A 200-MW coal-fired power plant consumes 1200 kg/min of coal to generate this el

Question

A 200-MW coal-fired power plant consumes 1200 kg/min of coal to generate this electricity. It is using coal with a 7.3% ash content (i.e., 7.3% of the coal mass becomes ash upon combustion), and pollution control strategies employed prevent 99.6% of this ash from being released to the atmosphere. Presume all other ash is contained to be collected for disposal.

(a) If the density of the ash is 650 kg/m^3, what volume of ash would be collected by the pollution control measures at this facility each month? Assume 1 month = 30 days.

(b) If the quantity of ash released from the power plant each minute were to be eventually despersed in 4,000,000 m^3 of air at 25 degrees Celsius and 1 atm, what would be its concentration expressed in ppm? Assume particulate matter to have a "molecular weight" of 29.0 g/mol.

Explanation / Answer

power plant using 1200 kg/min of coal which contains 7.3% of Ash. this implies; content of Ash in coal = (1200*7.3)/100 per minute Power plant uses ash content= 87.6 kg per minute only 99.6% ash released to the atmosphere and rest are collected for disposal. total mass of ash to be released = (87.6*99.6)/100 = 87.25 kg per minute (a) for this part; density of ash is given= 650 kg/m^3 we know that volume= mass/density mass of ash produces per minute =87.25 kg therefore, volume per month= 87.25*60*30/650 =241.62 m^3 (b) as per scale of concentration, volume concentration= volume of pollutant/(volume of pollutant + volume of air) volume of pollutant = 241.62 m^3 per month or = 0.134 m^3 per minute volume of air given= 4*10^6 m^3 therefore, from above formula , volume conc. = 0.134/(0.134+4*10^6) =3.35*10^(-8) concentration in ppm = 3.35*10^(-8) * 10^6 = .0335 ppm

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