Describing the tide at a point along an estuary (i.e., x = constant)

Transcription

1 MS 520 Part I: Estuarine Tides -- Carl Friedrichs A. Describing the observed tide in estuaries (Lecture 1) pp. 1-4 B. Scaling the dynamics of tides in estuaries (Lecture 1) pp C. Tides in short estuaries (Lecture 2) pp D. Tides is long, shallow, funnel-shaped estuaries (Lecture 2,3) pp E. Tides in long, deep, straight estuaries (Lecture 3) pp F. Tides in long, shallow, straight estuaries (Lecture 4) pp G. Tides in long estuaries, general equilibrium estuary (Lecture 4) pp H. Tides in intermediate length, deep, constant width estuaries (Lecture 5) pp I. Tides in intermediate length, constant width, arbitrary depth estuaries (Lecture 5) pp J. Summary (Lecture 5) p. 49 Describing the tide at a point along an estuary (i.e., x = constant) p.01 Tidal elevation at a point in space (at lowest order) = η(t) = a cos ωt Tidal velocity at a point in space (at lowest order) = u(t) = U cos (ωt - φ) a = tidal elevation amplitude (= 1/2 tidal range), U = tidal velocity amplitude ω = tidal frequency = 2π/T, T = tidal period T 12.4 hours for M 2 tide, ω 1.4 x 10-4 s -1 φ = phase of tidal velocity relative to tidal elevation. Greater φ means greater delay in waveform. 1 Elevation Velocity (flood positive) Tidal elevation (η/a) Tidal velocity (u/u) Time ω t/(2 π) In above example, φ = - π /2 = - 90 deg, i.e., velocity leads elevation by ~ 6 hours. (Caution -- sign depends on whether flood is positive or negative). This standing wave type of relative phase is common in short tidal estuaries. In a progressive wave, φ = 0, i.e., velocity and elevation are in phase. (If ebb is positive, φ = 180 deg.) Wind waves approaching a beach from offshore are progressive.

2 Phase of tidal velocity relative to tidal elevation in real estuaries: p deg = standing 0 deg = progressive ( measures exponential width convergence) (Friedrichs & Aubrey, JGR, 99: ) In real estuaries,(absolute) relative phase is between 0 and 90 deg and varies within individual estuaries. Example of real velocity - stage curve for a tidal marsh channel: p.03 x (Bayliss-Smith et al., ECSS, 9: ) X = observation site Tides in very shallow water can be highly distorted.

3 p.04 Describing the tide along an estuary (i.e., x varying in space) Tidal elevation and velocity in space (+x into estuary from sea) and time (at lowest order): η(x,t) = a(x) cos (ωt - kx) u(t) = U(x) cos (ωt - kx - φ) k = wave number = 2π/L t, L t = tidal wave length, c = ω/k = phase speed = L t /T k and φ can be functions of x, but are often approximated as constant. Ex. York River-ish: If c = 10 m/s, then η(x = 0) (mouth) and η(x = 40 km) (West Point) are: 1 Mouth West Point Tidal elevation (η/a) Time ω t/(2 π) kx = phase of tide relative to mouth of estuary. Phase kx (delay of wave form) increases into the estuary. p.05 Spatial variation in tidal amplitude and phase along real estuaries: a(x) L = 215 km, 95 km, 21 km kx (kx is either dimensionless (i.e., radians) or in degrees. π radians = 180 deg) (Friedrichs & Aubrey, JGR, 99: )

4 Dynamics of tides in estuaries: p.06 x = along-channel distance x b(x,t) = width of estuary w(x,t) = width of channel h(x,t) = cross-sectionally averaged depth of channel Cross-sectionally averaged momentum equation along a tidal estuary or embayment: u + u u x = g η x τ b ρh u = cross-sectionally averaged velocity τ b = bottom drag, ρ = water density Local acceleration + advective acceleration = pressure gradient + bottom friction (Note: Pressure gradient is barotropic -- i.e., ρ/ x is unimportant) Cross-sectionally integrated continuity equation: b η = x (whu) Rate of change in cross-sectional area of estuary = along-channel convergence in flux in channel (outside the channel the along-channel velocity is assumed to be negligible) p.07 Scaling Momentum u + u u x = g η x τ b ρh Local acceleration + advective acceleration = pressure gradient + bottom friction Bottom stress: τ b = ρ c d u u ρ c d {8/(3π)} U u Linearization of quadratic velocity formulation Drag coefficient: c d = 0.01 to depending on roughness scale Friction coefficient: r = c d {8/(3π)} U/h Relative importance of terms: O( ) indicates order of, i.e., approximate magnitude Local acceleration = u/ = / { U(x) cos (ωt - kx - φ) } O(ωU) Advective acceleration = u u/ x = u / x { U(x) cos (ωt - kx - φ) } O( U(dU/dx + ku) ) O( U(U/L U + ku) ), where L U is the length-scale over which U changes Pressure gradient = g η/ x = g / x {a(x) cos (ωt - kx) } O( g(da/dx + ka) ) O( g(a/l a + ka) ), where L a is the length-scale over which a changes Bottom friction = τ b /(ρh) = ru cos (ωt - kx - φ) O(rU) Bottom friction/local acceleration ru/(ωu) = r/ω Advective acceleration/local acceleration U(U/L U + ku)/(ωu) = U/(ω L U ) + ku/ ω = U/(ω L U ) + U/c

