Preliminary calculations: Lake morphometry Gas Transfer

Transcription

1 Preliminary calculations: Lake morphometry For all individual processes it will be necessary to know the depth of the epilimnion and hypolimnion. You have been given the following information: Epilimnion Hypolimnion Whole lake Surface area (m 2 ) 1.44x x x10 7 Mean depth (m) Max. depth (m) 86 Volume (m 3 ) 1.73x x10 8 There are two approaches to determining the volumes of the epilimnion and hypolimnion. The simplest approach is to assume that the depth of the thermocline (12 m) represents the mean depth of the epilimnion. Knowing the surface area and mean depth, one can calculate the volume of the epilimnion (1.73x10 8 m 3 ). By difference, one can calculate the volume of the hypolimnion (6.34x x10 8 = 4.61x10 8 m 3 ), and from the surface area and volume determine the mean depth of the hypolimnion (4.61x10 8 /1.36x10 7 = 33.9 m). The second approach is to use conic sections and assume that the epilimnion is a portion of a frustrum. The radius of the base of the frustrum is 2140 m, and the radius of the lower frustrum is 2080 m. The depth of the whole frustrum can be calculated as (2140m) (12m) ( m) = 428m. The volume of the larger frustrum can then be calculated as AH/3 = (1.44x10 7 m 2 ) (428m)/3 = 2.05x10 9 m 3. The depth of the smaller frustrum is simply = 416m, and its volume is (1.36x10 7 m 2 ) (416m)/3 = 1.89x10 9 m 3. The volume of the epilimnion is the difference between the volumes of the two frustra ( )x10 9 m 3 = 1.69x10 8 m 3. By difference, the volume of the hypolimnion can now be calculated as ( )x10 8 = 4.65x10 8 m 3. The mean depth of the hypolimnion can be calculated as V/A = 34.2m. The two methods yield similar results in this case. Although the hypolimnion gradually sinks during the summer, we will assume that the size of the epilimnion and hypolimnion remain constant during the summer to simplify the modeling exercise. 1. Gas Transfer Most of the parameterization for gas transfer was given to you in class (10/24). Based on the lecture notes, you should be able to write an equation for the O 2 flux into the water column that is a function of: Water temperature Wind speed Oxygen concentration in the epilimnion Assume an average windspeed of 10 m/s, and a temperature correction coefficient (θ) for gas transfer of

2 A. In calculating the gas transfer, do you need to consider both v a and v w (the mass transfer coefficients for air and water films) for the flux of oxygen? SOLUTION: For dissolution of oxygen into a lake, we need only consider the liquid film resistance. We may ignore air-film resistance; because the Henry's Law Constant is so small, the product of K H v a becomes negligibly small. Also, because of the large concentration in the atmosphere, negligible depletion will occur in the laminar boundary layer. B. Using the data in Table 2 below, calculate a temperature correction factor (θ) for the Henry's Law constant for oxygen. You must first calculate the missing values for the Henry's Law constant. Provide a graph showing that the values you predict using your temperature correction agree with the true values given in Table 2. SOLUTION: The missing values of K H are given in the table. Even though θ typically is used to correct the temperature dependence of kinetic rate constants, we may use a similar mathematical formulation to correct equilibrium constants. Rather than an activation energy, the thermodynamic value is the enthalpy of reaction ( H). F I ln K 2 DH F T2 - TI 1 K R TT HG 1 = KJ F H G HG I KJ 2 1 Therefore, θ may be defined as: q = exp DH RT T 1 2 KJ A plot of ln(k) vs. 1/T should yield a straight line with a slope of H/R. From the slope, the value of θ may be calculated as in the equation above using T values in the middle of the range encountered. 2

3 Table 2. Temperature dependence of Henry's Law Constant for oxygen. Temperature ( o C) Solubility (mg/l) * K H (mg L -1 atm -1 ) * -Solubility under air with 20% O ln(k H ) ln(k)= (1/t) r 2 = /T Figure 1. Temperature dependence of Henry's Law Constant for Oxygen. Note that absolute temperatures are used. The value of θ calculated from the slope (1808) equals The graph below documents that the predicted values (reference temperature of 4 o C) agree well with the true values given in Table 2. 3

