13-05-2021
Heat And Mass Transfer Cengel 4th Edition
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Hvac & R Research
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</title>n<meta http-equiv='Content-Type' content='text/html; charset='UTF-8'>n</head>nn<div><div><p><b>Chapter 3nSTEADY HEAT CONDUCTIONn</b></p>n<p><b>Steady Heat Conduction In Plane Wallsn</b></p>n<p><b>3-1C </b>(a) If the lateral surfaces of the rod are insulated, the heat transfer surface area of the cylindrical rod nis the bottom or the top surface area of the rod, . (b) If the top and the bottom surfaces of the rod are ninsulated, the heat transfer area of the rod is the lateral surface area of the rod, .n</p>n<p><b>3-2C</b> In steady heat conduction, the rate of heat transfer into the wall is equal to the rate of heat transfer out nof it. Also, the temperature at any point in the wall remains constant. Therefore, the energy content of the nwall does not change during steady heat conduction. However, the temperature along the wall and thus the nenergy content of the wall will change during transient conduction. n</p>n<p><b>3-3C </b>The temperature distribution in a plane wall will be a straight line during steady and one dimensional nheat transfer with constant wall thermal conductivity.n</p>n<p><b>3-4C </b>The thermal resistance of a medium represents the resistance of that medium against heat transfer.n</p>n<p><b>3-5C</b> The combined heat transfer coefficient represents the combined effects of radiation and convection nheat transfers on a surface, and is defined as <i>h</i>combined = <i>h</i>convection + <i>h</i>radiation. It offers the convenience of nincorporating the effects of radiation in the convection heat transfer coefficient, and to ignore radiation in nheat transfer calculations.n</p>n<p><b>3-6C</b> Yes. The convection resistance can be defined as the inverse of the convection heat transfer ncoefficient per unit surface area since it is defined as .n</p>n<p><b>3-7C </b>The convection and the radiation resistances at a surface are parallel since both the convection and nradiation heat transfers occur simultaneously.n</p>n<p><b>3-8C</b> For a surface of <i>A </i> at which the convection and radiation heat transfer coefficients are , the single nequivalent heat transfer coefficient is when the medium and the surrounding surfaces are at the same ntemperature. Then the equivalent thermal resistance will be .n</p>n<p><b>3-9C </b>The thermal resistance network associated with a five-layer composite wall involves five single-layer nresistances connected in series.n</p>n<p><b>3-10C</b> Once the rate of heat transfer is known, the temperature drop across any layer can be determined by nmultiplying heat transfer rate by the thermal resistance across that layer, n</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>1n-1</p>nn</div></div>n<div><div><p><b>3-11C </b>The temperature of each surface in this case can be determined fromn</p>n<p>where is the thermal resistance between the environment and surface i.n</p>n<p><b>3-12C </b>Yes, it is.n</p>n<p><b>3-13C</b> The window glass which consists of two 4 mm thick glass sheets pressed tightly against each other nwill probably have thermal contact resistance which serves as an additional thermal resistance to heat ntransfer through window, and thus the heat transfer rate will be smaller relative to the one which consists of na single 8 mm thick glass sheet.n</p>n<p><b>3-14C</b> Convection heat transfer through the wall is expressed as . In steady heat transfer, heat transfer rate nto the wall and from the wall are equal. Therefore at the outer surface which has convection heat transfer ncoefficient three times that of the inner surface will experience three times smaller temperature drop ncompared to the inner surface. Therefore, at the outer surface, the temperature will be closer to the nsurrounding air temperature.n</p>n<p><b>3-15C</b> The new design introduces the thermal resistance of the copper layer in addition to the thermal nresistance of the aluminum which has the same value for both designs. Therefore, the new design will be a npoorer conductor of heat.n</p>n<p><b>3-16C</b> The blanket will introduce additional resistance to heat transfer and slow down the heat gain of the ndrink wrapped in a blanket. Therefore, the drink left on a table will warm up faster.n</p>n<p><b>3-17</b> The two surfaces of a wall are maintained at specified temperatures. The rate of heat loss through the nwall is to be determined. n<b><i>Assumptions</i> 1</b> Heat transfer through the wall is steady since the surface temperatures remain constant at nthe specified values. <b>2 </b>Heat transfer is one-dimensional since any significant temperature gradients will nexist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivity is constant.n<b><i>Properties</i></b><i></i> The thermal conductivity is given to be <i>k</i> = 0.8 W/mF 0 D 7°C.n<b><i>Analysis</i></b>The surface area of the wall and the rate of heat loss through the wall aren</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>2n-2</p>nn</div></div>n<div><div><p><b>3-18 </b>The two surfaces of a window are maintained at specified temperatures. The rate of heat loss through nthe window and the inner surface temperature are to be determined.n<b><i>Assumptions</i> 1</b> Heat transfer through the window is steady since the surface temperatures remain constant nat the specified values. <b>2 </b>Heat transfer is one-dimensional since any significant temperature gradients will nexist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivity is constant. <b>4 </b>Heat transfer nby radiation is negligible.n<b><i>Properties</i></b><i></i> The thermal conductivity of the glass is given to be <i>k</i> = 0.78 W/mF 0 D 7°C.n<b><i>Analysis</i></b>The area of the window and the individual resistances aren</p>n<p>The steady rate of heat transfer through window glass is thenn</p>n<p>The inner surface temperature of the window glass can be determined fromn</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>3n-3</p>nn</div></div>n<div><div><p><b>3-19</b> A double-pane window consists of two 3-mm thick layers of glass separated by a 12-mm wide nstagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss through the nwindow and the inner surface temperature of the window are to be determined.n<b><i>Assumptions</i> 1</b> Heat transfer through the window is steady since the indoor and outdoor temperatures nremain constant at the specified values. <b>2 </b>Heat transfer is one-dimensional since any significant ntemperature gradients will exist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivities nof the glass and air are constant. <b>4 </b>Heat transfer by radiation is negligible.n<b><i>Properties</i></b><i></i> The thermal conductivity of the glass and air are given to be <i>k</i>glass = 0.78 W/mF 0 D 7°C and <i>k</i>air = n0.026 W/mF 0 D 7°C.n<b><i>Analysis </i></b><i></i> The area of the window and the individual resistances aren n</p>n<p>The steady rate of heat transfer through window glass then becomesn</p>n<p>The inner surface temperature of the window glass can be determined fromn</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>4n-4</p>nn</div></div>n<div><div><p><b>3-20</b> A double-pane window consists of two 3-mm thick layers of glass separated by an evacuated space. nFor specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner