On a cloudy summer day, a wind speed of 4.20 m/s was measured on a 10-m pole. Th
ID: 105249 • Letter: O
Question
On a cloudy summer day, a wind speed of 4.20 m/s was measured on a 10-m pole. The particulate matter (PM) concentration was found to be 1520 mu g/m^3 at a point 2 km downwind and 0.5 km perpendicular to the plume centerline from a coal-fired power plant. a. Determine the atmospheric stability category of the day. (indicate on the Eq. sheet) b. Estimate the windspeed (in m/s) at the effective stack height of 90 m. c. Determine sigma_x and sigma_z by the graphical method (show your work on the Eq. sheet). d. Determine sigma_y and sigma_z by the numerical method. e. Determine the particulate matter emission rate of the power plant in kg/s by using the following Table and the results in part d. (1 kg = 10^3 g = 10^9 mu g)Explanation / Answer
On a cloudy summer day, wind speed of 4.20 m/s was measured on a 10 m pole
Particulate matter concentration of 1520 mugm/meter cube at a point of 2 km downward and 0.5 km perpendicular to the plume centerline from coast-fired power plant.
Atmospheric stability of the Day
For x< 1 ; 440.8 + 1.941 + 9.27/3 = 150.7
For x>1 ; 459.7 + 2.094 + (-9.6) /3 = 150.7
Therefore the Effective Atmospheric stability of the Day is 150.7
The Wind speed in (m/s) at the effective stack height of 90m
u = u1 * (z/z1)p
Where,
u = desired but unknown wind speed, (us) ?
u1 = wind speed at known height, (u10)= 4.20 m/s
z = height where wind speed is unknown, h 90m
z1 = height where wind speed is known, 10m
p = exponent from t= 0.5
Therefore, u = u1 * (z/z1)p = 4.20*(90/10)to the power 0.5
U= 4.20*3 = 12.6 m/s
The emission rate for the particulate matter of power plant
E= C*Q* 10 to the power –6
Where 10 to the power –6 is conversion factor
E= 1520*1*10 to the power –6
E= 1.52 x 10 to the power –3 Mu gm
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