Chapter 3: Steady Heat Conduction. Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University
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1 Chapter 3: Steady Heat Conduction Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University
2 Objectives When you finish studying this chapter, you should be able to: Understand the concept of thermal resistance and its limitations, and develop thermal resistance networks for practical heat conduction problems, Solve steady conduction problems that involve multilayer rectangular, cylindrical, or spherical geometries, Develop an intuitive understanding of thermal contact resistance, and circumstances under which it may be significant, Identify applications in which insulation may actually increase heat transfer, Analyze finned surfaces, and assess how efficiently and effectively fins enhance heat transfer, and Solve multidimensional practical heat conduction problems using conduction shape factors.
3
4 Steady Heat Conduction in Plane Walls ) Considerable temperature difference between the inner and the outer surfaces of the wall (significant temperature gradient in the x direction). ) The wall surface is nearly isothermal. Steady one-dimensional modeling approach is justified.
5 Assuming heat transfer is the only energy interaction and there is no heat generation, the energy balance can be expressed as Zero for steady operation ate of heat transfer into the wall ate of heat transfer out of the wall - ate of change of the energy of the wall or in 0 dewall out 0 dt (3-) The rate of heat transfer through the wall must be constant ( Q ). cond wall, constant
6 Then Fourier s law of heat conduction for the wall can be expressed as Q dt, ka (W) dx (3-) cond wall emembering that the rate of conduction heat transfer and the wall area A are constant it follows dt/dxconstant the temperature through the wall varies linearly with x. Integrating the above equation and rearranging yields T T ka L cond, wall (W) (3-3)
7 Thermal esistance Concept- Conduction esistance Equation 3 3 for heat conduction through a plane wall can be rearranged as T T cond, wall (W) (3-4) wall Where wall is the conduction resistance expressed as wall L ka ( C/W) (3-5)
8 Analogy to Electrical Current Flow Eq. 3-5 is analogous to the relation for electric current flow I, expressed as I V V e (3-6) Heat Transfer Electrical current flow ate of heat transfer Electric current Thermal resistance Electrical resistance Temperature difference Voltage difference
9 Thermal esistance Concept- Convection esistance Thermal resistance can also be applied to convection processes. Newton s law of cooling for convection heat transfer rate ( ( ) can be rearranged as conv Qconv has Ts T T s T conv (W) (3-7) conv is the convection resistance conv ha s ( C/W) (3-8)
10 Thermal esistance Concept- adiation esistance The rate of radiation heat transfer between a surface and the surrounding ( 4 4 T ) s Tsurr rad εσ As Ts Tsurr hrad As ( Ts Tsurr ) (W) rad h A rad s ( K/W) (3-0) ( )( ) rad rad s surr s surr As ( Ts Tsurr ) rad (3-9) h εσ T + T T + T (W/m K) (3-)
11 Thermal esistance Concept- adiation and Convection esistance A surface exposed to the surrounding might involves convection and radiation simultaneously. The convection and radiation resistances are parallel to each other. When T surr T, the radiation effect can properly be accounted for by replacing h in the convection resistance relation by h combined h conv +h rad (W/m K) (3-)
12 Thermal esistance Network consider steady one-dimensional heat transfer through a plane wall that is exposed to convection on both sides. Under steady conditions we have or ate of heat convection into the wall h A T T T ( ), T L ate of heat conduction through the wall ( ) ka ha T T, (3-3) ate of heat convection from the wall
13 earranging and adding T T conv + T T wall T T conv T where,,,, T Q ( conv, + wall + conv,) Q total,, T T,, total (W) L total conv, wall conv, (C/W) ha ka ha (3-5) (3-6)
14 It is sometimes convenient to express heat transfer through a medium in an analogous manner to Newton s law of cooling as (3-8) Q UAΔT (W) where U is the overall heat transfer coefficient. Note that UA total ( C/K) (3-9)
15 Multilayer Plane Walls In practice we often encounter plane walls that consist of several layers of different materials. The rate of steady heat transfer through this two-layer composite wall can be expressed through Eq. 3-5 where the total thermal resistance is total conv, wall, wall, conv, L L ha ka ka ha (3-) + + +
16 Consider a.-m-high and -m-wide double-pane window consisting of two 3-mm-thick layers of glass (k 0.78 W/m C) separated by a -mm-wide stagnant air space (k 0.06 W/m C). Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface for a day during which the room is maintained at 4 C while the temperature of the outdoors is -5 C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h 0 W/m. C and h 5 W/m C, and disregard any heat transfer by radiation.
