dissolved fraction particulate fraction Resuspension

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1 CE4505 SURFACE WATER QUALITY PROJECT 2B. TORCH LAKE RECOVERY MODEL Due 10/29/09 The model that we will use for the next week is shown in Fig. 1 below. This model is a simplification of reality in two major respects. First, we have ignored several processes (biological uptake, decomposition) because they are very small relative to the remaining processes. Second, the lake is modeled here as a single, well-mixed box; seasonal stratification is ignored for the time being. Once we have this simple model working, we will add complexity back into the model. River inflows Precipitation System boundary Outflow TOTAL COPPER dissolved fraction particulate fraction Diffusion Resuspension Settling Sediments Figure 1. Box model of copper cycling to be used in initial modeling of copper in Torch Lake. A. The next step is to calibrate the model. Calibration is achieved by fitting measured data to the rate equation for each component process; the coefficients for the rate equations are chosen to yield the best fit between the measured and predicted concentrations. Each group is to choose two (2) of the five processes to calibrate; you will turn in these results next week. Come prepare to show the class how you calibrated one of the 2 processes. 1

2 A1. Settling The settling flux of copper is given by v s f particulate Cu Tutal A where v s is the settling velocity (m/d), f particulate is the fraction of total copper bound to particulate matter, Cu Total is the total Cu concentration in the water column, and A is the lake area. The objective of this calibration is to determine the value of the settling velocity. The settling velocity is calculated as the settling flux (g/m 2 d) divided by the concentration of suspended solids in the lake (C SS ). The settling flux is measured with sediment traps, cylinders suspended in the lake that catch the particles falling down through the water column. Given the data below (the sediment trap diameter is 10 cm), calculate the settling velocity appropriate for Torch Lake. Clearly indicate how your recommended value was calculated. Date of collection Total Average Annual Value Suspended Solids Conc. Mass collected Flux (g) g/m2d mg/l m/d 14-Jan Jan Feb Feb Mar Mar Apr Apr May May Jun Jun Jul Jul Jul Aug Aug Sep Sep Oct Oct Nov Nov Dec Dec Dec g/m2yr Settling velocity 2

3 A2. Resuspension In the sediments, there must be a mass balance on solids. The mass balance may be stated as: Burial = Settling Resuspension - Decomposition The settling flux may be calculated from the data in section B1 above. The sediment burial rate must be extrapolated from a few values measured in sediment cores to a lakewide average. The lake-wide average is calculated by weighting the individual core rates by the fraction of total lake area each represents. The necessary data are summarized in the table below. Mass of sediments to 1968 horizon Depth to 1968 horizon Burial Lake Water depth rate area (m) (g/m2) (cm) g/m2yr m E E E E E E+06 Decomposition is more difficult to calculate accurately. The rate of decomposition (g/m 2 yr) is equal to the inventory of organic carbon in the sediments times the decomposition rate constant, k (1/yr). D = I c k The value of k can be approximated by applying the first order rate law to the concentrations of organic carbon at the top (14%) and bottom (11%) of the layer of recent kt sediments: C = Coe Where t is the time in which the sediments accumulated (36 years). The inventory of organic carbon could either be measured or calculated by integrating the equation: dic = SC RC BC kic dt where S is the measured sedimentation rate of organic carbon, R is the resuspension rate of organic carbon, and B is the burial rate; all are in gc/m 2 yr. Integrating the last equation yields: ( SC BC RC) ( kt IC = 1 e k ) The trouble is that we do not know R C so we cannot calculate I C. We can estimate I C as the average organic carbon concentration in the recent sediments times the average mass of sediments in the recent layer of sediments. To calculate this average lake-wide inventory of recent sediments, we have to weight the values for our three cores by the areas of the lake that they represent. Use the data above to calculate values for k, D, and R C. Then verify your mass balance by showing that D = S-R-B. 3

4 A3. Dissolution It still must be determined whether the flux of dissolved Cu from the sediments is limited by the rate of diffusion or the rate of copper mineral dissolution. To make this evaluation, we will compare the rates of dissolution and diffusion. The copper mineral most likely to precipitate and dissolve is tenorite (CuO (s) ). The equilibrium constant for dissolution of tenorite is The ph of the lake was measured to be 7.9 in the epilimnion and 7.5 in the hypolimnion in fall The ph in the pore water is 7.7; what would be the equilibrium concentration of dissolved Cu 2+ in the pore water? CuO (s) + 2H + <==> Cu 2+ + H 2 O Does dissolved Cu 2+ represent a major fraction of dissolved Cu(II)? What other forms of dissolved Cu may be present? Based on calculations performed in CE4501, less than 1% of the dissolved copper exists as free Cu 2+. Use the following data to determine a rate constant (k) and value for f Cu according to the equation: dc * = kc ( f C Cu ) dt where C is the measured total Cu concentration, and C * is the equilibrium value for free Cu 2+ that you just calculated. Time Total Cu hr ug/l The rate constant that you determined is specific to the experimental conditions under which it was determined. The temperature in the sediments does not vary seasonally, and the measurements were made at the same temperature (4 o C). However, the rate does depend on the amount of solid surface area per unit volume of water. The mean grain size of the sediments is 8 μm and the particle density is 2.5 g/cm 3. In the experiment the concentration of solids was only 10 g/100 ml while the bulk density of the sediments is 0.75 g/cm 3. The measured rate constant must be multiplied by the ratio of the surface area/volume in the pore water to that in the experimental set-up. Calculate the rate constant for the porewater conditions; show and explain your calculations. 4

