On the Reynolds-Averaged Navier-Stokes Equations

Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 2 July 2019 doi:10.20944/preprints201907.0038.v1

On the Reynolds-averaged Navier-Stokes Equations

Bohua Sun1 1 School of Civil Engineering & Institute of Mechanics and Technology, Xi’an University of Architecture and Technology, Xi’an 710055, China [email protected]

This paper attempts to clarify an issue about the number of unknowns in the Reynolds-Averaged Navier-Stokes equations (RANS). The study discovers that all perspectives regarding the numbers of unknowns in the RANS stems from the misinterpretation of the Reynolds stress tensor. The current literature considers that the Reynolds stress tensor has six unknowns; however, this paper shows that the Reynolds stress tensor actually has only three unknowns, namely the three components of fluctuation velocity. The transport equation on both the Reynolds stress tensor and kinetic energy are discussed as well. As application, the turbulent Berger’s equation and Prandtl boundary layer equations are resolved. All solutions have no any adjustable parameters. The paper also propose a general algorithm for three dimensional turbulence flow. It is found that the magnitude of velocity fluctuations or turbulence are proportional to the flow pressure, which is a remarkable discovery.

Keywords: turbulence; the Reynolds stress tensor; turbulence closure problem

INTRODUCTION Navier–Stokes equations (RANS) as follows

ρ∇ · (u¯ ⊗ u¯) + ∇p¯ = µ∇2u¯ + ∇ · τ , (3) Turbulence is everywhere, controlling the drag on ∇ · cars, airplanes, and trains, whilst dictating the weath- u¯ = 0, (4) er through its influence on large-scale atmospheric and where dynamic viscosity µ, gradient operator ∇ = e ∇ , oceanic flows. Even solar flares are a manifestation of i i base vector in the i-coordinate e , and tensor product ⊗, turbulence since they are triggered by vigorous motions i and the Reynolds stress tensor is given by on the surface of the Sun [1–40]. ∫ The study of turbulence is a difficult subject owing to 1 t0+T τ = −ρu˜ ⊗ u˜ = −ρ lim (u˜ ⊗ u˜)dt, (5) its complex and forbidding mathematical descriptions, T →∞ T t0 which is often considered as the last unsolved problem in classical physics. In 1895 Reynolds published a seminal where T is the period of time over which the averaging work on turbulence [1], in which he proposed that flow takes place and must be sufficiently large to give mean- velocity u and pressure p are decomposed into its time- ingful averages and is needed to measure mean values averaged quantities, u¯,p ¯, and fluctuating quantities, u˜, depends on the accuracy desired. Reynolds stress is ap- p˜; thus, the Reynolds decompositions are: parent stress owing to the fluctuating velocity field u˜. Denoting kinematic viscosity ν = µ/ρ, the above equa- u = u¯(x) + u˜(x, t), (1) tions can be equivalently rewritten in conventional form: ∫ p(x, t) =p ¯(x) +p ˜(x, t), (2) · ∇ 1 ∇ 1 t+T · ∇ u¯ u¯ + ρ p¯ + limT →∞ T t (u˜ u˜)dt = ν∇2u¯, (6) where coordinates and time are (x, t). According to ∇ · Reynolds, u¯ represent a mean-motion at each point u¯ = 0 (7) and u˜ a motion at the same point relative to the mean-motion at the point. Therefore, the Reynolds HOW MANY UNKNOWNS ARE THERE IN THE called the u¯ mean-motion and u˜ relative-mean-motion RANS ？？？ [1]. The time-averaging∫ definition is given by u¯(x) = 1 t0+T limT →∞ u(x, t)dt, and time-fluctuation u˜ = u− T t0 All current literatures and textbooks believe that the u¯(x). Reynolds stresses τ has six unknowns (In the later, we will show that the Reynolds stresses have only three unknowns instead of six ones.). This understanding result- s a consensus of saying that there are 10 unknowns in REYNOLDS-AVERAGED NAVIER-STOKES Eqs.3, 4 and/or Eqs.35, 35. EQUATIONS If you carefully read Reynolds’ paper in 1895 [1], and will see that Reynolds did not discussed the number of With the decomposition the Navier-Stokes equa- unknowns in the RANS. However, from his presentation tion is then transformed into the Reynolds-averaged we are sure that he never considered the term ρu˜ ⊗ u˜ as