17 Assumptions Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivities of the glass and air are constant. 4 Heat transfer by radiation is negligible. Analysis A i o total 3 air (. m) ( m) 4. m, C/W h A (0 W/m. C)(.4 m ) L m glass C/W ka (0.78 W/m. C)(.4 m ) L 0.0 m 0.93 C/W ka (0.06 W/m. C)(.4 m ) T o, C/W o h A (5 W/m. C)(.4 m ) (0.006) conv conv conv, C/W conv, i Air 3 o T
18 The steady rate of heat transfer through window glass then becomes Q T T total [4 ( 5)] C C/W 4 W The inner surface temperature: T T T T conv, Q conv, 4 o C (4 W)(0.047 C/W) 9. C
19 EX.: The wall of a refrigerator is constructed of fiberglass insulation (k W/m C) sandwiched between two layers of -mm-thick sheet metal (k 5. W/m C). The refrigerated space is maintained at 3 C, and the average heat transfer coefficients at the inner and outer surfaces of the wall are 4 W/m C and 9 W/m C, respectively. The kitchen temperature on the outer surfaces of the refrigerator when the temperature of the outer surface drops to 0 C. Determine the minimum thickness of fiberglass insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces.
20 Q ho A( Troom T s, out ) ( 9 W/m. C) ( m )( 5 0) C45 W insulation / A T h room o T total + L k refrig T room metal T + refrig L k insulation + h i mm L mm i ins 3 o T room T refrig 45 W/m 9 ( 5 3) C 000. m L W/m. C 5. W/m. C W/m. C 4 W/m. C L m 0.45 cm
21 Thermal Contact esistance In reality surfaces have some roughness. When two surfaces are pressed against each other, the peaks form good material contact but the valleys form voids filled with air. As a result, an interface contains numerous air gaps of varying sizes that act as insulation because of the low thermal conductivity of air. Thus, an interface offers some resistance to heat transfer, which is termed the thermal contact resistance, c.
22 The value of thermal contact resistance depends on the surface roughness, material properties, temperature and pressure at the interface, type of fluid trapped at the interface. Thermal contact resistance is observed to decrease with decreasing surface roughness and increasing interface pressure. The thermal contact resistance can be minimized by applying a thermally conducting liquid called a thermal grease.
23 Q h A Δ T c int erface h c : thermal contact conductance (W/m.K), similar to convection heat transfer coefficient. See Table 3- & 3- Q/A h c (W/m. C) (3 7) ΔT int erface ΔT int erface c (m. C/W) (3 8) hc Q/A c : thermal contact resistance per unit area O: Δ Q T int erface int erface Where: int erface ha c : thermal contact resistance interface c A
24 EXAMPLE:Two 5-cm-diameter, 5 cm-long aluminum bars (k 76 W/m C) with ground surfaces are pressed against each other with a pressure of 0 atm. The bars are enclosed in an insulation sleeve and, thus, heat transfer from the lateral surfaces is negligible. If the top and bottom surfaces of the twobar system are maintained at temperatures of 50 C and 0 C, respectively, determine (a) the rate of heat transfer along the cylinders under steady conditions and (b) the temperature drop at the interface. Interface Bar Bar i interf o T T
25 The contact conductance at the interface of aluminum-aluminum plates for the case of ground surfaces and of 0 atm MPa pressure is: hc,400 W/m C (Table 3-). Analysis (a) C/W ha (, 400 W/m. C)[ (0.05 m) /4] interface c c π L 05. m ka (76 W / m. C)[ π(0.05 m) / 4] plate C/W T T (50 0) C Δ Δ 4.4 W + ( ) C/W total interface bar (b) Δ T Q (4.4 W)( C/W) 6.4 C interface interface
26 Generalized Thermal esistance Network The thermal resistance concept can be used to solve steady heat transfer problems that involve parallel layers or combined series-parallel arrangements. The total heat transfer of two parallel layers T T T T + + ( T T ) + total (3-9) + total total + (3-3)
27 Combined Series-Parallel Arrangement The total rate of heat transfer through the composite system T T total (3-3) where total 3 conv 3 conv + L L L 3 ; ; 3 ; conv ka ka ka 3 3 ha3 (3-33) (3-34)
28 EX.: Consider a 5-m-high, 8-m-long, and 0.-m-thick wall whose representative cross section is as given in the figure. The thermal conductivities of various materials used, in W/m C, are k A k F, k B 8, k C 0, k D 5, and k E 35. The left and right surfaces of the wall are maintained at uniform temperatures of 300 C and 00 C, respectively. Assuming heat transfer through the wall to be onedimensional, determine (a) the rate of heat transfer through the wall; (b) the temperature at the point where the sections B, D, and E meet; and (c) the temperature drop across the section F. Disregard any contact resistances at the interfaces.