4 Solubility (mg/l) Actual solubility Predicted Solubility Temperature C. Write out the full equation for the gas flux including all temperature corrections. SOLUTION: * W = J A= k ( C -C) V gas F HG gas I KJ aw KH V CA = -CW KH T d q gas u i F 02. atm I = V - C T -277 W H HG KH@ 277 q K KJ H In this equation, u is the windspeed (10 m/s), θ gas is the temperature correction for the gas transfer rate constant (given as 1.05 in the problem statement with a reference temperature of 25 o C), H is the epilimnion depth, V is the epilimnion volume, and θ KH is the temperature correction for the Henry's Law constant that we just calculated to be Given the K H values in the table above, we must use C w in mg/l. The wind correction above is based on the Wanninkhof formula (K aw in m/hr) given in Chapra (p. 381). Other wind-correction formulae are available (e.g., K aw = 0.015u α ) or the one given by Schwarzenbach et al. (Environmental Organic Chemistry, 1993, Wiley): K aw (m/hr) = u 2 D. To be sure that your flux is calculated properly, model the epilimnion of the lake as if it had gas transfer as the only flux in or out. Set the initial concentration in the water to zero, use a 1-d time step from 5/15-11/15, and use the water temperatures given in Table 1 below. Your model will be properly calibrated if your results match those shown in the graph below. 4

5 SOLUTION: This model may be incorporated into a spreadsheet using a first order Runge Kutta approximation as was done for Project 1. The input parameters that are needed include: θ - gas transfer 1.05 θ for Kh mg/l-atm u (m/s) 10 Mean depth (m) = 12 Surface area (m2) = 1.44E+07 It is necessary to interpolate to fill in the temperatures for all dates. The initial concentration is set to zero, and the only flux in the model is gas exchange. The results are shown in the graph below and are compared to the solubility of oxygen in the lake at the given temperatures. The Wanninkhof model has a faster mass transfer coefficient, and the model shows a faster reaeration of the lake. The model predicts that the lake would return to near equilibrium within about two weeks. In fall, the equilibrium concentration changes (due to cooling of the lake) faster than the reaeration resulting in a slight disequilibrium. Table. 1. Lake temperatures Date Epilimnetic temperature ( o C) Hypolimnetic temperature ( o C) May May June June July July August August Sept Sept Oct Oct Nov Nov

6 DO (mg/l) Equilibrium Conc. Schwarzenbach model 4/15/1998 6/4/1998 7/24/1998 9/12/ /1/ /21/1998 Date Chapra-Wanninkhof model Figure 1. Model of reaeration of lake epilimnion following complete deaeration. 6

7 Photosynthesis Photosynthesis and air-water exchange are the predominant sources of oxygen to most lakes. A summary of photosynthesis modeling in lakes is given on pages in your text book. (Note that there is an error in one equation in the text. The equation that reads P = r oc a ca k g a should have an H on the right hand side as does the equation above it.) You may use equation in combination with P = r oc a ca k g ah to calculate the oxygen production in L. Sempach. The coefficients that you need for this equation are summarized in Table 1 below. Because nitrogen is available in abundance in L. Sempach, you do not need to include the values for "n" or k sn in your model. Table 1. Summary of coefficients for L. Sempach. Parameter (units) Value I s (ly d -1 ) 300 PO 4 -P concentration (µg/l) 1 Chlorophyll a concentration (µg/l) 23 c ca (mg C/µg Chlor-a) 0.1 r oc (mg O 2 /mg C) 2.67 k 20 o C (d -1 ) 2 θ g (temperature correction factor) I a (ly d -1 ) 500 k ew (m -1 ) 0.1 N (inorganic suspended solids mg/l) 1 D (organic suspended solids mg/l) < 3 k sp or K m-p (half saturation constant µg/l) 3 H (mean depth of epilimnion, m) 12 f (fractional photoperiod) 0.63 (summer average) The equations above will allow you to calculate the production rate of oxygen for each day during the stratified period. Although chlorophyll and phosphate concentrations change during the period of stratification, these changes are minor in comparison with the changes in temperature. Therefore, we will use the average values of phosphate and chlorophyll concentrations given in Table 1, but you will correct for the changes in temperature. The seasonal progression of temperature is summarized in Table 2. Table 2. Temperatures in L. Sempach during the stratified period. Date Epilimnetic temperature ( o C) 3- Epilimnetic PO 4 Conc. (µg/l) May May June June July July August August

