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Thermodynamics HW Solutions 157

Course: PHY 4803, Fall 2010
School: UNF
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Word Count: 177

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2 Chapter Heat Conduction Equation T (r ) = 37,894r 2 + 98.34 ln r + 257.2 = 257.2 473.68r 2 + 98.34 ln r 4(20) The temperature at the center surface of the pipe is determined by setting radius r to be 17.5 cm, which is the average of the inner radius and outer radius. T (r ) = 257.2 473.68(0.175) 2 + 98.34 ln(0.175) = 71.2C 2-137 A spherical ball in which heat is generated uniformly is exposed to...

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2 Chapter Heat Conduction Equation T (r ) = 37,894r 2 + 98.34 ln r + 257.2 = 257.2 473.68r 2 + 98.34 ln r 4(20) The temperature at the center surface of the pipe is determined by setting radius r to be 17.5 cm, which is the average of the inner radius and outer radius. T (r ) = 257.2 473.68(0.175) 2 + 98.34 ln(0.175) = 71.2C 2-137 A spherical ball in which heat is generated uniformly is exposed to iced-water. The temperatures at the center and at the surface of the ball are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change any with time. 2 Heat transfer is one-dimensional., and there is thermal symmetry about the center point. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Properties The thermal conductivity is given to be k = 45 W/mC. Analysis The temperatures at the center and at the surface of the ball are determined directly from T s = T + & gr0 (2.6 10 6 W/m 3 )(0.15 m) = 0C + = 108.3C 3h 3(1200 W/m 2 .C) T0 = T s + & gr0 2 (2.6 10 6 W/m 3 )(0.15 m) 2 = 108.3C + = 325C 6k 6(45 W/m.C) 2-138 .... 2-141 Design and Essay Problems 2-80 D g h T
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UNF - PHY - 4803
Chapter 3 Steady Heat ConductionChapter 3STEADY HEAT CONDUCTIONSteady Heat Conduction In Plane Walls3-1C (a) If the lateral surfaces of the rod are insulated, the heat transfer surface area of the cylindrical rodis the bottom or the top surface area
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-11C The temperature of each surface in this case can be determined from&amp;&amp;Q = (T1 Ts1 ) / R1 s1 Ts1 = T1 (QR1 s1 )&amp;&amp;Q = (Ts2 T 2 ) / Rs2 2 Ts2 = T 2 + (QRs2 2 )where Ri is the thermal resistance between the environ
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-18 The two surfaces of a window are maintained at specified temperatures. The rate of heat loss throughthe window and the inner surface temperature are to be determined.Assumptions 1 Heat transfer through the window is
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-19 A double-pane window consists of two 3-mm thick layers of glass separated by a 12-mm widestagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss through thewindow and the inner sur
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-20 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 innersurface temperat
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-21&quot;GIVEN&quot;A=1.2*2 &quot;[m^2]&quot;L_glass=3 &quot;[mm]&quot;k_glass=0.78 &quot;[W/m-C]&quot;&quot;L_air=12 [mm], parameter to be varied&quot;T_infinity_1=24 &quot;[C]&quot;T_infinity_2=-5 &quot;[C]&quot;h_1=10 &quot;[W/m^2-C]&quot;h_2=25 &quot;[W/m^2-C]&quot;&quot;PROPERTIES&quot;k_air=conductivity
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-22E The inner and outer surfaces of the walls of an electrically heated house remain at specifiedtemperatures during a winter day. The amount of heat lost from the house that day and its its cost are to bedetermined.A
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-24 A power transistor dissipates 0.2 W of power steadily in a specified environment. The amount of heatdissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined.Assumpt
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-26 A person is dissipating heat at a rate of 150 W by natural convection and radiation to the surroundingair and surfaces. For a given deep body temperature, the outer skin temperature is to be determined.Assumptions 1
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-28E A wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermalresistance of the wall and its R-value of insulation are to be determined.Assumptions 1 Heat transfer through the w
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-29 The roof of a house with a gas furnace consists of 3-cm thick concrete that is losing heat to theoutdoors by radiation and convection. The rate of heat transfer through the roof and the money lostthrough the roof th
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-30 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat lossthrough that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to bedet
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-31 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat lossthrough that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to bedet
