Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 69
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(6.19)–(6.22), the two-dimensional steady-stateboundary layer equations are:∂u ∂v+=0∂x∂y(6.23)∂u∂ 2u∂u1 ∂p∂ 2u+v=−+ ν 2 = U Ux + ν 2∂x∂yρ ∂x∂y∂y∂T∂T∂pq ∂T 2µ ∂u 2u+v+= α 2 + βT u+∂x∂y∂y∂xρcp ∂yρcpu(6.24)(6.25)The boundary conditions for an impermeable surface areu(x,0) = v(x,0) = 0(6.26a)T (x,0) = T0 (x)(6.26b)u(x, ∞) = U (x)(6.26c)T (x, ∞) = T∞ (x)(6.26d)Equations (6.23)–(6.25) constitute a set of nonlinear partial differential equations.Under certain conditions similarity solutions can be found that allow conversion ofthis set into ordinary differential equations. The concept of similarity means thatBOOKCOMP, Inc. — John Wiley & Sons / Page 452 / 2nd Proofs / Heat Transfer Handbook / Bejan[452], (14)453HEAT TRANSFER FROM SINGLE OBJECTS IN UNIFORM FLOW123456789101112131415161718192021222324252627282930313233343536373839404142434445certain features (e.g., velocity profiles) are geometrically similar.
Analytically, thisamounts to combining the x and y spatial dependence on a single independent variable η. The velocity components u(x,y) and v(x,y) are expressed by a single nondimensional stram function f (η), and the temperature T (x,y) into a nondimensionaltemperature φ(η).Similarity for boundary layer flow follows from the observation that while theboundary layer thickness at each downstream location x is different, a scaled normaldistance η can be employed as a universal length scale. Presence of natural lengthscales (such as a finite-length, plate, cylinder, or sphere) generally precludes thefinding of similarity solutions.
Using the scaled distance, the similarity procedurefinds the appropriate normalized stream and temperature functions that are also validat all locations. Following Gebhart (1980), the similarity variables are defined asη(x,y) = yb(x)(6.27)ψ(x,y)vc(x)(6.28)T (x,y) − T∞ (x)φ(η) =T0 (x) − T∞ (x)(6.29)f (η) =Lines: 566 to 615d(x) = T0 (x) − T∞ (x)j (x) = T∞ (x) − Tref[453], (15)need to be determined. The transformed governing equations in terms of the foregoingnormalized variables are1 dc(x)1dc(x) c(x) db(x)f (η) +f (η)f (η) −+[f (η)]2b(x) dxb(x)dxb(x) dx1ρv 2 c(x)[(b(x)]3dp=0dx(6.30)φ (η)c(x) dd(x) 1 dc(x)+f (η)f (η) −f (η)φ(η)Prb(x) dxb(x)d(x) dx−c(x) dj (x) βT c(x) dp f (η) +f (η)b(x)d(x) dxρcp b(x)d(x) dx+ν2 [c(x)b(x)]2 q 1[f (η)]2 +=0cpd(x)d(x)[b(x)]2 kPrThe boundary conditions aref (0) = f (0) = 1 − φ(0) = φ(∞) = 0BOOKCOMP, Inc.
— John Wiley & Sons / Page 453 / 2nd Proofs / Heat Transfer Handbook / Bejan———2.14024pt PgVar———Long PagePgEnds: TEXwhere the allowable forms of b(x) and c(x) (defined later) and−[453], (15)(6.31)454123456789101112131415161718192021222324252627282930313233343536373839404142434445FORCED CONVECTION: EXTERNAL FLOWSand it is also noted thatdpdU= −ρUdxdxFor the momentum equation to be entirely a function of the independent variable η,1 dc(x)= C1b(x) dxc(x) db(x)= C2[b(x)]2 dxThis results in choices for the constants, C1 and C2 : kxke(C1 = C2 )c(x) = qk x(C1 = C2 )[454], (16)Lines: 615 to 681with C1 and C2 related by———q −1C2 =C1q0.08028pt PgVarU (x) ∝ x 2q−1[454], (16)———Normal PageFor the pressure gradient term to be independent of x, the free stream velocity * PgEnds: Ejectmust beAlso known as Falkner–Skan flow, this form arises in the flow past a wedge with anincluded angle βπ as seen in Fig.