5 A4. Diffusion A diffusive flux out of the sediments is maintained by dissolution of copper from the stamp sands into the pore waters. Dissolution maintains the copper concentration in the pore water at 800 mg/m 3. The diffusive flux may be estimated from Fick s Law as: 2 g g m ΔC ( 3 m ) J D 2 = eff myr yr Δzm ( ) where the effective diffusion coefficient (D eff ) equals the molecular diffusion coefficient for Copper at 4 o C (133 cm 2 /yr) times the porosity of the sediments (0.5 for stamp sands, 0.7 for the recent sediments). The concentration gradient may be calculated as the difference between concentrations in the lake and concentration in the pore water (800 mg/m 3 ) divided by the depth from the sediment surface to the stamp sand layer. In 1968, this depth was about 0.3 cm throughout the lake. As sediments have accumulated since that time, the depth to the stamp sands has increased at different rates in different regions of the lake. Sedimentation rates increase with increasing water depth. As for Part A2, we break the lake into 4 zones: zone 1 extends from the shore to 3 m depth, zone 2 extends from 3 to 12 m, zone 3 from 12 to 17 m, and zone 4 is all waters deeper than 17 m. The sedimentation rate in zone 1 is zero, so the stamp sands remain exposed to the lake in the littoral area, and the diffusive flux is not attenuated over time. In the other 3 zones, the diffusive flux becomes smaller because Cu must diffuse further as sediments accumulate. The lake-wide diffusion velocity (D eff /Δz) changes with time and is calculated as a weighted average of the values for all zones; the weighting factor is the fraction of the total lake area covered by each zone. Given the information in the table below, calculate the lake-wide diffusion velocity for 1968 and for Convert these values to first-order rate constants by dividing by the mean depth of the lake (15.2 m). Depth to 1968 Area (m2) frac. Area horizon cm Zone E % 0 Zone E % 5.5 Zone E % 5.75 Zone E %

6 A5. Hypolimnetic exchange Hypolimnetic exchange is the dispersive flux across the thermocline. Because the flux is driven by turbulence rather than molecular diffusion, the dispersion coefficient is identical for all ions, compounds and even non-molecular species such as heat. The usual method for determining the dispersion coefficient across a thermocline is to measure the mean epilimnetic and hypolimnetic temperatures over time, and from this information to calculate the heat flux and the dispersion coefficient. Dispersion can be modeled analogously to diffusion using Fick's Law. J = K dc dx Horizontal dispersion coefficients are much larger in lakes than are vertical dispersion coefficients (K z ). For transport through the thermocline, only vertical dispersion is important. The table below gives typical ranges for K Z. Region K Z (cm 2 sec -1 ) Lake epilimnion 0.1-1,000 Lake hypolimnion Thermocline As for diffusion, the dispersion coefficient (cm 2 /sec) can be transformed to a mass transfer coefficient or velocity (cm/sec) by dividing by the thickness (Δz) of the layer through which transport is occurring; in this case we would divide by the thickness of the thermocline. The thickness of the thermocline can be determined from the temperature profile. The first order rate constant will be the dispersion velocity divided by the depth of the epilimnion or hypolimnion. Heat also is transported by dispersion through the thermocline. From temperature profiles, one can calculate the rate of increase in heat content of the hypolimnion. Then, from Fick's Law, one can calculate the dispersion coefficient. F J K T T epi hypoi heat = zhg ΔzthermoclineKJ The flux will be proportional to the temperature gradient across the thermocline. Needless to say, both the thickness of the thermocline and the temperature gradient change with time. For this project, we will assume that K z remains constant throughout the summer. Thus, K z may be calculated as: Jheat ΔZthermocline Kz = Tepi Thypo The heat flux may be calculated as the change in heat content (calories) per unit time and surface area: bt2 T1g cwvρ w bt2 T1g cwhρ w Jheat = = A Δt Δt where c w is the heat capacity of water (~1 cal/g- o C), ρ w is the density of water, H is the mean depth of the hypolimnion, and T 1 and T 2 are the average hypolimnetic temperatures 6

7 at times t 1 and t 2. The area of the thermocline is the area of zone 1 plus zone 2 in section A4 above. From the data below plot the heat flux into the hypolimnion versus the temperature difference between epilimnion and hypolimnion. Calculate K z as the slope of this plot. (Be careful with units.) Provide a graph showing the data points and your regression line. Include the details of your calculations. Date Epilimnetic temperature ( o C) Hypolimnetic temperature ( o C) May May June June July July August August Sept Sept Oct Oct You will also need to know the thickness of the thermocline. This value may be determined from the temperature profile that you measured in September. Depth Temp. M oc

8 We will do questions B-D in class next week; we need the answers from part A to do them. You will turn in answers to questions B-D the following week with installment 3 of the project. B. Based on the first-order rate constants calculated in part A above, should we model the flux of dissolved Cu out of the sediments as diffusion or as dissolution? Explain your answer. C. Based on the model shown in Figure 1 and your answers to parts A & B, write the complete mass balance for Total Copper (i.e., dissolved plus particulate) in Torch Lake. D. Make a table in which you compare the first order rate constants for all of the processes in the model (Fig. 1) and also for hypolimnetic exchange. Which processes dominate copper cycling in Torch Lake. Calculate the eigenvalue and the residence time for copper in the lake. What is the time to 95% of steady state? Using the loadings for 1986 (table below) calculate the expected steady state concentration of copper. 8