nsurface temperature of the window are to be determined.n<b><i>Assumptions</i> 1</b> Heat transfer through the window is steady since the indoor and outdoor temperatures nremain constant at the specified values. <b>2 </b>Heat transfer is one-dimensional since any significant ntemperature gradients will exist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivities nof the glass and air are constant. <b>4 </b>Heat transfer by radiation is negligible.n<b><i>Properties</i></b><i></i> The thermal conductivity of the glass is given to be <i>k</i>glass = 0.78 W/mF 0 D 7°C.n<b><i>Analysis </i></b><i></i> Heat cannot be conducted through an evacuated space since the thermal conductivity of vacuum nis zero (no medium to conduct heat) and thus its thermal resistance is zero. Therefore, if radiation is ndisregarded, the heat transfer through the window will be zero. Then the answer of this problem is <b>zero n</b>since the problem states to disregard radiation.n<b><i>Discussion</i></b>In reality, heat will be transferred between the glasses nby radiation. We do not know the inner surface temperatures of nwindows. In order to determine radiation heat resistance we nassume them to be 5°C and 15°C, respectively, and take the nemissivity to be 1. Then individual resistances are n</p>n<p>The steady rate of heat transfer through window glass then becomesn</p>n<p>The inner surface temperature of the window glass can be determined fromn</p>n<p>Similarly, the inner surface temperatures of the glasses are calculated to be 15.2 and -1.2F 0 B 0C (we had nassumed them to be 15 and 5F 0 B 0C when determining the radiation resistance). We can improve the result nobtained by reevaluating the radiation resistance and repeating the calculations.n</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>5n-5</p>nn</div></div>n<div><div><p><b>3-21</b> n"GIVEN"nA=1.2*2 "[m^2]"nL_glass=3 "[mm]"nk_glass=0.78 "[W/m-C]"n"L_air=12 [mm], parameter to be varied"nT_infinity_1=24 "[C]"nT_infinity_2=-5 "[C]"nh_1=10 "[W/m^2-C]"nh_2=25 "[W/m^2-C]"n</p>n<p>"PROPERTIES"nk_air=conductivity(Air,T=25)n</p>n<p>"ANALYSIS"nR_conv_1=1/(h_1*A)nR_glass=(L_glass*Convert(mm, m))/(k_glass*A)nR_air=(L_air*Convert(mm, m))/(k_air*A)nR_conv_2=1/(h_2*A)nR_total=R_conv_1+2*R_glass+R_air+R_conv_2nQ_dot=(T_infinity_1-T_infinity_2)/R_totaln</p>n<p><b>Lair [mm] Q [W]n</b>2 307.8n4 228.6n6 181.8n8 150.9n</p>n<p>10 129n12 112.6n14 99.93n16 89.82n18 81.57n20 74.7n</p>n<p>2 4 6 8 10 12 14 16 18 20n50n</p>n<p>100n</p>n<p>150n</p>n<p>200n</p>n<p>250n</p>n<p>300n</p>n<p>350n</p>n<p><b>Lair [mm]n</b></p>n<p><b>Qn[Wn</b></p>n<p><b>]n</b></p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>6n-6</p>nn</div></div>n<div><div><p><b>3-22E </b>The inner and outer surfaces of the walls of an electrically heated house remain at specified ntemperatures during a winter day. The amount of heat lost from the house that day and its its cost are to be ndetermined.n<b><i>Assumptions</i> 1</b> Heat transfer through the walls is steady since the surface temperatures of the walls remain nconstant at the specified values during the time period considered. <b>2 </b>Heat transfer is one-dimensional since nany significant temperature gradients will exist in the direction from the indoors to the outdoors.<b> 3</b> Thermal nconductivity of the walls is constant.n<b><i>Properties</i></b><i></i> The thermal conductivity of the brick wall is given to be <i>k</i> = 0.40 Btu/hF 0 D 7ftF 0 D 7°F.n<b><i>Analysis </i></b><i></i>We consider heat loss through the walls only. The total heat transfer area isn</p>n<p>The rate of heat loss during the daytime isn nThe rate of heat loss during nighttime isn nThe amount of heat loss from the house that night will ben</p>n<p>Then the cost of this heat loss for that day becomesn</p>n<p><b>3-23</b> A cylindrical resistor on a circuit board dissipates 0.15 W of power steadily in a specified nenvironment. The amount of heat dissipated in 24 h, the surface heat flux, and the surface temperature of nthe resistor are to be determined.n<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat is transferred uniformly from all surfaces of the nresistor. n<b><i>Analysis </i></b><i></i>(<i>a</i>) The amount of heat this resistor dissipates during a 24-hour period isn</p>n<p>(<i>b</i>) The heat flux on the surface of the resistor isn n n(<i>c</i>) The surface temperature of the resistor can be determined fromn n</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>7n-7</p>nn</div></div>n<div><div><p><b>3-24</b> A power transistor dissipates 0.2 W of power steadily in a specified environment. The amount of heat ndissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined.n<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat is transferred uniformly from all surfaces of the ntransistor. n<b><i>Analysis </i></b><i></i>(<i>a</i>) The amount of heat this transistor dissipates during a 24-hour period isn</p>n<p>(<i>b</i>) The heat flux on the surface of the transistor isn n n(<i>c</i>) The surface temperature of the transistor can be determined fromn</p>n<p><b>3-25</b> A circuit board houses 100 chips, each dissipating 0.07 W. The surface heat flux, the surface ntemperature of the chips, and the thermal resistance between the surface of the board and the cooling nmedium are to be determined.n<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat transfer from the back surface of the board is nnegligible. <b>2 </b>Heat is transferred uniformly from the entire front surface. n<b><i>Analysis </i></b><i></i>(<i>a</i>) The heat flux on the surface of the circuit board is n</p>n<p>(<i>b</i>) The surface temperature of the chips isn n(<i>c</i>) The thermal resistance isn</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>8n-8</p>nn</div></div>n<div><div><p><b>3-26 </b>A person is dissipating heat at a rate of 150 W by natural convection and radiation to the surrounding nair and surfaces. For a given deep body temperature, the outer skin temperature is to be determined.n<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>The heat ntransfer coefficient is constant and uniform over the entire exposed nsurface of the person. <b>3 </b>The surrounding surfaces are at the same ntemperature as the indoor air temperature. <b>4 </b>Heat generation nwithin the 0.5-cm thick outer layer of the tissue is negligible.n<b><i>Properties</i></b><i></i> The thermal conductivity of the tissue near the skin is ngiven to be <i>k</i> = 0.3 W/mF 0 D 7°C.n<b><i>Analysis </i></b><i></i>The skin temperature can be determined directly fromn n</p>n<p><b>3-27</b> Heat is transferred steadily to the boiling water in an aluminum pan. The inner surface temperature of nthe bottom of the pan is given. The boiling heat transfer coefficient and the outer surface temperature of the nbottom of the pan are to be determined. n<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat transfer is one-dimensional since the thickness of nthe bottom of the pan is small relative to its diameter. <b>3 </b>The thermal conductivity of the pan is constant.n<b><i>Properties</i></b><i></i> The thermal conductivity of the aluminum pan is given to be <i>k</i> = 237 W/mF 0 D 7°C.n<b><i>Analysis </i></b><i></i>(<i>a</i>) The boiling heat transfer coefficient isn n n(<i>b</i>) The outer surface temperature of the bottom of the pan isn n</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>9n-9</p>nn</div></div>n<div><div><p><b>3-28E </b> A wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermal nresistance of the wall and its R-value of insulation are to be determined.n<b><i>Assumptions</i> 1</b> Heat transfer through the wall is one-dimensional. <b>2 </b>Thermal conductivities are constant.n<b><i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i>sheetrock = 0.10 Btu/hF 0 D 7ftF 0 D 7°F and <i>k </i>insulation = 0.020 nBtu/hF 0 D 7ftF 0 D 7°F.n<b><i>Analysis </i></b><i></i>(<i>a</i>) The surface area of the wall is not given and thus we consider na unit surface area (<i>A</i> = 1 ft2). Then the <i>R</i>-value of insulation of the wall nbecomes equivalent to its<i></i>thermal resistance, which is determined from.n n(<i>b</i>) Therefore, this is approximately a <b><i>R</i>-22</b> wall in English units.n</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>10n-10</p>nn</div></div>n<div><div><p><b>3-29 </b>The roof of a house with a gas furnace consists of 3-cm thick concrete that is losing heat to the noutdoors by radiation and convection. The rate of heat transfer through the roof and the money lost nthrough the roof that night during a 14 hour period are to be determined.n<b><i>Assumptions</i></b><i><b></b></i><b>1</b><i></i>Steady operating conditions exist. <b>2 </b>The emissivity and thermal conductivity of the roof are nconstant. n<b><i>Properties</i></b><i></i> The thermal conductivity of the concrete is given to be n<i>k </i>= 2 W/mF 0 D 7C. The emissivity of both surfaces of the roof is given nto be 0.9.n<b><i>Analysis </i></b><i></i>When the surrounding surface temperature is different nthan the ambient temperature, the thermal resistances network napproach becomes cumbersome in problems that involve radiation. nTherefore, we will use a different but intuitive approach.n</p>n<p>In steady operation, heat transfer from the room to the nroof (by convection and radiation) must be equal to the heat ntransfer from the roof to the surroundings (by convection and nradiation), that must be equal to the heat transfer through the roof nby conduction. That is,n</p>n<p>Taking the inner and outer surface temperatures of the roof to be <i>Ts,in </i> and <i>Ts,out </i>, respectively, the quantities nabove can be expressed asn</p>n<p> n</p>n<p>Solving the equations above simultaneously gives n nThe total amount of natural gas consumption during a 14-hour period isn</p>n<p>Finally, the money lost through the roof during that period isn</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>11n-11</p>nn</div></div>n<div><div><p><b>3-30 </b>An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss nthrough that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be ndetermined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves nwill be determined.n<b><i>Assumptions</i> 1</b> Heat transfer through the wall is steady and one-dimensional. <b>2 </b>Thermal conductivities are nconstant. <b>3 </b>The furnace operates continuously. <b>4 </b>The given heat transfer coefficient accounts for the nradiation effects.n<b><i>Properties</i></b><i></i> The thermal conductivity of the glass wool insulation is given to be <i>k</i> = 0.038 W/mF 0 D 7°C.n<b><i>Analysis </i></b><i></i>The rate of heat transfer without insulation isn</p>n<p> nIn order to reduce heat loss by 90%, the new heat transfer rate and nthermal resistance must ben nand in order to have this thermal resistance, the thickness of insulation must ben</p>n<p>Noting that heat is saved at a rate of 0.9F 0 B 41500 = 1350 W and the furnace operates continuously and thus n365F 0 B 424 = 8760 h per year, and that the furnace efficiency is 78%, the amount of natural gas saved per year nisn</p>n<p>The money saved isn</p>n<p>The insulation will pay for its cost of $250 inn</p>n<p>which is less than one year.n</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>12n-12</p>nn</div></div>n<div><div><p><b>3-31 </b>An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss nthrough that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be ndetermined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves nwill be determined.n<b><i>Assumptions</i> 1</b> Heat transfer through the wall is steady and one-dimensional. <b>2 </b>Thermal conductivities are nconstant. <b>3 </b>The furnace operates continuously. <b>4 </b>The given heat transfer coefficients accounts for the nradiation effects.n<b><i> Properties</i></b><i></i> The thermal conductivity of the expanded perlite insulation is given to be <i>k</i> = 0.052 W/mF 0 D 7°C.n<b><i>Analysis </i></b><i></i>The rate of heat transfer without insulation isn</p>n<p> nIn order to reduce heat loss by 90%, the new heat transfer rate and nthermal resistance must ben nand in order to have this thermal resistance, the thickness of insulation must ben</p>n<p>Noting that heat is saved at a rate of 0.9F 0 B 41500 = 1350 W and the furnace operates continuously and thus n365F 0 B 424 = 8760 h per year, and that the furnace efficiency is 78%, the amount of natural gas saved per year nisn</p>n<p>The money saved isn</p>n<p>The insulation will pay for its cost of $250 inn</p>n<p>which is less than one year.n</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>13n-13</p>nn</div></div>n<div><div><p><b>3-32</b> n"GIVEN"nA=2*1.5 "[m^2]"nT_s=80 "[C]"nT_infinity=30 "[C]"nh=10 "[W/m^2-C]"n"k_ins=0.038 [W/m-C], parameter to be varied"nf_reduce=0.90n</p>n<p>"ANALYSIS"nQ_dot_old=h*A*(T_s-T_infinity)nQ_dot_new=(1-f_reduce)*Q_dot_oldnQ_dot_new=(T_s-T_infinity)/R_totalnR_total=R_conv+R_insnR_conv=1/(h*A)nR_ins=(L_ins*Convert(cm, m))/(k_ins*A) "L_ins is in cm"n</p>n<p><b>kins [W/m.C] Lins [cm]n</b>0.02 1.8n</p>n<p>0.025 2.25n0.03 2.7n</p>n<p>0.035 3.15n0.04 3.6n</p>n<p>0.045 4.05n0.05 4.5n</p>n<p>0.055 4.95n0.06 5.4n</p>n<p>0.065 5.85n0.07 6.3n</p>n<p>0.075 6.75n0.08 7.2n</p>n<p>0.02 0.03 0.04 0.05 0.06 0.07 0.08n1n</p>n<p>2n</p>n<p>3n</p>n<p>4n</p>n<p>5n</p>n<p>6n</p>n<p>7n</p>n<p>8n</p>n<p><b>kins [W/m-C]n</b></p>n<p><b>L innsn[cmn</b></p>n<p><b>]n</b></p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>14n-14</p>nn</div></div>n<div><div><p><b>3-33E</b> Two of the walls of a house have no windows while the other two walls have 4 windows each. The nratio of heat transfer through the walls with and without windows is to be determined.n<b><i>Assumptions</i> 1</b> Heat transfer through the walls and the windows is steady and one-dimensional. <b>2 </b>Thermal nconductivities are constant. <b>3 </b>Any direct radiation gain or loss through the windows is negligible. <b>4 </b>Heat ntransfer coefficients are constant and uniform over the entire surface.n<b><i>Properties</i></b><i></i> The thermal conductivity of the glass is given to be <i>k</i>glass = 0.45 Btu/h.ftF 0 D 7°F. The R-value of the nwall is given to be 19 h.ft2F 0 D 7°F/Btu.n<b><i>Analysis </i></b><i></i>The thermal resistances through the wall without windows aren</p>n<p>The thermal resistances through the wall with windows aren nThen the ratio of the heat transfer through the walls with and without windows becomesn</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>15n-15</p>nn</div></div>n<div><div><p><b>3-34</b> Two of the walls of a house have no windows while the other two walls have single- or double-pane nwindows. The average rate of heat transfer through each wall, and the amount of money this household nwill save per heating season by converting the single pane windows to double pane