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Chapter 1. IntroductionChapter 1IntroductionHolman / Heat Transfer, 10th EditionHolman / Heat Transfer, 10th Edition1

Chapter 1. Introduction2Heat transfer is thermal energy transfer that is induced by atemperature difference (or gradient)Modes of heat transferConduction heat transfer: Occurs when a temperature gradientexists through a solid or a stationary fluid (liquid or gas).Convection heat transfer: Occurs within a moving fluid, orbetween a solid surface and a moving fluid, when they are atdifferent temperaturesThermal radiation: Heat transfer between two surfaces (that are notin contact), often in the absence of an intervening medium.Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction1‐1. Conduction Heat Transferqx T A xWhen the proportionality constant is inserted, Tqx kA x[1‐1]Holman / Heat Transfer, 10th Edition3

Chapter 1. IntroductionEnergy conducted in left face heat generated within element change in internal energy energy conducted out right faceThese energy quantities are given as follows:Energy in left face qx kA T xEnergy generated within element qA dx TChange in internal energy ρ cAdx τEnergy out right face qx dx T kA x x dx T T A k k x x x dx whereq energy generated per unit volume, W/m3c specific heat of material, J/kg CHolman / Heat Transfer, 10th Edition4

Chapter 1. Introductionρ density, kg/m3Combining the relations above gives kA T T T T qA dx ρ cAdx A k k x τ x x x T Tkqcρ or x x τqx q y qz qgen dx [1‐2]dE qx dx q y dy qz dz dτHolman / Heat Transfer, 10th Edition5

Chapter 1. Introduction Tqx k dy dz x T Tqx dx k k x x xq y k dx dz dx dy dz T y T T q y dy k k dy dx dz y y y Tqz k dx dy z T T qz dz k k dz dx dy z z z qgen q dx dy dzdE T ρ c dx dy dz τdτHolman / Heat Transfer, 10th Edition6

Chapter 1. Introduction7Figure 1‐2 Elemental volume for one‐dimensional heat conductionanalysis.Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction8Figure 1‐3 Elemental volume for three‐dimensional heat‐conductionanalysis: (a) cartesian coordinates; (b) cylindrical coordinates; (c)spherical coordinates.Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction T k x x T k y y T k z z9 T ρ qc τ [1‐3] For constant thermal conductivity, Equation (1‐3) is written 2T 2T 2T q 1 T 2 2 2 x y zk α τ [1 3a]Equation (1‐3a) may be transformed into either cylindrical orspherical coordinates by standard calculus techniques. The resultsare as follows:Cylindrical coordinates: 2T 1 T 1 2T 2T q 1 T 2 2 2 2r r r φk α τ [1 3b] r zHolman / Heat Transfer, 10th Edition

Chapter 1. Introduction10Spherical coordinates:1 21 T 1 2T q 1 T(rT ) 2 sin θ 2 222 θ r sin θ φr rr sin θ θ k α τ[1 3c]Steady‐state one‐dimensional heat flow (no heat generation):d 2T 02dx[1‐4]Steady‐state one‐dimensional heat flow in cylindrical coordinates(no heat generation):d 2T 1 dT 02[1‐5]drr drHolman / Heat Transfer, 10th Edition

Chapter 1. IntroductionSteady‐state one‐dimensional heat flow with heat sources:d 2T q 02[1‐6]dxkTwo‐dimensional steady‐state conduction without heat sources: 2T 2T 2 02 x y[1‐7]Holman / Heat Transfer, 10th Edition11

Chapter 1. Introduction1‐2. Thermal ConductivityFigure 1‐4Thermal conductivities of some typical gases .Holman / Heat Transfer, 10th Edition12

Chapter 1. IntroductionFigure 1‐5Thermal conductivities of some typical liquids.Holman / Heat Transfer, 10th Edition13

Chapter 1. IntroductionFigure 1‐6Thermal conductivities of some typical solids.Holman / Heat Transfer, 10th Edition14

Chapter 1. Introduction151‐3. Convection Heat TransferTo express the overall effect of convection, we use Newton’s lawof cooling:q hA (Tw T ) [1‐8]Figure 1‐7Convection heat transfer from a plate.Holman / Heat Transfer, 10th Edition

Chapter 1. IntroductionConvection Energy Balance on a Flow ChannelFigure 1-8 Convection in a channel.q m(ie ii )q mc p (Te Ti )q mc p (Te Ti ) hA(Tw, avg Tfluid , avg )[1 8a]m ρ umean AcHolman / Heat Transfer, 10th Edition16

Chapter 1. Introduction1-4. Radiation Heat Transferqemitted σ AT 4qnetexchangeA[1‐9] σ (T14 T24 )[1‐10]q Fε FGσ A (T14 T24 ) [1‐11]Radiation in an Enclosureq ε1σ A1 (T14 T24 )[1‐12]Holman / Heat Transfer, 10th Edition17

Chapter 1. Introduction1‐5. Dimensions and UnitsL lengthM massF forceτ timeT temperatureFor example, Newton’s second law of motion may be writtenForce time rate of change of momentumd (mv)F kdτF kmaF [1‐13]1magc[1‐14]Holman / Heat Transfer, 10th Edition18

Chapter 1. IntroductionSome typical systems of units are .1. 1‐pound force will accelerate a 1‐lb mass 32.17 ft/s 2 .2. 1‐pound force will accelerate a 1‐slug mass 1 ft/s 2 .3 1‐dyne force will accelerate a 1‐g mass 1 cm/s 2 .4. 1‐newton force will accelerate a 1‐kg mass 1 m/s 2 .5. 1‐kilogram force will accelerate a 1‐kg mass 9.806 m/s 2 .1.g c 32.17lb m ft/lb f s 22.g c 1slug ft/lb f s 223. g c 1g cm/dyn s4.g c 1kg m/N s 25.g c 9.806kg m m/kg f s 2Holman / Heat Transfer, 10th Edition19

