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On the Reynolds-Averaged Navier-Stokes Equations Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 2 July 2019 doi:10.20944/preprints201907.0038.v1 On the Reynolds-averaged Navier-Stokes Equations 1 Bohua Sun 1 School of Civil Engineering &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 ρr · (u¯ ⊗ u¯) + rp¯ = µr2u¯ + r · τ ; (3) Turbulence is everywhere, controlling the drag on r · 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 r = e r , 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]. Z 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 !1 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: R p(x; t) =p ¯(x) +p ~(x; t); (2) · r 1 r 1 t+T · r u¯ u¯ + ρ p¯ + limT !1 T t (u~ u~)dt = νr2u¯; (6) where coordinates and time are (x; t). According to r · 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-averagingR definition is given by u¯(x) = 1 t0+T limT !1 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 un- knowns 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 © 2019 by the author(s). Distributed under a Creative Commons CC BY license. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 2 July 2019 doi:10.20944/preprints201907.0038.v1 2 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 =6 v =6 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 Z 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 !1 T ly different perspectives. What we believe is that the Zt0 t +T Reynolds stress tensor τ has only three unknowns, name- 1 0 = lim vivjei ⊗ ejdt ly the velocity fluctuation components u0 (i = 1; 2; 3) due T !1 T i Zt0 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 !1 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 6 ⊗ 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 3 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 Z t +T 1 0 PROOF 2 = −ρ lim (~uiu~jei ⊗ ej)dt !1 T T t " Z0 # p 1 t0+T p Given two scalar functions,u ~ = 2U cos(!t) andv ~ = = −ρ lim u~iu~jdt ei ⊗ ej 2V cos(!t + θ), as independent unknowns, we can use T !1 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) .
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