature is measured\n", + "r = 0.0125 #[m] radius at which temperature is calculated\n", + "\n", + "\n", + "#CALCULATION\n", + "\t\n", + "\t# calculate the dimensionless heat-loss ratio for the infinite plate and \tinfinite cylinder which make up the multidimensional body\n", + "\t# for the plate we have \n", + "L = 0.05 \t\t# [m]\n", + "A = h*L/k \n", + "B = h**(2)*alpha*t/k**(2) \n", + "\t# from figure (4-14), for the plate, we read \n", + "Q_by_Q_o_plate = 0.22 \n", + "\n", + "\t# for the cylinder \n", + "r_o = 0.025 \t\t# [m]\n", + "\t# so we calculate \n", + "C = h*r_o/k \n", + "\t# and from figure(4-15) we have \n", + "Q_by_Q_o_cyl = 0.55 \n", + "\n", + "\t# the two heat ratios may be inserted in equation(4-22) to give \n", + "Q_by_Q_o_tot = Q_by_Q_o_plate+Q_by_Q_o_cyl*(1-Q_by_Q_o_plate) \n", + "c = 896 \t\t# [J/kg degree celsius] specific heat of aluminium\n", + "rho = 2707 \t\t# [kg/cubic meter] density of aluminium\n", + "\n", + "import math\n", + "\n", + "V = math.pi*r_o**(2)*l \t# [cubic meter]\n", + "Qo = rho*c*V*(Ti-Te) \t# [J]\n", + "Q = Qo*Q_by_Q_o_tot \t# [J] the actual heat loss in the 1-minute \n", + "\n", + "#RESULTS\n", + "\n", + "print \"The actual heat loss in the 1-minute:\",round((Q/1000),1),\" kJ\" " + ] + } + ], + "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 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_5_Principles_of_Convection.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_5_Principles_of_Convection.ipynb new file mode 100644 index 00000000..ec39bab5 --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_5_Principles_of_Convection.ipynb @@ -0,0 +1,552 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 Principles of Convection" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the increase in static pressure between sections 1 and 2 is: 61.88 kPa\n" + ] + } + ], + "source": [ + "#Example 5.1\n", + "# water flow in a diffuser \n", + "\n", + "#VARIABLE DECLARATION\n", + "Tw = 20 \t\t\t# [degree celcius] water temperature \n", + "m_dot = 8 \t\t\t# [kg/s] water flow rate \n", + "d1 = 0.03 \t\t\t#[m] diameter at section 1\n", + "d2 = 0.07 \t\t\t# [m] diameter at section 2\n", + "\n", + "#CALCULATION\n", + "import math\n", + "A1 = math.pi*d1**(2)/4 \t\t# [square meter] cross-sectional area at section 1\n", + "A2 = math.pi*d2**(2)/4 \t\t# [square meter] cross-sectional area at section 2\n", + "gc = 1 \t\t\t\t# [m/s**(2)] acceleration due to gravity\n", + "rho = 1000 \t\t\t# [kg/cubic m] density of water at 20 degree celcius\n", + "\n", + "\t# calculate the velocities from the mass-continuity relation\n", + "u1 = m_dot/(rho*A1) \t\t# [m/s]\n", + "u2 = m_dot/(rho*A2) \t\t# [m/s]\n", + "\t# the pressure difference is obtained by Bernoulli equation(5-7a)\n", + "p2_minus_p1 = rho*(u1**(2)-u2**(2))/(2*gc)\t # [Pa] \n", + "\n", + "#RESULTS\n", + "print\"the increase in static pressure between sections 1 and 2 is:\",round(p2_minus_p1/1000,2),\" kPa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The static temperature is: 529.0 K\n", + "Static pressure is: 0.5 MPa\n", + "Mach number is: 0.651\n" + ] + } + ], + "source": [ + "#Example Number 5.2\n", + "# isentropic expansion of air \n", + "\n", + "# Variable declaration\n", + "\n", + "Ta = 300.0+273.0\t \t\t# [K] air temperature\n", + "Pa = 0.7 \t\t\t\t# [MPa] pressure of air\n", + "u2 = 300 \t\t\t\t# [m/s] final velocity\n", + "gc = 1 \t\t\t\t\t# [m/s^(2)] acceleration due to gravity\n", + "Y = 1.4 \t\t\t\t# gama value for air \n", + "Cp = 1005 \t\t\t\t# [J/kg degree celsius]\n", + "\t#the initial velocity is small and the process is adiabatic. in terms of \t\ttemperature \n", + "\n", + "\n", + "#Calculation\n", + "\n", + "T2 = Ta-u2**(2)/(2*gc*Cp) \n", + "\n", + "#Result\n", + "print \"The static temperature is:\",round(T2,1),\"K\" \n", + "\n", + "\t# we may calculate the pressure difference from the isentropic relation \n", + "\n", + "p2 = Pa*((T2/Ta)**(Y/(Y-1))) \n", + "\n", + "print \"Static pressure is:\",round(p2,1),\"MPa\"\n", + "\n", + "\t# the velocity of sound at condition 2 is \n", + "a2 = (20.045*(T2**(0.5))) \t\t# [m/s] \n", + "\t#so that the mach no. is \n", + "M2 = u2/a2 \n", + "print \"Mach number is:\",round(M2,3) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The heat transfered in first case of the plate is 81.2 W\n", + "and the heat transfered in second case of the plate is: 114.8 W\n" + ] + } + ], + "source": [ + "#Example Number 5.4\n", + "#Calculate the heat transfereed in first 20 cm of the plate and the first 40 cm of the plate\n", + "\n", + "# Variable declaration\n", + "\n", + "\t# total heat transfer over a certain length of the plate is desired, so we \t\twish to calculate average heat transfer coefficients. \n", + "\t# for this purpose we use equations (5-44) and (5-45), evaluating the \t\tproperties at the film temperature :\n", + "Tp = 60+273.15 \t\t\t\t# [K] plate temperature \n", + "Ta = 27+273.15 \t\t\t\t# [K] air temperature\n", + "Tf = (Tp+Ta)/2 \t\t\t\t# [K]\n", + "u = 2 \t\t\t\t\t# [m/s] air velocity\n", + "\n", + "\t# from appendix A the properties are \n", + "\n", + "v = 17.36*(10**(-6)) \t\t\t# [square meter/s] kinematic viscosity\n", + "x1 = 0.2 \t\t\t\t# [m] distance from the leading edge of plate\n", + "x2 = 0.4 \t\t\t\t# [m] distance from the leading edge of plate\n", + "k = 0.02749 \t\t\t\t# [W/m K] heat transfer coefficient\n", + "Pr = 0.7 \t\t\t\t# prandtl number\n", + "Cp = 1006 \t\t\t\t# [J/kg K]\n", + "\n", + "\t# at x = 0.2m\n", + "\n", + "#Calculation\n", + "\n", + "Re_x1 =(u*x1/v) \t\t\t\t# reynolds number\n", + "Nu_x1 = 0.332*(Re_x1**(0.5))*(Pr**(0.333)) \t# nusselt number\n", + "hx1 = Nu_x1*k/x1 \t\t\t\t# [W/square meter K] \n", + "\n", + "\t# the average value of the heat transfer coefficient is twice this value, or\n", + "\n", + "h_bar1 = 2*hx1 \t\t\t\t\t# [W/square meter K] \n", + "\n", + "\t# the heat flow is \n", + "\n", + "A1 = x1*1 \t\t\t\t\t# [square meter] area for unit depth\n", + "q1 = h_bar1*A1*(Tp-Ta) \t\t\t\t# [W]\n", + "\n", + "\t# at x = 0.4m\n", + "\n", + "Re_x2 = u*x2/v \t\t\t\t\t# reynolds number\n", + "Nu_x2 = 0.332*Re_x2**(0.5)*Pr**(0.333) \t\t# nusselt number\n", + "hx2 = Nu_x2*k/x2 \t\t\t\t# [W/square meter K] \n", + "\n", + "\t# the average value of the heat transfer coefficient is twice this value, or\n", + "\n", + "h_bar2 = 2*hx2 \t\t\t\t\t# [W/square meter K] \n", + "\t# the heat flow is \n", + "A2 = x2*1 \t\t\t\t\t# [square meter] area for unit depth\n", + "q2 = h_bar2*A2*(Tp-Ta) \t\t\t\t# [W] \n", + "\n", + "#Result\n", + " \n", + "print\"The heat transfered in first case of the plate is\",round(q1,2),\"W\"\n", + "print\"and the heat transfered in second case of the plate is:\",round(q2,1),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average temperature difference along the plate is: 241.0 degree celsius\n", + "Temperature difference at the trailing edge is: 365.3 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 5.5\n", + "# Calculate the av temperature difference along the plate & Temperature diff at the trailing edge\n", + "\n", + "#Variable declaration\n", + "\n", + "u = 5 \t\t\t\t\t# [m/s] air velocity\n", + "l = 0.6 \t\t\t\t# [m] plate length\n", + "Ta = 27+273.15 \t\t\t\t# [K] temperature of airstream\n", + "\n", + "\t# properties should be evaluated at the film temperature, but we do not know \tthe plate temperature so for an initial calculation we take the properties at \tthe free-stream conditions of\n", + "\n", + "v = 15.69*10**(-6)\t \t\t#[square meter/s] kinematic viscosity\n", + "k = 0.02624 \t\t\t\t#[W/m deg celsius] heat transfer coefficient\n", + "Pr = 0.7 \t\t\t\t# prandtl number\n", + "Re_l = l*u/v \t\t\t\t# reynolds number\n", + "P = 1000 \t\t\t\t# [W] power of heater\n", + "qw = P/l**(2) \t\t\t\t# [W/square meter] heat flux per unit area \n", + "\n", + "\t# from equation (5-50) the average temperature difference is \n", + "\n", + "#Calculation\n", + "\n", + "Tw_minus_Tinf_bar = qw*l/(0.6795*k*(Re_l)**(.5)*(Pr)**(0.333)) \t # [degree celsius]\n", + "\n", + "\t# now, we go back and evaluate properties at \n", + "Tf = (Tw_minus_Tinf_bar+Ta+Ta)/2 \t# [degree celsius]\n", + "\n", + "\t# and obtain\n", + "\n", + "v1 = 28.22*10**(-6) \t\t\t# [square meter/s] kinematic viscosity\n", + "k1 = 0.035 \t\t\t\t# [W/m deg celsius] heat transfer coefficient\n", + "Pr1 = 0.687 \t\t\t\t# prandtl number\n", + "Re_l1 = l*u/v1 \t\t\t\t# reynolds number\n", + "Tw_minus_Tinf_bar1 = qw*l/(0.6795*k1*(Re_l1)**(0.5)*(Pr1)**(0.333)) #[degree celsius]\n", + "\n", + "\t# at the end of the plate(x = l = 0.6m) the temperature difference is obtained \tfrom equation (5-48) and (5-50) with the constant of 0.453\n", + "\n", + "Tw_minus_Tinf_x_equal_l = Tw_minus_Tinf_bar1*0.6795/0.453 \t# [degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print \"Average temperature difference along the plate is:\",round(Tw_minus_Tinf_bar),\" degree celsius\"\n", + "print \"Temperature difference at the trailing edge is:\",round(Tw_minus_Tinf_x_equal_l,1),\"degree celsius\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.7" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average value of the convection coefficient is 219.1 W/sq meter degree celsius\n", + " and the heat lost by the plate is 350.6 W\n" + ] + } + ], + "source": [ + "#Example Number 5.7\n", + "# Calculate