independent unknowns, while for some cases he even pro- w(x, t) = wjej = w1e1 + w2e2 + w3e3, we can defined a posed explicit expressions for the velocity fluctuation u˜, 3rd order tensor v ⊗ w as follows for instance, expressions in page 149 and Eq.(50) in page 158 of Reynolds paper [1], which means that he actually u ⊗ v ⊗ w = uiek ⊗ vkei ⊗ wjej = ukviwjek ⊗ ei ⊗ ej. take the velocity fluctuation u˜ as unknowns. Due to the (9) limited references, the author has no idea who was the first person proposed this 10-unknowns perception on the In general case of u ≠ v ≠ w, the tensor u ⊗ v ⊗ w RANS. is a general tensor with 27 components (elements), and Clearly, the RANS in Eqs.3, 4 has four independent have 9 unknowns, namely, u1, u2, u3, v1, v2, v3 and equations governing the mean velocity field; namely the w1, w2, w3, since the 27 components can be fully deter- three components of the Reynolds equation Eqs.3 togeth- mined by the 9 unknowns. er with the mean continuity equation Eq.4. However, If u = v = w, the tensor u⊗v⊗w is a 3rd order tensor these four equations contain more than four unknowns. with 27 components (elements), and have 3 unknowns, In addition to u¯ andp ¯, there are also the Reynolds stress- namely, u1, u2, u3, or v1, v2, v3 and/or w1, w2, w3, s- es τ , which results the Reynolds-averaged Navier-Stokes ince the 27 components can be fully determined by the 3 equations unclosed. The closure problem is being consid- unknowns. ered as the number one topic in turbulence, during the

past several decades, scientists and engineers have made Lemma 3 Giving a vector, v(x, t) = viei = v1e1 + many attempts towards solving the closure problem. v2e2 +v3e3, we can defined a 2nd order symmetric dyadic Regarding the RANS closure problem, the curren- tensor v ⊗ v and its mean value A(x) as follows t consensus is that there are six unknown components ∫ in the symmetric Reynolds stress tensor τ , namely 1 t0+T τ , τ , τ , τ , τ , τ . However, we have a complete- A(x) = lim v ⊗ vdt 11 12 13 22 23 33 T →∞ T ly different perspectives. What we believe is that the ∫t0 t +T Reynolds stress tensor τ has only three unknowns, name- 1 0 = lim vivjei ⊗ ejdt ly the velocity fluctuation components u′ (i = 1, 2, 3) due T →∞ T i ∫t0 to the fact that the Reynolds stress tensor is simply an 1 t0+T integration of a second order dyadic tensor of flow veloc- = lim [v1v1e1 ⊗ e1 + v1v2e1 ⊗ e2 T →∞ T ity fluctuations rather than a general symmetric tensor. t0 + v1v3e1 ⊗ e3 + v2v1e2 ⊗ e1

+ v2v2e2 ⊗ e2 + v2v3e2 ⊗ e3 LEMMAS + v3v1e3 ⊗ e1 + v3v2e3 ⊗ e2 + v v e ⊗ e ]dt (10) To support our statement, let’s us introduce some lem- 3 3 3 3 mas as follows where vivj = vjvi. Although A(x) has six independen- Lemma 1 Giving two vectors, v(x, t) = viei = v1e1 + t components,namely v1v1, v1v2, v1v3, v2v2, v2v3, v3v3, v2e2 + v3e3 and w(x, t) = wjej = w1e1 + w2e2 + w3e3, however it is clear there are only three independent quan- we can deﬁned a dyadic tensor v ⊗ w as follows [41] tities, namely, v1, v2, v3, in the A(x). It is because that the quantities v1v1, v1v2, v1v3, v2v2, v2v3, v3v3 can v ⊗ w = viei ⊗ wjej = viwjei ⊗ ej be fully determined by v1, v2, v3. = v1w1e1 ⊗ e1 + v1w2e1 ⊗ e2 + v1w3e1 ⊗ e3