8 Sept Sept Oct Oct Nov Nov A. Construct a one-box model for the epilimnetic dissolved oxygen in which photosynthesis is the only process occurring. Set the initial dissolved oxygen concentration at zero, and allow it to accumulate over the entire period of stratification. Provide the equations for your model, and a graph showing the oxygen concentration vs. time. You will have to interpolate to find the temperature on the days in between those listed in Table 2. Assume phosphate concentration remains constant throughout the period. SOLUTION: The equations needed for the model are: L NM c h OF HG 3- T f k k kh e -a e - [ PO ] 1 a0 4 g = o gt, = 20 C 3- e QP Km + [ PO4 ] J = C RQ Chlor k Chlor H O b g F 2 H G I K J g[ ] In these equations, k g is the first order growth rate constant (d -1 ) that is equal to an optimal value (k g,t=20 ) modified by temperature (θ=1.066), light availability, and nutrient availability. Multiplying the growth rate, k g, by the standing crop of algae (mg Chlor/m 3 ) and the mean depth will yield the growth of the algal population in terms of mass of chlorophyll per unit area and time. Because we are interested in the oxygen production, we must convert chlorophyll to oxygen. This conversion is done in two steps. First, chlorophyll is converted to carbon by using the ratio of C:Chlorophyll present in the algae, and then the carbon is converted to oxygen using the photosynthetic quotient (PQ). Photosynthetic quotients range from about 1 to 1.3. Results obtained using all of the values provided in the handout are shown in Figure 2a below. I KJ 8

9 D.O. (mg/l), Temp model DO, constant P Temperature 0 4/15/1998 6/4/1998 7/24/1998 9/12/ /1/ /21/1998 Date Figure 2a. Epilimnetic D.O. model with no processes other than photosynthesis. B. Now examine the effect that the rapid change in phosphorus concentration has on your model output. As for temperature, interpolate to find the daily phosphorus concentrations and incorporate these into your model. Graph the output for both models on the same page. SOLUTION: As shown in Fig. 2b, the high phosphorus concentrations in early summer result in a rapid phytoplankton growth and photosynthesis. This initial spurt in photosynthesis gives the lake approximately 10 mg/l more dissolved oxygen. 50 D.O. (mg/l) variable P 0 constant P 4/15/1998 6/4/1998 7/24/1998 9/12/ /1/ /21/199 8 Date Fig. 2b. Comparison of model with variable and constant P concentrations. 9

10 C. Next allow the photoperiod follow its actual sinusoidal pattern (f=0.5 on 3/21 and 9/21, and f max = 0.82 on 6/20) and, on the same graph as above, plot the output of your model. SOLUTION: For this model, all three parameters (temperature, phosphorus concentration, and photoperiod) vary with time and must have a separate column in the spreadsheet with the new value for each date. The photoperiod can be specified with the equation: F 2p f = + t -t ref I sin K J HG D.O. (mg/l) variable P constant P variable f 4/15/1998 6/14/1998 8/13/ /12/ /11/1998 Date Fig. 2c. Comparison of models with fixed and variable P concentration and photoperiod. The D.O. concentrations are calculated as above, but using the day-specific values for T,P, and f. The model output with this change is shown in figure 2c. The variable photoperiod allows oxygen production to exceed both of the models above through the time when the photoperiod is above the seasonal average (i.e., from 6/15 through 9/15). After that time, photosynthesis becomes limited by light, and the net rate is actually lower than in the previous models. These changes are shown more clearly in the graph of photosynthesis rate vs. time (Fig. 2d). 10

