Problem 2: (30 points) Re-do Example 11.12, with the following changes: a. (15 points) Plot the concentration of the pollutant 0.5 km downwind, along the centerline, as a function of stack height (from 15 m to 100 m, in 1 m intervals), assuming zero plume rise. Problem 2a must be submitted in Excel. b. (15 points) Estimate the class of stability for the atmosphere, and explain your reasoning and/or calculations. example/ 11.12 Use of the Gaussian Plume Dispersion Model A manufacturing process emits 2.4g of the pollutant SO, every minute. The stack height is 15 m and there is zero plume rise. Assume the wind speed is 3 m/s and the horizontal dispersion coefficient (ou) is 25 m and the vertical dispersion coefficient (o-) is 15 m for this situation. What is the concentration of the air pollutant 0.5 km downwind of the release along the centerline? What is the concentration at this same distance downwind but at a location 100 m to the side at ground level? ( 21 3)pl solution The problem is asking us to estimate the ground-level concentration of the pollutant in two locations. Both estimations are at ground level so we will use Equation 11.34. For the first question, we are estimating the concentration of the pollutant along the plume's centerline so the x.y.z coordinates are 500 m,0,0. Equation 11.34 can be written as: min 2.4g/min 60s (0m) C(500.0.0 (15 m2 exp exp # X 3 m/s x 25 m 15 m 2 x (25 m) 2 x (15 m) = 6.9 x 10-6 g/m x 10 ug/g 6.9 pg/m? What happens tothe ground-level concentration atthis locationifthestack height())isincreasedandweget some plume rise (Ah) so the value of H is increased to 50 m? Do the calculation yourself and note how the receptor located 0.5 km downwind along the center would now be exposed to a concentration of 0.014 mg/m? What happened? Remember also that we can use Equation 1135 to determine the x distance where the maximum concentration of the pollutant occurs. For the second question we are asked to estimate the SO2 concentration at x,y,z coordinates of 500, 100, 0 min 2.4g/min C(500,100.0) (100m 2 (15m2 exp TIX 3m/s x 25m x 15m 2 x (250) 2 x (157) 60s exp C - 2.3 10-9 'g/m x 100 Mg/8=0.0023 ug/m Note how the SO2 concentration in the y direction is much lower than the concentration along the centerline of plume because of dispersion. Problem 2: (30 points) Re-do Example 11.12, with the following changes: a. (15 points) Plot the concentration of the pollutant 0.5 km downwind, along the centerline, as a function of stack height (from 15 m to 100 m, in 1 m intervals), assuming zero plume rise. Problem 2a must be submitted in Excel. b. (15 points) Estimate the class of stability for the atmosphere, and explain your reasoning and/or calculations. example/ 11.12 Use of the Gaussian Plume Dispersion Model A manufacturing process emits 2.4g of the pollutant SO, every minute. The stack height is 15 m and there is zero plume rise. Assume the wind speed is 3 m/s and the horizontal dispersion coefficient (ou) is 25 m and the vertical dispersion coefficient (o-) is 15 m for this situation. What is the concentration of the air pollutant 0.5 km downwind of the release along the centerline? What is the concentration at this same distance downwind but at a location 100 m to the side at ground level? ( 21 3)pl solution The problem is asking us to estimate the ground-level concentration of the pollutant in two locations. Both estimations are at ground level so we will use Equation 11.34. For the first question, we are estimating the concentration of the pollutant along the plume's centerline so the x.y.z coordinates are 500 m,0,0. Equation 11.34 can be written as: min 2.4g/min 60s (0m) C(500.0.0 (15 m2 exp exp # X 3 m/s x 25 m 15 m 2 x (25 m) 2 x (15 m) = 6.9 x 10-6 g/m x 10 ug/g 6.9 pg/m? What happens tothe ground-level concentration atthis locationifthestack height())isincreasedandweget some plume rise (Ah) so the value of H is increased to 50 m? Do the calculation yourself and note how the receptor located 0.5 km downwind along the center would now be exposed to a concentration of 0.014 mg/m? What happened? Remember also that we can use Equation 1135 to determine the x distance where the maximum concentration of the pollutant occurs. For the second question we are asked to estimate the SO2 concentration at x,y,z coordinates of 500, 100, 0 min 2.4g/min C(500,100.0) (100m 2 (15m2 exp TIX 3m/s x 25m x 15m 2 x (250) 2 x (157) 60s exp C - 2.3 10-9 'g/m x 100 Mg/8=0.0023 ug/m Note how the SO2 concentration in the y direction is much lower than the concentration along the centerline of plume because of dispersion