+ v2w1e2 ⊗ e1 + v2w2e2 ⊗ e2 + v2w3e2 ⊗ e3 The lemma 3 actually states that any (time) averaging operation is just a method of data processing and will + v3w1e3 ⊗ e1 + v3w2e3 ⊗ e2 + v3w3e3 ⊗ e3. (8) not change the number of unknowns of the problem. It is clear that A(x) will be the Reynolds stress tensor ̸ ⊗ In general case of v = w, the dyadic tensor v w is a τ , if replacing v1, v2, v3 by the components of velocity general tensor with 9 components (elements),and have 6 ﬂuctuationsu ˜i (i = 1, 2, 3), respectively, namely, v1 = unknowns, namely, v1, v2, v3 and w1, w2, w3, since the u˜1, v2 =u ˜2, v3 =u ˜3. 9 components can be fully determined by the 6 unknowns. If v = w, the tensor v ⊗ w is a 2rd order tensor with 9 components (elements), and have 3 unknowns, namely, PROOF 1 v1, v2, v3 and/or w1, w2, w3, since the 9 components can be fully determined by the 3 unknowns. From the Lemma, we can see that the Reynolds stress Lemma 2 Giving three vectors, u(x, t) = uiei = u1e1 + tensor has only three unknown components. This can be u2e2 + u3e3 , v(x, t) = viei = v1e1 + v2e2 + v3e3 and proved easily as follows: the Reynolds stress tensor can Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 2 July 2019 doi:10.20944/preprints201907.0038.v1

be deﬁned by stress tensor has only three unknowns, namelyu ˜1, u˜2, u˜3. For two dimensional ﬂow, of course, the 2D Reynolds − ⊗ τ = ρu˜ u˜ stress tensor has only two unknowns, namelyu ˜1, u˜2.

= −ρu˜iei ⊗ u˜jej = −ρu˜iu˜jei ⊗ ej ∫ t +T 1 0 PROOF 2 = −ρ lim (˜uiu˜jei ⊗ ej)dt →∞ T T t [ ∫0 ] √ 1 t0+T √ Given two scalar functions,u ˜ = 2U cos(ωt) andv ˜ = = −ρ lim u˜iu˜jdt ei ⊗ ej 2V cos(ωt + θ), as independent unknowns, we can use T →∞ T t0 them to construct a 2nd order tensor or matrix with 4 2 2 = τijei ⊗ ej, (11) components as follows: (˜u) , u˜v,˜ v˜u,˜ (˜v) . Thus time averages give: and the ﬂuctuation velocity convective terms are given ∫ by: 1 t0+T lim (˜uu˜)dt = U 2 (14) T →∞ T t0 u˜ · ∇u˜ = u˜iei · [ek∂k ⊗ (˜ujej)] ∫ 1 t0+T = u˜ u˜ e · (e ⊗ e ) = u˜ u˜ (e · e )e lim (˜uv˜)dt = UV cos θ (15) i j,k i k j i j,k i k j T →∞ T ∫t0 = u˜iu˜j,kδikej = u˜iu˜j,iej t0+T ( ∫ ) 1 t+T lim (˜vu˜)d = UV cos θ (16) 1 T →∞ T t0 = lim u˜iu˜j,idt ej →∞ ∫ T T t 1 t0+T lim (˜vv˜)dt = V 2 (17) T →∞ = τij,iej, (12) T t0

