Question

In: Civil Engineering

The urban air above the city of Shanghai if the mixing height above the city is...

The urban air above the city of Shanghai if the mixing height above the city is 1210 m, the width of the box perpendicular to the wind is 105 m, the average annual wind speed is 15,400 m/hr, and the amount of sulfur dioxide discharged is 1375 × 106 lb/yr. Is this a conservative or non-conservative system? State all assumptions. Sketch a diagram to represent the system. Write the complete mass balance for SO2. Estimate the SO2 concentration in the air leaving the city in μg/m .
Hint: Determine the wind flow in m3/hr.

Expert Solution

avg wind speed = 15400/3600 = 4.27 m/s

wind flow = 1210*105*15400= 19.5657 m3/hr

SO2 discharge = 1375*106 lb/yr = 1375*106/365/24 = 16.63 lb/ hr = 117.74576437863671 mols/ hr

117.74576437863671 mols of SO2 = 22.4*117.745= 2637.50 litres/hr = 2.637 m3/ hr

concentration of SO2 in ppm = 1ppm of SO2= =

1 litre of SO2 has mass of = 64/22.4= 2.85 gm

1000litre or 1m3 = 2857.1gm

therefore 1 ppm of SO2 = 2857.1 gm/ 10^6 m3 =

1 ppm of SO2 = 2857.1micro gram / m3

0.0013441 ppm of SO2 = 3.840 microgram / metric cube

this a conservative system

assumption are standard temprature and pressure conditions

temperature of 273.15 K (0°C or 32°F)

and a pressure of 10⁵ Pascals formerly 1 atm

Ally Wells answered 1 year ago

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