A 425-m^3( 15.000 ft^3) basin filled with wastewater is to be aerated with 10 sp

Question

A 425-m^3( 15.000 ft^3) basin filled with wastewater is to be aerated with 10 spargers, each sparger using air at the rate of 7.08 x 10^-3 m^2/s. Find the time that is necessary to raise the dissolved oxygen level in the wastewater from 8 x 10^-2 to 2 x 10^_l mmol/L if the temperature of the water is 2X3 K and the depth of the water above the spargers is 3.2 m. The dissolved solids content will be low enough so that Henry's law will be obeyed, with a Henry's law constant of 3.27 x 10^4atm/mole fraction.

Explanation / Answer

given henry's law constant = 3.27 * 10^4 atm/mole fraction .

but 1 atm/k = 2*0.0825 mmol/lit=> 1atm = 2*0.0825 mmol-k/lit

therefore henry's law constant = 3.27 * 10^4 * 2*0.0825/ 32 mmol-k/lit

=> discharge through aerated sparger = rate of aeration * depth of water above sparger => 7.08 * 10^-3 * 3.2

=> 0.022656 m^3/s.

therefore time required to raise water level = (volume of basin * difference in oxygen level * temperature) / (discharge through aerated sparger * number of spargers * henry's constant )

=> time = (425 * ( 0.2-0.08)*283) / (0.22656 *10*3.27* 10^4 *2*0.0825 /32 )

=> time = 14433/ 382.0014 => 37.7825 sec.

=> therefore time required to raise water level = 37.7825 sec

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