11 1.2E+08 Rate (g O 2 /day) 1.0E E E E E+07 variable P constant P variable f 0.0E+00 4/15/1998 6/14/1998 8/13/ /12/ /11/1998 Date Fig. 2d. Photosynthesis rate vs. time for cases when only temperature varies, when phosphorus and temperature vary, and when photoperiod as well as temperature and phosphorus are varied seasonally. 11

12 3. Hypolimnetic Exchange The exchange of materials across the thermocline occurs via dispersion. Dispersion is faster than molecular diffusion, but it also is a random process. 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 ( z) 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. There are several methods for determining the value of the dispersion coefficient for a given lake. The method that we shall employ will use the heat budget of the 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 DzthermoclineKJ 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 DZthermocline Kz = Tepi - Thypo The heat flux may be calculated as the change in heat content (calories) per unit time and surface area: bt2 - T1g cwvr w bt2 - T1g cwhr w Jheat = = A Dt Dt 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 at times t 1 and t 2. A. 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. 12

13 (Be careful with units.) Provide a graph showing the data points and your regression line. Include in the appendix 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 Nov Nov SOLUTION First, there are two dimensions to be defined. The mean depth of the hypolimnion is defined as the volume divided by the surface area. We are given the surface area (1.36x10 7 m 2 ), and the volume may be calculated as the total lake volume (1.44x10 7 x44 = 6.34x10 8 m 3 ) minus the epilimnetic volume (1.44x10 7 x 12 = 1.73x10 8 m 3 ). Thus H H = 33.8 m. (Note that this is essentially equal to the lake mean depth minus the epilimnion depth.) We also will need the thickness of the thermocline. If the thermocline is defined as the depth of maximum temperature change, a graph of temperature change vs. depth can be used to define the thermocline thickness. The figure indicates that, although the zone of temperature change is 5-10 m thick, there is a 1-m layer of most rapid temperature change (the thermocline) occurring on both dates at m depth. The thickness of the thermocline was chosen as 5 m for the subsequent modeling. 13

14 Depth (m) Temperature Jun 15-Aug 6/15 dt 8/15 dt Fig. 3a. Temperature profiles indicate that the thermocline lies at 12.5 m in both June and August, and that the zone of most rapid temperature change increases from ~5 m thick in June to nearly 20 m in August. A thickness of 5 m is used in the calculations below. Heat flux (cal/m 2 d) 1.40E E E E E E E E Temperature difference ( o C) Fig. 3b. Heat flux into the hypolimnion as a function of the temperature gradient across the thermocline. The plot of heat flux vs. temperature gradient (Fig. 3b) shows a generally linear relationship with an r 2 of With 14 data points, this regression is statistically significant (P<0.01), and the slope of the regression divided by c w ρ w / z ( z = 5 m) equals the vertical dispersion coefficient (0.30 m 2 d -1 or cm 2 sec -1 ). The calculated dispersion coefficient is in the middle of the range typically observed in lakes. B. Just to demonstrate the speed of oxygen transfer across the hypolimnion, model the transfer of oxygen from the epilimnion into the hypolimnion using your calculated value of K z. Assume that the hypolimnion is anoxic initially, and that the epilimnion 14

15 has a dissolved oxygen concentration of 11 mg/l initally. Ignore all other oxygen sources and sinks. Determine how long it will take for the concentrations in the epilimnion and hypolimnion to equal each other. Provide a plot showing the concentrations in both epilimnion and hypolimnion. SOLUTION D.O. (mg/l) Hypolimnetic exchange Epilimnetic D.O. Hypolimnetic D.O Time (days) Fig. 3b. Exchange of oxygen across the thermocline of L. Sempach. Because the gas exchange is a dispersive process, it is driven by the gradient in concentration. All such processes act to erode the concentration gradient and equalize concentrations. With the low exchange coefficient calculated in part A (0.034 cm 2 /s), transfer of material across the thermocline is slow. It requires about 2 years to approach the steady state value when concentrations in epilimnion and hypolimnion will be equal. 15