5 Scaling Continuity b η = x (whu) p.08 Rate of change in cross-sectional area of estuary = along-channel convergence in flux in channel (outside the channel the along-channel velocity is assumed to be negligible) b(x,t) = width of estuary b LOW TIDE w b HIGH TIDE z = a z = 0 z = -a h(x,t) = cross-sectionally averaged depth of channel w(x,t) = width of channel < > = time-average b = amplitude of tidal estuary width change = (b HIGH TIDE - b LOW TIDE )/2 Change in x-sect area = b η/ = (<b> + ( b/a) η) η/ = <b> η/ + ( b/a) η η/ = Linear + Non-linear term O(b ω a + ( b/a) a ω a) = O(b ω a + ( b) ω a) Ratio of non-linear to linear term in change of x-sect area = b/b p.09 Scaling Continuity (cont.) b η = x (whu) Rate of change in cross-sectional area of estuary = along-channel convergence in flux in channel along-channel convergence in flux in channel x (whu) = w (<h> + η) u linear part 1 w w x + 1 h non-linear part h x + 1 u u x = w <h> u { 1 + (a/<h>)(η/a) } (1/L w + 1/L h + 1/L U + k) along-channel change in: width depth velocity velocity amplitude phase Ratio of non-linear to linear portion in along-channel flux convergence = a/h Ratio of along channel change in velocity phase to change in width = kl w, etc.

6 p.10 Estimating characteristic length-scales from data: (i) Estimating wave number, k: (Friedrichs & Aubrey, JGR, 99: ) Delaware L = 215 km Thames L = 95 km Tamar L = 21 km Phase = kx (First translate degrees or hours to radians) In Matlab: f = polyfit(x,phase,1); k = f(1) Result: Delaware k = 1/(58 km) ± 2% (one standard error) Thames k = 1/(70 km) ± 18% Tamar k = 1/(57 km) ± 14% p.11 (ii) For other characteristic length scales, fit observations to: F(x) = F o exp(±x/l F ) (such that L F is positive) Then log(f) = log(f o ) ± (1/L F ) x x/l x/l In Matlab: f = polyfit(x,log(f),1); 1/L F = abs(f(1)) (Friedrichs & Aubrey, JGR, 99: ) x/l L 1/L A 1/L b 1/L U Delaware 215 km (1/38) km -1 ± 3% (1/40) km -1 ± 3% (1/1700) km -1 ± 109% Thames 95 km (1/19) km -1 ± 4% (1/25) km -1 ± 3% (1/280) km -1 ± 39% Tamar 21 km (1/5.3) km -1 ± 4% (1/4.6) km -1 ± 9% (1/160) km -1 ± 109%

7 MS 520 Part I: Estuarine Tides -- Carl Friedrichs A. Describing the observed tide in estuaries (Lecture 1) pp. 1-4 B. Scaling the dynamics of tides in estuaries (Lecture 1) pp C. Tides in short estuaries (Lecture 2) pp D. Tides is long, shallow, funnel-shaped estuaries (Lecture 2,3) pp E. Tides in long, deep, straight estuaries (Lecture 3) pp F. Tides in long, shallow, straight estuaries (Lecture 4) pp G. Tides in long estuaries, general equilibrium estuary (Lecture 4) pp H. Tides in intermediate length, deep, constant width estuaries (Lecture 5) pp I. Tides in intermediate length, constant width, arbitrary depth estuaries (Lecture 5) pp J. Summary (Lecture 5) p. 49 p.12 Tides in short estuaries: Using continuity alone to solve for U: x x = L b η b η = x (whu) dx = x = L (whu) dx x x R.H.S. = ( w h u x = L w h u x ) = w h u No tidal flux at landward end of estuary L.H.S. = [ 1 + O{ kl + L/L a } ] η 0 x = L b dx x Errors due to along-estuary variation in (i) tidal phase (ii) tidal amplitude u(x,t) η 0 1 w h x = L b dx A b = estuarine surface area upstream of x A c = estuarine channel cross-sectional area at x x = η 0 A b A c with error O{ kl + L/L a } Small error if estuary is short relative to 1/k and L a