windows are to be ndetermined. n<b><i>Assumptions</i> 1</b> Heat transfer through the window is steady since the indoor and outdoor temperatures nremain constant at the specified values. <b>2 </b>Heat transfer is one-dimensional since any significant ntemperature gradients will exist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivities nof the glass and air are constant. <b>4 </b>Heat transfer by radiation is disregarded.n<b><i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i> = 0.026 W/mF 0 D 7°C for air, and 0.78 W/mF 0 D 7°C for nglass.n<b><i>Analysis </i></b><i></i>The rate of heat transfer through each wall can be determined by applying thermal resistance nnetwork. The convection resistances at the inner and outer surfaces are common in all cases.n</p>n<p><b>Walls without windows :n</b> nThen n</p>n<p><b>Wall with single pane windows:n</b></p>n<p>Then n</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>16n-16</p>nn</div></div>n<div><div><p><b>4th wall with double pane windows:n</b></p>n<p> nThen nThe rate of heat transfer which will be saved if the single pane windows are converted to double pane nwindows is n</p>n<p>The amount of energy and money saved during a 7-month long heating season by switching from single npane to double pane windows become n</p>n<p> Money savings = (Energy saved)(Unit cost of energy) = (19,011 kWh)($0.08/kWh) = <b>$1521n</b></p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>17n-17</p>nn</div></div>n<div><div><p><b>3-35</b> The wall of a refrigerator is constructed of fiberglass insulation sandwiched between two layers of nsheet metal. The minimum thickness of insulation that needs to be used in the wall in order to avoid ncondensation on the outer surfaces is to be determined.n<b><i>Assumptions</i> 1</b> Heat transfer through the refrigerator walls is steady since the temperatures of the food ncompartment and the kitchen air remain constant at the specified values. <b>2 </b>Heat transfer is one-ndimensional. <b>3 </b>Thermal conductivities are constant. <b>4 </b>Heat transfer coefficients account for the radiation neffects.n<b><i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i> = 15.1 W/mF 0 D 7°C for sheet metal and 0.035 W/nmF 0 D 7°C for fiberglass insulation.n<b><i>Analysis </i></b><i></i>The minimum thickness of insulation can be determined nby assuming the outer surface temperature of the refrigerator to be n10F 0 B 0C. In steady operation, the rate of heat transfer through the nrefrigerator wall is constant, and thus heat transfer between the nroom and the refrigerated space is equal to the heat transfer nbetween the room and the outer surface of the refrigerator. nConsidering a unit surface area,nUsing the thermal resistance network, heat transfer between the nroom and the refrigerated space can be expressed asn</p>n<p>Substituting,n</p>n<p>Solv ing for <i>L,</i> the minimum thickness of insulation is determined to ben<i>L </i>= 0.0045 m = <b>0.45 cmn</b></p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>18n-18</p>nn</div></div>n<div><div><p><b>3-36</b> n"GIVEN"nk_ins=0.035 "[W/m-C], parameter to be varied"nL_metal=0.001 "[m]"nk_metal=15.1 "[W/m-C], parameter to be varied"nT_refrig=3 "[C]"nT_kitchen=25 "[C]"nh_i=4 "[W/m^2-C]"nh_o=9 "[W/m^2-C]"nT_s_out=20 "[C]"n</p>n<p>"ANALYSIS"nA=1 "[m^2], a unit surface area is considered"nQ_dot=h_o*A*(T_kitchen-T_s_out)nQ_dot=(T_kitchen-T_refrig)/R_totalnR_total=R_conv_i+2*R_metal+R_ins+R_conv_onR_conv_i=1/(h_i*A)nR_metal=L_metal/(k_metal*A)nR_ins=(L_ins*Convert(cm, m))/(k_ins*A) "L_ins is in cm"nR_conv_o=1/(h_o*A)n</p>n<p><b>kins [W/m.C] Lins [cm]n</b>0.02 0.2553n</p>n<p>0.025 0.3191n0.03 0.3829n</p>n<p>0.035 0.4468n0.04 0.5106n</p>n<p>0.045 0.5744n0.05 0.6382n</p>n<p>0.055 0.702n0.06 0.7659n</p>n<p>0.065 0.8297n0.07 0.8935n</p>n<p>0.075 0.9573n0.08 1.021n</p>n<p><b>kmetal [W/m.C] Lins [cm]n</b>10 0.4465n</p>n<p>30.53 0.447n51.05 0.4471n71.58 0.4471n92.11 0.4471n112.6 0.4472n133.2 0.4472n153.7 0.4472n174.2 0.4472n194.7 0.4472n215.3 0.4472n235.8 0.4472n256.3 0.4472n276.8 0.4472n297.4 0.4472n</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>19n-19</p>nn</div></div>n<div><div><p>317.9 0.4472n338.4 0.4472n358.9 0.4472n379.5 0.4472n400 0.4472n</p>n<p>0.02 0.03 0.04 0.05 0.06 0.07 0.08n0.2n</p>n<p>0.3n</p>n<p>0.4n</p>n<p>0.5n</p>n<p>0.6n</p>n<p>0.7n</p>n<p>0.8n</p>n<p>0.9n</p>n<p>1n</p>n<p>1.1n</p>n<p><b>kins [W/m-C]n</b></p>n<p><b>L innsn[cmn</b></p>n<p><b>]n</b></p>n<p>0 50 100 150 200 250 300 350 400n0.4465n</p>n<p>0.4467n</p>n<p>0.4469n</p>n<p>0.4471n</p>n<p>0.4473n</p>n<p><b>kmetal [W/m-C]n</b></p>n<p><b>L innsn[cmn</b></p>n<p><b>]n</b></p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>20n-20</p>nn</div></div>n<div><div><p><b>3-37</b> Heat is to be conducted along a circuit board with a copper layer on one side. The percentages of nheat conduction along the copper and epoxy layers as well as the effective thermal conductivity of the nboard are to be determined.n<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat transfer is one-dimensional since heat transfer nfrom the side surfaces is disregarded <b>3 </b>Thermal conductivities are constant. <b> n<i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i> = 386 W/mF 0 D 7°C for copper and 0.26 W/mF 0 D 7°C for nepoxy layers.n<b><i>Analysis </i></b><i></i>We take the length in the direction of heat transfer to be <i>L </i>and the width of the board to be <i>w</i>. nThen heat conduction along this two-layer board can be expressed asn</p>n<p>Heat conduction along an “equivalent” board of thickness <i>t </i>= <i>t</i>copper<i></i>+ <i>t</i>epoxy<i></i> and thermal conductivity <i>k</i>eff ncan be expressed asn nSetting the two relations above equal to each other and solving for the effective conductivity givesn</p>n<p> nNote that heat conduction is proportional to <i>kt</i>. Substituting, the fractions of heat conducted along the ncopper and epoxy layers as well as the effective thermal conductivity of the board are determined to ben nandn</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>21n-21</p>nn</div></div>n<div><div><p><b>3-38E</b> A thin copper plate is sandwiched between two layers of epoxy boards. The effective thermal nconductivity of the board along its 9 in long side and the fraction of the heat conducted through copper nalong that side are to be determined.n<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat transfer is one-dimensional since heat transfer nfrom the side surfaces are disregarded <b>3 </b>Thermal conductivities are constant. <b> n<i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i> = 223 Btu/hF 0 D 7ftF 0 D 7°F for copper and 0.15 Btu/nhF 0 D 7ftF 0 D 7°F for epoxy layers.n<b><i>Analysis </i></b><i></i>We take the length in the direction of heat transfer to be <i>L </i>and the width of the board to be <i>w</i>. nThen heat conduction along this two-layer plate can be expressed as (we treat the two layers of epoxy as a nsingle layer that is twice as thick)n</p>n<p>Heat conduction along an “equivalent” plate of thick ness <i>t </i>= <i>t</i>copper<i></i>+ <i>t</i>epoxy<i></i> and thermal conductivity <i>k</i>eff ncan be expressed asn nSetting the two relations above equal to each other and solving for the effective conductivity givesn</p>n<p> nNote that heat conduction is proportional to <i>kt</i>. Substituting, the fraction of heat conducted along the copper nlayer and the effective