It is desired to remove trace amounts of benzene from water. A vessel that is 4

Question

It is desired to remove trace amounts of benzene from water. A vessel that is 4 m in diameter is to be used to strip benzene from water by bubbling air through the water at 298K. The water will flow at a continuous rate of 10 kg/s downward at 298 K. The water contains 830 ug/L benzene, and it is desired to remove 95% of the benzene with a water flowrate 25% higher than the required minimum. The benzene-water solution can be expected to follow Henry's law with m- 145. The bubbling apparatus is 2.75m in diameter at the bottom of the column with 1000 orifices, each 5 mm in diameter (a) Estimate the depth of water required for this separation (b) Estimate the power required for the air compressor if the mechanical efficiency is 45%.

Explanation / Answer

Stripping ratio, S=.00075*(A/H)

Normally, S=12

=>A/W, Air to Water flowrate = 16000/H

H=145

Water flowrate = 10 kg/s=10 litre/s (For water, 1 Kg=1 Litre)

Air flowrate = 16000/145*10=1103 litre/s

For 25% over design, Air flowrate = 1.25*1103=1379 litre/s

Concentration of benzene in water in = 830 ug/l

=>Mole fraction of bezene in water in = (830*10^-6/78)/(1000/18)=1.9*10^-7

After removing 95% benzene,

=>Mole fraction of bezene in water out = 1.9*10^-7*(1-.95)=9.5*10^-9

Number of transfer Units, NTU=ln(xin/xout)=ln(190/9.5)=3

For A/W=112, Height of Transfer unit = 4 ft

=> Height of column = NTU*HTU=3*4=12ft

b) Total Pressure drop in 12 ft water column = density*g*height of column = 1000*10*12/3.3=36363 Pa = 0.3 bar

Power required by air to overcome column Pressure drop = deltaP*Volumetric flowrate = 36363*1103/1000=40108 Watt

Compressor efficiency = 45%

=>Power required by Pump = 40108/.45=89128 W= 90 KW

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