8 p.13 Error from using continuity alone to solve for U: u(x,t) η 0 A b A c with error O{ kl + L/L a } (Friedrichs & Aubrey, JGR, 99: ) L 1/L a k kl + L/L a Delaware 215 km (1/570) km -1 ± 15% (1/58) km -1 ± 2% 4.0 ± 3% Thames 95 km (1/1500) km -1 ± 270% (1/70) km -1 ± 18% 1.4 ± 30% Tamar 21 km (1/88) km -1 ± 35% (1/57) km -1 ± 14% 0.6 ± 22% None of these systems is short because kl + L/L a are all too big. Small ratio depends on situation, but errors ~ 1/4 are often sufficiently small to neglect at lowest order and still gain useful insight into physics. p.14 Using continuity alone to solve for U in very short estuaries: O{ kl + L/L a } << 1 u(x,t) η 0 A b A c A b = estuarine surface area upstream of x A c = estuarine channel cross-sectional area at x At lowest order, η o (t) a o cos ωt, giving u(x,t) a o ω (sin ωt) { (A b (x,t)/a c (x,t) } Spatial variation in tidal velocity amplitude: U(x) a o ω A b (x)/a c (x) U(x) x = 0 x = L U(x) x = 0 x = L

9 p.15 Using continuity alone to solve for U in very short estuaries (cont.): O{ kl + L/L a } << 1 Spatial variation in tidal velocity amplitude: U(x) a o ω A b (x)/a c (x) U(x) x = 0 x = L U(x) x = 0 x = L u(t) in entrance of very short estuary: u(t) η 0 A b = η 0 A c A b (t) w h(t) Hypsometry p.16 Case (i): Estuarine hypsometry dominated by tidal variations in channel depth. Currents in channel strongest closer to low tide when depth is small. Case (ii): Hypsometry dominated by tidal variations in estuary surface area. Currents in channel strongest closer to high tide when surface area is large. (Pethick, 1980, ECSS, 11: )

10 MS 520 Part I: Estuarine Tides -- Carl Friedrichs A. Describing the observed tide in estuaries (Lecture 1) pp. 1-4 B. Scaling the dynamics of tides in estuaries (Lecture 1) pp C. Tides in short estuaries (Lecture 2) pp D. Tides is long, shallow, funnel-shaped estuaries (Lecture 2,3) pp E. Tides in long, deep, straight estuaries (Lecture 3) pp F. Tides in long, shallow, straight estuaries (Lecture 4) pp G. Tides in long estuaries, general equilibrium estuary (Lecture 4) pp H. Tides in intermediate length, deep, constant width estuaries (Lecture 5) pp I. Tides in intermediate length, constant width, arbitrary depth estuaries (Lecture 5) pp J. Summary (Lecture 5) p. 49 p.17 Tides in long estuaries: (1) Shallow, funnel-shaped estuaries -- lowest order solution Scaling Continuity b η = x (whu) Rate of change in cross-sectional area of estuary = along-channel convergence in flux in channel b(x,t) = width of estuary b = (b HIGH TIDE - b LOW TIDE )/2 b LOW TIDE w b HIGH TIDE z = a z = 0 z = -a h(x,t) = cross-sectionally averaged depth of channel w(x,t) = width of channel < > = time-average Change in x-sect area = b η/ = (<b> + ( b/a) η) η/ = Linear + Non-linear term O(b ω a + ( b) ω a) Ratio of non-linear to linear term in change of x-sect area = b/b For first-order case, assume b/b << 1 and w b Change in x-sect area: <b> η w η

11 p.18 (1) Shallow, funnel-shaped estuaries -- scaling continuity (cont.) along-channel convergence in flux in channel x (whu) = w (<h> + η) u 1 w w x + 1 h h x + 1 u u x linear part non-linear part = O( w <h> u { 1 + (a/<h>) } (1/L w + 1/L h + 1/L U + k) ) along-channel change in: width depth velocity velocity amplitude phase For first-order case, assume L w L b << L h, L U, 1/ k. Also assume a/h << 1 For first-order funnel shape, w b w o exp (-x/l b ) (1/w) w/ x = 1/(w o exp (-x/l b )) / x (w o exp (-x/l b )) = exp (x/l b ) (-1/L b ) exp (-x/l b )) = - 1/L b Then / x (whu) w <h> u (-1/L b ) - whu/l b Continuity: b η = x (whu) w η = whu η L b = hu L b p.19 (1) Shallow, funnel-shaped estuaries -- continuity (cont.) η = hu L b Change in cross-sectional area = Flux convergence due to width change Solve for velocity: u = (L b /h) η/ Elevation is defined to be: η = a cos (ωt - kx) Then velocity is: u = (L b /h) (-a ω) sin (ωt - kx) = a ω L b /h sin (ωt - kx) = U sin (ωt - kx) Tidal velocity amplitude: U = a ω L b /h Trig identity: u = U sin (ωt - kx) = U sin (kx - ωt) = U cos (kx - ωt + π/2) = U cos (ωt - kx - π/2) From earlier lecture, u is defined to be: u = U cos (ωt - kx + φ) Then velocity relative phase, φ = - π/2 = - 90 deg, i.e., Standing Wave relation

