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Friction and Ocean Turbulence Part II

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Friction and Ocean Turbulence Part II Friction and Ocean Turbulence Part II L. Goodman General Physical Oceanography MAR 555 School for Marine Sciences and Technology Umass-Dartmouth Unstratfied flow ! = !0 constant Example: Tidal flow over a mound Laminar Flow U Turbulent Flow H uL Reynolds Number Re = ; ! m2 u, L characteristic values of the mean flow, ! =10#6 sec For unstratfied flow " = "0 constant Turbulence occurs when Re > Rec ~ 3000 3 D Turbulence: Navier Stokes Equation (no gravity, no coriolis effect) Examples: tidal channel flow, pipe flow, river flow, bottom boundary layer) I II III IV 2 2 2 #ui #ui #ui #ui 1 #pi # ui # ui # ui +{u + v + w } = $ +! ( 2 + 2 + 2 ) #t #x #y #z "0 #x j #x #y #z I . Acceleration Homework: Use Dimensional analysis to show II . Advection (non-linear) that the Reynolds number, Re, goes as the ratio III. Dynamic Pressure of term II to the term IV. Interpret the Reynolds IV. Viscous Dissipation number criteria for the onset of turbulence. Flow past a circular cylinder as a function of Reynolds number From Richardson (1961). Laminar Cases Re =174 Re<1 Re = 20 Example : a toothpick Example: a finger moving at 1mm/s moving at 2cm/s Turbulent Cases Note: All flow at the same Reynolds number have the same streamlines. Flow past a 10cm diameter Re = 5,000 Re = 14,480 cylinder at 1cm/s looks the same as 10cm/s flow past a cylinder 1cm in diameter because in both cases Re = 1000. Re = 80000 Re = 1,000,000 Example: hand out of a car window moving at 60mph. Surface Wind Stress τ (Unstratified Boundary Layer Flow) Definition: Stress = force per unit area on a parallel surface ! ! air = # < uair 'wair ' > wind " air Air ! Water ! w = # < u 'w ' > " ! ! What is the relationship between and ? ! w ! air ! ! ! w =! air Empirical Formula for Wind Stress Drag Coefficient CD 2 ! = "CDU10 where U10 is the wind speed 10m above the water m 10#3 U <5 10 sec C = D m 2.5$10#3 U >5 10 sec Homework: Why do we use U 1 0 in the formula above? Why not use the wind speed right at the surface? Explain. Why do you think that the drag coefficient C D has a large change m in value when the wind speed (at 10m) increases above 5 ? sec Hint: what would cause a sudden increase in air turbulence with an increasing wind speed? Concept of Friction Velocity u* Definition ! = # < u 'w' >= (u*)2 " ! u* Characteristic velocity u* = of the turbulent eddies " Homework: Show that * u air = CD U10 * !air * * u w = u air = 3% of u air !w Example. If the wind at height of 10m over the ocean surface is 10 m/sec, calculate the stress at the surface on the air side and on the water side. Estimate the turbulent velocity on the air side and the water side. τ= ?, u*=? τ= ?, u*=? Since N .25 2 * ! air m m m u air = = = .5 U >5 C 2.5 10#3 kg sec 10 $ D = % "air 1.0 sec m3 2 kg #3 m 2 ! = "airCD (U10 ) = (1.0 3 )(2.5%10 )(10 ) N m sec .25 2 * ! w m m N u w = = = .016 .25 " kg sec ! = 2 =! air =! water w 1000 m m3 Vertical Turbulent Friction (Unstratified Boundary Layer Flows) Horizontal Equations (x, y) #u #u #u 1 #p 1 #! + u + v $ fv = $ ( )+ x #t #x #y " #x " #z #v #v #v 1 #p 1 #! y + u + v + fu = $ ( ) + dt #x #y " #y " #z Short hand form ! Du ! ! 