the heat lost by the plate\n", + "\n", + "# Variable declaration\n", + "\n", + "u = 1.2 \t\t\t# [m/s] oil velocity\n", + "l = 0.2 \t\t\t# [m] plate length as well as width (square) \n", + "To = 20+273.15 \t\t\t# [K] temperature of engine oil\n", + "Tu = 60+273.15 \t\t\t# [K] uniform temperature of plate \n", + "\t# First we evaluate the film temperature \n", + "T = (To+Tu)/2 \t\t\t# [K]\n", + "\t# and obtain the properties of engine oil are \n", + "rho = 876 \t\t\t# [kg/cubic meter] density of oil\n", + "v = 0.00024 \t\t\t# [square meter/s] kinematic viscosity\n", + "k = 0.144 \t\t\t# [W/m degree celsius] heat transfer coefficient\n", + "Pr = 2870 \t\t\t# prandtl number\n", + "\t# at the trailing edge of the plate the reynolds number is \n", + "\n", + "#Calculation\n", + "\n", + "Re = l*u/v \t\t\t# reynolds number\n", + "\n", + "\n", + "\n", + "\t# because the prandtl no. is so large we will employ equation(5-51) for the \t\tsolution. \n", + "\n", + "\t# we see that hx varies with x in the same fashion as in equation(5-44) , i.e. \thx is inversely proportional to the square root of x ,\n", + "\t# so that we get the same solution as in equation(5-45) for the average heat \t\ttransfer coefficient. \n", + "\n", + "\t# evaluating equation(5-51) at x = 0.2m gives\n", + "\n", + "Nux = (0.3387*(Re**(1.0/2.0))*(Pr**(1.0/3.0)))/((1+(0.0468/Pr)**(2.0/3.0))**(1.0/4.0))\n", + "\n", + "\n", + "hx = Nux*k/l \t\t\t# [W/sq m degree celsius] heat transfer coefficient\n", + "\n", + "\t# the average value of the convection coefficient is \n", + "\n", + "h = 2*hx \t\t\t# [W/square meter degree celsius] \n", + "\n", + "\t# so that total heat transfer is \n", + "\n", + "A = l**(2) \t\t\t# [square meter] area of the plate \n", + "q = h*A*(Tu-To) \t\t#[W] \n", + "\n", + "print \"Average value of the convection coefficient is\",round(h,1),\"W/sq meter degree celsius\"\n", + "print \" and the heat lost by the plate is\",round(q,1),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.8" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drag force exerted on the first 0.4 m of the plate is 5.45 mN\n" + ] + } + ], + "source": [ + "#Example Number 5.8\n", + "# Compute the drag force on the first 40 cm of the plate \n", + "\n", + "# Variable declaration\n", + "\n", + "\n", + "\t# data is used from example 5.4 \n", + "\t# we use equation (5-56) to compute the friction coefficient and then \t\tcalculate the drag force .\n", + "\t# an average friction coefficient is desired, so st_bar*pr**(2/3) = Cf_bar/2\n", + "\n", + "p = 101325 \t\t\t# [Pa] pressure of air\n", + "x = 0.4\t\t\t\t#[m] drag force is computed on first 0.4 m of the \t\t\t plate \n", + "R = 287 \t\t\t# []\n", + "Tf = 316.5 \t\t\t# [K]\n", + "u = 2 \t\t\t\t# [m/s] air velocity\n", + "Cp = 1006 \t\t\t# [J/kg K]\n", + "Pr = 0.7 \t\t\t# prandtl no.\n", + "rho = p/(R*Tf) \t\t\t# [kg/cubic meter] density at 316.5 K \n", + "h_bar = 8.698 \t\t\t# [W/square meter K] heat transfer coefficient\n", + "\n", + "\n", + "#Calculation\n", + "\t# for the 0.4m length\n", + "\n", + "st_bar = h_bar/(rho*Cp*u) \n", + "\n", + "\t# then from equation (5-56)\n", + "\n", + "Cf_bar = st_bar*Pr**(2.0/3.0)*2 \n", + "\n", + "\t# the average shear stress at the wall is computed from equation(5-52)\n", + "\n", + "tau_w_bar = Cf_bar*rho*u**(2)/2 \t# [N/square meter]\n", + "A = x*1 \t\t\t\t# [square meter] area per unit length \n", + "\n", + "\t# the drag force is the product of this shear stress and the area,\n", + "\n", + "D = tau_w_bar*A \t\t\t# [N] \n", + "\n", + "#Result\n", + "\n", + "print \"Drag force exerted on the first 0.4 m of the plate is\",round(D*1000,2),\"mN\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.9" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nul_bar is 2175.0\n", + "Heat transfer from plate is 2369.0 W\n" + ] + } + ], + "source": [ + "#Example Number 5.9\n", + "# Calculate the heat transfer from the plate\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101320.0 \t\t\t\t# [Pa] pressure of air\n", + "R = 287.0 \t\t\t\t# []\n", + "Ta = 20+273 \t\t\t\t# [K] temperature of air \n", + "u = 35 \t\t\t\t\t# [m/s] air velocity\n", + "L = 0.75 \t\t\t\t# [m] length of plate \n", + "Tp = 60+273 \t\t\t\t# [K] plate temperature \n", + "\n", + "\n", + "\t\t# we evaluate properties at the film temperature \n", + "\n", + "#Calculations\n", + "\n", + "Tf = (Ta+Tp)/2 \t\t\t\t# [K]\n", + "\n", + "\n", + "rho = (p/(R*Tf)) \t\t\t# [kg/cubic meter]\n", + "\n", + "\n", + "mu = 1.906*(10**(-5)) \t\t\t# [kg/m s] viscosity \n", + "k = 0.02723 \t\t\t\t# [W/m degree celsius]\n", + "Cp = 1007 \t\t\t\t# [J/kg K]\n", + "Pr = 0.7 \t\t\t\t# prandtl no.\n", + "\n", + "\t\t# the reynolds number is \n", + "\n", + "Rel = (rho*u*L)/mu \n", + "Rel=round(Rel)\n", + "\n", + "\t\t# and the boundary layer is turbulent because the reynolds number is \t\t\tgreater than 5*10**(5).\n", + "\t\t# therefore, we use equation(5-85) to calculate the average heat \t\t\ttransfer over the plate:\n", + "\n", + "Nul_bar = (Pr**(1.0/3.0))*(0.037*(Rel**(0.8))-871) \n", + "\n", + "print \"Nul_bar is\",round(Nul_bar)\n", + "A = L*1 \t\t\t\t# [square meter] area of plate per unit depth\n", + "h_bar = Nul_bar*k/L \t\t\t# [W/square meter degree celsius]\n", + "q = h_bar*A*(Tp-Ta) \t\t\t# [W] heat transfer from plate\n", + "\n", + "#Result\n", + "\n", + "print \"Heat transfer from plate is\",round(q),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 5.10" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Turbulent-boundary-layer thickness at the end of the plate from the leading edge of the plate is 16.5 mm\n", + "Turbulent-boundary-layer thickness at the end of the plate from the transition point at Re_crit = 5*10**(5) is 9.9 mm\n" + ] + } + ], + "source": [ + "#Example Number 5.10\n", + "# Calculate turbulent-boundary-layer thickness at the end of plate \n", + "\n", + "# Variable declaration\n", + "\n", + "\t\t# we have to use the data from example 5.8 and 5.9\n", + "Rel = 1.553*10**6 \t\t\t\t\t# from previous example\n", + "L = 0.75 \t\t\t\t\t\t# [m] length of plate\n", + "\t\t# it is a simple matter to insert this value in equations(5-91) and \t\t(5-95) \talong with\n", + "x = L \t\t\t\t\t\t\t# [m]\n", + "\t\t# turbulent-boundary-layer thickness are\n", + "\t\t# part a. from the leading edge of the plate \n", + "\n", + "#Calculation\n", + "\n", + "del_a = x*0.381*Rel**(-0.2) \t\t\t\t# [m] \n", + "\t\t# part b from the transition point at Recrit = 5*10**(5)\n", + "\n", + "del_b = x*0.381*Rel**(-0.2)-10256*Rel**(-1) \t\t# [m]\n", + "\n", + "#Result\n", + "\n", + "print \"Turbulent-boundary-layer thickness at the end of the plate from the leading edge of the plate is\",round(del_a*1000,1),\"mm\" \n", + "print \"Turbulent-boundary-layer thickness at the end of the plate from the transition point at Re_crit = 5*10**(5) is\",round(del_b*1000,1),\" mm\"" + ] + } + ], + "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 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_6_Empirical_and_Practical_Relations_for_Forced_Convection_Heat_Transfer.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_6_Empirical_and_Practical_Relations_for_Forced_Convection_Heat_Transfer.ipynb new file mode 100644 index 00000000..92b1940b --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_6_Empirical_and_Practical_Relations_for_Forced_Convection_Heat_Transfer.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 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_7_Natural_Convection_Systems.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_7_Natural_Convection_Systems.ipynb new file mode 100644 index 00000000..7793c6da --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_7_Natural_Convection_Systems.ipynb @@ -0,0 +1,794 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 Natural Convection Systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average wall temperature is 185.0 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 7.1\n", + "# constant heat flux from vertical plate \n", + "\n", + "# Variable declaration\n", + "\n", + "q_w = 800.0 \t\t\t# [W/square meter] radiant energy flux\n", + "H = 3.5 \t\t\t# [m] height of metal plate surface\n", + "W = 2 \t\t\t\t# [m] width of metal plate\n", + "Ta = 30 \t\t\t# [degree celsius] surrounding air temperature \n", + "\t# we treat this problem as one with constant heat flux on the surface since we \tdo not know the surface temperature, we must make an estimate for determining \t\tTf and the air properties.