12 p.20 (1) Shallow, funnel-shaped estuaries -- Example: Thames River Estuary Solution: η = a cos (ωt - kx), u = U sin (ωt - kx), U = a ω L b /h, φ - 90 deg Assumes: a/h << 1, b/b << 1, L w L b << L h, L U, 1/ k Values for Thames: a/h 0.24, b/b 0.17, L w 18 km, L b 25 km, L h 70 km, L U 280 km, 1/k 70 km Poorest approximations are L b /L h 0.36 << 1 and kl b 0.36 << 1 Solution predicts: Velocity relative phase φ - 90 deg Velocity amplitude U a ω L b /h (2.0 m)(1.4 x 10-4 s -1 )(25 x 10 3 m)/(8.5 m) = 0.82 m/s p.21 (1) Shallow, funnel-shaped estuaries -- Example: Thames River Estuary Poorest approximations are L b /L h 0.36 << 1 and kl b 0.36 << 1 (Hunt, 1963, Geophys. J. Roy. Astronom. Soc., 8: ) river ocean For example, the length scale for along-channel variation in depth, L h 70 km, is relatively short. Thus we don t expect the analytical predictions to be exact.

13 (1) Shallow, funnel-shaped estuaries -- Example: Thames River Estuary (cont.) p.22 Key features of the observed tidal velocity are similar to the predicted trend: 1) Observed U O(0.8 m/s) (within ~ 36%) 2) Velocity leads tidal elevation with a relative phase of about 90 deg. (Hunt, 1963, Geophys. J. Roy. Astronom. Soc., 8: ) p.23 Tides in long estuaries: (1) Shallow, funnel-shaped estuaries -- lowest order solution Scaling Momentum u + u u x = g η x r u Local acceleration + advective acceleration = pressure gradient + bottom friction Order of magnitude: ± O(ωU) ± O( U(U/L U + ku) ) ± O( g(a/l a + ka) ) ± O(rU) Advective acceleration/local acceleration (U/ ω)(1/l U + k) = {(0.8 m/s)/(1.4 x 10-4 s -1 )} {1/(280 km) + 1/(70 km)} = 0.10 << 1 Neglecting advective acceleration, momentum becomes: Plug in u = U sin (ωt - kx), η = a cos (ωt - kx) and get: u = g η x r u ω cos (ωt - kx) g a k sin (ωt - kx) + ru sin (ωt - kx) Because phase doesn t match other terms, nothing can balance this at lowest order. It must be a second order term. These two terms can cancel at lowest order i.e., If convergence in width dominates continuity (L w L b << L h, L U, 1/ k), it must follow that friction dominates acceleration (ω/r << 1) -- we ll confirm this in a bit. Reason -- If friction didn t dominate acceleration, rapid convergence would cause amplitude to increase along channel (creating small L U ), I.e., the estuary would be hypersynchronous.

14 p.24 Tides in long estuaries: (1) Shallow, funnel-shaped estuaries -- Scaling momentum (cont.) Lowest-order Momentum 0 = g η x r u 0 = pressure gradient + bottom friction 0 g a k sin (ωt - kx) + ru sin (ωt - kx) U = g a k/r From continuity we also had U = a ω L b /h, eliminating U and a gives: Phase speed c = ω/k = gh/(rl b ) Then for Thames (from rearranging relation above relation for c) ω/r = ω 2 L b /(kgh) = (1.4 x 10-4 s -1 ) 2 (25 km)/{(9.8 m/s 2 )(8.5 m)(1/70 km)} 0.4 which is consistent with assuming that kl b 0.36 << 1 p.25 Tides in long estuaries: (1) Shallow, funnel-shaped estuaries -- Gives 1st Order Wave Equation Lowest-order Momentum: 0 = g η x r u Lowest-order Continuity: η = hu L b Eliminate u and get: η + gh rl b η x = 0 or, equivalently η + c η x = 0 In physics, this is known as a 1st-Order Wave Equation because the derivatives are / and / x, not 2 / 2 and 2 / x 2, which would make it a 2nd-Order Wave Equation. A 1st-order wave equation allows propagation of waves in only one direction, in this case toward +x. Reflected waves (which propagate toward -x) are not allowed. Physically, reflected waves don t occur in this type of system because waves propagating toward -x have their energy spread by divergence of width as well as being damped by friction. Waves propagating toward +x are frictionally damped too, but convergence concentrates wave energy quickly enough to maintain an equilibrium, constant wave amplitude.