6.6. In this case,U (x) = C̄x mwherem=β2−βas indicated in potential flow theory. From the similarity requirement, the exponentq becomesq=m+11=22−βThen the pressure gradient term in eq. (6.30) becomes−11dp(C1 − C2 )3 mC̄2 x 2m−1=β=32ρν c(x)[b(x)] dxν2 k 4x 4q−3BOOKCOMP, Inc. — John Wiley & Sons / Page 454 / 2nd Proofs / Heat Transfer Handbook / BejanHEAT TRANSFER FROM SINGLE OBJECTS IN UNIFORM FLOW123456789101112131415161718192021222324252627282930313233343536373839404142434445455The arbitrary constants C1 and C2 may be chosen, without loss of generality, asC1 = 1andC2 = β − 1This results in1/22c(x) =Rexm+11/21 m+1b(x) =Rexx21/2y m+1η=Rexx21/22ψ(x,y) = νf (η)Rexm+1[455], (17)Lines: 681 to 735and for this choice of constants, the Falkner–Skan momentum equation and boundaryconditions become2f (η) + f (η)f (η) + 1 − f (η) β = 0———1.99225pt PgVar———Normal Page* PgEnds: Ejectf (η = 0) = f (η = 0) = 1 − f (η = ∞) = 0The often used (and much older) Blasius (1908) variables for flow past a flat plate(β = 0) are related to the Falkner–Skan variables η and f (η) asηB = 21/2 ηandf (ηB ) = 21/2 f (η)For similarity to hold, the energy equation must satisfy the conditionsc(x) dd(x)= C5b(x)d(x) dxν2 [c(x)b(x)]2[c(x)b(x)]2= K3= K3 C 6cp d(x)d(x)c(x) dj (x)= C7b(x)d(x) dx(6.32a)(6.32b)(6.32c)βT c(x) dpc(x) dp= K4= K4 C10ρcp b(x)d(x) dxb(x)d(x) dx(6.32d)11 q = F1 (η)2d(x)[b(x)] Pr k(6.32e)BOOKCOMP, Inc.
— John Wiley & Sons / Page 455 / 2nd Proofs / Heat Transfer Handbook / Bejan[455], (17)456123456789101112131415161718192021222324252627282930313233343536373839404142434445FORCED CONVECTION: EXTERNAL FLOWS6.4.3 Similarity Solutions for the Flat Plate at Uniform Temperature(m = 0)For the case of the flat plate, the similarity equations and boundary conditions reducetof (η) + f (η)f (η) = 0φ (η) + Pr · f (η)φ (η) = 0(6.33)f (0) = f (0) = 1 − f (∞) = 0(6.34)and1 − φ(0) = φ(∞) = 0andBoth the momentum and energy equations are ordinary differential equations in theform of two-point boundary value problems.
The momentum equation is solvedfirst because it is uncoupled from the energy equation. The velocity field is thensubstituted into the energy equation to obtain the temperature field and heat transfercharacteristics.The wall heat flux is obtained as∂T ∂ηq (x) = hx (T0 − T∞ ) = −k= −k(T0 − T∞ )φ (0)∂y y=0∂y= −k(T0 − T∞ )φ (0)1· Re1/2xx(6.35)hx x−φ (0)1/2= √ Re1/2x = F̄ (Pr)Rexk21/3Nux = 0.332Re1/2x · Pr(6.36)the surface-averaged heat transfer coefficient is determined as:6.4.41h dAs =L0Lhx dx = 2h(6.37)x=LSimilarity Solutions for a Wedge (m = 0)For a wedge at a uniform surface temperature, the expressions for the surface heatflux and the Nusselt number areBOOKCOMP, Inc.