16 4. Water column respiration As we discussed in class, the process that we are calling respiration in the water column is the combined respiration of algae, zooplankton, fish, bacteria, and all other organisms that live within the water column. The rate coefficients for this process are determined by incubating lake water in the dark and measuring the decrease in oxygen concentration over time. Data for such measurements from the epilimnion (made at two different temperatures) are provided in the table below together with results at one temperature from the hypolimnion. Table 1. Epilimnetic respiration measurements. * Time (hr) D.O. (mg/l) T=8 o C D.O. (mg/l) T=15 o C *The times given in the handout were incorrectly typed as 1/10 of the actual time. Table 2. Hypolimnetic respiration measurements (T=4 o C). Time (hr) D.O. (mg/l) A. Plot the data for each respiration measurement to determine the order of the reaction. Calculate the rate constant for each respiration measurement. SOLUTION Epilimnetic Respiration ln(d.o.) Time (hr) 8 oc 15 oc Fig. 4a. First order evaluation of epilimnetic respiration rates. 16

17 As shown in Fig. 4a, respiration in the epilimnion does not follow first order kinetics. The semi-log plot shows too much curvature to model this as first order. D.O. (mg/l) oc 15oC Linear (15oC) Linear (8 oc) Time (hr) Fig. 4b. Zero order evaluation of epilimnetic respiration. The plot of concentration vs. time (Fig. 4b) is more linear than is the plot of ln(concentration) vs. time. Therefore, this process can be modeled as a zero order reaction. The rate constants can be calculated as the slopes of the lines shown in Fig. 4b. These rate constants for the epilimnion are mg L -1 hr 8 o C and mg L -1 hr 15 o C. Hypolimnetic Respiration ln(do) (mg/l) Time (hrs) DO (mg/l) Fig. 4c. Comparison of first- and zero-order kinetic models for hypolimnetic respiration. Filled symbols and dashed line are the first-order model; open symbols and solid line are for the zero order model. From visual comparison alone, it is not as obvious which is the best model to use for hypolimnetic respiration. In such a case, we use statistics to determine the best model. The r 2 value for the zero order plot is and that for the regression of the first 17

18 order model is Because we know that the hypolimnion may go anoxic, it would be preferable to use the first order model. The zero order model would predict continued respiration even after all oxygen is consumed. The first order rate constant is hr -1. B. Determine the temperature correction coefficient for the epilimnetic measurements. SOLUTION: Because we are given information on rates at only two temperatures, it is not sensible to plot ln(k) vs. 1/T. Rather, we might as well use the equation: k1 T2- T1 = q k2 The rate constant for epilimnetic respiration at 8 o C is mg/l-hr and for 15 o C is mg/l-hr. Theta may be calculated as: F 1 I F k I 1 F 1 I. log( q) = log = log. HG - KJ HG KJ HG - K J F 0127 H G I. K J = T2 T1 k q = C. Write the rate law for the respiration reaction as determined from part A above. What is being ignored in this formulation? Can you suggest a method for improving the modeling of the reaction? SOLUTION: Epilimnetic: dc dt o mg = k o q = C Lhr T-15 C T-15 Hypolimnetic: dc dt = = 1 kc hr C The formulations as written assume that oxygen is the only substance limiting respiration rates. It is possible that carbon substrates are limiting or that dissolved nutrients limit bacterial growth and metabolism at least during some times of the year. Experiments could be conducted to see if respiration is stimulated by addition of carbon substrates or nutrients. It also is assumed that the temperature range is small enough to ignore in the hypolimnion; it would be an option to use the epilimetic value of theta. D. To demonstrate the effect of water column respiration on the dissolved oxygen in the water column, you will now model both the epilimnion and the hypolimnion as if respiration were the only process occurring. Set the initial oxygen concentration in both epilimnion and hypolimnion to 13 mg/l. Apply the temperature correction coefficient to the epilimnetic rates to account for the change in temperature over the summer. The lake temperatures are given in the table below. Provide the equations for your modeling and and a graph of the model output (dissolved oxygen concentration vs. time for epilimnion and hypolimnion). 18

