Question

Conventional flux in centrifuge?

Answer 1

The molecules sediment in centrifuge, which is rotating with angular velocity ωω. The chemical potential of molecule (considering potential associated with centrifuge force) is:

μi=μ0i+RTlnai−Miω2r2/2μi=μi0+RTln⁡ai−Miω2r2/2

∂μi∂r=0+RT∂aiai∂r−Miω2r∂μi∂r=0+RT∂aiai∂r−Miω2r

The total diffusion flux is:

Ji=16Cβ2νexp(−ΔdiffGRT)aiajRTγ(∂μj∂r−∂μi∂r),Ji=16Cβ2νexp⁡(−ΔdiffGRT)aiajRTγ(∂μj∂r−∂μi∂r),

where ii - stands for molecules and jj - solvent. If we assume that concentration of molecules are low (c→0c→0, ∂aj∂r=0∂aj∂r=0) one can get:

Ji=16Cβ2νexp(−ΔdiffGRT)(−∂aiγ∂r+aiMiω2rγRT)Ji=16Cβ2νexp⁡(−ΔdiffGRT)(−∂aiγ∂r+aiMiω2rγRT)

My problem is this: I know that conventional flux should be equaled to:

Jconvi=vci,Jiconv=vci,

where cici-concentration of molecules.

I do not understand how to reduce my equation and extract convectional part and so calledordinary part (J∼∂ci∂r)J∼∂ci∂r) of diffusion.

Answer 2

$\color{Black}{ \boxed{\bold{ \underline{Hey. Mate .Your .Answer}}}}$

$<b><u>$____________________________

✌️✌️${\mathbb{NICE.. QUESTION}}$✌️✌️

✔️✔️${\mathbb{CORRECT... ANSWER}}$✔️✔️

✔️✔️conventional membrane separation and CMS is presented and used to ..

. increase the permeate flux over the non-rotating case.


μi=μ0i+RTlnai−Miω2r2/2μi=μi0+RTln⁡ai−Miω2r2/2 ∂μi∂r=0+RT∂aiai∂r−Miω2r∂μi∂r=0+RT∂aiai∂r−Miω2r

The total diffusion flux is:

Ji=16Cβ2νexp(−ΔdiffGRT)aiajRTγ(∂μj∂r−∂μi∂r),Ji=16Cβ2νexp⁡(−ΔdiffGRT)aiajRTγ(∂μj∂r−∂μi∂r),

where ii - stands for molecules and jj - solvent. If we assume that concentration of molecules are low

(c→0c→0, ∂aj∂r=0∂aj∂r=0) one can get: Ji=16Cβ2νexp(−ΔdiffGRT)(−∂aiγ∂r+aiMiω2rγRT)Ji=16Cβ2νexp⁡(−ΔdiffGRT)(−∂aiγ∂r+aiMiω2rγRT)



$\color{Green}{ \boxed{\bold{ \underline{Hope .Help .You .Thanks}}}}$