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-32&quot;GIVEN&quot;A=2*1.5 &quot;[m^2]&quot;T_s=80 &quot;[C]&quot;T_infinity=30 &quot;[C]&quot;h=10 &quot;[W/m^2-C]&quot;&quot;k_ins=0.038 [W/m-C], parameter to be varied&quot;f_reduce=0.90&quot;ANALYSIS&quot;Q_dot_old=h*A*(T_s-T_infinity)Q_dot_new=(1-f_reduce)*Q_dot_oldQ_dot_ne
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-33E Two of the walls of a house have no windows while the other two walls have 4 windows each. Theratio of heat transfer through the walls with and without windows is to be determined.Assumptions 1 Heat transfer throug
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-34 Two of the walls of a house have no windows while the other two walls have single- or double-panewindows. The average rate of heat transfer through each wall, and the amount of money this householdwill save per heat
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction4th wall with double pane windows:RglassRiRairRwallRglassRoL wall R value2.31 m 2 C/W=== 0.033382 C/WkAA(20 4) 5(1.2 1.8)m 2Lglass0.005 m=== 0.002968 C/WkA(0.78 W/m 2 .C)(1.2 1.8)m 2L0.015 m= air =
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-35 The wall of a refrigerator is constructed of fiberglass insulation sandwiched between two layers ofsheet metal. The minimum thickness of insulation that needs to be used in the wall in order to avoidcondensation on
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-36&quot;GIVEN&quot;k_ins=0.035 &quot;[W/m-C], parameter to be varied&quot;L_metal=0.001 &quot;[m]&quot;k_metal=15.1 &quot;[W/m-C], parameter to be varied&quot;T_refrig=3 &quot;[C]&quot;T_kitchen=25 &quot;[C]&quot;h_i=4 &quot;[W/m^2-C]&quot;h_o=9 &quot;[W/m^2-C]&quot;T_s_out=20 &quot;[C]&quot;&quot;ANALYS
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction4000.44721.110.9L ins [ cm ]0.80.70.60.50.40.30.20 .020.030.040.050.060.070.08k ins [ W /m -C]0.4473L ins [ cm ]0.44710.44690.44670.4465050100150200250k m etal [ W /m -C]3-20300350400
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-37 Heat is to be conducted along a circuit board with a copper layer on one side. The percentages ofheat conduction along the copper and epoxy layers as well as the effective thermal conductivity of theboard are to be
UNF - PHY - 4803
Chapter 3 Steady Heat Conduction3-38E A thin copper plate is sandwiched between two layers of epoxy boards. The effective thermalconductivity of the board along its 9 in long side and the fraction of the heat conducted through copperalong that side are
UNF - PHY - 4803
Chapter 15 Steady Heat ConductionThermal Contact Resistance3-39C The resistance that an interface offers to heat transfer per unit interface area is called thermal contactresistance, Rc . The inverse of thermal contact resistance is called the thermal
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-46 Six identical power transistors are attached on a copper plate. For a maximum case temperature of85C, the maximum power dissipation and the temperature jump at the interface are to be determined.Assumptions 1 Stead
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-47 Two cylindrical aluminum bars with ground surfaces are pressed against each other in an insulationsleeve. For specified top and bottom surface temperatures, the rate of heat transfer along the cylinders andthe temp
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-48 A thin copper plate is sandwiched between two epoxy boards. The error involved in the total thermalresistance of the plate if the thermal contact conductances are ignored is to be determined.Assumptions 1 Steady op
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-52 A wall consists of horizontal bricks separated by plaster layers. There are also plaster layers on eachside of the wall, and a rigid foam on the inner side of the wall. The rate of heat transfer through the wall is
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-53&quot;GIVEN&quot;A=4*6 &quot;[m^2]&quot;L_brick=0.18 &quot;[m]&quot;L_plaster_center=0.18 &quot;[m]&quot;L_plaster_side=0.02 &quot;[m]&quot;&quot;L_foam=2 [cm], parameter to be varied&quot;k_brick=0.72 &quot;[W/m-C]&quot;k_plaster=0.22 &quot;[W/m-C]&quot;k_foam=0.026 &quot;[W/m-C]&quot;T_infinity
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction700600Q total [ W ]5004003002001001234567L foam [ cm ]3-298910
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-54 A wall is to be constructed of 10-cm thick wood studs or with pairs of 5-cm thick wood studs nailedto each other. The rate of heat transfer through the solid stud and through a stud pair nailed to each other,as wel
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-55 A wall is constructed of two layers of sheetrock spaced by 5 cm 12 cm wood studs. The spacebetween the studs is filled with fiberglass insulation. The thermal resistance of the wall and the rate of heattransfer thr
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-56E A wall is to be constructed using solid bricks or identical size bricks with 9 square air holes. Thereis a 0.5 in thick sheetrock layer between two adjacent bricks on all four sides, and on both sides of the wall.