\n", + "\t# an approximate value of h for free convection problems is \n", + "h = 10 \t\t\t\t# [W/square meter degree celsius]\n", + "dT = q_w/h \t\t\t# [degree celsius]\n", + "\t# then\n", + "Tf = (dT/2)+Ta \t\t\t# [degree celsius] approximately \n", + "\n", + "#Calculation\n", + "\n", + "\t# at Tf the properties of air are \n", + "v = 2.043*10**(-5) \t\t# [square meter/s]\n", + "k = 0.0295 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.7 \t\t\t# prandtl number\n", + "Beta = 1.0/(Tf+273) \t\t# [K**(-1)]\n", + "\t# from equation (7-30), with\n", + "x = 3.5 \t\t\t# [m]\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity \n", + "Gr_x = (g*Beta*q_w*x**(4))/(k*v**(2)) \n", + "\n", + "\n", + "\t# we may therefore use equation (7-32) to evaluate h_x\n", + "\n", + "h_x = (k*0.17*(Gr_x*Pr)**(1.0/4.0))/x \t# [W/square meter degree celsius]\n", + "\n", + "\t# in the turbulent heat transfer governed by equation (7-32), we note that\n", + "\t# Nu_x = h*x/k ~ (Gr_x)**(1/4) ~ x\n", + "\t# or h_x doest noy vary with x, and we may take this as the average value. the \tvalue of h\n", + "\n", + "\n", + "h = 5.41 \t\t\t# [W/square meter degree celsius]\n", + "\t# is less than the approximate value we used to estimate Tf, recalculating dT, \t we obtain\n", + "dT1 = q_w/h_x \t\t\t# [degree celsius]\n", + "\n", + "\n", + "\t# our new film temperature would be\n", + "Tf1 = Ta+dT1/2 \t\t\t# [degree celsius]\n", + "\t# at Tf the properties of air are\n", + "v1 = 2.354*10**(-5) \t\t# [square meter/s]\n", + "k1 = 0.0320 \t\t\t# [W/m degree celsius]\n", + "Pr1 = 0.695 \t\t\t# prandtl number\n", + "Beta1 = 1/(Tf1+273) \t\t# [K**(-1)]\n", + "\n", + "\t# then \n", + "Gr_x1 = (g*Beta1*q_w*x**(4))/(k1*v1**(2)) \n", + "\t# and h_x is caalculated from\n", + "h_x1 = (k1*0.17*(Gr_x1*Pr1)**(1.0/4.0))/x \t# [W/square meter degree celsius]\n", + "\n", + "\n", + "\t# our new temperature difference is calculated as \n", + "dT2 = q_w/h_x1 \t\t\t# [degree celsius]\n", + "\n", + "\n", + "\t# the average wall temperature is therefore\n", + "T_w_avg = dT2+Ta \t\t# [degree celsius]\n", + "\n", + "#Result\n", + "print \"The average wall temperature is\",round(T_w_avg),\"degree celsius\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer if the plate is 10 m wide is 9603.0 W\n" + ] + } + ], + "source": [ + "#Example Number 7.2\n", + "# heat transfer from isothermal vertical plate\n", + "\n", + "# Variable declaration\n", + "\n", + "H = 4.0 \t\t\t\t# [m] height of vertical plate\n", + "Tp = 60.0 \t\t\t# [degree celsius] plate temperature\n", + "Ta = 10.0 \t\t\t# [degree celsius] atmospheric temperature\n", + "\t# we first determine the film temperature as\n", + "Tf = (Tp+Ta)/2 \t\t\t# [degree celsius]\n", + "\t# the properties of interest are thus\n", + "v = 16.5*10**(-6) \t\t# [square meter/s]\n", + "k = 0.02685 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.7 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "\n", + "#Calculation\n", + "\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity \n", + "\t# and\n", + "Gr_into_Pr = (g*Beta*(Tp-Ta)*H**(3)*Pr)/(v**(2)) \n", + "\n", + "\t# we then may use equation (7-29) to obtain\n", + "Nu_bar_root = (0.825+(0.387*(Gr_into_Pr)**(1.0/6.0))/(1+(0.492/Pr)**(9.0/16.0))**(8.0/27.0) )\n", + "\n", + "\n", + "Nu_bar = (Nu_bar_root)**(2) \n", + "\t# the heat transfer coefficient is \n", + "h_bar = Nu_bar*k/H \t\t# [W/square meter degree celsius]\n", + "\n", + "\n", + "\t# the heat transfer is \n", + "A = H*10 \t\t\t# [square meter] for 10 m wide plate\n", + "q = h_bar*A*(Tp-Ta) \t\t# [W]\n", + "\n", + "\t# as an alternative, we could employ the simpler relation \n", + "Nu = 0.1*(Gr_into_Pr)**(1/3) \n", + "\n", + "#Result\n", + "print \"Heat transfer if the plate is 10 m wide is\",round(q),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "free-convection heat loss per unit length of heater is 443.0 W/m\n" + ] + } + ], + "source": [ + "#Example Number 7.3\n", + "# heat transfer from horizontal tube in water\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.02 \t\t\t# [m] diameter of heater\n", + "Ts = 38 \t\t\t# [degree celsius] surface temperature of heater\n", + "Tw = 27 \t\t\t# [degree celsius] water temperature\n", + "\t# the film temperature is \n", + "Tf = (Ts+Tw)/2 \t\t\t# [degree celsius]\n", + "\t# from appendix A the properties of water are \n", + "k = 0.630 \t\t\t# [W/m degree celsius] thermal conductivity\n", + "\t# and the following term is particularly useful in obtaining the product GrPr \t\tproduct when it is multiplied by d**(3)*DT\n", + "\t# g*Beta*rho**(2)*Cp/(mu*k) = 2.48*10**(10) [1/m**(3) degree celsius]\n", + "\n", + "K = 2.48*10**(10) \t\t# [1/m**(3) degree celsius]\n", + "Gr_into_Pr = K*(Ts-Tw)*d**(3) \n", + "\t\n", + "\t# using table 7-1 (page number -328), we get \n", + "\n", + "C = 0.53 \n", + "m = 1/4 \n", + "\t# so that\n", + "\n", + "Nu = C*(Gr_into_Pr)**(1.0/4.0) \n", + "h = Nu*k/d \t\t\t# [W/sq m deg C] convection heat transfer coefficient\n", + "\t# the heat transfer is thus\n", + "import math\n", + "\n", + "q_by_L = h*math.pi*d*(Ts-Tw) \t# [W/m]\n", + "print\"free-convection heat loss per unit length of heater is\",round(q_by_L),\"W/m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.4" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "#Example Number 7.4\n", + "# heat transfer from fine wire in air\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.00002 \t\t\t# [m] diameter of wire\n", + "L = 0.5 \t\t\t# [m] length of wire whose temperature is maintained\n", + "Ts = 54.0 \t\t\t# [degree celsius] surface temperature of wire \n", + "Pa = 101325.0 \t\t\t# [Pa] pressure of air\n", + "Ta = 0 \t\t\t\t# [degree celsius] temperature of air \n", + "\t# we first determine the film temperature as\n", + "Tf = (Ts+Ta)/2 \t\t\t# [degree celsius]\n", + "\t# the properties of interest are thus\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", + "Beta = 1/(Tf+273) \t\t# [K^(-1)]\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity \n", + "\t# and\n", + "Gr_into_Pr = (g*Beta*(Ts-Ta)*d**(3)*Pr)/(v**(2)) \n", + "\t# from table 7-1 we find\n", + "C = 0.675 \n", + "m = 0.058 \n", + "\t# so that\n", + "Nu_bar = C*(Gr_into_Pr)**(m) \n", + "h_bar = Nu_bar*k/d \t\t# [W/square meter degree celsius]\n", + "\t# the heat required is \n", + "import math\n", + "A = math.pi*d*L \t\t# [square meter] surface area of wire \n", + "q = h_bar*A*(Ts-Ta) \t\t# [W]\n", + "print \"Electric power to maintain the the wire temperature if the length is 0.5 m is\",round(q,3),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "free-convection heat loss per unit length is 1.5 kW/m\n" + ] + } + ], + "source": [ + "#Example Number 7.5\n", + "# heated horizontal pipe in air \n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.3048 \t\t\t# [m] diameter of pipe\n", + "Ts = 250.0 \t\t\t# [degree celsius] surface temperature of pipe \n", + "Ta = 15.0 \t\t\t# [degree celsius] temperature of air \n", + "\t# we first determine the Grashof-prandtl number product and then select the \tappropriate constants from table 7-1(page no.-328) for use with \tequation (7-25) \n", + "\t# the properties of air are evaluated at the film temperature:\n", + "Tf = (Ts+Ta)/2 \t\t\t# [degree celsius]\n", + "\t# the properties of interest are thus\n", + "v = 26.54*10**(-6) \t\t# [square meter/s]\n", + "k = 0.03406 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.687 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity\n", + "Gr_d_into_Pr = g*Beta*(Ts-Ta)*d**(3)*Pr/(v**(2)) \n", + "\t# from table 7-1 \n", + "C = 0.53 \n", + "m = 1.0/4.0 \n", + "Nu_d = C*(Gr_d_into_Pr)**(m) \n", + "h = Nu_d*k/d \t\t\t# [W/square meter degree celsius]\n", + "\t# the heat transfer per unit length is then calculated from \n", + "import math\n", + "\n", + "q_by_L = h*math.pi*d*(Ts-Ta) \t# [W/m]\n", + "print \"free-convection heat loss per unit length is\",round(q_by_L/1000,1),\"kW/m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer is 51.8 W\n" + ] + } + ], + "source": [ + "#Example Number 7.6 \n", + "# cube cooling in air\n", + "\n", + "# Variable declaration\n", + "\n", + "L = 0.2 \t\t\t\t# [m] side length of cube\n", + "Ts = 60 \t\t\t\t# [degree celsius] surface temperature of cube\n", + "Ta = 10 \t\t\t\t# [degree celsius] air temperature\n", + "\t# this is an irregular solid so we use the information in the last entry of \ttable 7-1(page no.-328) in the absence of a specific correlation for this \tgeometry. \n", + "\t# the properties were evaluated as\n", + "v = 17.47*10**(-6)\t\t\t # [square meter/s]\n", + "k = 0.02685 \t\t\t\t# [W/m degree celsius]\n", + "Pr = 0.70 \t\t\t\t# prandtl number\n", + "Beta = 3.25*10**(-3) \t\t\t# [K**(-1)]\n", + "g = 9.8 \t\t\t\t# [square meter/s] acceleration due to gravity \n", + "\t# the characteristic length is the distance a particle travels in the boundary \tlayer, which is L/2 along the bottom plus L along the side plus L/2 on the \ttop or\n", + "\n", + "\n", + "#Calculation\n", + "Gr_into_Pr = (g*Beta*(Ts-Ta)*(2*L)**(3)*Pr)/(v**(2)) \n", + "\t# from the last entry in table 7-1 we find\n", + "C = 0.52 \n", + "n = 1.0/4.0 \n", + "\t# so that\n", + "Nu = C*(Gr_into_Pr)**(n) \n", + "h_bar = Nu*k/(2*L) \t\t\t# [W/square meter degree celsius]\n", + "\t# the cube has six sides so the area is \n", + "A = 6*L**(2) \t\t\t\t# [square meter]\n", + "\t# the heat required is \n", + "q = h_bar*A*(Ts-Ta) \t\t\t# [W]\n", + "\n", + "#Result\n", + "print \"Heat transfer is\",round(q,1),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "heat transfer is 1.57 kW/m\n" + ] + } + ], + "source": [ + "#Example Number 7.7\n", + "# calculation with simplified relations \n", + "\n", + "#Variable declaration\n", + "\n", + "\t# this example is calculation of heat transfer with simplified relations for \texample (7.5) so we use the data of example 7.5\n", + "\n", + "d = 0.3048 \t\t\t# [m] diameter of pipe\n", + "Ts = 250 \t\t\t# [degree celsius] surface temperature of pipe \n", + "Ta = 15 \t\t\t# [degree celsius] temperature of air \n", + "\t# we first determine the Grashof-prandtl number product and then select the \tappropriate constants from table 7-1 for use with equation (7-25) \n", + "\t# the properties of air are evaluated at the film temperature:\n", + "\n", + "#Calculation\n", + "\n", + "Tf = (Ts+Ta)/2 \t\t\t\t# [degree celsius]\n", + "\t\t\t\t\t# the properties of interest are thus\n", + "v = 26.54*10**(-6)\t\t\t# [square meter/s]\n", + "k = 0.03406 \t\t\t\t# [W/m degree celsius]\n", + "Pr = 0.687 \t\t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t\t# [K**(-1)]\n", + "g = 9.8 \t\t\t\t# [square meter/s] acceleration due to gravity\n", + "\t# in example (7.5) we found that a rather large pipe with a substantial \t\ttemperature difference between the surface and air still had a GrPr product of \t1.57*10**(8)<10**(9), so laminar equation is selected from table 7-2(page \t\tno.