Question

A power plant, located close to the downtown of a developed city emits 1042 kg per hour of SO2 from stack with an effective height of 120 m. a) For a wind speed of 3 m/s (measured at the effective height) in an overcast winter morning, calculate the ground level SO2 concentration (in mg/m*) along the plume center line I km downwind. b) If the wind speed (3 m/s) had been measured at 20 m elevation instead of the effective height,what would the actual concentration of SO2 (in mg/m*) in the same location (in part a) at the night of the same day?

Answer 1

Part a:-

Emission of SO2 from stack, Q= 1042 kg/hr = 1042 * 1000 / (60 * 60) = 289.444 gm/sec

Wind speed measured at the effective height, u = 3m/s

Effective height of stack , H = 120 m

Concentration at X = 1km along centre line of plume means y =0 and x = 1km. This concentration is given by:

C(x,0) = ( Q / (π * u * σz *  σy )) * (e)^( -H2 / (2 *σz2)) gm/m3...............................(A)

where, C = the concentration of SO2 in gm/m3

Q = the pollutant emmision in gm/ sec

u = mean wind velocity in m/s

x and y = downwind and cross wind horizontal distances in m

σy = Plume's standard deviation in cross wind direction in m

σz = Plumes's standard deviation in vertical direction in m

H = effective height of stack in m

From chart of different atmospheric stabilities of Class C i.e. Slightly unstable conditions (as given overcast winter morning) for x = 1 km,

we get, σz = 75 and  σy = 110

Therefore, C (1,0) = (289.444 / ( 3.14 * 3 * 75 * 110 )) * (e)^ ( - (1202) / ( 2 * 752)) gm/m3

=> C (1,0) = 1.035 * 10^ (-3) gm/m3

=> C (1,0) = 1.035 mg/m3

So, the ground level SO2 concentration along the plume center line 1 km downwind is 1.035 mg/m3.

Part b :-

Wind velocity measured at 20 m elevation, u20 = 3m/s

Wind velocity at the effective height i.e. at 120 m = u120

u120 can be found out by equation => u2 = u1 ( z2 / z1 )p

where, p = empirical constant which depends on stability class and ux = wind velocity at elevation zx

Here, we have to found the concentration of SO2 in an overcast winter night in city, so, the stability class is D i.e. neutral. So for this class D, p = 0.25

=> u120 = u20 ( 120/20)0.25 = 3 * 1.5651 = 4.695 m/s

So, wind velocity , u measured at effective height of stack is 4.695 m/s

From chart of different atmospheric stabilities of Class D i.e. neutral conditions (as given overcast winter night with wind speed 4.695m/s) for x = 1 km, and y=0

we get, σz = 35 and  σy = 78

So, Actual concentration of SO2 in the same location (as in part a) at the night of the same day can be found out by:

C(x,0) = ( Q / (π * u * σz *  σy )) * (e)^( -H2 / (2 *σz2)) gm/m3

=> C (1,0) = (289.444 / ( 3.14 * 4.695 * 35 * 78 )) * (e)^ ( - (1202) / ( 2 * 352)) gm/m3

=> C (1,0) = 2.0138 * 10^ (-5) gm/m3

=> C (1,0) = 0.020138 mg/m3

Actual concentration of SO2 in the same location (as in part a) at the night of the same day is 0.020138 mg/m3.