1. 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)

2. Newport Harbor Back Bay

3. Newport Back Bay Salt Marsh

4. Newport Harbor Back Bay

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

6. Average ~ 1.3 m (4.5 ft)

7. Area = 62471 pix 2 = 3,500,000 m 2 = 1870 x 1870 m Average depth = 4 m (13ft) (range 2.5 m (8ft) to 6 m (20ft) Volume = 14,000,000 m 3 (8,700,000 m 3 – 21,000,000 m 3 ) Tidal range ± 1.3 m Volume = 4,500,000 m 3 Harbor Entrance = 220 m Depth = 6 m (20 ft) Cross-sectional area = 1320 m 2 Tidal cycle (high tide-low tide) ~ 6 hrs or 360 mins Flow rate (m 3 /min) at entrance = 12,600 m 3 /min Surface speed = 10 m/min (or 0.600 km/hr) Tidal Flow

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/m 2 = 100 biv/m 2 (range: 10 – 400 biv/m 2 ) Filtration rate range @ 10 biv/m 2 = 150 mL/min – 7,00 mL/min @ 100 biv/m 2 = 1,500 mL/min – 7,000 mL/min @ 400 biv/m 2 = 6,000 mL/min – 28,000 mL/min Filtration rate per tide @ 10 biv/m 2 = 0.15 L/min x 360 min = 54 L – 252 L @ 100 biv/m 2 = 540 L – 2520 L @ 400 biv/m 2 = 2160 L – 10,080 L Filtration rate per tide @ 10 biv/m 2 = 0.054 m 3 – 0.252 m 3 @ 100 biv/m 2 = 0.540 m 3 – 2.520 m 3 @ 400 biv/m 2 = 2.160 m 3 – 10.080 m 3

9. Area = 62471 pix 2 = 3,500,000 m 2 = 1870 x 1870 m Tidal range ± 1.3 m Volume = 4,500,000 m 3 Total Number of Bivalves in bay @ 10 biv/m 2 = 35,000,000 @ 100 biv/m 2 = 350,000,000 @ 400 biv/m 2 = 1,400,000,000 Filtration rate per tide @ 10 biv/m 2 = 0.054 m 3 – 0.252 m 3 @ 100 biv/m 2 = 0.540 m 3 – 2.520 m 3 @ 400 biv/m 2 = 2.160 m 3 – 10.080 m 3 Volume filtered per tide @ 10 biv/m 2 = 0.0540 x 3,500,000 = 190,000 m 3 to 880,000 m 3 (4 – 20 %) @ 100 biv/m 2 = 1,900,000 m 3 to 8,800,000 m 3 (40 - 200 %) @ 400 biv/m 2 = 7,500,000 m 3 to 29,000,000 m 3 (166 – 640 %) % of Newport Harbor Water Filtered per tide

10. San Diego Creek and Springs Newport Harbor Entrance Spatial Gradients of Variable Factors Nutrient distribution Bivalve habitat flow out = m 3 /min [Nutrient] = #/m 3 bivalve density (#/m 2 ) flow in = m 3 /min

11. Bivalve habitat (2D  #/m 2 )

12. San Diego Creek Newport Harbor Entrance Spatial Gradients of Factors Tidal changes in volume (  m 3 ) & surface area (  m 2 ) Relevance to habitat (#/m 2 )

13. Bivalve Density: Low, Medium, High ?

14. San Diego Creek Cyclical Changes Newport Harbor Entrance filtration = m 3 /min growth =  #/min tidal flow = ± m 3 /min turbidity = #/m 3

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

16. San Diego Creek Newport Harbor Entrance To simplify let’s make some assumptions …. i)Conservation of components (in = out) ii)Well mixed system iii)Volume and surface area = constant iv)1 principal factor determines phytoplankton growth  bivalve population Input = [In] x Flow in = #/m 3 x m 3 /min i.e. #/min Output = [Bay] x Flow out = #/m 3 x m 3 /min i.e. #/min

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

19. Population Growth Exponential growth – x t = x o (1+r) t or P (t) = P o (1+ growth rate) time interval – Solving over time (dx/dt) for changes in population size …. x = ae kt or P(t) = P ini. e (growth constant.time) Population grows rapidly for k>0

20. 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 –rP 2 /K carrying capacity (K) Exponential growth resource limits (feedback) P(t) = (K.P o.e rt )/(K+P o (e rt -1)) where lim P(t) = K t→∞ (r = frac. change/time)

21. 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.C bay (t))/dt = C in.Q – C bay (t).Q – K cons P(t).V ⇒ dC bay (t)/dt = Q/V(C in -C bay ( t )) – k cons.P( t ) ……. Eq. 1 where …. C = nutrient conc. (moles/m 3 ); Q = flow rate (m 3 /min) K cons = consumption rate (moles of nutrients/moles of PP.min)

22. Phytoplankton Growth Growth of Phytoplankton dP(t)/dt = k growth C bay (t)P(t)(1-P(t)/P max ) – k deg (V/min)P biv …. Eq. 2 k growth C bay (t) is growth rate term (r) P max is PP carrying capacity (K) – k deg (V/min)P biv is clearance rate of PP by bivalve population Turbidity Turb(t) = k opt.P(t) ….. Eq. 3 Turb(t) is interdependent with self-limiting growth from equation 2 and efficiency/size of bivalve population (clearance rate) k opt is a conversion constant

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

24. Parameter Exploration