-339). the heat transfer coefficient is given by \n", + "h = 1.32*((Ts-Ta)/d)**(1.0/4.0) \t\t# [W/square meter degree celsius]\n", + "\t# the heat transfer is then\n", + "import math\n", + " \n", + "q_by_L = h*math.pi*d*(Ts-Ta) \t\t# [W/m]\n", + "\n", + "#Result\n", + "\n", + "print \"heat transfer is\",round(q_by_L/1000,2),\"kW/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.8" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Free-convection heat transfer across the air space is 39.64 W\n", + "Radiation heat transfer across the air space is 15.37 W\n" + ] + } + ], + "source": [ + "#Example Number 7.8\n", + "# heat transfer across vertical air gap\n", + "\n", + "#Variable declaration\n", + "\n", + "L = 0.5 \t\t\t# [m] side length vertical square plate\n", + "d = 0.015 \t\t\t# [m] distance between plates\n", + "p = 101325.0 \t\t\t# [Pa] pressure of air\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "T1 = 100.0 \t\t\t# [degree celsius] temperature of first plate\n", + "T2 = 40.0 \t\t\t# [degree celsius] temperature of second plate\n", + "E = 0.2 \t\t\t# emissivity of both surfaces\n", + "\t\t# the properties of air is evaluated at the mean temperature\n", + "Tf = (T1+T2)/2 \t\t\t# [degree celsius]\n", + "rho = p/(R*(Tf+273)) \t\t# [Kg/m**(3)] density\n", + "k = 0.0295 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.70 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "mu = 2.043*10**(-5) \t\t# [Kg/m s] viscosity\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity\n", + "\t\t# the Grashof-prandtl number product is now calculated as \n", + "\n", + "Gr_into_Pr = (g*rho**(2)*Beta*(T1-T2)*(d)**(3)*Pr)/(mu**(2)) \n", + "\n", + "\t\t# we may now use eq(7-64) to calculate the effective thermal \t\tconductivity, with\n", + "L = 0.5 \t\t\t# [m]\n", + "deli=0.015\t\t\t# [m]\n", + "\t\t\t\t# and the constants taken from table 7-3:\n", + "\n", + "#Calculation\n", + "\n", + "Ke_by_K = 0.197*(Gr_into_Pr)**(1.0/4.0)*(L/deli)**(-1.0/9.0) \n", + "\t\t# the heat transfer may now be calculated with eq(7-54). the area is \n", + "A = L**(2) \t\t\t# [square meter]\n", + "q = Ke_by_K*k*A*(T1-T2)/deli \t# [W]\n", + " \t\t# the radiation flux is calculated with equation(7-67), taking \n", + "T1 = 373 \t\t\t# [K]\n", + "T2 = 313 \t\t\t# [K]\n", + "E1 = E \n", + "E2 = E \n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K**(4)]\n", + "q_A = sigma*(T1**(4)-T2**(4))/((1/E1)+(1/E2)-1) \t# [W/square meter]\n", + "q_rad = A*q_A \t\t\t\t\t\t# [W]\n", + "\n", + "#Result\n", + "print \"Free-convection heat transfer across the air space is\",round(q,2),\"W\" \n", + "print \"Radiation heat transfer across the air space is\",round(q_rad,2),\"W\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.9" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer across the air space is 10.34 W\n" + ] + } + ], + "source": [ + "#Example Number 7.9 \n", + "# heat transfer across horizontal air gap\n", + "\n", + "# Variable declaration\n", + "\n", + "a = 0.2 \t\t\t# [m] side length of plate\n", + "d = 0.01 \t\t\t# [m] seperation between two plates \n", + "p = 101325.0 \t\t\t# [Pa] pressure of air\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "T1 = 100.0 \t\t\t# [degree celsius] temperature of first plate\n", + "T2 = 40.0 \t\t\t# [degree celsius] temperature of second plate\n", + "\t# the properties are the same as given in example(7.8)\n", + "Tf = (T1+T2)/2 \t\t\t# [degree celsius]\n", + "rho = p/(R*(Tf+273)) \t\t# [Kg/m**(3)] density\n", + "k = 0.0295 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.70 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "mu = 2.043*10**(-5) \t\t# [Kg/m s] viscosity\n", + "g = 9.8 \t\t\t# [sq m/s] acceleration due to gravity\n", + "\t# the GrPr product is evaluated on the basis of the separating distance, so we \t have \n", + "\n", + "#Calculation\n", + "\n", + "Gr_into_Pr = (g*rho**(2)*Beta*(T1-T2)*(d)**(3)*Pr)/(mu**(2)) \n", + "\t# consulting table 7-3(page no.-344) we find\n", + "C = 0.059 \n", + "n = 0.4 \n", + "m = 0 \n", + "Ke_by_K = C*(Gr_into_Pr)**(n)*(a/d)**(m) \n", + "A = a**(2) \t\t\t# [square meter] area of plate \n", + "q = Ke_by_K*k*A*(T1-T2)/d \t# [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat transfer across the air space is\",round(q,2),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.10" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "3.47131600256\n", + "heat lost by the lower plate is 964.0 W\n" + ] + } + ], + "source": [ + "#Example Number 7.10\n", + "# heat transfer across water layer\n", + "\n", + "# Variable declaration\n", + "\n", + "L = 0.5 \t\t\t# [m] length of square plate\n", + "d = 0.01 \t\t\t# [m] seperation between square plates\n", + "T1 = 100\t\t\t# [degree F] temperature of lower plate\n", + "T2 = 80 \t\t\t# [degree F] temperature of upper plate\n", + "\n", + "dT=5.0*(T1-T2)/9.0\t\t\t#in Degree C\n", + "\t\t# we evaluate properties at mean temperature of 90 deg F and \t\t\t\tobtain, for water\n", + "k = 0.623 \t\t\t# [W/m degree celsus]\n", + "\t# and the following term is particularly useful in obtaining the product GrPr \n", + "\t# g*Beta*rho**(2)*Cp/(mu*k) = 2.48*10**(10) [1/m**(3) degree celsius]\n", + "\t# the Grashof-prandtl number product is now evaluated using the plate spacing \t\tof 0.01 m as the characterstic dimension\n", + "\n", + "#Calculation\n", + "\n", + "K = 2.48*10**(10.0) \t\t# [1/m**(3) degree celsius]\n", + "\n", + "Gr_into_Pr = K*(T1-T2)*(5.0/9.0)*d**(3.0) \n", + "\n", + "\t# now, using equation 7-64 and consulting table 7-3(page no.-344) we obtain\n", + "C = 0.13 \n", + "n = 0.3 \n", + "m = 0.0 \n", + "\t# therefore, equation (7-64) becomes\n", + "Ke_by_K = C*Gr_into_Pr**(n) \n", + "\n", + "\t# the effectve thermal conductivity is thus\n", + "ke = k*Ke_by_K \t\t\t# [W/m degree celsius]\n", + "\n", + "print ke\n", + "\t\t\t\t# and the heat transfer is\n", + "A = L**(2.0) \t\t\t# [square meter] area of plate\n", + "q = ke*A*(dT)/d \t# [W]\n", + "\n", + "#Result\n", + "print \"heat lost by the lower plate is \",round(q),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.11" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vacuum necessary for glass spacings of 1 cm is 13160.0 Pa\n", + "Vacuum necessary for glass spacings of 2 cm is 70154.0 Pa\n" + ] + } + ], + "source": [ + "#Example Number 7.11 \n", + "# reduction of convection in ar gap\n", + "\n", + "# Variable declaration\n", + "\n", + "Tm = 300.0 \t\t\t# [K] mean temperature of air\n", + "dT = 20.0 \t\t\t# [degree celsius] temperature difference\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "g = 9.8 \t\t\t# [m/s**(2)] acceleration due to gravity\n", + "p_atm = 101325.0 \t\t\t# [Pa] atmospheric pressure\n", + "\t# consulting table 7-13,we find that for gases, a value Grdeli_into_Pr<2000 is \tnecessary to reduce the system to one of pure \tconduction.\n", + "\t# at 300 K the properties of air are\n", + "k = 0.02624 \t\t\t# [W/m degree celsius]\n", + "Pr = 0.7 \t\t\t# prandtl no.\n", + "mu = 1.846*10**(-5) \t\t# [Kg/m s]\n", + "Beta = 1.0/300.0 \n", + "\n", + "\t# we have\n", + "Grdel_into_Pr = 2000.0 \n", + "\n", + "\t# Part A for spacing of 1cm\n", + "import math\n", + "deli = 0.01 \t\t\t# [m] spacing between plate\n", + "p = math.sqrt((Grdel_into_Pr*((R*Tm)**(2))*mu**(2))/(g*Beta*dT*deli**(3)*Pr)) \t# [Pa]\n", + "p=math.sqrt(7773/deli**3)\n", + "\n", + "\t# or vacuum\n", + "vacuum = p_atm-p \t\t# [Pa]\n", + "\n", + "\n", + "\t# Part B for spacing of 2cm\n", + "\n", + "deli1 = 0.02 \t\t\t# [m] spacing between plate\n", + "p1 = math.sqrt(Grdel_into_Pr*(R*Tm)**(2)*mu**(2)/(g*Beta*dT*deli1**(3)*Pr)) # [Pa]\n", + "p1=math.sqrt(7773/deli1**3)\n", + "\t# or vacuum\n", + "vacuum1 = p_atm-p1 \t\t# [Pa]\n", + "\n", + "\n", + "#Result\n", + "print \"Vacuum necessary for glass spacings of 1 cm is\",round(vacuum),\"Pa\" \n", + "print \"Vacuum necessary for glass spacings of 2 cm is\",round(vacuum1),\"Pa\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 7.13" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "heat transfer coefficient is 9.4 W/square meter degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 7.13\n", + "# combined free and forced convection with air\n", + "\n", + "# Variable declaration\n", + "\n", + "p = 101325.0 \t\t\t# [Pa] pressure of air\n", + "Ta = 27.0 \t\t\t# [degree celsius] temperature of air\n", + "d = 0.025 \t\t\t# [m] diameter of tube\n", + "u = 0.3 \t\t\t# [m/s] velocity of air\n", + "Tw = 140.0 \t\t\t# [degree celcius] temperature of tube wall\n", + "L = 0.4 \t\t\t# [m] length of tube\n", + "R = 287 \t\t\t# [] universal gas constant\n", + "\t# the properties of air are evaluated at the film temperature:\n", + "Tf = (Tw+Ta)/2 \t\t\t# [degree celcius]\n", + "\t# the properties of interest are thus\n", + "kf = 0.0305 \t\t\t# [W/m degree celcius]\n", + "Pr = 0.695 \t\t\t# prandtl number\n", + "Beta = 1/(Tf+273) \t\t# [K**(-1)]\n", + "g = 9.8 \t\t\t# [square meter/s] acceleration due to gravity\n", + "mu_f = 2.102*10**(-5) \t\t# [Kg/m s]\n", + "mu_w = 2.337*10**(-5) \t\t# [Kg/m s]\n", + "\n", + "#Calculations\n", + "\n", + "rho_f = p/(R*(Tf+273)) \t\t# [Kg/cubic meter]\n", + "\t# let us take the bulk temperature as 27 degree celsius for evaluating mu_b \tthen\n", + "mu_b = 1.8462*10**(-5) \t\t# [Kg/m s]\n", + "\t# the significant parameters are calculated as \n", + "Re_f = rho_f*u*d/mu_f \n", + "Gr = rho_f**(2)*g*Beta*(Tw-Ta)*d**(3)/mu_f**(2) \n", + "Z = Gr*Pr*d/L \t\t\t# constant\n", + "\t# according to figure(7-14)(page no.