— John Wiley & Sons / Page 456 / 2nd Proofs / Heat Transfer Handbook / Bejan——————Short Page* PgEnds: Eject[456], (18)where F̄ (Pr) is determined numerically and near Pr ≈ 1 is well approximated by0.332Pr1/3 so that,1h̄ =AsLines: 735 to 7990.86714pt PgVarThis results in the local Nusselt numberNux =[456], (18)HEAT TRANSFER FROM SINGLE OBJECTS IN UNIFORM FLOW1234567891011121314151617181920212223242526272829303132333435363738394041424344451q (x) = −k(T0 − T∞ )φ (0)xNux =m+1Rex24571/2hx x−φ (0)= √ [(m + 1)Rex ]1/2 = F̄ (m, Pr)Re1/2xk2(6.38)(6.39)For a wedge with a spatially varying surface temperature, eq. (6.32a)C5 =c(x) dd(x)b(x)d(x) dxyieldsb(x)m+1n1 dd(x)= C5= C5=d(x) dxc(x)2xx[457], (19)whereLines: 799 to 861m+1n ≡ C52———0.3433pt PgVar———Short Page* PgEnds: Ejectis a constant. This yieldsd(x) = [T0 (x) − T∞ ] = N x nwhere N is a constant arising from the integration for the surface temperature variation.
The energy equation is transformed to2nf (η)φ(η) = 0φ (η) + Pr f (η)φ (η) −m+1The resulting expressions for the heat flux and the local Nusselt number are1/2(m + 1)C̄x (2n+m−1)/2q (x) = −kφ (0)N2ν1/2m+1Nux = −φ (0)= F̄ (Pr, m, n)Re1/2Rexx2(6.40)(6.41)In the range of 0.70 ≤ Pr ≤ 10, Zhukauskas (1972, 1987) reports the correlation forthe data computed by Eckert:0.56ANu=1/2(2 − β)1/2RexwhereBOOKCOMP, Inc. — John Wiley & Sons / Page 457 / 2nd Proofs / Heat Transfer Handbook / Bejan(6.42)[457], (19)458123456789101112131415161718192021222324252627282930313233343536373839404142434445FORCED CONVECTION: EXTERNAL FLOWSβ=2mm+1andA = (β + 0.20)0.11 · Pr0.333+0.067β−0.026β2For the thermal boundary layer thickness to increase with x, two special cases ofthe foregoing solution corresponding to a flat plate (m = 0) are of interest (Fig. 6.7).These correspond respectively to a uniform heat flux surface and a line heat sourceat x = 0 (a line plume).
The heat flux can be written asq (x) = −kφ (0)NUν1/2x (2n−1)/2For the first condition (Fig. 6.7a), n = 21 . For the second condition (Fig. 6.7b), thetotal energy convected by the flow per unit length of the source is written asq (x) =∞[458], (20)ρcp u(T − T∞ ) dy0∞= νρcp c(x)d(x)Lines: 861 to 895f (η)φ(η) dη ∝ x(2n+1)/2(6.43)0For the convected energy to remain invariant with x, n must take on the value n = − 21 .Wedge Flow Limits With regard to wedge flow limits, numerical solutions to theFalkner–Skan equations have been obtained for −0.0904 ≤ m ≤ ∞, where the lowerlimit is set by the onset of boundary layer separation. In addition, the hydrodynamicboundary layer thickness isFigure 6.7 Two important cases for boundary layer flow at uniform free stream velocity.BOOKCOMP, Inc.
— John Wiley & Sons / Page 458 / 2nd Proofs / Heat Transfer Handbook / Bejan———0.24107pt PgVar———Normal Page* PgEnds: Eject[458], (20)HEAT TRANSFER FROM SINGLE OBJECTS IN UNIFORM FLOW1234567891011121314151617181920212223242526272829303132333435363738394041424344452νδ(x) = ηδ(m + 1)C̄1/2459x (1−m)/2where ηδ , the nondimensional thickness of the boundary layer, is bound at x = 0only for m ≤ 1. This requires that 1 ≥ m ≥ −0.0904. Additionally, the total energyconvected by the flow per unit width normal to the plane of flow is given by ∞q (x) = νρcp c(x)d(x)f (η)φ(η) dη ∝ Nx (2n+m+1)/20This provides the condition2n + m + 1≥026.4.5or n ≥ −m+12[459], (21)Prandtl Number EffectConsider the case of n = m = 0 first. In the limiting cases of Pr 1 and Pr 1(Fig. 6.8), the solution of the momentum equation can be approximated in closedform.