15 MS 520 Part I: Estuarine Tides -- Carl Friedrichs A. Describing the observed tide in estuaries (Lecture 1) pp. 1-4 B. Scaling the dynamics of tides in estuaries (Lecture 1) pp C. Tides in short estuaries (Lecture 2) pp D. Tides is long, shallow, funnel-shaped estuaries (Lecture 2,3) pp E. Tides in long, deep, straight estuaries (Lecture 3) pp F. Tides in long, shallow, straight estuaries (Lecture 4) pp G. Tides in long estuaries, general equilibrium estuary (Lecture 4) pp H. Tides in intermediate length, deep, constant width estuaries (Lecture 5) pp I. Tides in intermediate length, constant width, arbitrary depth estuaries (Lecture 5) pp J. Summary (Lecture 5) p. 49 p.26 Tides in long estuaries: (2) Deep, straight estuaries -- lowest order solution Scaling Continuity b η = x (whu) For first-order case, assume k >> 1/L w, 1/L h, 1/L a, 1/L U. Also assume a/h << 1, b/b << 1, w b L.H.S.: b η/ w / (a cos (ωt - kx)) = w ω a sin (ωt - kx) R.H.S.: / x (whu) w h u/ x w h / x (U cos (wt - kx + φ)) = w h k U sin (ωt - kx + φ) L.H.S. = R.H.S. : w ω a sin (ωt - kx) = w h k U sin (ωt - kx + φ) Then U = (a/h) (ω/k) φ = 0 deg Progressive wave relation

16 p.27 (2) Long, deep, straight estuaries -- Example: Chesapeake Bay Solution: η = a cos (ωt - kx), u = U cos (ωt - kx), φ = 0 deg Chesapeake Bay CBOS Buoy, mid-bay, x 170 km from mouth along-channel velocity u (m/s) η (meters) surface elevation Date in February 1999 Observed φ 0 to -30 deg, observed U 25 cm/s p.28 Tides in long estuaries: (2) Deep, straight estuaries -- lowest order solution Scaling Momentum u + u u x = g η x r u Neglect u u/ x relative to u/, which assumes (U/ω)(1/L U + k) << 1 Neglect ru relative to u/, which assumes r/ω << 1 Momentum becomes: u/ g η/ Plug in u = U cos (ωt - kx), η = a cos (ωt - kx) from continuity and get: ω U sin (ωt - kx) = g a k sin (ωt - kx), which gives U = gak/ω From continuity we had U = (a/h) (ω/k) Eliminating U gives: ω/k = (gh) 1/2 U = (a/h) (gh) 1/2

17 p.29 (2) Long, deep, straight estuaries -- Example: Chesapeake Bay Solution: η = a cos (ωt - kx), u = U cos (ωt - kx), U = (a/h) (gh) 1/2, φ = 0 deg, ω/k = (gh) 1/2 Assumes: 1/k << L w, L h, L a, L U. Also a/h << 1, b/b << 1, r/ω << 1 η (meters) Chesapeake Bay NOAA observed tidal elevations kx 57 deg 1 radian Windmill Point, Virginia x 70 km Solomons Island, Maryland x 150 km Date in February Wave number Tidal amplitude k = ( kx)/( x) = (1 radian)(80 km) 1/(80 km) a 0.25 m p.30 (2) Long, deep, straight estuaries -- Example: Chesapeake Bay Observed phase speed c = ω/k = (1.4 x 10-4 s -1 ) 2 (80 km) = 11 m/s c = ω/k = (gh) 1/2 gives effective depth h c 2 /g = (11.2 m/s) 2 /(9.8 m/s 2 ) = 12 m Non-linear/linear term a/h 0.02 << 1 Predicted amplitude of tidal velocity U (a/h) (ω/k) (0.02)(11 m/s) = 0.22 m/s Observed tidal velocity amplitude 0.25 m/s Friction/acceleration = r/ω c d U/(hω) (0.003) (0.25 m/s)/{(12 m)(1.4 x 10-4 s -1 )} = 0.44 Thus neglecting friction is a relatively poor assumption.