19 SOLUTION Water column respiration D.O. (mg/l) /15/1998 6/4/1998 7/24/1998 9/12/ /1/1998 Date Epilimnetic D.O. Hypolimne tic D.O. Fig. 4d. Model of water column respiration. As the model output demonstrates, the high rates of respiration in the epilimnion would completely deplete the dissolved oxygen if it were not resupplied by other processes. Respiration rates observed in the hypolimnion also are adequate to nearly deplete the hypolimnion of oxygen over the full stratification period. Date Epilimnetic temperature ( o C) Hypolimnetic temperature ( o C) May May June June July July August August Sept Sept Oct Oct Nov Nov

20 5. Sediment respiration Sediment respiration includes microbial decomposition as well as respiration by the micro- and macroinvertebrates that feed on the bacteria. Rates are measured by incubating sediment cores and measuring the rate of oxygen consumption (mg m -2 hr -1 ). Sediment temperatures, if not measured directly, are assumed to be the same as temperatures in the water column because most of the respiration occurs right at the sediment surface. A summary of lake temperatures is provided in Table 1 below. Because hypolimnetic temperatures do not change significantly, we will only apply a temperature correction to the epilimnetic rates. Table 1. Summary of lake temperatures during stratification. Date Epilimnetic temperature ( o C) Hypolimnetic temperature ( o C) May May June June July July August August Sept Sept Oct Oct Nov Nov In table 2 below are presented data from measurement of respiration rates at two temperatures for epilimnetic sediments, and table 3 presents data for hypolimnetic sediments at 4 o C. For both situations, a sediment core with a diameter of 6 cm was incubated in the dark with 500 ml of water above the sediments. The oxygen concentrations presented in the table were measured with an oxygen probe in the 500 ml of water above the sediments. Table 2. Epilimnetic sediment oxygen demand. Time (hr) O 2 (mg/l) T = 20 o C O 2 (mg/l) T = 10 o C

21 Table 3. Hypolimnetic sediment oxygen demand. Time (hr) O 2 (mg/l) T = 4 o C A. Determine the reaction order for each of the three experimental measurements. Calculate the rate constant for each of the three measurements. SOLUTION Respiration for all three experiments is modeled as a zero order reaction in Fig. 5a below. The data appear quite linear, and the high r 2 values (0.99) suggest that the zero order model is appropriate. D.O. (mg/l) Time (hr) Fig. 5a. Zero order model of sediment respiration rates. The uppermost line is the hypolimnetic experiment, and the two lower lines are the epilimnetic experiments. The rate constant for each experiment is calculated as the slope of the regression line. The slopes of the regression lines are in units of mg/l-hr. Multiplying these values by the volume of water above the cores (0.5 L) and dividing by the surface area of sediments in the cores (28.3 cm 2 ) will yield rate constants in units of mg/cm 2 -hr. The three rate constants in these units are: mg/cm 2 20 o C, mg/cm 2 10 o C, and mg/cm 2 hr in the hypolimnion. B. Determine the temperature correction coefficient (θ) for the epilimnetic sediment oxygen demand. SOLUTION: Because we are given information on rates at only two temperatures, it is not sensible to plot ln(k) vs. 1/T. Rather, we might as well use the equation: 21

22 k1 T2- T1 = q k2 The rate constant for epilimnetic respiration at 8 o C is mg/l-hr and for 15 o C is mg/l-hr. Theta may be calculated as: F 1 I F k I 1 log( q) = log HG T2 - T1KJ HG k 2KJ q = C. What assumptions are we making if we use the measured rate constants and merely correct them for seasonal temperature changes to obtain an estimate of the sediment oxygen demand during the stratified period? ANSWER We are assuming that the reaction rate depends on nothing other than the temperature. That means that the bacterial population is assumed to be constant and that neither oxygen nor carbon substrates ever limit the rate of respiration. D. Construct a model for dissolved oxygen in both the epilimnion and hypolimnion in which sediment respiration is the ONLY process in the model. Set the initial concentration of dissolved oxygen at mg/l in both epilimnion and hypolimnion. Use the morphometric data for Lake Sempach, and calculate how the dissolved oxygen concentrations will decrease with time. Provide all the necessary equations for your model and present a graph showing the dissolved oxygen concentration vs. time in both the epilimnion and hypolimnion. Sediment respiration 15 D.O. (mg/l) /15/1998 6/14/1998 8/13/ /12/ /11/1998 Time Epilimnion Hypolimnion The model equation (identical for epilimnion and hypolimnion but with appropriate values for each layer) is: 22