Subsequently, the energy equation can be solved. For Pr 1, the velocitycomponents u and U are approximately equal throughout the thermal boundary layer.This results in f (η) ≈ 1 or f (η) ≈ η + K. For an impermeable wall, K = 0 andthe energy equation simplifies toφ (η) + Pr · ηφ (η) = 0This can be integrated to yieldφ (η) = e−η1 +C2whereη21 =η2 · Pr2Figure 6.8 Laminar flat plate boundary layer flow at limiting Prandtl numbers.BOOKCOMP, Inc. — John Wiley & Sons / Page 459 / 2nd Proofs / Heat Transfer Handbook / BejanLines: 895 to 945———0.66211pt PgVar———Normal PagePgEnds: TEX[459], (21)460123456789101112131415161718192021222324252627282930313233343536373839404142434445FORCED CONVECTION: EXTERNAL FLOWSand with the boundary conditions1 = φ(0) = φ(∞) = 0another integration providesφ(η) = 1 − erf(η1 )where2erf(η1 ) = √π η1e−v dv20[460], (22)From this the local Nusselt number is determined asφ (0)2Nux = − √ Re1/2x = √π2PrRex21/21/2= 0.565Re1/2x · Pr(6.44)Lines: 945 to 1016———and for Pr 1, the nondimensional stream function near the wall is expressed asf (0) f (0)f (0) 2 f (0) 3f (η) =+η+η +η + ···0!1!2!3!(6.45)However, f (0) = f (0) = 0 and the momentum equation shows that f (0) = 0.This results inf (η) =0.332f (0) 2η = √ η22!2f (0) =√ √2 fB (0) = 0.332 2(6.46)upon usingwhere the subscript B refers to the Blasius variables defined in Section 6.4.2.
Theenergy equation then becomesφ (η) +0.332Pr 2 η φ (η) = 0√2(6.47)This can be integrated and evaluated at the surface to yieldφ (0)− √ = 0.339Pr1/321/3Nux = 0.339Re1/2x · PrBOOKCOMP, Inc. — John Wiley & Sons / Page 460 / 2nd Proofs / Heat Transfer Handbook / Bejan(6.48a)(6.48b)5.77745pt PgVar———Normal PagePgEnds: TEX[460], (22)461HEAT TRANSFER FROM SINGLE OBJECTS IN UNIFORM FLOW1234567891011121314151617181920212223242526272829303132333435363738394041424344456.4.6Incompressible Flow Past a Flat Plate with Viscous DissipationUsing the Blasius normalized variables for the flow and then defining the normalizedtemperature asθ(η) =2cp (T − T∞ )U2the energy equation and thermal boundary conditions becomeθ (η) + 21 Pr · f (η)θ (η) + 2Pr · f (η)2 = 0(6.49)andθ(η = ∞) = 0and θ(η = 0) =2cp (T0 − T∞ )= θoU2(constant)(6.50)The solution to eq.
(6.49), which is a nonhomogeneous equation, consists of thesuperposition of a homogeneous part:[461], (23)Lines: 1016 to 1085———4.72226pt PgVarθH = C1 φ(η) − C2———Normal Page* PgEnds: Ejectand a particular solutionθP = θAW (η)[461], (23)where the subscript AW refers to an adiabatic wall condition.The governing equations and boundary conditions for these areθAW + 21 Pr · f θAW + 2Pr(f )2 = 0(6.51)φ + 21 Pr · f φ = 0(6.52)andθAW (0) = θAW (∞) = 0andφ(0) = 1 − φ(∞)(6.53)For the boundary conditions for the complete problem to be satisfied,θ(∞) = 0 = θAW (∞) + C1 φ(∞) + C2which leads to C1 = −C2 andθ(0) =2cp (TAW − T0 )2cp (T0 − T∞ )= θAW (0) + C1 φ(0) + C2 =+ C2U2U2orBOOKCOMP, Inc. — John Wiley & Sons / Page 461 / 2nd Proofs / Heat Transfer Handbook / Bejan462123456789101112131415161718192021222324252627282930313233343536373839404142434445FORCED CONVECTION: EXTERNAL FLOWSC2 = −C1 =2cp (T0 − TAW )U2The solution for θAW (ηB ) is obtained numerically by solving eq.
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