1 1 # ! + f $u = % & p + ! Dt " h " #z ! ! ! D # ! where = % < u 'w' > & = + u '& " Dt #t Vertical Equation Hydrostatic condition "p = #!g "z Role of Bottom Stress ExampleExample : Steady: Steady State State, Channel Narrow flow Channel with a constantflow ( f =0), surface constant slope surface , α. (No slope wind) , α, no wind. Surface Stress Stress ζ Surface ! = 0 z = D Flow Direction Why? z p = !0 g(D # z +" ) Bottom Stress ! = $"0 g# D Bottom x z = 0 1 #p 1 # $p ${!0 g(D % z %" )} 0= $ + ! but = = %!0 g# x x "0 #x "0 #z $ $ #! $! = %"0 g#(D % z) Note " = < 0 #x Typical Values |! |%1cm /(1km to 10km) =10$5 to 10$6 & for D =10m & Friction velocity on the bottom is " u* = = g |! | D #0 & u* = (1$ 3)cm / sec Homework: For the example just given, using | ! | = 1 0 " 6 ; D = 1 0 m estimate the turbulent velocity at the surface, at the mid depth, and on the bottom. Where is the turbulence a maximum? Homework: Suppose in the example above the surface was completely flat, α = 0 ,but m there was wind with a speed of U 1 0 = 1 0 at a height of 10 m above the surface. sec Calculate the stress and estimate the turbulent velocity: (a) at the surface; (b) at mid depth; ( c ) on the bottom. Relating Stress to Velocity Viscous (molecular) stress in boundary layer flow Low Reynolds Number Flow Note: Viscous Stress is proportional to shear. %u m2 ! = " # # =10$6 molecular viscoisty 0 %z sec Turbulence Case: Eddy Viscosity Assumption #u ! = " k k eddy viscosity 0 #z Note. At a fixed boundary u = 0 because of molecular friction. In general k = k(z). Mixing Length Theory: Modeling k k = u!l l a characteristic length , u ! a characteristic velocity of the turbulence Back to constant surface slope example where we found that ! = $"0 g#(D $ z) α z = D z = 0 If we use the eddy viscosity assumption with constant k Note we have used the fact that $u 2 k g (D z) $(z) $(z ) ! = "0 = %"0 # % & =1 & = 2z $z $z $z g# z z u = % (D % ) 2 ! k 2 & (u*) = = %g" D #0 (u*)2 z = z(1% ) k 2D u* = bottom friction velocity m2 Homework: For the example just given using | á |= 10-6; D = 10m k = 10-4 sec estimate the mean velocity at the surface, at the mid depth, and on the bottom. Log Layer Note: in the previous example near the bottom, τ independent of z Bottom Boundary Layer ! " constant z Eddy size " to distance from bottom # k = !u * z $u ! = "0 (u*) = "0k $z z ! ln( ) $u u * z 1 % = 0 = $z # z !z z # = .4, Von Karman's constant u * z u = ln( ) z0 the roughness parameter # z0 Homework: The wind blows along a coastal channel of depth 40 meters with a speed of 10 m/sec. If the drag coefficient between the water and air is found to be 2.5 x 10-3, (a) calculate the stress induced by the wind on the water surface; (b) calculate the in air friction velocity, the in water friction velocity; © what is the stress on the bottom of the water? (d) what is the stress at a depth of 10m, 20 ,30? (d) what is the shear ! u at the surface, at a depth of 10m, on the bottom? ! z Assume that the flow acts as a log “layer “ layer in the lower 5 m of the m2 water, above which k = 1 0 -3 . sec Vertical Distribution of Turbulence Mixed Layer/Surface Ekman Layer (turbulent) Pycnocline (Intermittently turbulent) Bottom Ekman Layer (turbulent) Bottom Boundary Layer (turbulent log layer) Explain what produces turbulence in the: (a) mixed layer; (b) the bottom boundary layer; ( c) in the pycnocline. Turbulence in the Pycnocline Velocity Shear !u g #" N = $ ! !z " #z ρ Gradient Richardson Number Turbulence occurs when 2 N 1 Rig = Ri !u 2 g < ( ) 4 !z 1 Billow clouds showing a Kelvin-Helmholtz (Ri < ) g 4 instability at the top of a stable atmospheric boundary layer. Photography copyright Brooks Martner, NOAA Environmental Technology Laboratory. Depth(m) Distance (m) 1 Turbulence Observed in an internal solitary wave resulting in (Rig < ) Goodman and Wang (JMS, 2008) 4 kg m Homework: " = 22 u = .7 z = 0 m A certain marginal sea of depth 1000m has ! m3 sec kg m a density and velocity profile as shown in the figure " = 22 u = .6 z =50m below with values given in the table to the right. ! m3 sec kg m Assume that the density and velocity vary linearly " = 22.1 u = .1 z =100m within each “layer”. Calculate and sketch the N 2 ! m3 sec !u kg m and profiles. Where should there be turbulence? " = 26 u = .05 z =800m !z ! m3 sec kg m " = 26 u = 0 z =1000m ! m3 sec z =0 m z =50 m z =100 m " ! u z =800 m z =1000 m.
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