23 c h q V dc F k mg T T = sed area m ref dt H G I 2 - md K J. ( ) 2 As shown in the model output above, concentrations in both epilimnion and hypolimnion are slowly depleted by this process. E. 6. Oxidation of reduced substances Because inadequate oxygen is present in the hypolimnion to fully oxidize all of the organic matter, anaerobic bacteria will use other electron acceptors (Mn(IV), Fe(III), NO 3 -, SO 4 2- ) as well as resort to fermentation. The result of these processes is the production of reduced substances including Mn 2+, Fe 2+, H 2 S, and CH 4. The first three of these substances can react chemically with oxygen, and all four can be oxidized catalytically by bacteria. For all four of the substances (with the possible exception of Fe), it is thought that the microbially catalyzed reaction is most important because rates of microbial catalysis are much faster than are the rates of the abiotic oxidations. The concentrations of the four substrates in the hypolimnion are not constant during the stratified period. At the onset of stratification, the concentrations of all substrates are very close to zero. Concentrations of all substrates increase during the stratified period because the rate of production exceeds the rate of oxidation. Rates of production of the four reduced substances have been measured and are reported in the table below. Because we are interested in the consumption of oxygen during the oxidation of these substrates, we must consider the stoichiometry of the oxidation reactions. 2Mn 2+ + O 2 + 2OH - 2MnO 2 + 2H + (Mn:O 2 = 2) 4Fe 2+ + O H 2 O 4Fe(OH) 3 + 8H + (Fe:O 2 = 4) H 2 S + 2O 2 SO H + (H 2 S:O 2 = 0.5) CH 4 + 2O 2 CO 2 + 2H 2 O (CH 4 :O 2 = 0.5) This comparison points out that sulfide oxidation is the major reaction in L. Sempach. Table 1. Rates of production of reduced substrates in L. Sempach. Substrate Production rate (mmole m -2 d -1 ) Oxygen equivalent rate (mmole m -2 d -1 ) Mn Fe H 2 S CH Each of the four substrates is oxidized by a different group of bacteria. Hence for each substrate there exists a different value of K m and V max. However, because sulfide oxidation accounts for >90% of the oxygen consumption, we shall ignore all bacterial groups except the sulfide oxidizing bacteria. In Lake Sempach, sulfide is oxidized by a dense layer of Beggiotoa, a large, white bacterium that grows on the sediment surface. Realize that the bacteria need both oxygen and sulfide for this reaction; there is a halfsaturation constant for both substrates. The rate law is: 1 do [ F IF I 2] [ HS 2 ] [ O2 ] - = Vmax B 2 dt K + [ H S] K [ O ] HG KJ HG + M-H S 2 M-O KJ 23

24 where B is the concentration of bacteria and V max is expressed in moles per bacterium. (Those of you who have had microbiology will realize that we have included the yield coefficient into the value of V max.) As you have heard in other classes, typically (although not necessarily) only one substrate is limiting at any point in time. If only one substrate is limiting, the other one must be present in concentrations much larger than K M, and, therefore, one of the terms in parentheses is equal to one. We will assume that oxygen is the limiting substrate. As discussed in class, the Monod kinetic model can be simplified to either a first order or zero order expression depending on whether the substrate concentration is large or small relative to the half saturation constant (K m ). In the case of oxygen, the halfsaturation constant for Beggiotoa is relatively high (50 µm). Concentrations of oxygen near the sediment surface are almost always lower than this value. Reported values of V max are about 20 pmole bacterium -1 hr -1 at 25 o C. The density of bacteria is estimated to be 10 5 cm -2 on the sediment surface in L. Sempach. Again, because oxygen consumption is dominated by sulfide oxidation, we shall consider the temperature dependence only of this microbial process. Rates of sulfide oxidation by Beggiotoa were measured in the laboratory at two temperatures. Concentrations of sulfide were kept above 20 µm and oxygen was kept at saturated concentrations, so the rates measured are essentially equal to V max (times the bacterial density). From the results of these measurements (see Table 2 below) you will determine the temperature coefficient for sulfide oxidation. Table 2. Experimental measurement of sulfide oxidation rates at two temperatures. Time (minutes) [H 2 S] 20 o C [H 2 S] 10 o C A. From the information in Table 2, determine the temperature correction coefficient, θ, for microbial sulfide oxidation. What is the value of the coefficient when expressed as Q 10? Provide the appropriate graphs and calculations in the appendix to support your answer. SOLUTION Sulfide oxidation is zero order with respect to sulfide concentration; the plot of concentration vs. time is linear as evidenced by the high r 2 values. The rate constants may be calculated from the slopes to be µm hr -1 at 20 o C and µm hr -1 at 10 o C. From the two rate constants, θ, the temperature correction factor, may be calculated as follows: 1 logbqg F k I 1 = log T - T k 2 1 HG 2 KJ 24