18 p.31 (Carter & Pritchard, 1988) Tidal amplitude (roughly) constant, suggesting balance between channel convergence increasing amplitude and frictional dissipation decreasing amplitude, i.e., kl b ω/r L b (ω/r)(1/k) (0.44) -1 (80 km) 180 km (Seems reasonable ) In lower bay, width increases landward, leading to rapid decrease in amplitude with distance into estuary. (Higher amplitude on east side is 2-D Kelvin wave more on this later.) p.32 Tides in long estuaries: (2) Deep straight estuaries -- Gives 2nd Order Wave Equation Lowest-order Momentum: u = g η x 2 u x = g 2 η x 2 Lowest-order Continuity: η = h u x 2 η 2 = h 2 u x Eliminate 2 u and get: 2 η gh 2 η or, equivalently x 2 x = 0 2 η 2 2 c 2 2 η x 2 = 0 In physics, this is known as a 2nd-Order Wave Equation because the derivatives are 2 / 2 and 2 / x 2. A 2nd-order wave equation allows propagation of waves in both +x and -x. Reflected waves (which propagate toward -x) are allowed. Physically, reflected waves can occur in this type of system because the mild rate of channel width change does not change cause rapid divergence of reflected wave energy. Likewise, weak friction has a mild impact on reflected wave amplitude. (In the Chesapeake Bay, width convergence and friction are still sufficient to damp most of the wave reflected off the head of the bay.)

19 MS 520 Part I: Estuarine Tides -- Carl Friedrichs A. Describing the observed tide in estuaries (Lecture 1) pp. 1-4 B. Scaling the dynamics of tides in estuaries (Lecture 1) pp C. Tides in short estuaries (Lecture 2) pp D. Tides is long, shallow, funnel-shaped estuaries (Lecture 2,3) pp E. Tides in long, deep, straight estuaries (Lecture 3) pp F. Tides in long, shallow, straight estuaries (Lecture 4) pp G. Tides in long estuaries, general equilibrium estuary (Lecture 4) pp H. Tides in intermediate length, deep, constant width estuaries (Lecture 5) pp I. Tides in intermediate length, constant width, arbitrary depth estuaries (Lecture 5) pp J. Summary (Lecture 5) p. 49 p.33 Tides in long estuaries: (3) Shallow, straight estuaries Continuity b η = wh u x Assumes a/h << 1, b/b << 1, k >> 1/L w 1/L h. Does not assume k >> 1/L a or 1/L U Momentum 0 = g η x r u Assumes ω/r << 1 / x {Momentum} to eliminate u in continuity gives η = wgh br 2 η x 2 Or, equivalently, η = D 2 η x 2 D = wgh br In physics, this is known as a Diffusion Equation, with D = Diffusion Coefficient Solution η = a o exp(-kx) cos (ωt - kx), u = U o exp(-kx) cos (ωt - kx - π/4), φ = - 45 deg U o = (a o /h) (b/w) (ωd) 1/2 Phase speed c = ω/k = (2ωD) 1/2

20 p.34 (3) Long, shallow, straight estuaries -- Example: The Fleet, England (Robinson et al., 1983, ECSS, 16: ) 3.3 hrs 1.7 rad 5 km k = kx/x = (1.7 rad)/(5 km) 1/2.9 km b 390 m, w 130 m, h 0.7 m, a o 0.7 m c = ω/k (1.4 x 10-4 s -1 )(2.9 km) = 0.4 m/s D = c 2 /(2ω) 570 m 2 /s, ω/r = ωdb/(wgh) 0.04 Thus frictional dominance is a very good assumption. Is it reasonable to neglect nonlinearity? NO: a/h (0.5 m)/(0.7 m) 0.7 p.35 Tidal asymmetry in shallow, straight estuaries: Tidal phase speed c = ω/k = (2ωD) 1/2 = 2ωwgh br 1/2 Friction parameter r = 8 3π c d U h c = 3 π ωwg h 2 4 c d Ub 1/2 c ~ h/(b 1/2 ) = channel depth estuary width 1/2 If the channel is much deeper at high tide than at low tide, high tide will propagate into the estuary faster. If the estuary is much wider at high tide than at low tide, low tide will propagate into the estuary faster. This results in tidal asymmetries which favor flood or ebb dominance.

21 p.36 Tidal asymmetry in shallow, straight estuaries: Phase speed c ~ h/(b 1/2 ) = channel depth estuary width 1/2 Case (i) h HT /(b HT ) 1/2 > h LT /(b LT ) 1/2 so C HT > C LT Flood Dominance Case (ii) h HT /(b HT ) 1/2 < h LT /(b LT ) 1/2 so C HT < C LT Ebb Dominance p.37 Tidal asymmetry in shallow, straight estuaries (cont.): Tidal Phase Speed c ~ (h 2 /b) 1/2 h 2 /b = (<h> + η ) 2 /(<b> + b (η/a) ) <h 2 /b> (1 + 2(a/h)(η/a) )/(1 + ( b/b)(η/a) ) <h/b> (1 + 2(a/h)(η/a) ( b/b)(η/a) ) = <h/b> (1 + γ (η/a) ) γ = 2(a/h) ( b/b) = asymmetry parameter If γ > 0, tidal changes in depth dominate changes in width Then high tide propagates faster than low tide flood dominance If γ < 0, tidal changes in width dominate changes in depth Then low tide propagates faster than high tide ebb dominance