where the Reynolds stress tensor in index format τij is In the definition√ of the Reynolds stress√ tensor, θ = 0, we defined by haveu ˜ = 2U cos(ωt) andv ˜ = 2V cos(ωt) and their ∫ averaging 1 t0+T ∫ t +T τij = τji = −ρ lim u˜iu˜jdt. (13) 1 0 T →∞ 2 T t0 lim (˜uu˜)dt = U (18) T →∞ T ∫t0 It is clear that any τij can be calculated by fluctuation ve- 1 t0+T locity componentsu ˜ , u˜ , u˜ , which means that the τ lim (˜uv˜)dt = UV (19) 1 2 3 ij T →∞ T t0 are dependent on the componentsu ˜1, u˜2, u˜3. In other ∫ 1 t0+T words, the componentsu ˜1, u˜2, u˜3 are real unknowns. It lim (˜vu˜)dt = UV (20) is the key point why we say that the Reynolds stress ten- T →∞ T ∫t0 sor τ has 6 unknowns is a big mistake in the turbulence 1 t0+T modelling. lim (˜vv˜)dt = V 2 (21) T →∞ T Therefore, the formulation in Eq.11 reveals that the t0 Reynolds stress tensor τ = −ρu˜ ⊗ u˜ can be fully cal- Therefore, the Reynolds stress tensor is given by culated by three independent components of fluctuation − 2 ⊗ ⊗ velocityu ˜1, u˜2, u˜3. In other words, the Reynolds stress τ = ρ(U e1 e1 + UV e1 e2 tensor has only three unknowns rather than six ones. 2 + VUe2 ⊗ e1 + V e2 ⊗ e2). (22) It means that the averaging technique is just a mathematical processing, which can provide us a mean value, The above process shows that,u ˜ or U, andv ˜ or V , are however, can’t change the number of unknowns of the the independent unknowns, with them we can get the problem. Reynolds stress tensor τ . The misinterpretation regarding the numbers of un- If keeping the θ, namely, then we are talking about known components of all literature may stem from con- the time correlation (autocorrelations) at the same point. sidering the Reynolds stress tensor as a general 2nd or- The correlation between the same (Greek autos = self der symmetric tensor with six independent components. or same) fluctuating quantity measured at two different However, the Reynolds stress tensor is not an arbitrary times (at the same point in space) is not itself very rele- 2nd order tensor, actually each of its component is made vant to the behaviour of turbulence and its measurement by the bi-product of fluctuation velocity components, requires a time delay mechanism (usually a tape recorder which means that the Reynolds stress tensor is an dyadic with movable heads or a digital sample-and-delay sys- tensor of the velocity fluctuation. The unknown compo- tem). nents to construct the dyadic tensor are the three compo- In the same way, you can define the space correla- nents of fluctuation velocity u˜. Therefore, the Reynolds tion, however, all literatures and textbooks said that the Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 2 July 2019 doi:10.20944/preprints201907.0038.v1

Reynolds stress tensor is the one-point or single-point Therefore, the Reynolds tress tensor can be also given by velocity fluctuation correlation. Therefore, θ = 0. ∫ Even not setting θ = 0, the Reynolds stress still has 1 t0+T τ = −ρ lim σdt three independent unknowns, namely the three compo- T →∞ T nents of velocity fluctuations. This is the key point of ∫t0 t0+T this article. The conclusion of saying that the RANS has 1 2 2 2 = −ρ lim [(˜u1) + (˜u2) + (˜u3) ]ê2 ⊗ eˆ2dt. T →∞ only 7 independent unknowns is still correct. T t0 (33)

PROOF 3 It is proved once again that both the tensor σ and the Reynolds stress tensor τ are the only function of To support this obvious statement from other point of u˜1, u˜2, u˜3. As a summary, the number of unknowns of view, let’s study the problem by tensor representation the Reynolds stress tensor is listed in the Table II: theory and write the Reynolds stress tensor as follows: ∫ 1 t0+T τ = −ρ lim σdt, (23) TABLE I: Numbers of unknowns in the Reynolds stress T →∞ T t0 Current literatures This paper Numbers 6 3 ⊗ ⊗ ⊗ where σ = u˜(x, t) u˜(x, t) =u ˜iei u˜jej =u ˜iu˜jei ej. Unknowns τ11, τ12, τ13, τ22, τ23, τ33 u˜1, u˜2, u˜3 The characteristic equation of σ is given by

3 2 λ − I1λ + I2λ − I3 = 0. (24) Although the Reynolds-averaged Navier-Stokes (RAN- S) equations are unclosed, however, the four-equations 2 2 2 2 2 2 Note tr(σ) = (˜u1) +(˜u2) +(˜u3) and tr(σ) = tr(σ ), RANS in Eqs.3 and 4 contain only 7 unknowns instead thus of 10 ones. The list of unknowns in the RANS are sum- 2 2 2 marized in the below Table II: I1 = tr(σ) = (˜u1) + (˜u2) + (˜u3) (25) 1 I = I = [(tr(σ)2 − tr(σ2)] = 0 (26) 2 2 2 TABLE II: Numbers of unknowns in the RANS I3 = det(σ) = εijk(u1uj)(u2uj)(u3uj) Current literature This paper 2 2 2 Numbers 10 7 = (˜u1u˜2u˜3) − (˜u1u˜2u˜3) − (˜u1u˜2u˜3) Unknowns u¯1, u¯2, u¯3 u¯1, u¯2, u¯3 2 2 2 + (˜u1u˜2u˜3) + (˜u1u˜2u˜3) − (˜u1u˜2u˜3) p¯ p¯ = 0 (27) τ11, τ12, τ13, τ22, τ23, τ33 u˜1, u˜2, u˜3