thermal conductivity of the plate are determined to ben nandn</p>n<p><b>Chapter 3 <i>Steady Heat Conductionn</i></b></p>n<p>22n-22</p>nn</div></div>n</body></html>','canEdit':false,'canDelete':false,'canReport':false,'userVote':null,'previewLimit':3,'advEnabled':true,'totalVotes':0,'title':'u00c7engel - Solution Heat and Mass Transfer 2th Ed - heat chap03-001, Manuais, Projetos, Pesquisas de Engenharia Mecu00e2nica','isPremiumEnabled':false,'hasQuizcardSet':null}'><div><div><div><div><main><div><div><div><span><span>Pré-visualização</span><span>3 páginas / 22</span></span></div><div><div><div><div><div><div></div><div></div></div></div></div></div></div></div></div><div><div><div></div><div></div><div></div></div></div><div><div><div><div><div><div><div><div><p><b>Chapter 3STEADY HEAT CONDUCTION</b></p><p><b>Steady Heat Conduction In Plane Walls</b></p><p><b>3-1C </b>(a) If the lateral surfaces of the rod are insulated, the heat transfer surface area of the cylindrical rod is the bottom or the top surface area of the rod, . (b) If the top and the bottom surfaces of the rod are insulated, the heat transfer area of the rod is the lateral surface area of the rod, .</p><p><b>3-2C</b> In steady heat conduction, the rate of heat transfer into the wall is equal to the rate of heat transfer out of it. Also, the temperature at any point in the wall remains constant. Therefore, the energy content of the wall does not change during steady heat conduction. However, the temperature along the wall and thus the energy content of the wall will change during transient conduction. </p><p><b>3-3C </b>The temperature distribution in a plane wall will be a straight line during steady and one dimensional heat transfer with constant wall thermal conductivity.</p><p><b>3-4C </b>The thermal resistance of a medium represents the resistance of that medium against heat transfer.</p><p><b>3-5C</b> The combined heat transfer coefficient represents the combined effects of radiation and convection heat transfers on a surface, and is defined as <i>h</i>combined = <i>h</i>convection + <i>h</i>radiation. It offers the convenience of incorporating the effects of radiation in the convection heat transfer coefficient, and to ignore radiation in heat transfer calculations.</p><p><b>3-6C</b> Yes. The convection resistance can be defined as the inverse of the convection heat transfer coefficient per unit surface area since it is defined as .</p><p><b>3-7C </b>The convection and the radiation resistances at a surface are parallel since both the convection and radiation heat transfers occur simultaneously.</p><p><b>3-8C</b> For a surface of <i>A </i> at which the convection and radiation heat transfer coefficients are , the single equivalent heat transfer coefficient is when the medium and the surrounding surfaces are at the same temperature. Then the equivalent thermal resistance will be .</p><p><b>3-9C </b>The thermal resistance network associated with a five-layer composite wall involves five single-layer resistances connected in series.</p><p><b>3-10C</b> Once the rate of heat transfer is known, the temperature drop across any layer can be determined by multiplying heat transfer rate by the thermal resistance across that layer, </p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>1-1</p></div></div><div><div><p><b>3-11C </b>The temperature of each surface in this case can be determined from</p><p>where is the thermal resistance between the environment and surface i.</p><p><b>3-12C </b>Yes, it is.</p><p><b>3-13C</b> The window glass which consists of two 4 mm thick glass sheets pressed tightly against each other will probably have thermal contact resistance which serves as an additional thermal resistance to heat transfer through window, and thus the heat transfer rate will be smaller relative to the one which consists of a single 8 mm thick glass sheet.</p><p><b>3-14C</b> Convection heat transfer through the wall is expressed as . In steady heat transfer, heat transfer rate to the wall and from the wall are equal. Therefore at the outer surface which has convection heat transfer coefficient three times that of the inner surface will experience three times smaller temperature drop compared to the inner surface. Therefore, at the outer surface, the temperature will be closer to the surrounding air temperature.</p><p><b>3-15C</b> The new design introduces the thermal resistance of the copper layer in addition to the thermal resistance of the aluminum which has the same value for both designs. Therefore, the new design will be a poorer conductor of heat.</p><p><b>3-16C</b> The blanket will introduce additional resistance to heat transfer and slow down the heat gain of the drink wrapped in a blanket. Therefore, the drink left on a table will warm up faster.</p><p><b>3-17</b> The two surfaces of a wall are maintained at specified temperatures. The rate of heat loss through the wall is to be determined. <b><i>Assumptions</i> 1</b> Heat transfer through the wall is steady since the surface temperatures remain constant at the specified values. <b>2 </b>Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivity is constant.<b><i>Properties</i></b><i></i> The thermal conductivity is given to be <i>k</i> = 0.8 W/mF 0 D 7°C.<b><i>Analysis</i></b>The surface area of the wall and the rate of heat loss through the wall are</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>2-2</p></div></div><div><div><p><b>3-18 </b>The two surfaces of a window are maintained at specified temperatures. The rate of heat loss through the window and the inner surface temperature are to be determined.<b><i>Assumptions</i> 1</b> Heat transfer through the window is steady since the surface temperatures remain constant at the specified values. <b>2 </b>Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivity is constant. <b>4 </b>Heat transfer by radiation is negligible.<b><i>Properties</i></b><i></i> The thermal conductivity of the glass is given to be <i>k</i> = 0.78 W/mF 0 D 7°C.<b><i>Analysis</i></b>The area of the window and the individual resistances are</p><p>The steady rate of heat transfer through window glass is then</p><p>The inner surface temperature of the window glass can be determined from</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>3-3</p></div></div><div><div><p><b>3-19</b> A double-pane window consists of two 3-mm thick layers of glass separated by a 12-mm wide stagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined.<b><i>Assumptions</i> 1</b> Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. <b>2 </b>Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivities of the glass and air are constant. <b>4 </b>Heat transfer by radiation is negligible.<b><i>Properties</i></b><i></i> The thermal conductivity of the glass and air are given to be <i>k</i>glass = 0.78 W/mF 0 D 7°C and <i>k</i>air = 0.026 W/mF 0 D 7°C.<b><i>Analysis </i></b><i></i> The area of the window and the individual resistances are </p><p>The steady rate of heat transfer through window glass then becomes</p><p>The inner surface temperature of the window glass can be determined from</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>4-4</p></div></div><div><div><p><b>3-20</b> A double-pane window consists of two 3-mm thick layers of glass separated by an evacuated space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined.<b><i>Assumptions</i> 1</b> Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. <b>2 </b>Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivities of the glass and air are constant. <b>4 </b>Heat transfer by radiation is negligible.<b><i>Properties</i></b><i></i> The thermal conductivity of the glass is given to be <i>k</i>glass = 0.78 W/mF 0 D 7°C.<b><i>Analysis </i></b><i></i> Heat cannot be conducted through an evacuated space since the thermal conductivity of vacuum is zero (no medium to conduct heat) and thus its thermal resistance is zero. Therefore, if radiation is disregarded, the heat transfer through the window will be zero. Then the answer of this problem is <b>zero </b>since the problem states to disregard radiation.<b><i>Discussion</i></b>In reality, heat will be transferred between the glasses by radiation. We do not know the inner surface temperatures of windows. In order to determine radiation heat resistance we assume them to be 5°C and 15°C, respectively, and take the emissivity to be 1. Then individual resistances are </p><p>The steady rate of heat transfer through window glass then becomes</p><p>The inner surface temperature of the window glass can be determined from</p><p>Similarly, the inner surface temperatures of the glasses are calculated to be 15.2 and -1.2F 0 B 0C (we had assumed them to be 15 and 5F 0 B 0C when determining the radiation resistance). We can improve the result obtained by reevaluating the radiation resistance and repeating the calculations.</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>5-5</p></div></div><div><div><p><b>3-21</b> 'GIVEN'A=1.2*2 '[m^2]'L_glass=3 '[mm]'k_glass=0.78 '[W/m-C]'L_air=12 [mm], parameter to be varied'T_infinity_1=24 '[C]'T_infinity_2=-5 '[C]'h_1=10 '[W/m^2-C]'h_2=25 '[W/m^2-C]'</p><p>'PROPERTIES'k_air=conductivity(Air,T=25)</p><p>'ANALYSIS'R_conv_1=1/(h_1*A)R_glass=(L_glass*Convert(mm, m))/(k_glass*A)R_air=(L_air*Convert(mm, m))/(k_air*A)R_conv_2=1/(h_2*A)R_total=R_conv_1+2*R_glass+R_air+R_conv_2Q_dot=(T_infinity_1-T_infinity_2)/R_total</p><p><b>Lair [mm] Q [W]</b>2 307.84 228.66 181.88 150.9</p><p>10 12912 112.614 99.9316 89.8218 81.5720 74.7</p><p>2 4 6 8 10 12 14 16 18 2050</p><p>100</p><p>150</p><p>200</p><p>250</p><p>300</p><p>350</p><p><b>Lair [mm]</b></p><p><b>Q[W</b></p><p><b>]</b></p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>6-6</p></div></div><div><div><p><b>3-22E </b>The inner and outer surfaces of the walls of an electrically heated house remain at specified temperatures during a winter day. The amount of heat lost from the house that day and its its cost are to be determined.<b><i>Assumptions</i> 1</b> Heat transfer through the walls is steady since the surface temperatures of the walls remain constant at the specified values during the time period considered. <b>2 </b>Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors.<b> 3</b> Thermal conductivity of the walls is constant.<b><i>Properties</i></b><i></i> The thermal conductivity of the brick wall is given to be <i>k</i> = 0.40 Btu/hF 0 D 7ftF 0 D 7°F.<b><i>Analysis </i></b><i></i>We consider heat loss through the walls only. The total heat transfer area is</p><p>The rate of heat loss during the daytime is The rate of heat loss during nighttime is The amount of heat loss from the house that night will be</p><p>Then the cost of this heat loss for that day becomes</p><p><b>3-23</b> A cylindrical resistor on a circuit board dissipates 0.15 W of power steadily in a specified environment. The amount of heat dissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined.<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat is transferred uniformly from all surfaces of the resistor. <b><i>Analysis </i></b><i></i>(<i>a</i>) The amount of heat this resistor dissipates during a 24-hour period is</p><p>(<i>b</i>) The heat flux on the surface of the resistor is (<i>c</i>) The surface temperature of the resistor can be determined from </p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>7-7</p></div></div><div><div><p><b>3-24</b> A power transistor dissipates 0.2 W of power steadily in a specified environment. The amount of heat dissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined.<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat is transferred uniformly from all surfaces of the transistor. <b><i>Analysis </i></b><i></i>(<i>a</i>) The amount of heat this transistor dissipates during a 24-hour period is</p><p>(<i>b</i>) The heat flux on the surface of the transistor is (<i>c</i>) The surface temperature of the transistor can be determined from</p><p><b>3-25</b> A circuit board houses 100 chips, each dissipating 0.07 W. The surface heat flux, the surface temperature of the chips, and the thermal resistance between the surface of the board and the cooling medium are to be determined.<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat transfer from the back surface of the board is negligible. <b>2 </b>Heat is transferred uniformly from the entire front surface. <b><i>Analysis </i></b><i></i>(<i>a</i>) The heat flux on the surface of the circuit board is </p><p>(<i>b</i>) The surface temperature of the chips is (<i>c</i>) The thermal resistance is</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>8-8</p></div></div><div><div><p><b>3-26 </b>A person is dissipating heat at a rate of 150 W by natural convection and radiation to the surrounding air and surfaces. For a given deep body temperature, the outer skin temperature is to be determined.<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>The heat transfer coefficient is constant and uniform over the entire exposed surface of the person. <b>3 </b>The surrounding surfaces are at the same temperature as the indoor air temperature. <b>4 </b>Heat generation within the 0.5-cm thick outer layer of the tissue is negligible.<b><i>Properties</i></b><i></i> The thermal conductivity of the tissue near the skin is given to be <i>k</i> = 0.3 W/mF 0 D 7°C.<b><i>Analysis </i></b><i></i>The skin temperature can be determined directly from </p><p><b>3-27</b> Heat is transferred steadily to the boiling water in an aluminum pan. The inner surface temperature of the bottom of the pan is given. The boiling heat transfer coefficient and the outer surface temperature of the bottom of the pan are to be determined. <b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat transfer is one-dimensional since the thickness of the bottom of the pan is small relative to its diameter. <b>3 </b>The thermal conductivity of the pan is constant.<b><i>Properties</i></b><i></i> The thermal conductivity of the aluminum pan is given to be <i>k</i> = 237 W/mF 0 D 7°C.<b><i>Analysis </i></b><i></i>(<i>a</i>) The boiling heat transfer coefficient is (<i>b</i>) The outer surface temperature of the bottom of the pan is </p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>9-9</p></div></div><div><div><p><b>3-28E </b> A wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermal resistance of the wall and its R-value of insulation are to be determined.<b><i>Assumptions</i> 1</b> Heat transfer through the wall is one-dimensional. <b>2 </b>Thermal conductivities are constant.<b><i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i>sheetrock = 0.10 Btu/hF 0 D 7ftF 0 D 7°F and <i>k </i>insulation = 0.020 Btu/hF 0 D 7ftF 0 D 7°F.