# 1M3-3-2. Mass Transfer from a Pipe Wall. Pure water at 26.1°C is flowing at a velocity of0.0305 m/s ina tube having an inside diameter of 6.35 mm. The tube is 1.829 m longwith the last 1.22 m having the walls coated with benzoic acid. Assuming that thevelocity profile is fully developed, calculate the average concentration of benzoicacid at the outlet. Use the physical property data from Example 21.3-2. [Hint: First,calculate the Reynolds number Duplu. Then, calculate NReNse(D/L) (T/4), which isthe same as WIDABPL.Ans.(CA-CAO)/(CAi-CAO) = 0.0744, cA 2.193 x 103 kg mol/m.3.31.3- represented within +30 % by JpSmJD0.036N2mental data for liquids are correlated within about +40% by the f-50000 (L2):Re,L-0.5Using Eq. (21.3-29),0.007580.99(1.700 x 10Jp 0.99N05JD0.99N0.5Re,LRe.LThe definition of Jp from Eq. (21.3-5) iske(Ne(21.3-5)MPLE 21.3-2.Mass Transfer from a Flat Platee volume of pure water at 26.1°C is flowing parallel to a flat plate Solving for ke k= Jpv(Nc)2. Substituting known values and solving,ic acid, where L 0.244 m in the direction of flow. The water velm/s. The solubility of benzoic acid in water is 0.02948 kg mol/m2.5.85 x 10-6 mS-2/3k0.00758 0.0610Sy of benzoic acid is 1.245 x 10 m2/s. Calculate the mass-transfer In this case, diffusion is for A through nondiffusing B, so k, in Eq. (21.1-10)and the flux NAon: Since the solution is quite dilute, the physical properties of wfrom Appendix A.2 can be used:should be used:(e-CA)= k (eA-C2)Х вм(21.1-10)ACubp8.71 x 10 Pa s1.0 and keke Also, cA1 = 2.948 x 10-2 ks0 (large volume of fresh water). Substituting intoSince the solution is very dilute, XBMmol/m3 (solubility) and CA2Eq. (21.1-10),P 996 kgm3kgmolт2.948 x 100NA k.(CAI-CA2) = |5.85 x 10-6m3m2DAR 1.245 x 10S-91.726 x 10-7 kgmolS.m2hmidt number is8.71 x 10 Pa sNo702Sckg996m3pDd mtem2AB1.245 x 10-9Mass transfer for flow past single spheres. For flow past single spheres and for v= D,uplu, where v is the average velocity in the empty test section before the sphrwood number, which is k'D,/DAB, should approach a value of 2.0. This can beEq. (19.1-24), which was derived for a stagnant medium. Rewriting Eq. (19.-24.,where Dp is the sphere diameter,ynolds number iskg(0.244m 0.0610Lup996m21.700 x 102DABNACA-CA)= k(CA- CA2)8.71x 10 Pa -sRemass-transfer coefficient ke, which is k' for a dilute solution, is then2DАВDPIrranging,ort Processes: Momentum, Heat, and MassP N 2.0DАВShourse, natural convection effects could increase k..

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Asked Oct 22, 2019
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Step 1

Physical property data to be used from example 21.3-2 are:

Step 2

Given data to be used further:

Step 3

Calculate the value of S...

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