29 Analysis (a) The representative surface area is: A m T T 3 A 4 B L ka C L ka A B 0.0m 0.04 C/W ( W/m. C)(0. m ) L 0.05 m 0.06 C/W ka C (0 W/m. C)(0.04 m ) 0.05 m 0.6 C/W (8 W/m. C)(0.04 m ) D E F L ka L ka L ka D E F (5 W/m. 0.m o C)(0.06 m 0.m (35 W/m. C)(0.06 m 0.06 m ( W/m. C)(0. m 0. C/W ) 0.05 ) o C/W 0.5 C/W )
30 mid, mid, total 5 T T total 6 mid, mid, (300 00) C C/W mid, mid, C/W 0.05 C/W C/W 57 W (for a 0. m m section) Then steady rate of heat transfer through entire wall becomes Q total (57 W) (5 m)(8 m) 0. m W (b) The total thermal resistance between left surface and the point where the sections B, D, and E meet is total mid +, C/W T T T T Q total 300 C (57 W)(0.065 C/W) total 63 C
31 (c) The temperature drop across the section F can be determined from ΔT ΔT Q F (57 W)(0.5 C/W) F 43 C
32 Heat Conduction in Cylinders Consider the long cylindrical layer Assumptions: the two surfaces of the cylindrical layer are maintained at constant temperatures T and T, no heat generation, constant thermal conductivity, one-dimensional heat conduction. Fourier s law of heat conduction Q cond cyl dt, ka (W) dr (3-35)
33 dt cond, cyl ka (W) (3-35) dr Separating the variables and integrating from rr, where T(r )T, to rr, where T(r )T r T cond, cyl dr kdt (3-36) A r r T T Substituting A πrl and performing the integrations give T T cond, cyl π Lk (3-37) ln r / r ( ) Since the heat transfer rate is constant T T Qcond, cyl (3-38) cyl
34 Sphere: (3-40) (3-4)
35 Thermal esistance with Convection Steady one-dimensional heat transfer through a cylindrical or spherical layer that is exposed to convection on both sides T T,, total ( ) (3-3) + + total conv, cyl conv, ln ( r / r ) πrl h πlk πrl h + + ( ) (3-43) (3-44) Sphere
36 Multilayered Cylinders Steady heat transfer through multilayered cylindrical or spherical shells can be handled just like multilayered plane. The steady heat transfer rate through a three-layered composite cylinder of length L with convection on both sides is expressed by Eq. 3-3 where: total conv, cyl, cyl,3 cyl,3 conv, ln ( r / r ) ln ( r / r ) ln ( r / r ) πrl h πlk πlk πlk πrl h ( ) (3-46) ( ) 3 4
37 EX.: Steam at 30 C flows in a stainless steel pipe (k5 W/m C) whose inner and outer diameters are 5 cm and 5.5 cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation (k W/m C). Heat is lost to the surroundings at 5 C by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of 5 W/m C. Taking the heat transfer coefficient inside the pipe to be 80 W/m C, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.