-354), the mixed convection flow regime is \tencountered. thus we must use equation(7-77).\n", + "\t# The graetz number is calculated as \n", + "Gz = Re_f*Pr*d/L \n", + "\t# and the numerical calculation for equation(7-77) becomes\n", + "Nu = 1.75*(mu_b/mu_w)**(0.14)*(Gz+0.012*(Gz*Gr**(1.0/3.0))**(4.0/3.0))**(1.0/3.0) \n", + "\t# the average heat transfer coefficient is calculated as \n", + "h_bar = Nu*kf/d \t\t# [W/square meter degree celsius]\n", + "\n", + "#Result\n", + "\n", + "print \"heat transfer coefficient is\",round(h_bar,2),\"W/square meter degree celsius\"" + ] + } + ], + "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 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_8_Radiation_Heat_Transfer.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_8_Radiation_Heat_Transfer.ipynb new file mode 100644 index 00000000..97eeda1a --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_8_Radiation_Heat_Transfer.ipynb @@ -0,0 +1,1006 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 Radiation Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total energy absorbed = 52.75 kW\n", + "total energy transmitted= 58.2 kW\n" + ] + } + ], + "source": [ + "#Example Number 8.1\n", + "# transmission and absorption in a gas plate\n", + "\n", + "# Variable declaration\n", + "\n", + "T = 2000+273 \t\t\t\t# [K] furnace temperature \n", + "L = 0.3 \t\t\t\t# [m] side length of glass plate\n", + "t1 = 0.5 \t\t\t\t# transmissivity of glass betweenb/n lambda1 \t\t\t\t\tto lambda2\n", + "lambda1 = 0.2 \t\t\t\t# [micro m] \n", + "lambda2 = 3.5 \t\t\t\t# [micro m] \n", + "E1 = 0.3 \t\t\t\t# emissivity of glass upto lambda2 \n", + "E2 = 0.9 \t\t\t\t# emissivity of glass above lambda2\n", + "t2 = 0 \t\t\t\t\t# transmissivity of glass except in the range \t\t\t\t\tof lambda1 to lambda2\n", + "\n", + "#Calculation\n", + "\n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K**(4)]\n", + "A = L**(2) \t\t\t\t# [square meter] area of glass plate\n", + "\t# calculating constants to use table 8-1(page no.-379-380)\n", + "K1 = lambda1*T \t\t\t\t# [micro m K]\n", + "K2 = lambda2*T \t\t\t\t# [micro m K]\n", + "\t# from table 8-1\n", + "Eb_0_lam1_by_sigmaT4 = 0 \n", + "Eb_0_lam2_by_sigmaT4 = 0.85443 \n", + "Eb = sigma*T**(4) \t\t\t# [W/square meter]\n", + "\t# total incident radiation is \n", + "\t# for 0.2 micro m to 3.5 micro m\n", + "TIR = Eb*(Eb_0_lam2_by_sigmaT4-Eb_0_lam1_by_sigmaT4)*A \t# [W]\n", + "TRT = t1*TIR \t\t\t\t# [W]\n", + "RA1 = E1*TIR \t\t\t\t# [W] for 0<lambda<3.5 micro m\n", + "RA2 = E2*(1-Eb_0_lam2_by_sigmaT4)*Eb*A # [W] for 3.5 micro m <lambda< infinity \n", + "TRA = RA1+RA2 \t\t\t\t# [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Total energy absorbed =\",round(TRA/1000,2),\"kW\" \n", + "print \"total energy transmitted=\",round(TRT/1000,1),\"kW\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "net radiant heat exchange between the two plates is 18.33 kW\n" + ] + } + ], + "source": [ + "#Example Number 8.2\n", + "# heat transfer between black surfaces\n", + "\n", + "# Variable declaration\n", + "\n", + "L = 1 \t\t\t\t\t# [m] length of black plate\n", + "W = 0.5 \t\t\t\t# [m] width of black plate\n", + "T1 = 1000+273 \t\t\t\t# [K] first plate temperature\n", + "T2 = 500+273 \t\t\t\t# [K] second plate temperature\n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "\t\t# the ratios for use with figure 8-12(page no.-386) are\n", + "Y_by_D = W/W \n", + "X_by_D = L/W \n", + "\t\t# so that \n", + "F12 = 0.285 \t\t\t\t# radiation shape factor \n", + "\t\t# the heat transfer is calculated from\n", + "q = sigma*L*W*F12*(T1**(4)-T2**(4)) \n", + "print \"net radiant heat exchange between the two plates is\",round(q/1000,2),\"kW\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Shape factor between the open ends of the cylinder is 0.0768\n" + ] + } + ], + "source": [ + "#Example Number 8.3\n", + "# shape-factor algebra for open ends of cylinder\n", + "\n", + "# Variable declaration\n", + "\n", + "d1 = 0.1 \t\t\t\t# [m] diameter of first cylinder\n", + "d2 = 0.2 \t\t\t\t# [m] diameter of second cylinder\n", + "L = 0.2 \t\t\t\t# [m] length of cylinder\n", + "\t\t# we use the nomenclature of figure 8-15(page no.-388) for this \t\tproblem and designate the open ends as surfaces 3 and 4.\n", + "\t\t# we have \n", + "L_by_r2 = L/(d2/2) \n", + "r1_by_r2 = 0.5 \n", + "\t\t# so from figure 8-15 or table 8-2(page no.-389) we obtain\n", + "F21 = 0.4126 \n", + "F22 = 0.3286 \n", + "\t\t# using the reciprocity relation (equation 8-18) we have\n", + "\n", + "#Calculation\n", + "F12 = (d2/d1)*F21 \n", + "\t\t# for surface 2 we have F12+F22+F23+F24 = 1.0\n", + "\t\t# and from symmetry F23 = F24 so that\n", + "F23 = (1-F21-F22)/2 \n", + "F24 = F23 \n", + "\t\t# using reciprocity again,\n", + "import math\n", + "\n", + "A2 = math.pi*d2*L \t\t\t# [m**2]\n", + "A3 = math.pi*(d2**2-d1**2)/4 \t\t# [m**2]\n", + "F32 = A2*F23/A3 \n", + "\t\t# we observe that F11 = F33 = F44 = 0 & for surface 3 F31+F32+F34 =1.0\n", + "\t\t# so, if F31 can be determined, we can calculate the desired quantity \t\tF34. for surface 1 F12+F13+F14 = 1.0\n", + "\t\t# and from symmetry F13 = F14 so that\n", + "F13 = (1-F12)/2 \n", + "F14 = F13 \n", + "\t\t# using reciprocity gives\n", + "A1 = math.pi*d1*L \t\t\t# [square meter]\n", + "F31 = (A1/A3)*F13 \n", + "\t\t# then \n", + "F34 = 1-F31-F32 \n", + "\n", + "#Result\n", + "print \"Shape factor between the open ends of the cylinder is\",F34 " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Shape factor between the top surface and the side is 0.52\n", + "Shape factor between the side and itself is 0.398\n" + ] + } + ], + "source": [ + "#Example Number 8.4\n", + "# shape-factor algebra for truncated cone\n", + "\n", + "#Variable declaration\n", + "\n", + "d1 = 0.1 \t\t\t# [m] diameter of top of cone\n", + "d2 = 0.2 \t\t\t# [m] diameter of bottom of cone\n", + "L = 0.1 \t\t\t# [m] height of cone\n", + "\t\t#we employ figure 8-16(page no.-390) for solution of this problem and \t\ttake the nomenclature as shown, designating the top as surface 2,\n", + "\t\t# the bottom as surface 1, and the side as surface 3. thus the desired \t\tquantities are F23 and F33. we have \n", + "Z = L/(d2/2) \n", + "Y = (d1/2)/L \n", + "\t\t# thus from figure 8-16(page no.-390) \n", + "F12 = 0.12 \n", + "\t\t# from reciprcity(equatin 8-18)\n", + "\n", + "import math\n", + "A1 = math.pi*d2**(2)/4 \t\t# [square meter]\n", + "A2 = math.pi*d1**(2)/4 \t\t# [square meter]\n", + "F21 = A1*F12/A2 \n", + "\t\t#and\n", + "F22 = 0 \n", + "\t\t# so that \n", + "F23 = 1-F21 \n", + "\t\t# for surface 3 F31+F32+F33 = 1, so we must find F31 and F32 in order \t\tto evaluate F33. since F11 = 0 we have\n", + "F13 = 1-F12 \n", + "\t\t# and from reciprocity \n", + "A3 = math.pi*((d1+d2)/2)*((d1/2-d2/2)**(2)+L**(2))**(1.0/2.0) \t# [square meter]\n", + "\t\t# so from above equation\n", + "F31 = A1*F13/A3 \n", + "\t\t# a similar procedure is applies with surface 2 so that \n", + "F32 = A2*F23/A3 \n", + "\t\t# finally from above equation \n", + "F33 = 1-F32-F31 \n", + "print \"Shape factor between the top surface and the side is\",F23 \n", + "print \"Shape factor between the side and itself is\",round(F33,3) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of F12 is 0.424\n", + "Value of F13 is 0.262\n", + "Value of F11 is 0.313\n" + ] + } + ], + "source": [ + "#Example Number 8.5\n", + "# shape-factor algebra for cylindrical reflactor\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.6 \t\t\t\t# [m] diameter of long half-circular cylinder\n", + "L = 0.2 \t\t\t\t# [m] length of square rod\n", + "\t\t# we have given figure example 8-5(page no.-397) for solution of this \t\t\tproblem and take the nomenclature as shown, \n", + "\t\t# from symmetry we have \n", + "F21 = 0.5 \n", + "F23 = F21 \n", + "\t\t# in general, F11+F12+F13 = 1. to aid in the analysis we create the \t\t\tfictious surface 4 shown in figure example 8-5 as dashed line.