Chapter 1. Introduction1.2.3.4.5.20lb f ftlb f ftdyn cm 1ergN m 1joule(J)kg f m 9.806JIn addition, we may use the units of energy that are based onthermal phenomena:1 Btu will raise 1 lb m of water 1 F at 68 F .1 cal will raise 1 g of water 1 C at 20 C .1 kcal will raise 1 kg of water 1 C at 20 C .Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction21Some conversion factors for the various units of work and energyare1Btu 778.16lb f ft1Btu 1055J1kcal 4182J1lb f ft 1.356J1Btu 252calW gmgc[1‐15]kinW/m ChinW/m 2 C1N 1kg m/s 2[1‐16]Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction22Example 1‐1 Conduction Through Copper PlateOne face of a copper plate 3 cm thick is maintained at 400 C ,and the other face is maintained at 100 C . How much heat istransferred through the plate?SolutionFrom Appendix A, the thermal conductivity for copper is 370W/m C at 250 C . From Fourier’s lawqdT kAdxIntegrating givesqΔT (370)(100 400) k 3.7MW/m 2 [1.173 106 Btu/h ft 2 ] 2ΔxA3 10Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction23Example 1‐2 Convection CalculationAir at 20 C blows over a hot plate 50 by 75 cm maintained at2250 C . The convection heat‐transfer coefficient is 25 W/m C .Calculate the heat transfer.SolutionFrom Newton’s law of coolingq hA (Tw T ) (25)(0.50)(0.75)(250 20) 2.156kW[7356Btu/h]Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction24Example 1‐3 Multimode Heat TransferAssuming that the plate in Example 1‐2 is made of carbon steel(1%) 2 cm thick and that 300 W is lost from the plate surfaceby radiation, calculate the inside plate temperature.SolutionThe heat conducted through the plate must be equal to the sum ofconvection and radiation heat losses:qcond qconv qrad kAΔT 2.156 0.3 2.456kWΔx( 2456)(0.02)ΔT 3.05 C[ 5.49 F](0.5)(0.75)(43)where the value of k is taken from Table 1‐1. The inside platetemperature is thereforeTi 250 3.05 253.05 CHolman / Heat Transfer, 10th Edition

Chapter 1. Introduction25Example 1‐4 Heat Source and ConvectionAn electric current is passed through a wire 1 mm in diameterand 10 cm long. The wire is submerged in liquid water atatmospheric pressure, and the current is increased until the waterboils. For this situation h 5000W/m C , and the water temperaturewill be 100 C . How much electric power must be supplied to thewire to maintain the wire surface at 114 C ?Solution2The total convection loss is given by Equation (1‐8): q hA (Tw T )For this problem the surface area of the wire isA π dL π (1 10 3 )(10 10 2 ) 3.142 10 4 m 2The heat transfer is thereforeq (5000W/m 2 C)(3.142 10 4 m 2 )(114 100) 21.99W[75.03Btu/h] and thisis equal to the electric power that must be applied.Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction26Example 1‐5 Radiation Heat TransferTwo infinite black plates at 800 C and 300 C exchange heat byradiation. Calculate the heat transfer per unit area.SolutionEquation (1‐10) may be employed for this problem, so we findimmediatelyq/ A σ (T14 T24 ) (5.669 10 8 )(10734 5734 ) 69.03kW/m 2 [21, 884Btu/h ft 2 ]Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction27Example 1‐6 Total Heat Loss by Convection and RadiationA horizontal steel pipe having a diameter of 5 cm is maintainedat a temperature of 50 C in a large room where the air and walltemperature are at 20 C . The surface emissivity of the steel maybe taken as 0.8. Using the data of Table 1‐3, calculate the totalheat lost by the pipe per unit length.SolutionThe total heat loss is the sum of convection and radiation. FromTable 1‐3 we see that an estimate for the heat‐transfer coefficient2for free convection with this geometry and air is h 6.5W/m C .The surface area is π dL , so the convection loss per unit length isq/L]conv h(π d )(Tw T ) (6.5)(π )(0.05)(50 20) 30.63W/mThe pipe is a body surrounded by a large enclosure so theradiation heat transfer can be calculated from Equation (1‐12).Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction28With T1 50 C 323 K and T2 20 C 293 K , we haveq/L]rad ε1 (π d1 )σ (T14 T24 ) (0.8)(π )(0.05)(5.669 10 8 )(3234 2934 ) 25.04W/mThe total heat loss is thereforeq/L]tot q/L]conv q/L]rad 30.63 25.04 55.67W/mIn this example we see that the convection and radiation are aboutthe same. To neglect either would be a serious mistake.Holman / Heat Transfer, 10th Edition

Chapter 1. Introduction1‐6. SummarydT kA hA (Tw T ) Fε FGσ A (Tw4 Ts4 )dy wallTs temperature of surroundingsTw surface temperatureT fluid temperatureHolman / Heat Transfer, 10th Edition29

Chapter 1. IntroductionFigure 1‐9 Combination of conduction, convection, and radiationheat transfer.Holman / Heat Transfer, 10th Edition30

Holman / Heat Transfer, 10th Edition Heat transfer is thermal energy transfer that is induced by a temperature difference (or gradient) Modes of heat transfer Conduction heat transfer: Occurs when a temperature gradient exists through a solid or a stationary fluid (liquid or gas). Convection heat transfer: Occurs within a moving fluid, or