22 Numerical solution (Friedrichs & Aubrey, 1988) p.38 Analytical solution for γ = 0 Intertidal Storage Volume Volume in Channels 2a b hb (1 - b/b) γ < 0 γ > 0 = a/h γ = 0 when b/b = 2a/h, i.e., when Intertidal Storage Volume Volume in Channels 4(a/h) 2 (1-2a/h) γ = 2(a/h) ( b/b) = asymmetry parameter is a simpler, more physics-based parameter for distinguishing between flood- vs. ebb-dominant (shallow) tidal estuaries. MS 520 Part I: Estuarine Tides -- Carl Friedrichs A. Describing the observed tide in estuaries (Lecture 1) pp. 1-4 B. Scaling the dynamics of tides in estuaries (Lecture 1) pp C. Tides in short estuaries (Lecture 2) pp D. Tides is long, shallow, funnel-shaped estuaries (Lecture 2,3) pp E. Tides in long, deep, straight estuaries (Lecture 3) pp F. Tides in long, shallow, straight estuaries (Lecture 4) pp G. Tides in long estuaries, general equilibrium estuary (Lecture 4) pp H. Tides in intermediate length, deep, constant width estuaries (Lecture 5) pp I. Tides in intermediate length, constant width, arbitrary depth estuaries (Lecture 5) pp J. Summary (Lecture 5) p. 49

23 Tides in long estuaries: (4) General equilibrium estuary (Ex. James River) Look for solution of form η = a cos (ωt - kx), u = U cos (ωt - kx + φ) Assume equilibrium channel so that a, U nearly constant in space Also assume k and/or 1/L b >> 1/L h, 1/L a, 1/L U. Also assume a/h << 1, b/b << 1, w b ~ exp(-x/l b ) Do NOT assume k >> 1/L w, 1/L b and do NOT assume r/ω << 1 Continuity: Change in x-sect area = divergence in flux due to width and phase Momentum: Acceleration = Pressure gradient + Friction Plug in form for η and u and after some algebra you find lowest order solution: General equilibrium estuary Shallow, strongly Funnel-shaped Deep, Constant width ω/r O(1), kl b O(1) ω/r << 1, kl b << 1 ω/r >> 1, kl b >> 1 U ω a L b h {1 + ( k L b ) 2 } 1/2 U (a/h) ω L b U (a/h) ( ω/k) tan φ = 1/(kL b ) φ = 90 deg φ = 0 deg Tides in long estuaries: (4) General equilibrium estuary (cont.) Next allow η and U to change slowly along estuary So that a(x) ~ U(x) ~ exp (µkx), µ = amplitude growth factor If µ > 0, a and U increase into estuary ( hyper-synchronous ) If µ < 0, a and U decrease into estuary ( hypo-synchronous ) Scaling requires µ << kl b, ω/r Perturbation approach similar to analysis of asymmetry factor yields µ = ω/r kl b For pure equilibrium estuary (a, U constant), ω/r = kl b c = ω/k = rl b Recall from previous notes: Shallow, strongly Funnel-shaped Deep, Constant width c = gh/(rl b ) c = (gh) 1/2 Relation requires c = gh/(rl b ) = rl b = (gh) 1/2 for all equilibrium estuaries! For µ 0, perturbation solution gives c = (gh) 1/2 (ω/r)/(kl b )

24 Tides in intermediate length estuaries: (1) Deep, constant width, constant depth Governing equation from last lecture: Solution: 2 η 2 c 2 2 η x 2 = 0 η = a cos (ωt - k(x-l)) + a cos (ωt + k(x-l)) u = U cos (ωt - k(x-l)) U cos (ωt + k(x-l)), 2 u 2 c 2 2 u x 2 = 0 Incident wave + Reflected wave Apply boundary condition : η(x = 0) = a o cos ωt Substitute in, apply trig identities, get η = a o cos k(l-x) cos kl cos ωt u = (gh) 1/2 a o sin k(l-x) h cos kl sin ωt vs. general form η = a obs cos (ωt - k obs x), u = U obs cos (ωt - k obs x + φ) ω/k = (gh) 1/2 and U = (a/h)(ω /k) above, but k k obs. Similarly a a obs, U U obs. Above relations predict k obs = 0, i.e., c obs = ω/k obs is predicted to be infinite. Also a obs and U obs change in x, and φ = 0 ( standing wave relation ). The situation where these solutions would apply at lowest order should have (gh) 1/2 /c obs << 1 and L/L a = O(1). If L/L a << 1, then use the short estuary solution. Intermediate length, deep, constant width, constant depth Ex. Gulf St. Vincent, Australia Bowers & Lennon, 1990, ECSS, 30: amplitude (meters) phase (deg) 120 km k obs = (kl obs /L) (0.26 rad)/(120 km) = 1/(460 km) c obs = ω/k obs (1.4 x 10-4 s -1 )(460 km) = 64 m/s (gh) 1/2 /c obs {(9.8 m/s 2 )(25 m)} 1/2 /(64 m/s) = 0.24 L a /L a/a obs (0.3 m)/(0.4 m) = 0.75

