Thermodynamics, Heat Transfer, And Fluid Flow. V.2. Heat Transfer (776131), страница 4

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The thickness is 1 in. and the thermal conductivityis 0.12 Btu/hr-ft-°F. Compute the temperature difference across the material.Figure 1Rev. 0Conduction Through a SlabPage 7HT-02CONDUCTION HEAT TRANSFERHeat TransferSolution:Using Equation 2-4:Q̇ ∆T k A  ∆x Solving for ∆T:∆T ∆x Q̇ k ABtu   1 ft1000 hr   12 Btu 20.12 1 fthr ft °F ∆T694° FExample:A concrete floor with a conductivity of 0.8 Btu/hr-ft-°F measures 30 ft by 40 ft with athickness of 4 inches. The floor has a surface temperature of 70°F and the temperaturebeneath it is 60°F.

What is the heat flux and the heat transfer rate through the floor?Solution:Using Equations 2-1 and 2-4:Q̇Q̇A ∆T k  ∆x Btu   10° F 0.8 hr ft ° F   0.333 ft 24HT-02Btuhr ft 2Page 8Rev. 0Heat TransferCONDUCTION HEAT TRANSFERUsing Equation 2-3:Q̇ ∆T k A  ∆x Q̇ A24 Btu  (1200 ft 2)hr ft 2 28,800BtuhrEquivalent Resistance MethodIt is possible to compare heat transfer to current flow in electrical circuits. The heat transfer ratemay be considered as a current flow and the combination of thermal conductivity, thickness ofmaterial, and area as a resistance to this flow.

The temperature difference is the potential ordriving function for the heat flow, resulting in the Fourier equation being written in a formsimilar to Ohm’s Law of Electrical Circuit Theory. If the thermal resistance term ∆x/k is writtenas a resistance term where the resistance is the reciprocal of the thermal conductivity divided bythe thickness of the material, the result is the conduction equation being analogous to electricalsystems or networks.

The electrical analogy may be used to solve complex problems involvingboth series and parallel thermal resistances. The student is referred to Figure 2, showing theequivalent resistance circuit. A typical conduction problem in its analogous electrical form isgiven in the following example, where the "electrical" Fourier equation may be written asfollows.∆TRthQ̇=(2-6)Q̇= Heat Flux ( Q̇ /A) (Btu/hr-ft2)∆T= Temperature Difference (oF)Rth= Thermal Resistance (∆x/k) (hr-ft2-oF/Btu)where:Rev. 0Page 9HT-02CONDUCTION HEAT TRANSFERHeat TransferFigure 2Equivalent ResistanceElectrical AnalogyExample:A composite protective wall is formed of a 1 in.

copper plate, a 1/8 in. layer of asbestos,and a 2 in. layer of fiberglass. The thermal conductivities of the materials in units ofBtu/hr-ft-oF are as follows: kCu = 240, kasb = 0.048, and kfib = 0.022. The overalltemperature difference across the wall is 500°F. Calculate the thermal resistance of eachlayer of the wall and the heat transfer rate per unit area (heat flux) through the compositestructure.Solution:RCu∆xCuRasbkCu 1 ft 1 in  12 in Btu240hr ft °F0.000347HT-02hr ft 2 °FBtu∆xasbRfibkasb 1 ft 0.125 in  12 in Btu0.048hr ft °F0.2170hr ft 2 °FBtuPage 10∆xfibkfib 1 ft 2 in  12 in Btu0.022hr ft °F7.5758hr ft 2 °FBtuRev. 0Heat TransferQ̇ACONDUCTION HEAT TRANSFER(Ti(RCuTo )RasbRfib)500°F(0.00034764.20.21707.5758)hr ft 2 °FBtuBtuhr ft 2Conduction-Cylindrical CoordinatesHeat transfer across a rectangular solid is the most direct application of Fourier’s law.

Heattransfer across a pipe or heat exchanger tube wall is more complicated to evaluate. Across acylindrical wall, the heat transfer surface area is continually increasing or decreasing. Figure 3is a cross-sectional view of a pipe constructed of a homogeneous material.Figure 3Rev. 0Cross-sectional Surface Area of a Cylindrical PipePage 11HT-02CONDUCTION HEAT TRANSFERHeat TransferThe surface area (A) for transferring heat through the pipe (neglecting the pipe ends) is directlyproportional to the radius (r) of the pipe and the length (L) of the pipe.A = 2πrLAs the radius increases from the inner wall to the outer wall, the heat transfer area increases.The development of an equation evaluating heat transfer through an object with cylindricalgeometry begins with Fourier’s law Equation 2-5.Q̇ ∆T k A  ∆r From the discussion above, it is seen that no simple expression for area is accurate. Neither thearea of the inner surface nor the area of the outer surface alone can be used in the equation.

Fora problem involving cylindrical geometry, it is necessary to define a log mean cross-sectionalarea (Alm).AlmAouterAinner(2-7)A ln  outer  Ainner Substituting the expression 2πrL for area in Equation 2-7 allows the log mean area to becalculated from the inner and outer radius without first calculating the inner and outer area.Alm2 π router L2 π rinner L2 π rL outerln  2 π rinner L rrinner outer2 π L  ln router rinner This expression for log mean area can be inserted into Equation 2-5, allowing us to calculate theheat transfer rate for cylindrical geometries.HT-02Page 12Rev. 0Heat TransferQ̇CONDUCTION HEAT TRANSFER ∆T k Alm  ∆r rri k 2 π L  o ln ro ri Q̇T o r oTi ri 2 π k L (∆T)ln (ro / ri)(2-8)where:L= length of pipe (ft)ri= inside pipe radius (ft)ro= outside pipe radius (ft)Example:A stainless steel pipe with a length of 35 ft has an inner diameter of 0.92 ft and an outerdiameter of 1.08 ft.

