Problem:SO2is released from a 90m stack at a rate of 800 g/s. During the release period the average wind velocity at 10m is 4 m/sec and the atmospheric stability is class B. The observed plume rise for the stack is 60m. Claculate the ground level concentration at: (i), 3000 m downwind and 100 m crosswind, and (ii), 3000 m downwind and 500 m crosswind.Solution:Step 1:Determine the values of the dispersion coefficients,yandz, from the appropriate graphsfor x = 3000 (Appendix A).y= 417 m

z= 363 m

Step 2:Determine the effective stack height by adding stack height to plume riseH= 90 m + 60m

= 150 m

Step 3:Determine the windspeed at effective stack height.u150

u10=(z150

z10)p

u150=(4 m/sec)(150 m

10 m)0.15=6.0 m/sec

Step 4:Compute concentration using Gaussian dispersion model at (i) y=100 m, and (II) y = 500 m.(i)C=Q

uyzexp (-H2

2z2)exp (-y2

2y2)

C=800 g/sec

(6 m/sec)(417 m)(363 m)exp ((-150 ft)2

2(363 m)2)exp ((-100 ft)2

2(417 m)2)

C=2.5 x 10-4g

m3=250g

m3

(ii)C=Q

uyzexp (-H2

2z2)exp (-y2

2y2)

C=800 g/sec

(6 m/sec)(417 m)(363 m)exp ((-150 ft)2

2(363 m)2)exp ((-500 ft)2

2(417 m)2)

C=1.25 x 10-4g

m3=125g

m3

Exercise:At what wind speed would the ground level concentration at 100 m be 70g/m3?(21.4 m/sec)

Problem:The traffic on the highway can be considered a line source. The I-75 traffic density along a certain stretch is 6,000 vehicles per hour with an average speed of 50 mph. The average hydrocarbon emission rate for each vehicle can be taken as 0.03 g/s. Assume it is 2:30 p.m. on a bright, sunny day and the wind is blowing perpendicular to a portion of the highway at 6 m/s. Calculate the total hydrocarbon concentration at a point 500 m downwind.Solution:Step 1:Begin by calculating how many vehicles are crossing through one meter.6000 vehicles/hour

50 miles/hour(1 mile

1600 m)=0.075vehicles

m

Step 2:Calculate the emission rate, qq=0.075 vehicles

m(0.03 g

s(vehicles))=2.25 x 10-3g

s(m)

Step 3:Determine the stability class. For a clear, sunny day with speed of 6 m/s, the stability classis C.Step 4:Assuming the area is rural,zcan be calculated from Brigg's formula:z=0.08(x)(1+0.0002x)

=0.08(500 m)(1+0.0002(500 m))

=38.13 m

Step 5:Calculate concentration at 500 m.C=2q

(2)(z)()

C=2(2.25 x 10-4g/s(m))

(2)(38.13 m)(6 m/s)

C=8 x 10-6g

m3=8g

m3

Exercise:After a sale on radar dectors, the average speed of the vehicles increases to 90 mph. What is the new hydrocarbon concentration at a point 500 m downwind?(4.36g/m3)