25 Intermediate length, deep, constant width, constant depth: Nodes and Antinodes k(l-x) = 0 k(l-x) = π/2 a(x) ~ cos k(l-x), U(x) ~ sin k(l-x) There is an maximum or antinode in elevation amplitude (and a minimum or node in velocity amplitude) at k(l-x) = 0. If the system is long enough, there will be a node in elevation (and an antinode in elevation) at k(l-x) = π/2. An antinode occurs where the incident and reflected waves reinforce each other. A node occurs where the incident and reflected waves cancel each other. (from Thurman, 1981) Intermediate length, deep, constant width, constant depth: Quarter wave resonance a ~ U ~ 1/(cos kl) If kl = π/2, then cos kl = 0 and a and U become infinitely large. This happens because when kl = π/2, the incident and reflected waves for elevation cancel at the mouth, i.e., the node for elevation occurs at the mouth and the ratio of elevation at the mouth relative to the inner estuary is zero. However, the tide at the estuary mouth is constrained to equal the non-zero ocean tidal amplitude, a o. Thus the tide within the estuary becomes infinitely large in comparison. Of course the estuarine tidal amplitude never actually becomes infinite, since infinitely strong currents would generate intense bed friction, damping tidal amplitude within the tidal estuary (and violating the assumption that r/ω << 1). This phenomena is termed quarter wave resonance because when kl = π/2, L = (π/2)(1/k) = (π/2)l tide /(2π) = L tide /4 Ex. Bay of Fundy, h 100 m. Resonant length = T (gh) 1/2 /4 (12.4 x 3600 sec){(9.8 m/s 2 )(100 m)} 1/2 /4 = 350 km

26 (from Prandle & Rahman, 1980) Tides in intermediate length estuaries: (2) constant width, constant depth, arbitrary depth Governing equation becomes: 2 η 2 + r η gh 2 η x 2 = 0 Properties of both a second-order wave equation and a diffusion equation. Reflected and incident wave, but with wave decaying as it propagates Solution: η = a e -µkx cos (ωt - k(x-l)) + a e µkx cos (ωt + k(x-l)) u = U e -µkx cos (ωt - k(x-l) + φ) U e µkx cos (ωt + k(x-l) + φ) Damped reflection: Incident wave + Reflected wave In this case µ is a positive number and always causes the incident and reflected waves to decay with distance as they propagate. Substitute in, apply b.c.s, get µ = ( 1 + (ω/r) 2 ) -1/2, µ = 0 for frictionless case, µ = 1 for strong friction η/η o = (1/2) 1/2 (cos 2kx + cosh 2µkx) 1/2, U = (a/h)(ω/k)(1 + µ 2 ) 1/2, tan φ = µ 2 gh = (ω/k) 2 (1 + (r/ω) 2 ) ) 1/2 { 1 + (1 + (r/ω) 2 ) -1/2 }

27 Damped reflection (cont): (Bowers & Lennon, 1990, ECSS, 30: 17-34)

28

29 SUMMARY OF TIDES IN ESTUARIES CASE MOMENTUM Continuity Assumptions U φ(deg) c Short None All terms kl, L/L a << 1 aωa b /Ac - 90 infinite (Swash Bay) Shallow, Friction + η/ + kl b << 1, aωl b /h - 90 gh/(rl b ) Funnel Press. Grad. width change ω/r << 1, kl a >> 1 (Thames) Deep, Accel. + η/ + kl b >> 1, (a/h)(gh) 1/2 0 (gh) 1/2 Straight, Press. Grad. phase change ω/r >> 1, kl a >> 1 Long (Chesapeake) Long, Friction + η/ + ω/r << 1, (a o /h)(b/w)(ωd) e -kx - 45 (2ωD) 1/2 Straight, Press. Grad. phase change + kl b >> 1 D = (wgh)/(br) Shallow amplitude change kl a = O(1) (Fleet) General Accel. + η/ + kl a >> 1. walb. 0 to gh/(rl b ) "Equilibrium" Press. Grad. phase change + ω/r kl b h {1+(kLb) 2 } 1/2-90 = (gh) 1/2 Estuary + Friction width change (James) Intermediate Accel. + η/ + L/L a = O(1) (gh) 1/2 (a/h) sin k(l-x) - 90 infinite length, deep, Press. Grad. phase change +. cos kl straight amplitude change (St. Vincent) Intermediate Accel. + η/ + see notes see notes - 45 to - 90 see notes length & depth, Press. Grad. phase change + straight + Friction amplitude change (Long Island)