Chapter 3: Steady Heat Conduction Yoav Peles Department of Mechanical, Aerospace and Nuclear Engineering Rensselaer Polytechnic Institute Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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. 3.1 Steady Heat Conduction in Plane Walls 1) Considerable temperature difference between the inner and the outer surfaces of the wall (significant temperature gradient in the x direction). 2) The wall surface is nearly isothermal. Steady one-dimensional modeling approach is justified. • 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 Rate of heat transfer into the wall or - Rate of heat transfer out of the wall = Rate of change of the energy of the wall =0 P (3-1) dE Q in − Q out = wall = 0 dt The rate of heat transfer through the wall must be constant ( Q cond ,wall = constant ). For the control volume (CV): [入] - [出] + [生成量] = [增加量] 無 穩態 • Then Fourier’s law of heat conduction for the wall can be expressed as dT Qcond , wall = −kA dx (W) (3-2) • Remembering that the rate of conduction heat transfer and the wall area A are constant it follows dT/dx=constant the temperature through the wall varies linearly with x. • Integrating the above equation and rearranging yields T1 − T2 Qcond , wall = kA L (W) (3-3) Thermal Resistance ConceptConduction Resistance • Equation 3–3 for heat conduction through a plane wall can be rearranged as T1 − T2 Qcond , wall = Rwall (W) (3-4) • Where Rwall is the conduction resistance expressed as Rwall L = kA ( D C/W) (3-5) Analogy (類比) to Electrical Current Flow • Eq. 3-5 is analogous to the relation for electric current flow I, expressed as V1 − V2 I= Re (3-6) Heat Transfer Electrical current flow Rate of heat transfer ÅÆ Electric current Thermal resistance ÅÆ Electrical resistance Temperature difference ÅÆ Voltage difference Thermal Resistance ConceptConvection Resistance • Thermal resistance can also be applied to convection processes. • Newton’s law of cooling for convection heat transfer rate ( Q conv = hAs (Ts − T∞ )) can be rearranged as Ts − T∞ Qconv = Rconv (W) (3-7) • Rconv is the convection resistance Rconv 1 = hAs (D C/W) (3-8) Thermal Resistance ConceptRadiation Resistance • The rate of radiation heat transfer between a surface and the surrounding Ts − Tsurr 4 4 Qrad = εσ As Ts − Tsurr = hrad As (Ts − Tsurr ) = (W) Rrad ( Rrad = hrad 1 hrad As ) (3-9) (K/W) (3-10) Q rad 2 2 2 = = εσ Ts + Tsurr (Ts + Tsurr ) (W/m ⋅ K) As (Ts − Tsurr ) ( ) (3-11) Thermal Resistance Concept - Radiation and Convection Resistance • A surface exposed to the surrounding might involves convection and radiation simultaneously. • The convection and radiation resistances are parallel to each other. • When Tsurr≈T∞, the radiation effect can properly be accounted for by replacing h in the convection resistance relation by hcombined = hconv+hrad (W/m2K) (3-12) Thermal Resistance Network • consider steady one-dimensional heat transfer through a plane wall that is exposed to convection on both sides. Rate of heat convection into the wall Q = h1 A (T∞ ,1 − T1 ) T1 − T2 = kA L = h2 A (T2 − T∞ ,2 ) (3-13) = Rate of heat conduction through the wall = Rate of heat convection from the wall Rearranging and adding ⎧T∞ ,1 − T1 = Q ⋅ Rconv ,1 ⎪ + ⎨T1 − T2 = Q ⋅ Rwall ⎪ ⋅R − = T T Q 2 ∞ ,2 conv ,2 ⎩ T∞ ,1 − T∞ ,2 = Q ( Rconv ,1 + Rwall + Rconv ,2 ) = Q ⋅ Rtotal where T∞,1 − T∞,2 Q= Rtotal Rtotal = Rconv ,1 + Rwall + Rconv ,2 (W) (3-15) 1 L 1 D = + + ( C/W) h1 A kA h2 A (3-16) • It is sometimes convenient