Therefore, the characteristic equation of σ in Eq.23 has three roots as follows In the future, turbulence modelling will be focus on velocity fluctuationsu ˜1, u˜2, u˜3 instead of the Reynolds λ1 = 0, λ2 = I1, λ3 = 0. (28) stress τij. Advantage of modelling the velocity fluctuationsu ˜1, u˜2, u˜3 is obvious, since it reduce the 6 compo- From spectra representation theory of a symmetric ten- nents of τij into 3 components of velocity fluctuations, sor, we can rewrite the tensor σ as follows and from experimental point of view, the components of velocity fluctuations are easy to be measured than the ⊗ ⊗ ⊗ σ = λ1eˆ1 eˆ1 + λ2eˆ2 eˆ2 + λ3eˆ3 eˆ3 Reynolds stress tensor. = λ2eˆ2 ⊗ eˆ2, (29)

where the base eî is an eigenvector of σ, which can be determined by eigenvalue equation TRANSPORT EQUATION OF REYNOLDS STRESS TENSOR σ · eˆ = 2keˆ· eˆ (30) 1 2 2 With the above new understanding on the number of where the fluctuation kinetic energy k = 2 [(˜u1) +(˜u2) + 2 (˜u3) ], we find unknowns. In the following, we will present complete new perception on the higher-order correction of the Reynolds 1 stress tensor. ê2 = √ u˜ℓeℓ (31) 2k In order to get more information about the Reynolds Hence we have stress tensor, Reynolds [1] derived velocity fluctuation equations, and reformulated in index-tensorial form by 2 2 2 σ = [(˜u1) + (˜u2) + (˜u3) ]ê2 ⊗ eˆ2, (32) Chou [4]. Here we present the equations in a bold-face Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 2 July 2019 doi:10.20944/preprints201907.0038.v1

tensorial form as follows means thatu ˜1, u˜2, u˜3 are unknowns. Similarly, ∇ · ⊗ ⊗ ⊗ ∇ ∂p˜ ρu˜,t + ρ (u¯ u˜ + u˜ u¯ + u˜ u˜) + p˜ (34) u˜ ⊗ (∇p˜) = u˜ e ⊗ e ∫ i ∂x i j 2 1 t+T j = µ∇ u˜ + ρ limT →∞ ∇ · (u˜ ⊗ u˜)dt, ∫ T t 1 t0+T ∂p˜ ∇ · u˜ = 0. (35) = [ lim (˜ui )dt]ei ⊗ ej. (39) T →∞ T t0 ∂xj According to the author’s investigations [38–40], Eqs.3, ∂p˜ The mean value of u˜i can be calculated byp ˜ and 3, 3, and 3 form a closed integral-differential equations ∂xj system, in which there are 8 equations with 8 unknowns u˜1, u˜2, u˜3. u¯, u˜, p,¯ p˜. In the same way, we can