4.4 Buoyant plume from a steady heat source

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1 1 Notes on 1.63 Advanced Envionmental Fluid Mechanics Instucto: C. C. Mei, Octobe 25, buoyplum.tex 4.4 Buoyant plume fom a steady heat souce [Refeence]: Gebhat, et. al. (Jalluia, Mahajan, Saammakia), Buoyancy-induced Flows and Tanspot, 1988, Hemisphee Publishing Copoation Let T = T T = tempeatue vaiation whee T i nfty is a constant (no ambient statification). Fo a stong enough heat souce, we expect bounday laye behavio, À x,uàv, p = The bounday laye equations ae (u) x + (v) u u x + v u = gβ(t T )+ ν u T x + v T = k The centeline = is an axis of symmety, = (4.4.1) T u (4.4.2) (4.4.3) Fa outside the plume v = u = T = (4.4.4) u andt T, ( T ) (4.4.5) Rewite (4.4.3) as (u T ) x + (v T ) = (u T ) x T (u) x + (v) + (v T ) = k T (4.4.6)

2 2 afte using continuity. Now integating the last equation fom =to = Z x = k2π 2πu Td+2π d T (v T ) d theefoe Z 2πu Td+2πv T x Using the bounday conditions, we get o =2πk T = = (4.4.7) 2πu Td = constant Note that 2πd uρc T = ate of buoyancy flux = ate of heat flux = Q(given ate of heat elease at x =) theefoe, Q = This is a bounday condition. Let the steam function ψ be defined by 2πdρuC T (4.4.8) u = ψ, v = ψ x (4.4.1) is automatically satisfied. Fom the momentum equation: 1 ψ 1 2 ψ x 1 ψ x 1 ψ = gβ T + ν " 1 # ψ (4.4.9) (4.4.1) Fom the enegy equation 1 ψ T x 1 ψ T x = k T (4.4.11) and fom the buoyancy flux condition Q =2πρC d 1 ψ T (4.4.12)

3 3 Ty a similaity solution with the one-paamete tansfomation Fom (4.4.1), fom (4.4.11) and fom (4.4.12) Fom these thee equations we get x λ a x, = λ b, ψ = λ c ψ, T = λ d T λ 2c 4b a = λ 2c 4b a = λ d = λ c 4b (4.4.13) λ c+d 2b a = λ d 2b (4.4.14) λ c+d = 1 (4.4.15) c a =1, b a = 1 2, d a = 1. We leave it as an execise to show that the similaity vaiable can be taken to be and the similaity solutions to be Afte much algeba, and noting we get fom (4.4.1) and fom (4.4.11). η = 1 x, 1/2 Befoe integating, let us nomalize : It follows fom (4.4.18) that η = x 1/2 (4.4.16) ψ = xf (η), and T = x 1 G(η) (4.4.17) η x = 1 2 x = 1 1 3/2 2 x 1/2 x = η 2x F νf + (F ν)+gβηg = (4.4.18) η νγ α 3 F + γ α 3 k(ηg ) +(FG) = (4.4.19) η = α η, F = γ F, G = σḡ. (4.4.2) F (γ F ν)+gβασ ηḡ = (4.4.21) η

4 4 whee pime denotes d/d η. Settingγ = ν and ν 2 α 3 = gβασ which elates σ and α, σ = ν2 (4.4.22) gβα 4 we get F F + ( η F 1) + ηḡ = (4.4.23) Simila nomalization of (4.4.19) gives. which can be simplified to.whee kασ α 2 ( ηḡ ) + γσ α ( F Ḡ) = (4.4.24) ( ηḡ ) + P( F Ḡ) = (4.4.25) P = ν k = Pandtl Numbe (4.4.26) Fo wate ν =1 2 cm 2 /s, k =1.42cm 2 /s, hence P =7. Foaiν =.145cm 2 /s, k =.22cm 2 /s, hence P =.75. We now integate (4.4.25)to give ηḡ + P F Ḡ =constant Since ψ(x, ) =, we must have F () = ; the constant above is zeo. Equation (4.4.27) can be witten ηḡ + P F Ḡ = (4.4.27) Ḡ Ḡ = P F η, d ln Ḡ F o = P d η η Z η ln Ḡ = P F d η +constant η Z η Ḡ( η) =Ḡ() exp F P η d η (4.4.28) Substituting Eqn. (4.4.28) into Eqn. (4.4.23), the esulting equation fo F must be integated numeically. Now let us find the bounday condtions fo F o F.

5 5 Eqn. (4.4.8 ) becomes Z Q 2πρC = d Theefoe, Let us choose so that 1 ψ G(η) x = d x1/2 F G x = d η F Ḡ = Q 2πρCνσ dη(f G)=νσ d η( F Ḡ) (4.4.29) (4.4.3) Q 2πρCνσ = 1 (4.4.31) d η F Ḡ = 1 (4.4.32) is the bounday condition fo F and Ḡ. Now (4.4.31) defines σ, thescaleofg. Note that lage Q implies lage σ and smalle α. Thus a stonge heat souce leads to a geate centeline tempeatue and a thinne plume. Also, u as hence u = 1 ψ = F η = ν F α 2 η, The adial velocity is, in geneal v = 1 ψ x = 1 F η F 2 as η η Since we must have, Clealy v as η, F () =. F ( η) = as η (4.4.33) The numeical esults by Mollendof & Gelhat, 1974, ae shown in Figs , fo vaious Pandtl numbes. A schlieian photogaph due to Gebhat (copied fom Van Dyke An Album of Fluid Motion) ishowninfiguefig:plumevd. Remak: u = 1 ψ = F µ = x η F 1 x 1/2 Along the centeline u(x, ) = ³ F = constant depending on P η. Why? Buoyancy acceleation is counteacted by entainment.

6 6 Remak: Let the adius of the plume be a which vaies as a x 1/2 This is consistent with the behavio that u x,and T x 1,since On the othe hand the mass flux ate is and the momentum flux ate is a 2 u T = Q ua 2 x u 2 a 2 x hence both appoach zeo at the souce. Thus a plume is the esult of enegy souce, not of mass o momentum.

7 Figue 4.4.1: Velocity and tempeatue pofiles in a themal plume 7

8 8 Figue 4.4.2: A 2D themal plume fom a line heat souce. Gebhat, Pea and Schoo 197, Fom Van Dyke, photo by

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