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction(b) The thermal resistance network and the individual thermal resistances if the wall is constructed ofbricks with air holes areR2RiR1R3T1R6R4R7T 2R5Aairholes = 9(1.25 / 12) (1.25 / 12) = 0.0977 ft 2Abricks =
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-57 A composite wall consists of several horizontal and vertical layers. The left and right surfaces of thewall are maintained at uniform temperatures. The rate of heat transfer through the wall, the interfacetemperatu
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-58 A composite wall consists of several horizontal and vertical layers. The left and right surfaces of thewall are maintained at uniform temperatures. The rate of heat transfer through the wall, the interfacetemperatu
UNF - PHY - 4803
Chapter 15 Steady Heat ConductionRtotal = R1 + Rmid ,1 = 0.04 + 0.025 = 0.065 C/WThen the temperature at the point where The sections B, D, and E meet becomes&amp; T T T = T QR&amp;Q= 11total = 300C (571 W)(0.065 C/W) = 263CRtotal(c) The temperature drop
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-60 A coat is made of 5 layers of 0.1 mm thick cotton fabric separated by 1.5 mm thick air space. The rateof heat loss through the jacket is to be determined, and the result is to be compared to the heat loss througha
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-61 A kiln is made of 20 cm thick concrete walls and ceiling. The two ends of the kiln are made of thinsheet metal covered with 2-cm thick styrofoam. For specified indoor and outdoor temperatures, the rate ofheat trans
UNF - PHY - 4803
Chapter 15 Steady Heat ConductionL_wall=0.2 &quot;[m], parameter to be varied&quot;k_concrete=0.9 &quot;[W/m-C]&quot;T_in=40 &quot;[C]&quot;T_out=-4 &quot;[C]&quot;L_sheet=0.003 &quot;[m]&quot;L_styrofoam=0.02 &quot;[m]&quot;k_styrofoam=0.033 &quot;[W/m-C]&quot;h_i=3000 &quot;[W/m^2-C]&quot;&quot;h_o=25 [W/m^2-C], parameter to be
UNF - PHY - 4803
Chapter 15 Steady Heat Conductionho [W/m2.C]5101520253035404550Qtotal [W]55515720958010084817879279013291776930509406594894160000140000Q total [ W ]12000010000080000600000 .080.120.160.20.240.280.32L w all [ m ]95000
UNF - PHY - 4803
Chapter 15 Steady Heat ConductionAssumptions 1 Steady operating conditions exist. 2 Heat transfer through the plate is one-dimensional. 3Thermal conductivities are constant.Properties The thermal conductivities are given to be k = 0.10 Btu/hftF for epo
UNF - PHY - 4803
Chapter 15 Steady Heat ConductionHeat Conduction in Cylinders and Spheres3-64C When the diameter of cylinder is very small compared to its length, it can be treated as anindefinitely long cylinder. Cylindrical rods can also be treated as being infinite
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-67 A spherical container filled with iced water is subjected to convection and radiation heat transfer at itsouter surface. The rate of heat transfer and the amount of ice that melts per day are to be determined.Assum
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-68 A steam pipe covered with 3-cm thick glass wool insulation is subjected to convection on its surfaces.The rate of heat transfer per unit length and the temperature drops across the pipe and the insulation are tobe
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-69&quot;GIVEN&quot;T_infinity_1=320 &quot;[C]&quot;T_infinity_2=5 &quot;[C]&quot;k_steel=15 &quot;[W/m-C]&quot;D_i=0.05 &quot;[m]&quot;D_o=0.055 &quot;[m]&quot;r_1=D_i/2r_2=D_o/2&quot;t_ins=3 [cm], parameter to be varied&quot;k_ins=0.038 &quot;[W/m-C]&quot;h_o=15 &quot;[W/m^2-C]&quot;h_i=80 &quot;[W/m
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction200310180300160290280120270100260802506040123456t ins [ cm ]3-4678924010 T in s [ C ]Q [W ]140
