1 / 24

Newport Beach Harbor/Back Bay Bivalve Restoration Project

Newport Beach Harbor/Back Bay Bivalve Restoration Project. Computational modeling approaches to ecosystems Rationalize interactions Predictive ? Concepts Variables Parameter estimates Models Bath tub (simple) Cyclical (dynamic) Spatial (non-uniform, complex). Back Bay. Newport Harbor.

zoie
Télécharger la présentation

Newport Beach Harbor/Back Bay Bivalve Restoration Project

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Newport Beach Harbor/Back BayBivalve Restoration Project • Computational modeling approaches to ecosystems • Rationalize interactions • Predictive ? • Concepts • Variables • Parameter estimates • Models • Bath tub (simple) • Cyclical (dynamic) • Spatial (non-uniform, complex)

  2. Back Bay Newport Harbor

  3. Newport Back Bay Salt Marsh

  4. Back Bay Newport Harbor

  5. Newport Beach Harbor Harbor Shoreline = 16 km Area = 62471 pix2 =3,500,000 m2 = 1870 x 1870 m Average depth = 4 m (13ft) (range 2.5 m (8ft) to 6 m (20ft) Volume = 14,000,000 m3 (8,700,000 m3 – 21,000,000 m3) Upper Newport Bay Low tide channels = 18407 pix2= 1,000,000 m2 = 1015 x 1015 m Average depth = 2.5 m Volume = 2,500,000 m3 High tide area = +10340 pix2= 579,040 m2 = 761 x 761 m Average depth = 0.5 m Volume = 290,000 m3 Flood plain = +32798 pix2 = 1,800,000 m2 = 1355 x 1355 m Average depth = 0.5 m Volume = 900,000 m3

  6. Average ~ 1.3 m (4.5 ft)

  7. Tidal Flow Area = 62471 pix2 =3,500,000 m2 = 1870 x 1870 m Average depth = 4 m (13ft) (range 2.5 m (8ft) to 6 m (20ft) Volume = 14,000,000 m3 (8,700,000 m3 – 21,000,000 m3) Tidal range ± 1.3 m Volume = 4,500,000 m3 Harbor Entrance = 220 m Depth = 6 m (20 ft) Cross-sectional area = 1320 m2 Tidal cycle (high tide-low tide) ~ 6 hrs or 360 mins Flow rate (m3/min) at entrance = 12,600 m3/min Surface speed = 10 m/min (or 0.600 km/hr)

  8. Bivalve Statistics Typical oyster filtration rate = 22 - 100 L/day = 0.92 – 4.2 L/hr = 15 – 70 mL/min Average size of oyster bivalve = 5 cm No of of bivalves/m2 = 100 biv/m2 (range: 10 – 400 biv/m2) Filtration rate range @ 10 biv/m2= 150 mL/min – 7,00 mL/min @ 100 biv/m2 =1,500 mL/min – 7,000 mL/min @ 400 biv/m2 = 6,000 mL/min – 28,000 mL/min Filtration rate per tide @ 10 biv/m2= 0.15 L/min x 360 min = 54 L – 252 L @ 100 biv/m2 = 540 L – 2520 L @ 400 biv/m2 = 2160 L – 10,080 L Filtration rate per tide @ 10 biv/m2 = 0.054 m3 – 0.252 m3 @ 100 biv/m2 = 0.540 m3 – 2.520 m3 @ 400 biv/m2 = 2.160 m3 – 10.080 m3

  9. % of Newport Harbor Water Filtered per tide Area = 62471 pix2 =3,500,000 m2 = 1870 x 1870 m Tidal range ± 1.3 m Volume = 4,500,000 m3 Total Number of Bivalves in bay @ 10 biv/m2 = 35,000,000 @ 100 biv/m2 = 350,000,000 @ 400 biv/m2 = 1,400,000,000 Filtration rate per tide @ 10 biv/m2 = 0.054 m3 – 0.252 m3 @ 100 biv/m2 = 0.540 m3 – 2.520 m3 @ 400 biv/m2 = 2.160 m3 – 10.080 m3 Volume filtered per tide @ 10 biv/m2 = 0.0540 x 3,500,000 = 190,000 m3 to 880,000 m3 (4 – 20 %) @ 100 biv/m2 = 1,900,000 m3 to 8,800,000 m3 (40 - 200 %) @ 400 biv/m2 = 7,500,000 m3 to 29,000,000 m3 (166 – 640 %)

