Mathematics 108 Professor Alan H. Stein November 10, 2004 SOLUTIONS 1. (10 points) Consider a one-dimensional diffusion situation where a gas is of mass 15 grams is released in the center of a thin tube. Assume a diffusion constant D = 0.1 centimeters per second. Calculate the concentration of the gas at the center and at locations 0.5, 1, 1.5, 2, 5 and 10 centimeters from the center at times 1, 2, 5 and 10 seconds after the gas is released. Putting the data into a spreadsheet based on the one-dimensional diffusion spreadsheet on the class home page yields the following concentrations, given in grams per centimeter. One Dimensional Diffusion Mass M = 15 Diffusion Const D = 0.1 Distance Time Concentration 0.5 1 7.162296173 0.5 2 6.922361951 0.5 5 5.280979901 0.5 10 3.975052985 1 1 1.098373692 1 2 2.71083589 1 5 3.629560868 1 10 3.295434671 1.5 1 0.048259172 1.5 2 0.568222071 1.5 5 1.942763935 1.5 10 2.410991509 2 1 0.000607493 2 2 0.06375275 2 5 0.809864498 2 10 1.556653115 5 1 9.61792 10 27 5 2 2.5367 10 13 5 5 2.23008 10 5 5 10 0.008168566 10 1 3.5716 10 108 10 2 4.88834 10 54 10 5 1.15419 10 21 10 10 5.87657 10 11 Page 1 of 6
Page 2 of 6 2. (20 points) Suppose 1000 kilograms of a soluble material are spilled into the center of a large shallow lake and gradually diffuses out into the lake. Suppose the diffusion constant in the north/south directions is 0.4 square meters per minute while the diffusion constant in the east/west directions is 0.8 square meters per minute. (a) Find the concentrations at a point 10 meters north and 15 meters east of the center 10, 20, 30 and 60 minutes after the material is spilled. Putting the data into the two-dimensional diffusion spreadsheet yields the following concentrations, given in kilograms per square meter. Two Dimensional Diffusion Mass M = 1000 Diffusion Constants D1 = 0.4 D2 = 0.8 Distances x = 10 y = 15 Concentration C = 2.40017 10 5 Time Concentration 10 C = 2.40017 10 5 20 C = 0.009187521 30 C = 0.056031988 60 C = 0.25629195 (b) Find the concentrations at a point 15 meters north and 10 meters east of the center 10, 20, 30 and 60 minutes after the material is spilled. Putting the data into the two-dimensional diffusion spreadsheet yields the following concentrations, given in kilograms per square meter. Two Dimensional Diffusion Mass M = 1000 Diffusion Constants D1 = 0.4 D2 = 0.8 Distances x = 15 y = 10 Concentration C = 2.40017 10 5 Time Concentration 10 C = 4.82813 10 7 20 C = 0.001303068 30 C = 0.015238718 60 C = 0.133656832
Page 3 of 6 3. (20 points) A chemical plant emits a volatile organic compound into the air at the rate of 100 grams per second from a stack 100 meters above ground. Given conditions, the effective stack height is actually 130 meters. The average wind speed is 9 meters per second and the wind is constant. (a) Find the average concentration at a point directly in line with the center of the plume at a point 1, 000 meters downwind if the stability class is C. Putting the data in the Plume Spreadsheet, we get the following: Calculation of Concentration Ratio C/Q and Concentration C from an Elevated Release Using the Gaussian Plume Model Note: See README file on installation disk for suggestions on enhancing the utility of this spreadsheet program.?-?-?-?-? Input Data?-?-?-?-?-? Wind Speed (meters/sec) = 9 Down Wind Distance (meters) = 1000 Stack Height (meters) = 130 Horiz Dist from Plume Center (meters) = 0 Receptor Elevation (meters) = 130 Stability Class = c Source term Q (quantity per second) 100 *-*-*-*-*-*-*-* Calculated Results *-*-*-*-*-*-*-*-* Vertical Dist from Plume Center Line = 0 SigmaY (meters)= 105.9605283 SigmaZ (meters)= 60.74697882 Concentration Ratio C/Q (sec/m 3 ) = 2.75E 06 Concentration C (quantity /m 3 )= 2.75E 04 Note: Here we enter the stack height as 130, since that is the effective stack height, the horizontal distance from the plume center as 0 and the receptor elevation as 130, since the center of the plume is 130 meters is that high. We conclude the concentration is 2.75 10 4 grams per cubic meter. In the other parts, we will only note the changes in what was entered from the preceding part. (b) Find the average concentration at a point 10 meters above and 15 meters to the side of the center of the plume at a point 1, 000 meters downwind if the stability class is C. Changes: Horiz Dist from Plume Center (meters) = 15 Receptor Elevation (meters) = 140 (since it s 10 meters above the center of the plume.