<b><i>Analysis </i></b><i></i>(<i>a</i>) The surface area of the wall is not given and thus we consider a unit surface area (<i>A</i> = 1 ft2). Then the <i>R</i>-value of insulation of the wall becomes equivalent to its<i></i>thermal resistance, which is determined from. (<i>b</i>) Therefore, this is approximately a <b><i>R</i>-22</b> wall in English units.</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>10-10</p></div></div><div><div><p><b>3-29 </b>The roof of a house with a gas furnace consists of 3-cm thick concrete that is losing heat to the outdoors by radiation and convection. The rate of heat transfer through the roof and the money lost through the roof that night during a 14 hour period are to be determined.<b><i>Assumptions</i></b><i><b></b></i><b>1</b><i></i>Steady operating conditions exist. <b>2 </b>The emissivity and thermal conductivity of the roof are constant. <b><i>Properties</i></b><i></i> The thermal conductivity of the concrete is given to be <i>k </i>= 2 W/mF 0 D 7C. The emissivity of both surfaces of the roof is given to be 0.9.<b><i>Analysis </i></b><i></i>When the surrounding surface temperature is different than the ambient temperature, the thermal resistances network approach becomes cumbersome in problems that involve radiation. Therefore, we will use a different but intuitive approach.</p><p>In steady operation, heat transfer from the room to the roof (by convection and radiation) must be equal to the heat transfer from the roof to the surroundings (by convection and radiation), that must be equal to the heat transfer through the roof by conduction. That is,</p><p>Taking the inner and outer surface temperatures of the roof to be <i>Ts,in </i> and <i>Ts,out </i>, respectively, the quantities above can be expressed as</p><p>Solving the equations above simultaneously gives The total amount of natural gas consumption during a 14-hour period is</p><p>Finally, the money lost through the roof during that period is</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>11-11</p></div></div><div><div><p><b>3-30 </b>An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves will be determined.<b><i>Assumptions</i> 1</b> Heat transfer through the wall is steady and one-dimensional. <b>2 </b>Thermal conductivities are constant. <b>3 </b>The furnace operates continuously. <b>4 </b>The given heat transfer coefficient accounts for the radiation effects.<b><i>Properties</i></b><i></i> The thermal conductivity of the glass wool insulation is given to be <i>k</i> = 0.038 W/mF 0 D 7°C.<b><i>Analysis </i></b><i></i>The rate of heat transfer without insulation is</p><p> In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be and in order to have this thermal resistance, the thickness of insulation must be</p><p>Noting that heat is saved at a rate of 0.9F 0 B 41500 = 1350 W and the furnace operates continuously and thus 365F 0 B 424 = 8760 h per year, and that the furnace efficiency is 78%, the amount of natural gas saved per year is</p><p>The money saved is</p><p>The insulation will pay for its cost of $250 in</p><p>which is less than one year.</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>12-12</p></div></div><div><div><p><b>3-31 </b>An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves will be determined.<b><i>Assumptions</i> 1</b> Heat transfer through the wall is steady and one-dimensional. <b>2 </b>Thermal conductivities are constant. <b>3 </b>The furnace operates continuously. <b>4 </b>The given heat transfer coefficients accounts for the radiation effects.<b><i> Properties</i></b><i></i> The thermal conductivity of the expanded perlite insulation is given to be <i>k</i> = 0.052 W/mF 0 D 7°C.<b><i>Analysis </i></b><i></i>The rate of heat transfer without insulation is</p><p> In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be and in order to have this thermal resistance, the thickness of insulation must be</p><p>Noting that heat is saved at a rate of 0.9F 0 B 41500 = 1350 W and the furnace operates continuously and thus 365F 0 B 424 = 8760 h per year, and that the furnace efficiency is 78%, the amount of natural gas saved per year is</p><p>The money saved is</p><p>The insulation will pay for its cost of $250 in</p><p>which is less than one year.</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>13-13</p></div></div><div><div><p><b>3-32</b> 'GIVEN'A=2*1.5 '[m^2]'T_s=80 '[C]'T_infinity=30 '[C]'h=10 '[W/m^2-C]'k_ins=0.038 [W/m-C], parameter to be varied'f_reduce=0.90</p><p>'ANALYSIS'Q_dot_old=h*A*(T_s-T_infinity)Q_dot_new=(1-f_reduce)*Q_dot_oldQ_dot_new=(T_s-T_infinity)/R_totalR_total=R_conv+R_insR_conv=1/(h*A)R_ins=(L_ins*Convert(cm, m))/(k_ins*A) 'L_ins is in cm'</p><p><b>kins [W/m.C] Lins [cm]</b>0.02 1.8</p><p>0.025 2.250.03 2.7</p><p>0.035 3.150.04 3.6</p><p>0.045 4.050.05 4.5</p><p>0.055 4.950.06 5.4</p><p>0.065 5.850.07 6.3</p><p>0.075 6.750.08 7.2</p><p>0.02 0.03 0.04 0.05 0.06 0.07 0.081</p><p>2</p><p>3</p><p>4</p><p>5</p><p>6</p><p>7</p><p>8</p><p><b>kins [W/m-C]</b></p><p><b>L ins[cm</b></p><p><b>]</b></p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>14-14</p></div></div><div><div><p><b>3-33E</b> Two of the walls of a house have no windows while the other two walls have 4 windows each. The ratio of heat transfer through the walls with and without windows is to be determined.<b><i>Assumptions</i> 1</b> Heat transfer through the walls and the windows is steady and one-dimensional. <b>2 </b>Thermal conductivities are constant. <b>3 </b>Any direct radiation gain or loss through the windows is negligible. <b>4 </b>Heat transfer coefficients are constant and uniform over the entire surface.<b><i>Properties</i></b><i></i> The thermal conductivity of the glass is given to be <i>k</i>glass = 0.45 Btu/h.ftF 0 D 7°F. The R-value of the wall is given to be 19 h.ft2F 0 D 7°F/Btu.<b><i>Analysis </i></b><i></i>The thermal resistances through the wall without windows are</p><p>The thermal resistances through the wall with windows are Then the ratio of the heat transfer through the walls with and without windows becomes</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>15-15</p></div></div><div><div><p><b>3-34</b> Two of the walls of a house have no windows while the other two walls have single- or double-pane windows. The average rate of heat transfer through each wall, and the amount of money this household will save per heating season by converting the single pane windows to double pane windows are to be determined. <b><i>Assumptions</i> 1</b> Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. <b>2 </b>Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. <b>3 </b>Thermal conductivities of the glass and air are constant. <b>4 </b>Heat transfer by radiation is disregarded.<b><i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i> = 0.026 W/mF 0 D 7°C for air, and 0.78 W/mF 0 D 7°C for glass.<b><i>Analysis </i></b><i></i>The rate of heat transfer through each wall can be determined by applying thermal resistance network. The convection resistances at the inner and outer surfaces are common in all cases.</p><p><b>Walls without windows :</b> Then </p><p><b>Wall with single pane windows:</b></p><p>Then </p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>16-16</p></div></div><div><div><p><b>4th wall with double pane windows:</b></p><p> Then The rate of heat transfer which will be saved if the single pane windows are converted to double pane windows is </p><p>The amount of energy and money saved during a 7-month long heating season by switching from single pane to double pane windows become </p><p> Money savings = (Energy saved)(Unit cost of energy) = (19,011 kWh)($0.08/kWh) = <b>$1521</b></p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>17-17</p></div></div><div><div><p><b>3-35</b> The wall of a refrigerator is constructed of fiberglass insulation sandwiched between two layers of sheet metal. The minimum thickness of insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces is to be determined.