38 i o T T A A i o πd L π( m) ( m) 057. m i πd L π( m) ( m) m o i o total h A i o i i pipe o insulation h A + (80 W/m (5 W/m + + ln( r3 / r πk L. C)(0.57 m ) ln(5.75 /.75) C/W π (0.038 W/m. C)(m). C)(0.36m 0.08 C/W ) ln( r / r ) ln(.75 /.5) C/W πk L π (5 W/m. C)(m) o C/W ) C/W
39 ( ) Q T T 30 5 C C / W 93.9 W total The temperature drops across the pipe and the insulation are ΔT ΔT pipe insulation Q ( W)(0.000 C / W) C pipe Q ( 939. W)(3.089 C / W) 90 C insulation
40 Critical adius of Insulation Adding more insulation to a wall or to the attic always decreases heat transfer. Adding insulation to a cylindrical pipe or a spherical shell, however, is a different matter. Adding insulation increases the conduction resistance of the insulation layer but decreases the convection resistance of the surface because of the increase in the outer surface area for convection. The heat transfer from the pipe may increase or decrease, depending on which effect dominates.
41 A cylindrical pipe of outer radius r whose outer surface temperature T is maintained constant. The pipe is covered with an insulator (k and r ). Convection heat transfer at T and h. The rate of heat transfer from the insulated pipe to the surrounding air can be expressed as T T T T + r r ins conv ln ( ) / + πlk h( πr L) (3-37)
42 The variation of the heat transfer rate with the outer radius of the insulation r is shown in the figure. The value of r at which Q reaches a maximum is determined by d dr 0 Performing the differentiation and solving for r yields k rcr, cylinder (m) (3-50) h Thus, insulating the pipe may actually increase the rate of heat transfer instead of decreasing it.
43 Sphere: EX.: A -mm-diameter and 0-m-long electric wire is tightly wrapped with a -mm-thick plastic cover whose thermal conductivity is k 0.5 W/m C. Electrical measurements indicate that a current of 0 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a medium at T h 30 C with a heat transfer coefficient of h 4 W/m C, determine the temp. at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.
44 W VI conv plastic total e h A ln( r o o / r πkl conv + ( 8 V)(0 A) (4 W/m ln( /) 80 W π(0.5 W/m. C)(0 m) plastic T. C)[ π(0.004 m)(0 m)] C/W C/W plastic C/W conv T Then the interface temperature becomes T T T T total + Q total The critical radius of plastic insulation is r cr h k 0.5 W/m. C 4 W/m. C 30 C + (80 W)(0.405 C/W) m 6.5 mm 6.4 C Doubling the thickness of the plastic cover will increase the outer radius of the wire to mm, which is less than the critical radius of insulation. Therefore, doubling the thickness of plastic cover will increase the rate of heat loss and decrease the interface temperature.
45 Heat Transfer from Finned Surfaces Newton s law of cooling ( ) Q ha T T conv s s Two ways to increase the rate of heat transfer: increasing the heat transfer coefficient, increase the surface area fins Fins are the topic of this section. Finned Surfaces: extended surfaces attached to the surface area by exposing a larger surface area to convection & radiation to enhance heat transfer
46 Fin Equation Under steady conditions, the energy balance on this volume element can be expressed as ate of heat conduction into the element at x ate of heat conduction from the element at x+δx + ate of heat convection from the element or cond, x cond, x+δx + conv where ( )( ) conv h pδx T T Substituting and dividing by Δx, we obtain cond, x+δx cond, x + hp ( T T ) 0 Δx (3-5)
47 Taking the limit as Δx 0 gives dq cond dx + hp T T ( ) 0 From Fourier s law of heat conduction we have cond Substitution of Eq into Eq gives d dx kac dx (3-53) dt (3-54) dt kac hp T T dx ( ) 0 (3-55)
48 For constant cross section and constant thermal conductivity Where Temp. Excess θ d θ m θ 0 (3-56) dx T T m ; Equation 3 56 is a linear, homogeneous, second-order differential equation with constant coefficients. The general solution of Eq is mx θ ( x) Ce + C e (3-58) mx hp ka C and C are constants whose values are to be determined from the boundary conditions at the base and at the tip of the fin. c
49 Boundary Conditions Several boundary conditions are typically employed: At the fin base Specified temperature boundary condition, expressed as: θ(0) θ b T b -T θ b At the fin tip. Specified temperature. Infinitely Long Fin 3. Adiabatic tip 4. Convection (and combined convection and radiation). :Temp. Excess at Fin Base