\n", + "\t\t# for this surface \n", + "F41 = 1.0 \n", + "\t\t# now, all radiation leaving surface 1 will arrive either at 2 or at \t\t\t3. likewise,this radiation will arrive at the imaginary surface 4, so \t\t\tthat F41 = F12+F13 say eqn a\n", + "\t\t# from reciprocity\n", + "#Calculation\n", + "\n", + "import math\n", + "\n", + "A1 =math.pi*d/2 \t\t\t# [square meter]\n", + "A4 = L+2*math.sqrt(0.1**(2)+L**(2)) \t# [square meter]\n", + "A2 = 4*L \t\t\t\t# [square meter]\n", + "\t\t# so that \n", + "F14 = A4*F41/A1 \t\t\t# say eqn b\n", + "\t\t# we also have from reciprocity\n", + "F12 = A2*F21/A1 \t\t\t# say eqn c\n", + "\t\t# combining a,b,c, gives\n", + "F13 = F14-F12 \n", + "\t\t# finally\n", + "F11 = 1-F12-F13 \n", + "\n", + "#Result\n", + "\n", + "print \"Value of F12 is \",round(F12,3) \n", + "print \"Value of F13 is \",round(F13,3) \n", + "print \"Value of F11 is \",round(F11,3) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of the insulated surface is 599.4 K\n", + "Heat lost by the surface at 1000K is 8.229 kW\n" + ] + } + ], + "source": [ + "#xample Number 8.7\n", + "# surface in radiant balance\n", + "\n", + "# Variable declaration\n", + "\n", + "w = 0.5 \t\t\t# [m] width of plate \n", + "L = 0.5 \t\t\t# [m] length of plate\n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "\t# from the data of the problem \n", + "T1 = 1000 \t\t\t# [K] temperature of first surface\n", + "T2 = 27+273 \t\t\t# [K] temperature of room\n", + "A1 = w*L \t\t\t# [square meter] area of rectangle\n", + "A2 = A1 \t\t\t# [square meter] area of rectangle\n", + "E1 = 0.6 \t\t\t# emissivity of surface 1\n", + "\t#eq(8-41) may not be used for the calculation because \tone of the heat-exchanging surfaces is not convex. The radiation \t\t\tnetwork is shown in figure example 8-7(page no.-404) where surface 3 is the \t\troom and surface 2 is the insulated surface. n\tJ2 \"floats\" in the network and is determined from the overall radiant balance.\n", + " \n", + "\t# from figure 8-14(page no.-387) the shape factors are \n", + "F12 = 0.2 \n", + "F21 = F12 \n", + "\t# because\n", + "F11 = 0 \n", + "F22 = 0 \n", + "F13 = 1-F12 \n", + "F23 = F13 \n", + "\t# the resistances are \n", + "R1 = (1-E1)/(E1*A1) \n", + "R2 = 1/(A1*F13) \n", + "R3 = 1/(A2*F23) \n", + "R4 = 1/(A1*F12) \n", + "\t# we also have\n", + "Eb1 = sigma*T1**(4) \t\t# [W/square meter]\n", + "Eb3 = sigma*T2**(4) \t\t# [W/square meter]\n", + "J3 = Eb3 \t\t\t# [W/square meter]\n", + "\t# the overall ckt is a series parallel arrangement and the heat transfer is \n", + "R_equiv = R1+(1/((1/R2)+1/(R3+R4))) \n", + "q = (Eb1-Eb3)/R_equiv \t\t # [W]\n", + "\t# this heat transfer can also be written as q = (Eb1-J1)/((1-E1)/(E1*A1))\n", + "\t# inserting the values \n", + "J1 = Eb1-q*((1-E1)/(E1*A1)) \t # [W/square meter]\n", + "\t# the value of J2 is determined from proportioning the resistances between J1 \t\tand J3, so that \n", + "\t# (J1-J2)/R4 = (J1-J3)/(R4+R2)\n", + "J2 = J1-((J1-J3)/(R4+R2))*R4 \t # [W/square meter]\n", + "Eb2 = J2\t\t\t # [W/square meter]\n", + "\t# finally, we obtain the temperature of the insulated surface as\n", + "T2 = (Eb2/sigma)**(1.0/4.0) \t # [K]\n", + "\n", + "print \"Temperature of the insulated surface is\",round(T2,1),\"K\" \n", + "print \"Heat lost by the surface at 1000K is\",round(q/1000,3),\"kW\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.8" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net radiant exchange is 798.0 W\n" + ] + } + ], + "source": [ + "#Example Number 8.8 \n", + "# open hemisphere in large room\n", + "\n", + "# Variable declaration\n", + "\n", + "d = 0.3 \t\t\t# [m] diameter of hemisphere\n", + "T1 = 500+273 \t\t\t# [degree celsius] temperature of hemisphere\n", + "T2 = 30+273 \t\t\t# [degree celsius] temperature of enclosure \n", + "E = 0.4 \t\t\t# surface emissivity of hemisphere\n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)] constant\n", + "\t\n", + "\t# in the given figure example 8-8(page no.-407) we take the inside of the \t\tsphere as surface 1 and the enclosure as surface 2.\n", + "\t# we also create an imaginary surface 3 covering the opening.\n", + "\t# then the heat transfer is given by\n", + "#Calculation\n", + "Eb1 = sigma*T1**(4) \t\t# [W/square meter]\n", + "Eb2 = sigma*T2**(4) \t\t# [W/square meter]\n", + "\n", + "import math\n", + "A1 = 2*math.pi*(d/2)**(2) \t# [square meter] area of surface 1\n", + "\t# calculating the surface resistance \n", + "R1 = (1-E)/(E*A1) \n", + "\t# since A2 tends to 0 so R2 also tends to 0\n", + "R2 = 0 \n", + "\t# recognize that all of the radiation leaving surface 1 which \twill \teventually arrive at enclosure 2 will also hit the imaginary surface \t\t\t3(F12=F13). we also recognize that A1*F13 = A3*F31. but \n", + "F31 = 1.0 \n", + "A3 = math.pi*(d/2)**(2) \t# [square meter]\n", + "F13 = (A3/A1)*F31 \n", + "F12 = F13 \n", + "\t# then calculating space resistance \n", + "R3 = 1/(A1*F12) \n", + "\t# we can claculate heat transfer by inserting the quantities in eq (8-40):\n", + "q = (Eb1-Eb2)/(R1+R2+R3) \t# [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Net radiant exchange is\",round(q),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.9" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For emissivity of 0.2 the value of effective emissivity is 0.467\n", + " For emissivity of 0.5 the value of effective emissivity is 0.738\n", + "For emissivity of 0.8 the value of effective emissivity is 0.907\n" + ] + } + ], + "source": [ + "#Example Number 8.9\n", + "# effective emissivity of finned surface\n", + "\n", + "# Variable declaration\n", + "\n", + "\t# for unit depth in the z-dimension we have \n", + "A1 = 10 \t\t\t# [square meter]\n", + "A2 = 5 \t\t\t\t# [square meter]\n", + "A3 = 60 \t\t\t# [square meter]\n", + "\t# the apparent emissivity of the open cavity area A1 is given by \tequation(8-47) as \n", + "\t# Ea1 = E*A3/[A1+E*(A3-A1)]\n", + "\t# for const surface emissivity the emitted energy from the total area A1+A2 is\n", + "\t# e1 = Ea1*A1+E*A2*Eb\n", + "\t# and the energy emitted per unit area for that total area is \n", + "\t# e_t = [(Ea1*A1+E*A2)/(A1+A2)]*Eb\n", + "\t# the coeff of Eb is the effective emissivity, E_eff of the combination of the \tsurface and open cavity. inserting \n", + "\t# above equations gives the following values\n", + "\n", + "\n", + "\n", + "#Calculation & Results\n", + "\n", + "\t# for E = 0.2\n", + "\n", + "E = 0.2 \n", + "Ea1 = E*A3/(A1+E*(A3-A1)) \n", + "E_eff = ((Ea1*A1+E*A2)/(A1+A2)) \n", + "\n", + "print \"For emissivity of 0.2 the value of effective emissivity is\",round(E_eff,3) \n", + "\n", + "\t# for E = 0.5\n", + "\n", + "E = 0.5 \n", + "Ea1 = E*A3/(A1+E*(A3-A1)) \n", + "E_eff = ((Ea1*A1+E*A2)/(A1+A2))\n", + " \n", + "print \" For emissivity of 0.5 the value of effective emissivity is \",round(E_eff,3) \n", + "\n", + "\t# for E = 0.8\n", + "\n", + "E = 0.8 \n", + "Ea1 = E*A3/(A1+E*(A3-A1)) \n", + "E_eff = ((Ea1*A1+E*A2)/(A1+A2)) \n", + "\n", + "print \"For emissivity of 0.8 the value of effective emissivity is \",round(E_eff,3) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.10" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat tranfer is reduced by 93.2 percent\n" + ] + } + ], + "source": [ + "#Example Number 8.10\n", + "# heat transfer reduction with parallel plate shield\n", + "\n", + "# Variable declaration\n", + "\n", + "E1 = 0.3 \t\t\t# emissivity of first plane\n", + "E2 = 0.8 \t\t\t# emissivity of second plane\n", + "E3 = 0.04 \t\t\t# emissivity of shield\n", + "\n", + "#Calculation\n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "\t\t# the heat transfer without the shield is given by \n", + "\t\t# q_by_A = sigma*(T1**4-T2**4)/((1/E1)+(1/E2)-1) = \t\t0.279*sigma*(T1**4-T2**4)\n", + "\t\t# T1 is temp of 1st plane & T2 is temperature of second plane\n", + "\t\t# the radiation network for the problem with the shield in place is \t\t\tshown in figure (8-32) (page no.-410). \n", + "\t\t# the resistances are \n", + "R1 = (1-E1)/E1 \n", + "R2 = (1-E2)/E2 \n", + "R3 = (1-E3)/E3 \n", + "\t\t# the total resistance with the shield is \n", + "R = R1+R2+R3 \n", + "\t\t# and the heat transfer is \n", + "\t\t# q_by_A = sigma*(T1**4-T2**4)/R = 0.01902*sigma*(T1**4-T2**4)\n", + "\n", + "#Result\n", + "\n", + "print \"Heat tranfer is reduced by\",round((((0.279-0.01902)/0.279)*100),1),\"percent\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.11" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of the outer cylinder is 716.0 K\n", + "Total heat lost by inner cylinder is 2826.0 W\n" + ] + } + ], + "source": [ + "#Example Number 8.11\n", + "# open cylindrical shield in large room\n", + "\n", + "# Variable declaration\n", + "\n", + "\t# two concentric cylinders of example(8.3) have \n", + "T1 = 1000 \t\t\t# [K] \n", + "E1 = 0.8 \n", + "E2 = 0.2 \n", + "T3 = 300 \t\t\t# [K] room temperature \n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "\t# refer to figure example 8-11(page no.-413) for radiation network\n", + "\t# the room is designed as surface 3 and J3 = Eb3, because the room is very \t\tlarge,(i.e.its surface is very small) \n", + "\t# in this problem we must consider the inside and outside of surface 2 and \t\tthus have subscripts i and o to designate the respective quantities. \n", + "\t# the shape factor can be obtained from example 8-3 as\n", + "F12 = 0.8253 \n", + "F13 = 0.1747 \n", + "F23i = 0.2588 \n", + "F23o = 1.0 \n", + "\t# also\n", + "\n", + "#Calculations\n", + "import math\n", + "A1 = math.pi*0.1*0.2 \t\t# [square meter] area of first cylinder\n", + "A2 = math.pi*0.2*0.2 \t\t# [square meter] area of second cylinder\n", + "Eb1 = sigma*T1**4 \t\t# [W/square meter]\n", + "Eb3 = sigma*T3**4 \t\t# [W/square meter]\n", + "\t# the resistances may be calculated as \n", + "R1 = (1-E1)/(E1*A1) \n", + "R2 = (1-E2)/(E2*A2) \n", + "R3 = 1/(A1*F12) \n", + "R4 = 1/(A2*F23i) \n", + "R5 = 1/(A2*F23o) \n", + "R6 = 1/(A1*F13) \n", + "\t# the network could be solved as a series-parallel circuit to obtain the heat \t\ttransfer, butwe will need the radiosities anyway, so we setup three nodal\t\tequations to solve for J1,J2i, and J2o.