<b><i>Assumptions</i> 1</b> Heat transfer through the refrigerator walls is steady since the temperatures of the food compartment and the kitchen air remain constant at the specified values. <b>2 </b>Heat transfer is one-dimensional. <b>3 </b>Thermal conductivities are constant. <b>4 </b>Heat transfer coefficients account for the radiation effects.<b><i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i> = 15.1 W/mF 0 D 7°C for sheet metal and 0.035 W/mF 0 D 7°C for fiberglass insulation.<b><i>Analysis </i></b><i></i>The minimum thickness of insulation can be determined by assuming the outer surface temperature of the refrigerator to be 10F 0 B 0C. In steady operation, the rate of heat transfer through the refrigerator wall is constant, and thus heat transfer between the room and the refrigerated space is equal to the heat transfer between the room and the outer surface of the refrigerator. Considering a unit surface area,Using the thermal resistance network, heat transfer between the room and the refrigerated space can be expressed as</p><p>Substituting,</p><p>Solv ing for <i>L,</i> the minimum thickness of insulation is determined to be<i>L </i>= 0.0045 m = <b>0.45 cm</b></p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>18-18</p></div></div><div><div><p><b>3-36</b> 'GIVEN'k_ins=0.035 '[W/m-C], parameter to be varied'L_metal=0.001 '[m]'k_metal=15.1 '[W/m-C], parameter to be varied'T_refrig=3 '[C]'T_kitchen=25 '[C]'h_i=4 '[W/m^2-C]'h_o=9 '[W/m^2-C]'T_s_out=20 '[C]'</p><p>'ANALYSIS'A=1 '[m^2], a unit surface area is considered'Q_dot=h_o*A*(T_kitchen-T_s_out)Q_dot=(T_kitchen-T_refrig)/R_totalR_total=R_conv_i+2*R_metal+R_ins+R_conv_oR_conv_i=1/(h_i*A)R_metal=L_metal/(k_metal*A)R_ins=(L_ins*Convert(cm, m))/(k_ins*A) 'L_ins is in cm'R_conv_o=1/(h_o*A)</p><p><b>kins [W/m.C] Lins [cm]</b>0.02 0.2553</p><p>0.025 0.31910.03 0.3829</p><p>0.035 0.44680.04 0.5106</p><p>0.045 0.57440.05 0.6382</p><p>0.055 0.7020.06 0.7659</p><p>0.065 0.82970.07 0.8935</p><p>0.075 0.95730.08 1.021</p><p><b>kmetal [W/m.C] Lins [cm]</b>10 0.4465</p><p>30.53 0.44751.05 0.447171.58 0.447192.11 0.4471112.6 0.4472133.2 0.4472153.7 0.4472174.2 0.4472194.7 0.4472215.3 0.4472235.8 0.4472256.3 0.4472276.8 0.4472297.4 0.4472</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>19-19</p></div></div><div><div><p>317.9 0.4472338.4 0.4472358.9 0.4472379.5 0.4472400 0.4472</p><p>0.02 0.03 0.04 0.05 0.06 0.07 0.080.2</p><p>0.3</p><p>0.4</p><p>0.5</p><p>0.6</p><p>0.7</p><p>0.8</p><p>0.9</p><p>1</p><p>1.1</p><p><b>kins [W/m-C]</b></p><p><b>L ins[cm</b></p><p><b>]</b></p><p>0 50 100 150 200 250 300 350 4000.4465</p><p>0.4467</p><p>0.4469</p><p>0.4471</p><p>0.4473</p><p><b>kmetal [W/m-C]</b></p><p><b>L ins[cm</b></p><p><b>]</b></p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>20-20</p></div></div><div><div><p><b>3-37</b> Heat is to be conducted along a circuit board with a copper layer on one side. The percentages of heat conduction along the copper and epoxy layers as well as the effective thermal conductivity of the board are to be determined.<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat transfer is one-dimensional since heat transfer from the side surfaces is disregarded <b>3 </b>Thermal conductivities are constant. <b><i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i> = 386 W/mF 0 D 7°C for copper and 0.26 W/mF 0 D 7°C for epoxy layers.<b><i>Analysis </i></b><i></i>We take the length in the direction of heat transfer to be <i>L </i>and the width of the board to be <i>w</i>. Then heat conduction along this two-layer board can be expressed as</p><p>Heat conduction along an “equivalent” board of thickness <i>t </i>= <i>t</i>copper<i></i>+ <i>t</i>epoxy<i></i> and thermal conductivity <i>k</i>eff can be expressed as Setting the two relations above equal to each other and solving for the effective conductivity gives</p><p> Note that heat conduction is proportional to <i>kt</i>. Substituting, the fractions of heat conducted along the copper and epoxy layers as well as the effective thermal conductivity of the board are determined to be and</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>21-21</p></div></div><div><div><p><b>3-38E</b> A thin copper plate is sandwiched between two layers of epoxy boards. The effective thermal conductivity of the board along its 9 in long side and the fraction of the heat conducted through copper along that side are to be determined.<b><i>Assumptions</i> 1</b> Steady operating conditions exist. <b>2 </b>Heat transfer is one-dimensional since heat transfer from the side surfaces are disregarded <b>3 </b>Thermal conductivities are constant. <b><i>Properties</i></b><i></i> The thermal conductivities are given to be <i>k</i> = 223 Btu/hF 0 D 7ftF 0 D 7°F for copper and 0.15 Btu/hF 0 D 7ftF 0 D 7°F for epoxy layers.<b><i>Analysis </i></b><i></i>We take the length in the direction of heat transfer to be <i>L </i>and the width of the board to be <i>w</i>. Then heat conduction along this two-layer plate can be expressed as (we treat the two layers of epoxy as a single layer that is twice as thick)</p><p>Heat conduction along an “equivalent” plate of thick ness <i>t </i>= <i>t</i>copper<i></i>+ <i>t</i>epoxy<i></i> and thermal conductivity <i>k</i>eff can be expressed as Setting the two relations above equal to each other and solving for the effective conductivity gives</p><p> Note that heat conduction is proportional to <i>kt</i>. Substituting, the fraction of heat conducted along the copper layer and the effective thermal conductivity of the plate are determined to be and</p><p><b>Chapter 3 <i>Steady Heat Conduction</i></b></p><p>22-22</p></div></div></div></div></div></div></div></div><div><div><div><div></div></div></div></div></main><div><div><div><div></div></div></div></div><div><div><div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div></div></div></div><div><div><div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div><div><div><div><div><div><div><div></div></div><div><div></div></div></div></div></div></div></div></div></div></div></aside></div></div></div></div></div></div></div></div></body>
- Solutions-manual-for-heat-and-mass-transfer-fundamentals-and-applications-5th-edition-by-cengel.pdf for the Solutions Manual for Heat and Mass Transfer Fundamentals and Applications 5th Edition by Cengel I found from them both solutions manual/test bank from search box.
- I'm in desperate need of this solution manual. If you have an earlier edition I'll take that too. I really need help here. 'Heat and Mass Transfer: Fundamentals & Applications' 4th ed by Cengel and Ghajar.
Heat And Mass Transfer Cengel 5th Edition Pdf
With complete coverage of the basic principles of heat transfer and a broad range of applications in a flexible format, Heat and Mass Transfer: Fundamentals and Applications, by Yunus Cengel and Afshin Ghajar provides the perfect blend of fundamentals and applications.
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