50 A wl+ wt fin wl Ac wt Ab wt p t+ w w π Afin D +πdl 4 πdl π Ac Ab D 4 p πd
51 Infinitely Long Fin (T fin tip T h ) For a sufficiently long fin the temperature at the fin tip approaches the ambient temperature Boundary condition: θ(l )T(L)-T 0 When x so does e mx C x0: e mx C θ b The temperature distribution: T( x) T mx x hp / kac e e T T b heat transfer from the entire fin dt ka hpka T T c c b dx x 0 ( ) T (3-60) (3-6)
52 Adiabatic Tip Boundary condition at fin tip: dθ dx x L 0 (3-63) After some manipulations, the temperature distribution: T( x) T cosh m L x T T cosh ml b ( ) heat transfer from the entire fin dt ka hpka T T tanh ml c c b dx x 0 ( ) (3-64) (3-65)
53 Convection (or Combined Convection and adiation) from Fin Tip A practical way of accounting for the heat loss from the fin tip is to replace the fin length L in the relation for the insulated tip case by a corrected length defined as L c L+A c /p (3-66) For rectangular and cylindrical fins L c is L c,rectangular L+t/ L c,cylindrical L+D/4
54 Fin Efficiency To maximize the heat transfer from a fin the temperature of the fin should be uniform (maximized) at the base value of T b In reality, the temperature drops along the fin, and thus the heat transfer from the fin is less To account for the effect we define a fin efficiency Q fin η fin fin,max Actual heat transfer rate from the fin Ideal heat transfer rate from the fin if the entire fin were at base temperature or fin ηfin fin,max ηfinhafin( Tb T ) (3-69)
55 Fin Efficiency For constant cross section of very long fins: η long, fin fin,max fin b ( ) ( ) fin hpkac Tb T kac ha T T L hp ml (3-70) For constant cross section with adiabatic tip: η adiabatic, fin fin hpkac Tb T tanh al ha T T fin,max fin b ( ) ( ) tanh ml ml (3-7)
56
57
58 Fin Effectiveness The performance of the fins is judged on the basis of the enhancement in heat transfer relative to the no-fin case. The performance of fins is expressed ε in terms of the fin effectiveness ε fin defined as fin ε fin ε ε fin fin ha T T fin fin no fin b b < > Does not affect Insulation Enhancing ( ) (3-7) Heat transfer rate from the fin of base area A b Heat transfer rate from the surface of area A b
59 emarks regarding fin effectiveness The thermal conductivity k of the fin material should be as high as possible. It is no coincidence that fins are made from metals. The ratio of the perimeter to the cross-sectional area of the fin p/a c should be as high as possible. The use of fins is most effective in applications involving a low convection heat transfer coefficient. The use of fins is more easily justified when the medium is a gas instead of a liquid and the heat transfer is by natural convection instead of by forced convection.
60 Overall Effectiveness An overall effectiveness for a finned surface is defined as the ratio of the total heat transfer from the finned surface to the heat transfer from the same surface if there were no fins. fin ε fin, overall no fin ( ) unfin + η fin Afin h A ha no fin (3-76)
61 Proper Length of a Fin An important step in the design of a fin is the determination of the appropriate length of the fin once the fin material and the fin cross section are specified. The temperature drops along the fin exponentially and asymptotically approaches the ambient temperature at some length.
62 EX.: Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 80 o C. Circular aluminum alloy 04-T6 fins (k 86 W/m C) of outer diameter 6 cm and constant thickness mm are attached to the tube. The space between the fins is 3 mm, and thus there are 50 fins per meter length of the tube. Heat is transferred to the surrounding air at T h 5 o C, with a heat transfer coefficient of 40 W/m oc. Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.
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64
65 Heat Transfer in Common Configurations Many problems encountered in practice are two- or three-dimensional and involve rather complicated geometries for which no simple solutions are available. An important class of heat transfer problems for which simple solutions are obtained encompasses those involving two surfaces maintained at constant temperatures T and T. The steady rate of heat transfer between these two surfaces is expressed as QSk(T - T ) (3-79) S is the conduction shape factor, which has the dimension of length.
66 Table 3-7
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