UNIVERSITY OF OSLO. Faculty of Mathematics and Natural Sciences. MEK4300/9300 Viscous flow og turbulence

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1 UNIVERSITY OF OSLO Faculty of Mathematics and Natural Sciences Examination in: Day of examination: Friday 15. June 212 Examination hours: This problem set consists of 5 pages. Appendices: Permitted aids: MEK43/93 Viscous flow og turbulence None Rottmann: Matematische Formelsamlung, certified calculator Please make sure that your copy of the problem set is complete before you attempt to answer anything. Problem 1 Turbulence Splitting of v and p v = v+v, p = p+p. The continuity and Navier-Stokes equations Averaging then gives v =, t +v v = 1 ρ p+ν 2 v v =, t +v v = 1 ρ p+ν 2 v. We need to work more on the convective term only: and v v = vv, vv = vv+vv +v v+v v = vv+v v, since the terms which are linear in the fluctuations are nihilated by the averaging. Insertion in the averaged momentum equation then gives where t +v v = 1 ρ p+ 1 ρ τ, τ = µ v+ v )+S, S = ρv v. The term S = {S ij } = { ρu i u j }, which stems from the convective term, is re-interpreted as the Reynolds stress tensor. Continued on page 2.)

2 Examination in MEK43/93, Friday 15. June 212 Page 2 Problem 2 Gravity driven viscous flow 2a We align the z-axis vertically downwards and employ cylindrical coordinates. Due to the symmetry and informaton given in the text we assume v = v r r)i r +v z r)k. From the formula sheets we then find the continuity equation 1 rv r ) =. r r while the components of the Navier-Stokes equation become { r i r : v r r = 1 p 1 ρ r +ν r ) r v } r r r r r 2, z k : v r r = 1 p ρ z +g + ν r ) z. r r r At the cylinder we have the no-slip condition v z a) = v r a) =. On the free surface, r = b we have the dynamic condition for the stress which means zero shear stress. 2b The continuity equation gives p = i r P = p i r, v r = A r, which corresponds to a line source at r =. The boundary condition v r a) = the implies that v r is zero throughout the fluid. To find an expression for the stress at the surface we start with ) v = i r r +k v z r)k = dv z z dr i rk. The stress tensor then becomes P = pi +µ v+ v ) = pi +µ dv z dr i rk+ki r ), and the condition at r = b p i r = i r P = pi r +µ dv z dr k, which imply p = p and dvz dr = at the surface. Continued on page 3.)

3 Examination in MEK43/93, Friday 15. June 212 Page 3 The i r component of the momentum equation now simply states p r =, which imply p = pz). Combined with p = p at r = b this gives p = p throughout the fluid. Then the equation set for v r becomes = g + ν d r dv ) z, r dr dr dv v z a) =, zb) dr =. Integration of the momentum equation r dv z dr = g 2ν r2 +C, v z = g 4ν r2 +Clnr/a)+D, where C and D are integration constants. The boundary conditions give g 4ν a2 +D =, g 2ν b+ C b =. The expresssion for v z then becomes v z = g a 2 r 2 +2b 2 lnr/a) ). 4ν 2c Since the downward acceleration is zero the drag per height must equal the weight of the fluid per height. Also, since the kinetec energy is constant the rate of loss of potential energy per height must equal the dissipation per height. Problem 3 Boundary layer 3a We assume a constant density. Then conservation of mass implies conservation of volume. There is no transport of volume through either ii) or iii). For boundary iv) we have a transport into the volume per width) Q iv) = HU. At ii) we have Y Q ii) = ux, y)dy, where the minus sign appears because fluid is leaving the volume through ii) when u >. Since y = H is outside the boundary layer the flux Q ii) may be rewritten Q ii) Continued on page 4.) Y ux,y)dy H udy = udy Uδ.

4 Examination in MEK43/93, Friday 15. June 212 Page 4 Zero net influx then implies = Q ii) +Q ii) = QH ux,y)dy Uδ = U u)dy Uδ. Rearrangement then gives δ = 1 U U u)dy 1 u ) dy. U Comment: This calculation is asymptotically valid for large H, as indicated by the occasional. 3b The boundary layer equations in this case read u x + y =, u u x +v u y = ν 2 u y 2. At the plate we have a no-slip slip condition ux,) =, vx,) =, for x, while the boundary layer flow is matched to the outer flow through lim u = U. y There is no matching condition on v. The above set is simplified in relation to the Navier-Stokes equations in the following manners 3c The pressure is adapted from the outer solution and is treated as known and constant through the boundary layer. For the Blasius profile the pressure is constant in the outer solution, hence no pressure gradient in the equation. Since the pressure is not unknown the y component of the momentum equation may be disregarded. The ν 2 u/ 2 x term is neglected, since variations along the boundary layer are small than the transverse ones. To find an expression for v it is slightly simpler to introduce the stream function, ψ, than to employ the continuity equation directly ψ y = u ψ = Uf η)dy = Ufη)+ψ x). Continued on page 5.)

5 Examination in MEK43/93, Friday 15. June 212 Page 5 We may choose ψ = at the streamline y = and an f such that f) =. Then ψ = Uf and v = ψ x = U ηf f ). substitution into the momentum equation implies ) Uf U y 2 f + U ηf f ) U f = νu 2f, which is re-arranged into U ff = f. ν Since the first factor on the left hand side dependens only on x while the restdepends only on η) itmust beaconstant. Thisimplies =const. and = B x. B is arbitrary and conveniently chosen such that = 2νx/U. The equation for f then becomes Boundary conditions are 3d f +ff = f. f) =, f ) =, f ) = 1. The End From point a) the boundary layer thickness is given by δ = 1 u ) dy = U 1 f η) ) dη = lim η fη)). η From the graph and table in figure 2 η = appears to be way outside the boundary layer and η fη) = for this η value. Hence 2νx νx δ = U = U. The shear stress is τ w = µ u) y = µu ρµu f ) =.33 x. The drag then is D τ w dx =.66 ρµux. Results are not valid near the leading edge δ is singular, for instance) and the solution becomes unstable for large Re x = Ux/ν.

Candidates must show on each answer book the type of calculator used. Log Tables, Statistical Tables and Graph Paper are available on request.

UNIVERSITY OF EAST ANGLIA School of Mathematics Spring Semester Examination 2004 FLUID DYNAMICS Time allowed: 3 hours Attempt Question 1 and FOUR other questions. Candidates must show on each answer book