to express heat transfer through a medium in an analogous manner to Newton’s law of cooling as Q = UA∆T (W) (3-18) • where U is the overall heat transfer coefficient. • Note that 1 UA = Rtotal D ( C/K) (3-19) Multilayer Plane Walls • The rate of steady heat transfer through a two-layer composite wall can be expressed through Eq. 3-15 where the total thermal resistance is Rtotal = Rconv ,1 + Rwall ,1 + Rwall ,2 + Rconv ,2 1 L1 L2 1 = + + + h1 A k1 A k 2 A h2 A (3-22) 3.2 Thermal Contact Resistance • 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, Rc. • 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. 3.3 Generalized Thermal Resistance 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 ⎛ 1 T1 − T2 T1 − T2 1 ⎞ + = (T1 − T2 ) ⎜ + ⎟ Q = Q1 + Q2 = R1 R2 ⎝ R1 R2 ⎠ 1 (3-29) Rtotal ⎛ 1 R1 R2 1 1 ⎞ = ⎜ + ⎟ → Rtotal = Rtotal ⎝ R1 R2 ⎠ R1 + R2 (3-31) Combined Series-Parallel Arrangement The total rate of heat transfer Through the composite system T1 − T∞ Q= Rtotal (3-32) where Rtotal = R12 + R3 + Rconv R1R2 = + R3 + Rconv (3-33) R1 + R2 L3 L1 L2 1 R1 = ; R2 = ; R3 = ; Rconv = k1 A1 k2 A2 k3 A3 hA3 (3-34) 3.4 Heat Conduction in Cylinders Consider the long cylindrical layer Assumptions: – the two surfaces of the cylindrical layer are maintained at constant temperatures T1 and T2, – no heat generation, – constant thermal conductivity, – one-dimensional heat conduction. Fourier’s law of heat conduction dT Qcond ,cyl = − kA dr (W) (3-35) dT (3-35) Qcond ,cyl = − kA (W) dr Separating the variables and integrating from r=r1, where T(r1)=T1, to r=r2, where T(r2)=T2 r2 ∫ Q cond ,cyl dr = − T2 ∫ kdT (3-36) A T =T1 Substituting A =2πrL and performing the integrations give T1 − T2 (3-37) Qcond ,cyl = 2π Lk ln ( r2 / r1 ) r = r1 Since the heat transfer rate is constant T1 − T2 Qcond ,cyl = Rcyl (3-38) Thermal Resistance with Convection Steady one-dimensional heat transfer through a cylindrical or spherical layer that is exposed to convection on both sides T∞ ,1 − T∞ ,2 Q= Rtotal (3-32) where Rtotal = Rconv ,1 + Rcyl + Rconv ,2 = ln ( r2 / r1 ) 1 1 = + + ( 2π r1L ) h1 2π Lk ( 2π r2 L ) h2 (3-43) 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. 3-32 where: Rtotal = Rconv ,1 + Rcyl ,1 + Rcyl ,3 + Rcyl ,3 + Rconv ,2 = ln ( r2 / r1 ) ln ( r3 / r2 ) ln ( r4 / r3 ) 1 1 = + + + + 2π Lk 2 2π Lk3 ( 2π r1L ) h1 2π Lk1 ( 2π r2 L ) h2 (3-46) 3.5 Critical Radius of Insulation (略) • 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. • A cylindrical pipe of outer radius r1 whose outer surface temperature T1 is maintained constant. • The pipe is covered with an insulator (k and r2). • Convection heat transfer at T∞ and h. • The rate of heat transfer from the insulated pipe to the surrounding air can be expressed as T1 − T∞ T1 − T∞ Q= = (3-37) Rins + Rconv ln ( r2 / r1 ) 1 + 2π Lk h ( 2π r2 L ) • The variation of the heat transfer rate with the outer radius of the insulation r2 is shown in the figure. • The value of r2 at which Q reaches a maximum is determined by dQ =0 dr2 • Performing the differentiation and solving for r2 yields rcr ,cylinder k = h (m) (3-50) • Thus, insulating the pipe may actually increase the rate of heat transfer instead of decreasing it. 3.6 Heat Transfer from Finned Surfaces • Newton’s law of cooling Q conv = hAs (Ts − T∞ ) • Two ways to increase the rate of heat transfer: – increasing the heat transfer coefficient, – increase the surface area Æfins • Fins are the topic of this section. Fin Equation Under steady conditions, the