An organic contaminant with a molecular mass of 280 g/mol has to be removed from

Question

An organic contaminant with a molecular mass of 280 g/mol has to be removed from an

aqueous medium. The concentration of this contaminant is 5 g/l. The membrane

selected to perform its removal has an internal diameter of 1 mm and a length of 60 cm.

The filtration process is inside-out. Optimal operating conditions were defined as:

transmembrane pressure of 10 bar, cross-flow velocity of 1.5 m/s and temperature of 25

.C. Under these conditions, a permeate flux of 15 l/(m2.h) was obtained. It was also

observed an apparent rejection of the contaminant of 90%.

a) Determine the mass transfer coefficient for the backward diffusion

process

b) Determine the concentration of solute at the membrane surface and the

polarization modulus.

Additional information:

Please consider as valid the following mass transfer correlation:

Sh = kdi/D= 0, 04 Re0,75 Sc0,33

with: Re = (p v d) /u ; Sc = u/(pD)

p = 103 kg/m3; u = 1.1X 10-3 Pa·s; D = 4.2 X 10-10 m2/s

1 bar = 1 X 105 Pa pi = R T Cmolar R = 0,082 bar l/(mol·K)

Explanation / Answer

Answer:

I have recalculated the values of the constants provided in the question

a) Mass transfer coefficient for back diffusion process

Step 1: we calculate the Mass transfer coefficient

Kc= Sh x D/ ( dia of membrane)

=(0.04X 4.2 X 10-10 )/ 0.001

= 1.68X 10-8 m/s

Step 2:Concentration of solute at membrane surface

the flux of solute

N = permeate flux X (difference in pressure)

= 15 X (10)

=150 l/m2 h

= 4.167 X 10-5 m/s

j = K ( diff. in concentration)

= 8.85 mol/l

hence concentration of solute at the membrane surface = 8.85 mol/l

Step 3:

Polarization factor:

tranfer fluk (J)X reject factor/ mass transfer coeff.

= 0.2

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