25 Filling in the rate constants and temperatures yields a value for θ of Theta is related to Q 10 by the expression: Q 10 = θ 10. Therefore, the Q 10 for sulfide oxidation is H2S Conc. (um) Sulfide Oxidation y = x R 2 = y = x R 2 = oC 10oC Regression (20oC) Regression (10oC) Time (hr) Fig. 6a. Sulfide concentration vs. Time. Because this plot is linear, it may be concluded that the reaction is zero order. The slopes for the lines give the rate(s) and rate constants at each temperature. B. Given the value for K m and the oxygen concentration in the lake, which simplification would you propose for the Monod rate law? Calculate the value of the rate constant. Transform this areal rate constant into a volumetric rate constant that can be used in the mass balance for the hypolimnion. SOLUTION: When K m is large relative to the substrate concentration, the Monod equation can be simplified to a first-order rate law in which the rate constant is equal to V max /K m. For oxygen concentrations below about 1 mg/l, the concentration would be smaller than the value of K m (50 µm or 1.6 mg/l). Thus, we can use the first-order simplification at very low oxygen concentrations. The value of V max is given as 20 pmole bacterium -1 hr -1 (at 25 o C), and the density of bacteria is given as 10 5 cm -2. Thus the rate constant may be calculated as: V F pmole I bacteria L mole k = = x cm K HG m bacterium hr K J F H G max 5 I 11 2 cm K J F H G I -6 m c h 11 50mmole pmole 42. x10 L The value of the first-order rate constant is hr -1. However, at oxygen concentrations above 3 mg/l, we would have to use the zero order simplification in which the rate constant is equal to V max. Converting to the proper units yields: 11 2 F pmole 5 bacteria 136. x10 cm mg mg k = Vmax = bacterium hr cm 42. x10 L 10 9 pmole Lhr HG I K J F H G I K J F H G I K J F H G I K J = KJ F H G I K J F H G I K J 25

26 C. Formulate the expression for oxygen consumption in the hypolimnion that will be utilized in the oxygen mass balance. This expression must include the temperature correction coefficient. SOLUTION: For oxygen concentrations greater than 1 mg/l we could use the following equation: V dc T-T =-2k q ref V dt For lower oxygen concentrations we must either switch to the first-order rate constant with the expression below, or we must include a logic step to stop the reaction when the concentration reaches zero mg/l. V do dt 2 =-2k O q 2 T-T ref V D. To demonstrate the rate of oxygen consumption due to oxidation of reduced substances, model the hypolimnion as if this was the only process occurring. Set the initial oxygen concentration in the hypolimnion at 13 mg/l. The concentration should decrease with time because you will allow no inputs to the hypolimnion. Provide the equation that you use and a graph of the model output (Oxygen concentration vs. time). SOLUTION: The model output indicates that oxygen concentrations decrease linearly in a zero-order fashion until concentrations are below 1 mg/l. At that point, the reaction is modeled as a first-order reaction, and rates taper to zero as the concentration decreases. Sulfide Oxidation D.O. (mg/l) /15/1998 6/14/1998 8/13/ /12/ /11/1998 Date 26