energy balance on this volume element can be expressed as Rate of heat Rate of heat conduction into = conduction from the the element at x element at x+∆x + Rate of heat convection from the element Q cond , x = Q cond , x +∆x + Q conv where Q conv = h ( p∆x )(T − T∞ ) Substituting and dividing by ∆x, we obtain Q cond , x +∆x − Q cond , x ∆x + hp (T − T∞ ) = 0 (3-52) Taking the limit as ∆x → 0 gives dQ cond + hp (T − T∞ ) = 0 dx (3-53) From Fourier’s law of heat conduction we have dT Qcond = − kAc dx (3-54) Substitution of Eq. 3-54 into Eq. 3–53 gives d ⎛ dT ⎞ ⎜ kAc ⎟ − hp (T − T∞ ) = 0 dx ⎝ dx ⎠ (3-55) For constant cross section and constant thermal conductivity Where d 2θ 2 −m θ =0 2 dx θ = T − T∞ (3-56) hp ; m= kAc • Equation 3–56 is a linear, homogeneous, second-order differential equation with constant coefficients. • The general solution of Eq. 3–56 is θ ( x) = C1e mx + C2 e − mx (3-58) • C1 and C2 are constants whose values are to be determined from the boundary conditions at the base and at the tip of the fin. Boundary Conditions Several boundary conditions are typically employed at the fin tip : • At the fin base – Specified temperature boundary condition, expressed as: θ(0)= θb= Tb-T∞ (a) Infinitely Long Fin (Tfin tip = T) • 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 emx→∞ C1=0 • @ x = 0: emx = 1 C2= θb • The temperature distribution: T ( x) − T∞ −x − mx =e =e Tb − T∞ hp / kAc (3-60) • heat transfer from the entire fin dT Q = − kAc dx = hpkAc (Tb − T∞ ) x =0 (3-61) (b) Adiabatic Tip • Boundary condition at fin tip: dθ dx =0 x=L (3-63) • After some manipulations, the temperature distribution: T ( x) − T∞ cosh m ( L − x ) = Tb − T∞ cosh mL (3-64) • heat transfer from the entire fin dT Q = − kAc dx = hpkAc (Tb − T∞ ) tanh mL x =0 (3-65) (c & d) Convection (or Combined Convection and Radiation) 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 Lc=L+Ac/p (3-66) • For rectangular and cylindrical fins Lc is • Lc,rectangular=L+t/2 • Lc,cylindrical =L+D/4 Fin Efficiency • To maximize the heat transfer from a fin the temperature of the fin should be uniform (maximized) at the base value of Tb • 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 Actual heat transfer rate from the fin η fin = = Q Ideal heat transfer rate from the fin fin ,max or if the entire fin were at base temperature Q fin = η finQ fin ,max = η fin hAfin (Tb − T∞ ) (3-69) Fin Efficiency • For constant cross section of very long fins: ηlong , fin Q fin = Q = fin ,max hpkAc (Tb − T∞ ) hAfin (Tb − T∞ ) 1 kAc 1 (3-70) = = L hp mL • For constant cross section with adiabatic tip: ηadiabatic , fin Q fin = Q fin ,max = hpkAc (Tb − T∞ ) tanh aL hAfin (Tb − T∞ ) tanh mL = mL (3-71) 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 in terms of the fin effectiveness εfin defined as Heat transfer rate ε fin Q fin Q fin = = = Qno fin hAb (Tb − T∞ ) (3-72) from the fin of base area Ab Heat transfer rate from the surface of area Ab Remarks 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/Ac 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. 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. Q fin ε fin ,overall = Q no fin = h ( Aunfin + η fin Afin ) hAno fin (3-76) 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. 3.7 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 T1 and T2. • The steady rate of heat transfer between these two surfaces is expressed as Q=Sk(T1=T2) (3-79) • S is the conduction shape factor, which has the dimension of length. Table 3-7