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-70 A 50-m long section of a steam pipe passes through an open space at 15C. The rate of heat loss fromthe steam pipe, the annual cost of this heat loss, and the thickness of fiberglass insulation needed to save 90perc
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-71 An electric hot water tank is made of two concentric cylindrical metal sheets with foam insulation inbetween. The fraction of the hot water cost that is due to the heat loss from the tank and the paybackperiod of t
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-72&quot;GIVEN&quot;L=2 &quot;[m]&quot;D_i=0.40 &quot;[m]&quot;D_o=0.46 &quot;[m]&quot;r_1=D_i/2r_2=D_o/2&quot;T_w=55 [C], parameter to be varied&quot;T_infinity_2=27 &quot;[C]&quot;h_i=50 &quot;[W/m^2-C]&quot;h_o=12 &quot;[W/m^2-C]&quot;k_ins=0.03 &quot;[W/m-C]&quot;Price_electric=0.08 &quot;[\$/kWh]&quot;
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction4035f HeatLoss [ %]3025201510540506070T w [ C]3-508090
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-73 A cold aluminum canned drink that is initially at a uniform temperature of 3C is brought into a roomair at 25C. The time it will take for the average temperature of the drink to rise to 10C with and withoutrubber i
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-74 A cold aluminum canned drink that is initially at a uniform temperature of 3C is brought into a roomair at 25C. The time it will take for the average temperature of the drink to rise to 10C with and withoutrubber i
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-75E A steam pipe covered with 2-in thick fiberglass insulation is subjected to convection on its surfaces.The rate of heat loss from the steam per unit length and the error involved in neglecting the thermalresistance
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-76 Hot water is flowing through a 3-m section of a cast iron pipe. The pipe is exposed to cold air andsurfaces in the basement. The rate of heat loss from the hot water and the average velocity of the water inthe pipe
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-77 Hot water is flowing through a 15 m section of a copper pipe. The pipe is exposed to cold air andsurfaces in the basement. The rate of heat loss from the hot water and the average velocity of the water inthe pipe a
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-78E Steam exiting the turbine of a steam power plant at 100F is to be condensed in a large condenserby cooling water flowing through copper tubes. For specified heat transfer coefficients, the length of thetube requir
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-79E Steam exiting the turbine of a steam power plant at 100F is to be condensed in a large condenser bycooling water flowing through copper tubes. For specified heat transfer coefficients and 0.01-in thick scalebuild
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-80E&quot;GIVEN&quot;T_infinity_1=100 &quot;[F]&quot;T_infinity_2=70 &quot;[F]&quot;k_pipe=223 &quot;[Btu/h-ft-F], parameter to be varied&quot;D_i=0.4 &quot;[in]&quot;&quot;D_o=0.6 [in], parameter to be varied&quot;r_1=D_i/2r_2=D_o/2h_fg=1037 &quot;[Btu/lbm]&quot;h_o=1500 &quot;[Btu/h
UNF - PHY - 4803
Chapter 15 Steady Heat ConductionDo[in]0.50.5250.550.5750.60.6250.650.6750.70.7250.750.7750.80.8250.850.8750.90.9250.950.9751Ltube [ft]11541153115211511151115011491149114811481148114711471147114611461146114611451
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction11801175L tube [ ft]117011651160115511501145050100150200250300350400k pipe [ Btu/h-ft-F]1155.0L tube [ ft]1152.51150.01147.51145.00 .50.60.70.8D o [ in]3-600.91
UNF - PHY - 4803
Chapter 15 Steady Heat Conduction3-81 A 3-m diameter spherical tank filled with liquid nitrogen at 1 atm and -196C is exposed toconvection and radiation with the surrounding air and surfaces. The rate of evaporation of liquid nitrogenin the tank as a r