still do the fourth order and After some tensor algebra we arrive at an important higher orders as in [4] and textbooks. However, there is equation on the Reynolds stress tensor as follows no more unknowns can be created by any order equation about the Reynolds stress tensor. ∂τ + u¯ · ∇τ = −τ · ∇u¯ − ∇u¯ · τ ∂t TRANSPORT EQUATION OF TURBULENCE ∇ ⊗ ∇ ⊗ ∇ ∇ ⊗ +2µI : u˜ u˜ + u˜ ( p˜) + ( p˜) u˜ KINETIC ENERGY +µ∇2τ + ρ∇ · (u˜ ⊗ u˜ ⊗ u˜). (36) If we do contraction operation for index i and j in In current literature, it is believe that Eq.36 has 31 un- the Eq.36, we have following transport equation for the knowns. However, we have complete different opinion. turbulence kinetic energy k We believe the Reynolds stress equation Eq. 36 has only ∂k 7 unknowns. Specifically, accoutring for all symmetries, ρ( + u¯ · ∇k) = τ : ∇u¯ − µ∇u˜ : ∇u˜ we have the following listed in the Table III. ∂t 1 +µ∇2k − ∇ · (p˜u˜) − ρ∇ · [(u˜ · u˜)u˜] (40) 2 TABLE III: Numbers of unknowns in the Reynolds stress e- where the kinetic energy k = − 1 τ = 1 u˜ u˜ = 1 u˜ · u˜. quation 2 kk 2 k k 2 The number of unknowns in the kinetic energy equation Current literature This paper Numbers 31 7 Eq.40 is listed in the Table IV. Unknowns u¯ (3) u¯1, u¯2, u¯3 u˜ ⊗ u˜ ⊗ u˜ (10) p˜ TABLE IV: Numbers of unknowns in the kinetic energy equa- I : ∇u˜ ⊗ ∇u˜ (6) u˜1, u˜2, u˜3 tion u˜ ⊗ (∇p˜) (6) Current literature This paper τ (6) Numbers 21 7 Unknowns k (3) u˜1, u˜2, u˜3 The above statement can be easily proved. For in- u¯ (3) u¯1, u¯2, u¯3 stance, τ (6) p˜ ∇u˜ : ∇u˜ (3) p˜u˜ (3) u˜ ⊗ u˜ ⊗ u˜ = u˜iu˜ju˜kei ⊗ ej ⊗ ek ∫ [(u˜ · u˜)u˜] (3) 1 t0+T = [ lim (˜uiu˜ju˜k)dt]ei ⊗ ej ⊗ ek, (37) T →∞ T t0 In order to support the above understanding about the number of unknowns, let’s present some examples in and the following. The aim of doing these exercise are only for demonstration purpose: to show general algorithm I : ∇u˜ ⊗ ∇u˜ = (ek ⊗ ek):(∇u˜ ⊗ ∇u˜) to find turbulence solutions. We will not going to give ⊗ ∇ ⊗ ∇ · ∇ ⊗ · ∇ = (ek ek):( u˜ u˜) = (ek u˜) (ek u˜) complete solutions here, so that we can stay on our key