Page 4 of 6 We obtain a concentration of 2.68 10 4 grams per cubic meter. (c) Find the average concentration at a point directly in line with the center of the plume at a point 1, 500 meters downwind if the stability class is C. Changes: Horiz Dist from Plume Center (meters) = 0 Receptor Elevation (meters) = 130 Down Wind Distance (meters) = 1500 We obtain a concentration of 1.34 10 4 grams per cubic meter. (d) Find the average concentration at a point 10 meters above and 15 meters to the side of the center of the plume at a point 1, 500 meters downwind if the stability class is C. Changes: Horiz Dist from Plume Center (meters) = 15 Receptor Elevation (meters) = 140 We obtain a concentration of 1.32 10 4 grams per cubic meter. (e) Find the average concentration at a point directly in line with the center of the plume at a point 1, 000 meters downwind if the stability class is E. We have the same input as for (a) except the Stability Class = e. We obtain a concentration of 2.12 10 3 grams per cubic meter. (f) Find the average concentration at a point 10 meters above and 15 meters to the side of the center of the plume at a point 1, 000 meters downwind if the stability class is E. We have the same input as for (b) except the Stability Class = e. We obtain a concentration of 1.70 10 3 grams per cubic meter. (g) Find the average concentration at a point directly in line with the center of the plume at a point 1, 500 meters downwind if the stability class is E. We have the same input as for (c) except the Stability Class = e. We obtain a concentration of 1.17 10 3 grams per cubic meter. (h) Find the average concentration at a point 10 meters above and 15 meters to the side of the center of the plume at a point 1, 500 meters downwind if the stability class is E. We have the same input as for (d) except the Stability Class = e. We obtain a concentration of 1.02 10 3 grams per cubic meter.
Page 5 of 6 4. (10 points) Describe two examples of two-dimensional diffusion. 5. (10 points) Explain the difference between physical stack height and effective stack height, why there can be a difference and why the difference can be important. 6. (5 points) Suppose that exhaust gases are leaving a vertical smoke stack at a constant rate while the wind at the top of the stack is blowing at 15 miles per hour. What will the effect be on the effective stack height if the wind calms down to 5 miles per hour? Explain the effect qualitatively; obviously, you do not have sufficient information to quantify the effect. The effective stack height will increase. 7. (10 points) Suppose a mass of 15 grams is released at the center of a long, thin tube and teh diffusion constant for the materials involved is D = 0.1 cm 2 /sec. Find the concentration at a point 5 centimeters from the center 10 seconds after the mass is released. You may use the one-dimensional diffusion formula C = M e x2 4Dt. 4πDt C = 15 4π 0.1 10 e 52 4 0.1 10 8.16856586 10 3, so the concentration will be approximately 8.16856586 10 3 grams per centimeter. 8. (10 points) Consider the attached table defining atmospheric stability classes. Determine the stability class (A, B, C, D or E) under each of the following conditions. Justify your answer. (a) It s a bright, sunny day with virtually no wind. Stability Class A, since the incoming solar radiation will be strong with a surface wind < 4.5 miles per hour. (b) It s a heavy, overcast day with occasional showers and winds between 8 10 miles per hour. Stability Class B, by Note 1. (c) It s a fairly clear evening, with about 15% cloud cover and winds between 8 10 miles per hour. Stability Class E, since by Note 4 Night refers to evening as well, there s less than 3/8 cloud cover and winds between 8 10 miles per hour fit in the class of winds between 6.7 11.2 miles per hour.
Page 6 of 6 9. (5 points) Consider the Gaussian Plum Model: C = Q u 1 [ e y 2 1 2σy 2 σ y 2π σ z 2π e (z H) 2 2 ] + e (z+h)2 2. and e (z+h)2 2, in the right- In general terms, explain the presence of two terms, e (z H) 2 2 most factor. The term e (z+h) 2 2 accounts for a gas bouncing off the ground. Table 3-1 Definitions of Atmospheric Stability Classes Surface wind speed Day Night miles per meters per Incoming solar radiation Thinly overcast or 3/8 cloud hour second strong moderate slight 4/8 cloud cover024