  10. Spatial Gradients of Variable Factors • Nutrient distribution • Bivalve habitat San Diego Creek and Springs flowin = m3/min [Nutrient] = #/m3 Newport Harbor Entrance flowout = m3/min bivalve density (#/m2)

  11. Bivalve habitat (2D  #/m2)

  12. Spatial Gradients of Factors • Tidal changes in volume (Dm3) & surface area (Dm2) • Relevance to habitat (#/m2) San Diego Creek Newport Harbor Entrance

  13. Bivalve Density: Low, Medium, High ?

  14. Cyclical Changes San Diego Creek tidal flow = ± m3/min growth = D#/min Newport Harbor Entrance turbidity = #/m3 filtration = m3/min

  15. Stochastic Events • Weather • Surge run-off San Diego Creek Newport Harbor Entrance

  16. To simplify let’s make some assumptions …. • Conservation of components (in = out) • Well mixed system • Volume and surface area = constant • 1 principal factor determines phytoplankton growth  bivalve population San Diego Creek Input = [In]xFlowin = #/m3 xm3/min i.e. #/min Newport Harbor Entrance Output = [Bay] xFlowout = #/m3 xm3/min i.e. #/min

  17. Bivalve Ecosystem Sunlight Fresh water flow • Nutrients (N, P, Fe) turbidity % dieoff PP carrying capacity (max population size) salinity Phytoplankton Biv carrying capacity (max population size) turbidity Salt water (tidal) food chain Bivalves % dieoff toxicity Pollutants

  18. Population Growth • Exponential growth • xt = xo(1+r)t or P(t) = Po(1+ growth rate)time interval • Solving over time (dx/dt) for changes in population size …. x = aektorP(t) = Pini . e(growthconstant.time) Population grows rapidly for k>0

  19. Logistic growth (Verhulst-Pearl equation) • Initial stage of growth is approximately exponential; then, as saturation begins, growth slows, and, at maturity, net growth stops • P(t) = 1/(1+e-t) • Solving over time …. dP(t)/dt = P(t) . (1-P(t)) • For biological systems where rate of reproduction is proportional to both the existing population and the amount of available resources  self-limiting growth of population • dP/dt = rP(1-P/K) where r = growth rate and K is the carrying capacity • note early exponential growth depends on +rP; later, competition for food/space etc. is due to larger term –rP2/K carrying capacity (K) resource limits (feedback) P(t) = (K.Po.ert)/(K+Po(ert-1)) where limP(t) = K t→∞ Exponential growth (r = frac. change/time)

  20. What is limiting for phytoplankton (PP) growth ? • Nutrient sources (N, P, Fe) or sunlight • Are these constant, variable or variable + periodic (seasonal) • Will these variables change the growth rate (r), carrying capacity (K) or both ? • How is increasing [PP] related to turbidity  reduces sunlight  reduces growth ? Nutrient Level Change in Bay d(V.Cbay(t))/dt= Cin.Q – Cbay(t).Q – KconsP(t).V ⇒ dCbay(t)/dt = Q/V(Cin-Cbay(t)) – kcons.P(t) ……. Eq. 1 where …. C = nutrient conc. (moles/m3); Q = flow rate (m3/min) Kcons = consumption rate (moles of nutrients/moles of PP.min)

  21. Phytoplankton Growth Growth of Phytoplankton dP(t)/dt = kgrowthCbay(t)P(t)(1-P(t)/Pmax) – kdeg(V/min)Pbiv…. Eq. 2 kgrowthCbay(t) is growth rate term (r) Pmax is PP carrying capacity (K) – kdeg(V/min)Pbivis clearance rate of PP by bivalve population Turbidity Turb(t) = kopt.P(t) ….. Eq. 3 • Turb(t) is interdependent with self-limiting growth from equation 2 • and efficiency/size of bivalve population (clearance rate) • kopt is a conversion constant

  22. Bivalve Population Growth Can write similar equations for bivalve population changes….. • Dependent on [PP] (measured as ∞ turbidity) • Bivalve Kbiv (carrying capacity) is dependent on [PP] and time-varying variables (e.g. seasons) • Bivalve growth rate (rbiv) is dependent on filtration rate (efficiency) and [PP] (resource) What factors might affect efficiency ? • Habitat (spatial differences: tidal flats v. channels) • Pollution (impaired growth)

  23. Parameter Exploration

More Related