A stack was sampled for 5 minutes. The gas velocity (at stack conditions) throug

Question

A stack was sampled for 5 minutes. The gas velocity (at stack conditions) through the sampler was 15.0 m/s and the area of the sample probe was 4 square cm. The net mass of particulates collected was 0.40 grams.

a. Calculate the concentration of particulates in the stack in micrograms per cubic meter.

b. The temperature in the stack is 350 degrees F and the pressure is 725 mm Hg. If the total flow rate of gas out of the stack is 200 standard cubic meters per second, calculate the total mass emissions of particulates out of the stack in kg/day.

Explanation / Answer

a) flow rate of gas = 15*(4*10^-4) = 6*10^-3 m^3/s...... therefore volume collected in 5 min = 6*10^-3 * 5* 60 = 1.8 m^3..... therefore concentration = 0.4g/1.8 m^3 = (0.4*10^6 micro gram)/( 1.8 m^3) = 7.2*10^5 microgram /m^3..........................

b) concentration = 0.72 g/m^3.... therefore total mass emission of particulates out of the stacke in kg/day = (200m^3/hr * 24hr/day)*(0.72g/m^3 * 0.001 kg/g) = 3.456 kg/day

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