Sulfur dioxide and carbon dioxide emissions estimates: Emissions from the Neal p
ID: 513556 • Letter: S
Question
Sulfur dioxide and carbon dioxide emissions estimates: Emissions from the Neal power plant can be can be estimated using the parameters below: m^dot_coal = 100 kg/s = coal mass flow rate P_s = 0.0025 sulfur fraction of coal P_coal ash = 0.03 = fraction of coal converted to ash P_air = 0.15 = fraction of excess air used to combust coal P_S ash = 0.05 = fraction of sulfur captured in ash Using the parameters provided and reasonable assumptions, calculate the following: [3]The average daily mass (kg/d) of sulfur dioxide released from the stack. [4] The amount of oxygen consumed per day in reactions with coal expressed in both mass (kg/d) and volume (m^3/d). The total volume of air used in the combustion process each day (m^3/d) [5] The concentration (ppmv) of SO_2 in the stack. [5] The mass of CO_2 emitted each year (kg/y).Explanation / Answer
a) Average daily mass (kg/d) of sulfur dioxide released from the stack:
In 1 sec mass of coal flow = 100 kg
Sulfur fraction of coal = 0.0025
fraction of sulfur converted to sulfur dioxide = 1- PS ash = (1 - 0.05) = 0.95
Sulfur dioxide released from the stack in 1 sec =
In 1 sec mass of coal flow * Sulfur fraction of coal * fraction of sulfur converted to sulfur dioxide = 100 * 0.0025 * 0.95 = 0.2375
Average daily mass (kg/d) of sulfur dioxide released from the stack =
Sulfur dioxide released from the stack in 1 sec * 60 sec * 60 mins * 24 hours
0.2375 * 60 * 60 * 24 = 2052 kg/d
c). Total volume of the air used in the combustion process each day:
Coal mass flow in 1 sec = 100 Kg
Coal mass flow in 1 day = 100 Kg * 60 sec * 60 min * 24 hours
= 8,640,000
Total volume of the air used in the combustion process in 1 day = Fraction of air used in the process * coal mass flow in 1 day
= 0.15 * 8,640, 000
= 1,296,000 m3/d
d). Fraction of Sulfur converted to SO2 = ( 1 - PS ash) = ( 1- 0.05) = 0.95
Concentration of SO2 in the stack = mass of coal * Sulfur fraction of the coal * Fraction of sulfur converted to SO2
= 100 Kg * 0.0025 * 0.95
= 0.2375 Kg
= 0.2375 * 106 mg
= 237500 mg or 237500 ppmv
e) mass of CO2 emitted in 1 second = mcoal * Pair
= 100 * 0.15
= 15 Kg / s
mass of CO2 emitted in 1 year = mass of CO2 emitted in 1 second * 60 sec * 60 min * 24 hours * 365 days
= 15 * 60 * 60 * 24 * 365
= 4,730,400,00
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