= (∇ku˜) ⊗ (∇ku˜) = (∇ku˜iei) ⊗ (∇ku˜jej) focus, namely, to determine the number of unknowns in the Reynolds-averaged Navier-Stokes equations. ∂u˜ ∂u˜ = (∇ u˜ )(∇ u˜ )e ⊗ e = i j e ⊗ e k i k j i j ∂x ∂x i j ∫ k k t0+T 1 ∂u˜i ∂u˜j EXAMPLE 1: ONE DIMENSIONAL BERGER’S = [ lim ( )dt]ei ⊗ ej. EQUATION T →∞ T t0 ∂xk ∂xk (38) One dimensional Navier-Stokes equation, namely,

∂u˜i ∂u˜j Bergers’s equation It is clear that mean value of u˜iu˜ju˜k and can ∂xk ∂xk be calculated by the velocity ﬂuctuationsu ˜1, u˜2, u˜3. It u,t + uu,x = νu,xx + S(x, t), (41) Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 2 July 2019 doi:10.20944/preprints201907.0038.v1

6 √ where the source term S(x, t) is given by 1, u,xx√(0, 0) = 1, and obtain ω = κ and κ = 1+ν+ 1+ν − 1. S(x, t) = cos(x + t) + ν sin(x + t) 2ν2 Hence, the 1st order solution is obtained B √ + B cos(x + t) tanh(A − x) √ 2ν 1 + ν u(x, t) = 1 + ν tanh(A1 − x) + sin(x + t) cos(x + t) 2ν 1 √ 1 1 +√ sin( κt + κx). (49) − B2 sin(x + t)[1 − tanh2(A − Bx)], (42) 2ν 2ν 2κ where the trigonometric tangent function is defined as (1) The 1st order turbulence velocity u1st is different sinh(x) ex−e−x from the exact solution of the Bergers equation in Eq.41. tanh(x) = = − , and A, B are constant. cosh(x) ex+e x The exact solution of Berger’s equation is given by Substituting the Reynolds velocity decomposition u = u¯(x)+u ˜(x, t) into Eq.41, after time-averaging, we can get B u (x, t) = B tanh(A − x) + sin(x + t). (50) its RANS as follows exact 2ν → ∞ u¯u¯,x + u˜u˜,x = νu¯,xx + S,¯ (43) (2) In particular, when x , their asymptotic be- havior are totally different. The exact solution is going namely to be uexact = B + sin(x + t), due to the property of ∫ limx→∞ tanh(x) → 1, while the turbulence solution 1 t0+T u¯u¯,x + lim u˜u˜,xdt = νu¯,xx + S,¯ (44) √ √ T →∞ T 1 t0 u(x, t) = 1 + ν + √ sin( κt + κx).. (51) 2κ ¯ 1 where the mean source term S = 2 . Obviously, the Eq.44 has only two unknowns, one is the mean velocityu ¯ and (3) The turbulence kinetic energy is given by k = another is velocity fluctuationu ˜, as stated in the previous 1/(2κ) part. For a better approximation,√ the velocity fluctuations Since we have only one integral-differential equation can be assumed asu ˜(x, t) = 2a(x) sin(ωt). Eq.44, therefore, the Eq.44 is not closed. To solve the Eq.44, we have to model either the mean fieldu ¯ or fluc- EXAMPLE 2: TURBULENT BOUNDARY LAYER tuationu ˜. FLOW OF SEMI-INFINITE SURFACE We can use the steady solution ofu ¯u¯,x = νu¯,xx as the 1st order approximate solution of mean field. Hence, the Here velocity components u = u , v = u , directions 1st order exact solution is given by 1 2 1 → x, 2 → y are adopted respectively. B According to Prandtl’s boundary layer concept [2, 42], u¯(x) = B tanh(A − 1 x), (45) 1 1 2ν the two dimensional Navier-Stokes equations can be sim- plified to following boundary layer equations where the constants A1,B1 will be determined later. For fluctuation fieldu ˜, since its meanu ˜ must be zero, dU ∫ u,t + uu,x + vu,y = U + µu,yy, (52) 1 t0+T namely, limT →∞ udt˜ ≡ 0, hence, we can take the dx T t0 1st order velocity fluctuation proposed by Reynolds [1] as u,x + v,y = 0, (53) an approximate solation √ where U(x) is incoming velocity, U = const. for infinite u˜(x, t) = 2a sin(ωt + κx), (46) surface, hence U,x = 0. In duct and boundary layer flow, we can approximate where a, frequency ω and wave number κ are to be de- with excellent accuracy a simpler turbulent boundary termined. layer equations Substituting both Eqs.45 and 46 into Eq.44, we obtain dU ρ(¯uu¯ +v ¯u¯ ) = U + τ + τ , (54) 1 ,x ,y dx xx,x xy,y a = √ (47) 2κ u¯,x +v ¯,y = 0, (55) Therefore, we have the 1st order velocity field as follows where total stress ∫ t0+T B1 √1 1 u(x, t) = B1 tanh(A1 − x) + sin(ωt + κx). (48) τxx = −ρu˜u˜ = −ρ lim u˜udt,˜ (56) 2ν 2κ T →∞ T t0 ∫ t0+T For initial-boundary conditions u(0, 0) = 1, u (0, 0) = 1, 1 √ ,x τxy = µu¯,y − ρu˜v˜ = µu¯,y − ρ lim u˜vdt.˜ (57) 1 →∞ we get B = 1 + ν, tanh(A ) = √ u (0, 0) = T T t 1 1 1+ν ,t 0 Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 2 July 2019 doi:10.20944/preprints201907.0038.v1

To solve the above equations, we firstly omit the fluctu- For three dimensional turbulent flow, we propose fol- ation term and reduce Eqs.54 and 55 into followings lowing algorithm: (1) Find solution of corresponding equations of steady ρ(¯uu¯,x +v ¯u¯,y) = µu¯,yy, (58) flow u¯:

u¯,x +v ¯,y = 0, (59) ∇ · (u¯ ⊗ u¯) = ν∇2u¯. (67) Blasius [42] has obtained their solutions as follows (2) Calculate pressure (source term)p ¯: u¯ = Uf ′(ξ), (60) √ ∇2p¯ = −ρ∇ · (u¯ · ∇u¯). (68) 1 νU v¯ = (ξf ′ − f), (61) 2 x (3) Propose an approximate solution for velocity fluctu- √ ations u˜: where ξ = y U , U is incoming velocity, f ′ = df , and √ νx dξ u˜ = 2C(x) cos(ωt). (69) function f(ξ) is the solution of equation ff ′′ + 2f ′′′ = 0 ′ ′ ∞ under conditions f(0) = f (0) = 0, f ( ) = 1. (4) Calculate the Reynolds stress tensor τ : For fluctuation fieldu ˜ andv ˜, since their mean must be ∫ vanish, hence, we can take the 1st order velocity fluctu- 1 t+T τ = −ρ lim (u˜ ⊗ u˜)dt (70) ation suggested by Reynolds [1] as their solutions →∞ T T t √ − ⊗ u˜ = 2c cos(φ), (62) = ρ(C C). √ 1 v˜ = 2c2 cos(φ), (63) (5) Determine the coefficient vector an: Substitute u¯, p,¯ τ into the RANS, hence we have where phase function φ = ωt + κxx + κyy, and constants 1 c1, c2. ∇ · (u¯ ⊗ u¯) − ν∇2u¯ = − ∇p¯ − ∇ · (C ⊗ C). (71) Substituting Eqs. 60,61,62 and 63 into Eq. 54, we ρ obtain Since u¯ is the solution of ∇ · (u¯ ⊗ u¯) = ν∇2u¯, therefore, dU (c2)κ + (c c )κ = U = 0. (64) the left side of the above equation is zero, hence 1 x 1 2 y dx 1 ∇ · (C ⊗ C + p¯I) = 0, (72) to determine c1, c2, we need mass conversation law of ρ fluctuation fieldu ˜,x +v ˜,y = 0, namely κx = −κy = κ, find the c1 = c2 = C, hence, we can write down velocity This relation reveals that the fluctuation magnitude is field as follows proportional to pressure, the turbulence can not main- √ tained without continua source supplying. ′ u = Uf (ξ) + 2C cos(ωt + κx − κy), (65) (6) Calculate pressure fluctuationp ˜ √ 1 νU √ v = (ξf ′ − f) + 2C cos(ωt + κx − κy), (66) ∇2p˜ = −ρ∇ · [u¯ · ∇u˜ + u˜ · ∇u¯ + u˜ · ∇u˜] − τ . (73) 2 x where frequency ω, wave number κ and constant C can (7) Calculate total pressure p: be determined by 3 initial-boundary conditions. p =p ¯ +p. ˜ (74) In√ the future, for a better approximation,√ we can take u˜ = 2c1(x, y) cos(ωt) andv ˜ = 2c2(x, y) cos(ωt). (8) Calculate total velocity field u We must point out here that the one and/or two dimensional Navier-Stokes model can not represent a real u = u¯ + C(x) cos ωt. (75) turbulence, since turbulence motion is always three dimensional [7]. In the similar way, you can get higher order solutions.

EXAMPLE 3: THREE DIMENSIONAL CONCLUSIONS TURBULENT FLOW SOLUTION ALGORISM In summary, this article has re-visited a fundamental Although there are diﬀerent views about turbulence, problem in turbulence analysis, namely, how many un- there is a consensus that the deterministic Navier-Stokes knowns are there in the Reynolds stress? This study not equation probably contains all information, which is rel- only clariﬁed the number of unknowns in the formula- evant to turbulence [18]. tions of the Reynolds-averaged Navier-Stokes equation, Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 2 July 2019 doi:10.20944/preprints201907.0038.v1

but also discovered the number of unknowns is much less [11] M. Lesieur, Turbulence in Fluids. 2nd ed. Kluwer, Dor- than the conventional accoutring. drecht (1990). As application of this study, we have resolved the Berg- [12] D.C. Wilcox, Turbulence Modeling for CFD. D C W In- er’s equation and turbulent Pranddtl boundary layer e- dustries (1993). [13] S.B. Pope, Turbulent Flows. Cambridge University Press, quations. All solutions have no any adjustable param- Cambridge (2000). eters. It is found that the Reynolds stress is mainly [14] P.A. Davidson, Turbulence. Oxford University Press, depends on the source term (ie. flow pressure), which Oxford (2004). means that the turbulence can not maintained if there is [15] B. Hof, Experimental Observation of Nonlinear Traveling no source supplying. Waves in Turbulent Pipe Flow. Science 305, 1594 (2004). Although the Reynolds-Averaged Navier-Stokes- [16] G. Falkovich and K.R. Sreenivasan, Lessons from hydro- dynamic turbulence. 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