\n", + "\t# we sum the currents into each node and set them equal to zero:\n", + "\t# node J1: (Eb1-J1)/R1+(Eb3-J3)/R6+(J2i-J1)/R3 = 0\n", + "\t# node J2i: (J1-J2i)/R3+(Eb3-J2i)/R4+(J2o-J2i)/(2*R2) = 0\n", + "\t# node J2o: (Eb3-J2o)/R5+(J2i-J2o)/(2*R2) = 0\n", + "\t# these equations can be solved by matrix method and the solution is \n", + "J1 = 49732 \t\t\t# [W/square meter]\n", + "J2i = 26444 \t\t\t# [W/square meter]\n", + "J2o = 3346 \t\t\t# [W/square meter]\n", + "\t# the heat transfer is then calculated from\n", + "q = (Eb1-J1)/((1-E1)/(E1*A1)) \t# [W]\n", + "\t# from the network we see that\n", + "Eb2 = (J2i+J2o)/2\t\t # [W/square meter]\n", + "\t# and \n", + "T2 = (Eb2/sigma)**(1.0/4.0) \t# [K]\n", + "\t# if the outer cylinder had not been in place acting as a \"shield\" the heat \t\tloss from cylinder 1 could have been calculated from equation(8-43a) as \n", + "q1 = E1*A1*(Eb1-Eb3)\t\t # [W]\n", + "\n", + "#Result\n", + "\n", + "print \"Temperature of the outer cylinder is\",round(T2),\"K\" \n", + "print \"Total heat lost by inner cylinder is\",round(q1),\"W\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.12" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat-transfer rate between the two planes is 5621.0 W/square meter\n", + "Temperature of the gas is 592.4 K\n", + "Ratio of heat-transfer with presence of gas to without presence of gas is 0.97\n" + ] + } + ], + "source": [ + "#Example Number 8.12\n", + "# network for gas radiation between parallel plates\n", + "\n", + "# Variable declaration\n", + "\n", + "T1 = 800 \t\t\t\t# [K] temperature of first plate \n", + "E1 = 0.3 \t\t\t\t# emissivity\n", + "T2 = 400 \t\t\t\t# [K] temperature of second plate\n", + "E2 = 0.7 \t\t\t\t# emissivity\n", + "Eg = 0.2 \t\t\t\t# emissivity of gray gas\n", + "tg = 0.8 \t\t\t\t# transmissivity of gray gas \n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K**(4)]\n", + "\t# the network shown in figure 8-39(page no.-419) applies to this problem. all \t\tthe shape factors are unity for large planes and the various resistors can be \t\tcomputed on a unit area basis as \n", + "\n", + "#Calculation\n", + "F12 = 1 \n", + "F1g = 1 \n", + "F2g = F1g \n", + "R1 = (1-E1)/E1 \n", + "R2 = (1-E2)/E2 \n", + "R3 = 1/(F12*(1-Eg)) \n", + "R4 = 1/(F1g*Eg) \n", + "R5 = 1/(F2g*Eg) \n", + "Eb1 = sigma*T1**(4) \t\t\t# [W/square meter]\n", + "Eb2 = sigma*T2**(4) \t\t\t# [W/square meter]\n", + "\n", + "\t# the equivalent resistance of the center \"triangle\" is \n", + "\n", + "R = 1/((1/R3)+(1/(R4+R5))) \n", + "\n", + "\t# the total heat transfer is then \n", + "\n", + "q_by_A = (Eb1-Eb2)/(R1+R2+R) \t\t# [W/square meter]\n", + "\n", + "\t# heat transfer would be given by equation (8-42):\n", + "\n", + "q_by_A1 = (Eb1-Eb2)/((1/E1)+(1/E2)-1) \t# [W/square meter]\n", + "\n", + "\t# the radiosities may be computed from q_by_A = (Eb1-J1)*(E1/(1-E1)) = \t(J2-Eb2)*(E2/(1-E2))\n", + "\n", + "J1 = Eb1-q_by_A*((1-E1)/E1) \t\t# [W/square meter]\n", + "J2 = Eb2+q_by_A*((1-E2)/E2) \t\t# [W/square meter]\n", + "\n", + "\t# for the network Ebg is just the mean of these values\n", + "\n", + "Ebg = (J1+J2)/2 \t\t\t# [W/square meter]\n", + "\n", + "\t# so that the temperature of the gas is\n", + "Tg = (Ebg/sigma)**(1.0/4.0) \t\t# [K]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat-transfer rate between the two planes is\",round(q_by_A),\"W/square meter\" \n", + "print \"Temperature of the gas is\",round(Tg,1),\"K\" \n", + "print \"Ratio of heat-transfer with presence of gas to without presence of gas is\",round(q_by_A/q_by_A1,2) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.13" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Apparent emissivity of covered opening is 0.6269\n", + "If there were no cover present, the value of Ea(apparent emissivity) would be 0.8571\n" + ] + } + ], + "source": [ + "#Example Number 8.13\n", + "# cavity with transparent cover \n", + "\n", + "#Variable declaration\n", + "\n", + "E1 = 0.5 \t\t\t# emissivity of rectangular cavity\n", + "t2 = 0.5 \t\t\t# transmissivity\n", + "rho2 = 0.1 \t\t\t# reflectivity\n", + "E2 = 0.4 \t\t\t# emissivity\n", + "\t# from example 8-9 we have\n", + "\t# per unit depth in the z direction we have \n", + "\n", + "#Calculation\n", + "A1 = 60\n", + "A2 = 10.0 \n", + "\t# we may evaluate K from equation(8-96a)\n", + "K = E1/(t2+(E2/2))\n", + "\n", + "\n", + "\t# the value of Ea is then computed from equation (8-96) as \n", + "Ea = (t2+(E2/2))*K/((A2/A1)*(1-E1)+K) \n", + "print \"Apparent emissivity of covered opening is \",round(Ea,4) \n", + "\t# when no cover present, the value of Ea would be given by eq (8-47) as\n", + "Ea1 = E1*A1/(A2+E1*(A1-A2)) \n", + "\n", + "#Result\n", + "print \"If there were no cover present, the value of Ea(apparent emissivity) would be \",round(Ea1,4) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.14" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radiation lost through the quartz window to a room temperature of 30 degree celsius is 112171.0 W/square meter\n", + "With no windows at all, the heat transfer would be 148397.0 W/square meter\n" + ] + } + ], + "source": [ + "#Example Number 8.14\n", + "# Transmitting and reflecting system for furnace opening\n", + "\n", + "# Variable declaration\n", + "\t\n", + "T1 = 1000+273 \t\t\t# [K] temperature of furnace\n", + "lamda = 4.0 \t\t\t# [micro meter]\n", + "\t\t#for 0 < lamda < 4 micro meter\n", + "t1 = 0.9 \n", + "E1 = 0.1 \n", + "rho1 = 0 \n", + "\t\t#for 4 micro meter < lamda < infinity \n", + "t2 = 0 \n", + "E2 = 0.8 \n", + "rho2 = 0.2 \n", + "sigma = 5.669*10**(-8) \t\t# [W/square meter K**(4)]\n", + "T3 = 30+273\t\t \t# [K] room temperature\n", + "\t\t# because the room is large it may be treated as a blackbody.\n", + "\t\t# analyze the problem by calculating the heat transfer for each \t\t\twavelength band and then adding them together to obtain the total. the \t\tnetwork for each band is a modification of figure 8-57 \n", + "A1 = 1.0 \t\t\t# [square meter]\n", + "A2 = 1.0 \t\t\t# [square meter]\n", + "A3 = 1.0 \t\t\t# [square meter]\n", + "F12 = 1.0 \n", + "F13 = 1.0 \n", + "F32 = 1.0 \n", + "\t\t# the total emissive powers are \n", + "\n", + "#Calculation\n", + "\n", + "Eb1 = sigma*T1**(4) \t\t# [W/square meter]\n", + "Eb3 = sigma*T3**(4) \t\t# [W/square meter]\n", + "\t\t# to determine the fraction of radiation in each wavelength band\n", + "lamba_into_T1 = lamda*T1 \t# [micro meter K]\n", + "lamba_into_T3 = lamda*T3 \t# [micro meter K]\n", + "\t\t# consulting table 8-1, we find \n", + "Eb1_0_to_4 = 0.6450*Eb1 \t# [W/square meter]\n", + "Eb3_0_to_4 = 0.00235*Eb3 \t# [W/square meter]\n", + "Eb1_4_to_inf = (1-0.6450)*Eb1 \t# [W/square meter]\n", + "Eb3_4_to_inf = (1-0.00235)*Eb3 \t# [W/square meter]\n", + "\t\t# apply these numbers to the network for the two wavelengths bands, \t\t\twith unit areas.\n", + "\n", + "\t\t# 0 < lamda < 4 micro meter band:\n", + "R1 = 1/(F13*t1) \n", + "R2 = 1/(F32*(1-t1)) \n", + "R3 = 1/(F12*(1-t1)) \n", + "R4 = rho1/(E1*(1-t1)) \n", + "\t\t# the net heat transfer from the network is then \n", + "R_equiv_1 = 1/(1/R1+1/(R2+R3+R4)) \n", + "q1 = (Eb1_0_to_4-Eb3_0_to_4)/R_equiv_1 # [W/square meter]\n", + "\n", + "\t\t# 4 micro meter < lamda < infinity band:\n", + "R2 = 1/(F32*(1-t2)) \n", + "R3 = 1/(F12*(1-t2)) \n", + "R4 = rho2/(E2*(1-t2)) \n", + "\t\t# the net heat transfer from the network is then \n", + "\t\t# R1 is infinity\n", + "R_equiv_2 = R2+R3+R4*2 \n", + "q2 = (Eb1_4_to_inf-Eb3_4_to_inf)/R_equiv_2 # [W/square meter]\n", + "\n", + "\t\t# the total heat loss is then \n", + "q_total = q1+q2 \t\t\t# [W/square meter]\n", + "\t\t# with no windows at all, the heat transfer would have been the \t\t\tdifference in blackbody emissive powers,\n", + "Q = Eb1-Eb3 \t\t\t\t# [W/square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Radiation lost through the quartz window to a room temperature of 30 degree celsius is\",round(q_total),\"W/square meter\" \n", + "\n", + "print \"With no windows at all, the heat transfer would be\",round(Q),\"W/square meter\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.20" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radiation equilibrium temperature for the plate exposed to solar flux if the surface is coated with :\n", + "White paint is 39.5 degree celsius\n", + "Flat black lacquer is 104.8 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 8.20\n", + "# solar-environment equilibrium temperatures \n", + "\n", + "# Variable declaration\n", + "\n", + "q_by_A_sun = 700 \t\t\t# [W/m**(2)] solar flux\n", + "T_surr = 25+273 \t\t\t# [K] surrounding temperature\n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K**(4)]\n", + "\t# at radiation equilibrium the netenergy absorbed from sun must equal the \tlong-wavelength radiation exchange with the surroundings,or\n", + "\t# (q_by_A_sun)*alpha_sun = alpha_low_temp*sigma*(T**4-T_surr**4) (a)\n", + "\n", + "\t# case (a) for white paint\n", + "\n", + "\t# for white paint we obtain from table 8-4\n", + "\n", + "#Calculation\n", + "\n", + "alpha_sun = 0.12 \n", + "alpha_low_temp = 0.9 \n", + "\t# so that equation (a) becomes\n", + "T = ((q_by_A_sun)*alpha_sun/(alpha_low_temp*sigma)+T_surr**(4))**(1.0/4.0)\t # [K]\n", + "\n", + "\n", + "#Result\n", + "\n", + "print\"Radiation equilibrium temperature for the plate exposed to solar flux if the surface is coated with :\"\n", + "print \"White paint is\",round(T-273,1),\"degree celsius\" \n", + "\n", + "\t# case (b) for flat black lacquer we obtain\n", + "\n", + "alpha_sun = 0.96 \n", + "alpha_low_temp = 0.95 \n", + "\t# so that equation (a) becomes\n", + "T = ((q_by_A_sun)*alpha_sun/(alpha_low_temp*sigma)+T_surr**(4))**(1.0/4.0) \t# [K]\n", + "\n", + "print \"Flat black lacquer is\",round(T-273,1),\"degree celsius\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 8.23" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the true air temperature is 28.6 degree celsius\n" + ] + } + ], + "source": [ + "#Example Number 8.23\n", + "# temperature measurement error caused by radiation \n", + "\n", + "# Variable declaration\n", + "\n", + "E = 0.9 \t\t\t\t# emissivity of mercury-in-glass thermometer \n", + "Tt = 20+273 \t\t\t\t# [K] temperature indicated by thermometer \n", + "Ts = 5+273 \t\t\t\t# [K] temperature of walls\n", + "sigma = 5.669*10**(-8) \t\t\t# [W/square meter K^(4)]\n", + "h = 8.3 \t\t\t\t# [W/sq m] heat transfer coefficient for \t\t\t\t\tthermometer\n", + "\t# we employ equation(8-113) for the solution: h*(Tinf-Tt) =sigma*E*(Tt^4-Ts^4)\n", + "\t# inserting the values in above equation \n", + "\n", + "Tinf = sigma*E*(Tt**4-Ts**4)/h+Tt \t# [K]\n", + "print\"the true air temperature is\",round(Tinf-273,1),\"degree celsius\" " + ] + } + ], + "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 +} diff --git a/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_9_Condensation_and_Boiling_Heat_Transfer.ipynb b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_9_Condensation_and_Boiling_Heat_Transfer.ipynb new file mode 100644 index 00000000..977f0e4b --- /dev/null +++ b/Heat_Transfer_(In_SI_Units)_by_J_P_Holman/Chapter_9_Condensation_and_Boiling_Heat_Transfer.ipynb @@ -0,0 +1,308 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 Condensation and Boiling Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of reynolds no. is 49.7 so the laminar assumption was correct\n", + "The heat transfer is 2368.0 w\n", + "Total mass flow condensate is 3.78 kg/h\n" + ] + } + ], + "source": [ + "#Example Number 9.1\n", + "# condensation on vertical plate\n", + "\n", + "#Variable declaration\n", + "\n", + "\t# we have to check the reynolds no. to that film is laminar or turbulent\n", + "Tf = (100+98)/2 \t\t\t# [degree celsius]\n", + "Tw = 98 \t\t\t\t# [degree celsius]\n", + "RHOf=960 \t\t\t\t# [kg/cubic meter] \n", + "MUf=2.82*10**(-4) \t\t\t# [kg/m s]\n", + "Kf=0.68 \t\t\t\t# [W/m degree celsius]\n", + "g=9.81 \t\t\t\t\t# [m/s**(2)]\n", + "L=0.3 \t\t\t\t\t# [m]\n", + "\t# RHOf(RHOf-RHOv)~RHOf**(2)\n", + "\t# let us assume laminar film condensate \n", + "Tsat=100 \t\t\t\t# [degree celsius]\n", + "Tg=100 \t\t\t\t\t# [degree celsius]\n", + "Hfg=2255*10**(3) \t\t\t# [J/kg]\n", + "\n", + "#Calculation\n", + "\n", + "hbar=0.943*((RHOf**(2)*g*Hfg*Kf**(3)/(L*MUf*(Tg-Tw)))**(0.25)) \t# [W/sq m deg celsius]\n", + "h=hbar \t\t\t\t\t# [W/square meter degree celsius]\n", + "\t# checking reynolds no. with equation(9-17)\n", + "Ref=4*h*L*(Tsat-Tw)/(Hfg*MUf) \n", + "\n", + "print \"Value of reynolds no. is\",round(Ref,1),\"so the laminar assumption was correct\" \n", + "\t# the heat transfer is now calculated from \n", + "A=0.3*0.3 \t\t\t\t# [square meter]\n", + "q=hbar*A*(Tsat-Tw) \t\t\t# [W]\n", + "mdot=q/Hfg \t\t\t\t# [kg/h]\n", + "\n", + "#Result\n", + "\n", + "print \"The heat transfer is\",round(q),\"w\" \n", + "mdot=mdot*3600 \t\t\t\t# [kg/h]\n", + "print \"Total mass flow condensate is\",round(mdot,2),\"kg/h\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total surface area is 3.99 square meter/m\n", + "Heat transfer is 100.1 kW/m\n", + "Total mass flow of condensate is 159.8 kg/h\n" + ] + } + ], + "source": [ + "#Example Number 9.2\n", + "# condensation on tube tank\n", + "\n", + "# Variable declaration\n", + "\n", + "\t# the condensate properties are obtained from previous example\n", + "\t# replacing L by n*d\n", + "Tw=98 \t\t\t\t# [degree celsius]\n", + "RHOf=960 \t\t\t# [kg/cubic meter] \n", + "MUf=2.82*10**(-4) \t\t# [kg/m s]\n", + "Kf=0.68 \t\t\t# [W/m degree celsius]\n", + "g=9.81 \t\t\t\t# [m/s^(2)]\n", + "Tsat=100 \t\t\t# [degree celsius]\n", + "Tg=100 \t\t\t\t# [degree celsius]\n", + "Hfg=2255*10**(3) \t\t# [J/kg]\n", + "d=0.0127 \t\t\t# [m]\n", + "n=10 \n", + "\n", + "#Calculation\n", + "\n", + "hbar=0.725*((RHOf**(2)*g*Hfg*Kf**(3)/(n*d*MUf*(Tg-Tw)))**(0.25)) # [W/sq m deg C]\n", + "\t# total surface area is \n", + "n=100 \n", + "Al=n*22*d/7 \t\t\t# [square meter]\n", + "print \"Total surface area is\",round(Al,2),\"square meter/m\" \n", + "\t# so the heat transfer is \n", + "Ql=hbar*Al*(Tg-Tw)\t\t # [W]\n", + "\n", + "#Result \n", + "\n", + "print \"Heat transfer is\",round(Ql/1000,2),\"kW/m\" \n", + "\t# total mass flow of condensate is then \n", + "mdotl=Ql/Hfg \t\t\t# [kg/h]\n", + "mdotl=mdotl*3600\t\t# [kg/h]\n", + "print \"Total mass flow of condensate is\",round(mdotl,1),\"kg/h\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer in a 1.0 m length of tube is 2810.0 W/m\n" + ] + } + ], + "source": [ + "#Example Number 9.4\n", + "# Flow boiling\n", + "# Variable declaration\n", + "\n", + "p =0.5066\t\t\t# [MPa] pressure of water \n", + "d = 0.0254 \t\t\t# [m] diameter of tube \n", + "Tw = 10.0 \t\t\t# [degree celsius]\n", + "\t\t\t\t# for calculation we use equation (9-45), noting that \n", + "dT = 10.0 \t\t\t# [degree celsius]\n", + "\t\t\t\t# the heat transfer coefficient is calculated as \n", + "\n", + "import math\n", + "h = 2.54*Tw**(3)*math.exp(p/1.551) \t# [W/square meter degree celsius]\n", + "\n", + "\t\t# the surface area for a 1-m length of tube is \n", + "L = 1 \t\t\t\t# [m]\n", + "import math\n", + "A = math.pi*d*L \t\t# [square meter]\n", + "\n", + "\t\t# so the heat transfer is \n", + "q = h*A*dT \t\t\t# [W/m]\n", + "print \"Heat transfer in a 1.0 m length of tube is\",round(q),\"W/m\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat flux obtained is 22.8 kW/square meter\n", + "If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is 23.0 percent\n" + ] + } + ], + "source": [ + "#Example Number 9.5\n", + "# water boiling in a pan \n", + "\n", + "# Variable declaration\n", + "\n", + "p = 0.101\t\t\t\t# [MPa] pressure of water \n", + "dT_x = 8 \t\t\t\t# [degree celsius]\n", + "p1 = 0.17 \t\t\t\t# [MPa] given operating pressure\n", + "\t# we will use the simplified relation of table 9-13(page no.-506) for the \t\testimates.we do not know the value of q_by_A and so must choose one of the two \trelation for a horizontal surface from the table\n", + "\t# we anticipate nucleate boiling, so choose\n", + "h = 5.56*dT_x**(3) \t\t\t# [W/square meter degree celsius]\n", + "\t\t\t\t\t# and the heat flux is \n", + "q_by_A = h*dT_x \t\t\t# [W/square meter]\n", + "\t# for operation as a pressure cooker we obtain the value of h from \t\tequation(9-44)\n", + "\n", + "#Calculation\n", + "\n", + "hp = h*(p1/p)**(0.4) \t\t\t# [W/square meter degree celsius]\n", + "\t# the corresponding heat flux is \n", + "q_by_A1 = hp*dT_x \t\t\t# [W/square meter]\n", + "\n", + "#Result\n", + "\n", + "print \"Heat flux obtained is\",round(q_by_A/1000,1),\"kW/square meter\" \n", + "per_inc = 100*(q_by_A1-q_by_A)/q_by_A \n", + "\n", + "print \"If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is\",round(per_inc),\"percent\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exa 9.6" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thus the heat transfers 14.0 times the heat of a pure copper rod with a substantial temperature gradient\n" + ] + } + ], + "source": [ + "#Example Number 9.6\n", + "# heat-flux comparisons \n", + "\n", + "# Variable declaration\n", + "\n", + "Tw = 200.0 \t\t\t\t# [degree celsius] water temperature \n", + "L = 0.08 \t\t\t\t# [m] length of solid copper bar\n", + "dT = 100.0 \t\t\t\t# [degree C] temp differential in copper bar\n", + "\t#using the data of table 9-4(page no.-508)\n", + "\t# the heat flux per unit area is expressed as q_by_A = -k*del_T/dx\n", + "\t# from table A-2(page no.-) the thermal conductivity of copper is \n", + "k = 374.0 \t\t\t\t# [W/m degree celsius]\n", + "\n", + "#Calculation\n", + "\n", + "q_by_A = -k*(-dT)/L \t\t\t# [W/square meter]\n", + "\n", + "\t# from table 9-4(page no.-508) the typical axial heat flux for a water heat \t\tflux for a water heat pipe is \n", + "q_by_A_axial = 0.67 \t\t\t# [kW/csquare meter]\n", + "q_by_A = q_by_A/(1000*10**(4.0)) \t# [kW/csquare meter]\n", + "time=q_by_A_axial/q_by_A\n", + "\n", + "#Result\n", + "\n", + "print \"Thus the heat transfers \",round(time),\"times the heat of a pure copper rod with a substantial temperature gradient\" " + ] + } + ], + "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" + } + }, + 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