The temperature of the inner surface of the pipe is 122oF and thetemperature of the outer surface is 118oF. The thermal conductivity of the stainless steelis 108 Btu/hr-ft-oF.Calculate the heat transfer rate through the pipe.Calculate the heat flux at the outer surface of the pipe.Solution:Q̇2 π k L (ThTc)ln (ro/ri)Btu 6.28 108 (35 ft) (122°Fhr ft °F 0.54 ftln0.46 ft5.92 x 105Rev.

0118°F)BtuhrPage 13HT-02CONDUCTION HEAT TRANSFERQ̇Heat TransferQ̇AQ̇2 π ro LBtuhr2 (3.14) (0.54 ft) (35 ft)5.92 x 1054985Btuhr ft 2Example:A 10 ft length of pipe with an inner radius of 1 in and an outer radius of 1.25 in has anouter surface temperature of 250°F. The heat transfer rate is 30,000 Btu/hr. Find theinterior surface temperature. Assume k = 25 Btu/hr-ft-°F.Solution:Q̇2 π k L (ThTc )ln (ro / ri )Solving for Th:ThQ̇ ln (ro / ri)2 π k LTcBtu   1.25 in 30,000 lnhr  1 in Btu 2 (3.14) 25(10 ft)hr ft °F 250°F254°FThe evaluation of heat transfer through a cylindrical wall can be extended to include a compositebody composed of several concentric, cylindrical layers, as shown in Figure 4.HT-02Page 14Rev.

0Heat TransferCONDUCTION HEAT TRANSFERFigure 4Rev. 0Composite Cylindrical LayersPage 15HT-02CONDUCTION HEAT TRANSFERHeat TransferExample:A thick-walled nuclear coolant pipe (ks = 12.5 Btu/hr-ft-°F) with 10 in. inside diameter(ID) and 12 in. outside diameter (OD) is covered with a 3 in. layer of asbestos insulation(ka = 0.14 Btu/hr-ft-oF) as shown in Figure 5. If the inside wall temperature of the pipeis maintained at 550°F, calculate the heat loss per foot of length. The outside temperatureis 100°F.Figure 5HT-02Pipe Insulation ProblemPage 16Rev.

0Heat TransferCONDUCTION HEAT TRANSFERSolution:Q̇L2π (Tin r  ln  2  r   1 ksTo )r  ln  3   r2  ka 2π (5500F ln  6 in  5 in  12.5 Btuhr ft oF971100 oF) 9 in ln  6 in Btu0.14hr ft oFBtuhr ftSummaryThe important information in this chapter is summarized below.Conduction Heat Transfer Summary•Conduction heat transfer is the transfer of thermal energy by interactions betweenadjacent molecules of a material.•Fourier’s Law of Conduction can be used to solve for rectangular and cylindricalcoordinate problems.•Heat flux ( Q̇ ) is the heat transfer rate ( Q̇ ) divided by the area (A).•Heat conductance problems can be solved using equivalent resistance formulasanalogous to electrical circuit problems.Rev.

0Page 17HT-02CONVECTION HEAT TRANSFERHeat TransferCONVECTION HEAT TRANSFERHeat transfer by the motion and mixing of the molecules of a liquid or gas iscalled convection.EO 1.9Given the formula for heat transfer and the operatingconditions of the system, CALCULATE the rate of heattransfer by convection.ConvectionConvection involves the transfer of heat by the motion and mixing of "macroscopic" portions ofa fluid (that is, the flow of a fluid past a solid boundary).

The term natural convection is usedif this motion and mixing is caused by density variations resulting from temperature differenceswithin the fluid. The term forced convection is used if this motion and mixing is caused by anoutside force, such as a pump. The transfer of heat from a hot water radiator to a room is anexample of heat transfer by natural convection.

The transfer of heat from the surface of a heatexchanger to the bulk of a fluid being pumped through the heat exchanger is an example offorced convection.Heat transfer by convection is more difficult to analyze than heat transfer by conduction becauseno single property of the heat transfer medium, such as thermal conductivity, can be defined todescribe the mechanism. Heat transfer by convection varies from situation to situation (upon thefluid flow conditions), and it is frequently coupled with the mode of fluid flow.

In practice,analysis of heat transfer by convection is treated empirically (by direct observation).Convection heat transfer is treated empirically because of the factors that affect the stagnant filmthickness:Fluid velocityFluid viscosityHeat fluxSurface roughnessType of flow (single-phase/two-phase)Convection involves the transfer of heat between a surface at a given temperature (Ts) and fluidat a bulk temperature (Tb). The exact definition of the bulk temperature (Tb) varies dependingon the details of the situation. For flow adjacent to a hot or cold surface, Tb is the temperatureof the fluid "far" from the surface. For boiling or condensation, Tb is the saturation temperatureof the fluid. For flow in a pipe, Tb is the average temperature measured at a particular crosssection of the pipe.HT-02Page 18Rev. 0Heat TransferCONVECTION HEAT TRANSFERThe basic relationship for heat transfer by convection has the same form as that for heat transferby conduction:Q̇h A ∆TQ̇= rate of heat transfer (Btu/hr)h= convective heat transfer coefficient (Btu/hr-ft2-°F)A= surface area for heat transfer (ft2)(2-9)where:∆T = temperature difference (°F)The convective heat transfer coefficient (h) is dependent upon the physical properties of the fluidand the physical situation.

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