Question

A hollow-fiber dialysis unit is one design that provides a large surface area between blood and dialysate to permit efficient mass exchange. Total blood flow rates in the device are typically 200 mL/min, similar to flow within the kidneys. A typical hollow-fiber unit is 30 cm long with 250 fibers, each of diameter 200 µm. The fluid density is 1.05 g/cm³ and the viscosity is 0.03 g/cm-s. Assuming that blood behaves as a Newtonian fluid, determine the entrance length for each fiber and assess whether it is small relative to the length of the unit.

Answer 1

I will tell you how to start this, and from there, try to do it yourself:

We are asking for the entrance length. The assumption is important, because it basically tells us we can use the maths we know, without having to account for variable change and intermolecular forces. Basically, if we didn't have that assumption, smaller forces come into play making the problem much harder than it needs to be.

Straight away we are going to change the blood flow rate from 200 (mL min^-1) to (cm^3 s^-1). 1mL = 1cm^3. We divide the number by 60 to go from min^-1 to s^-1. So it's 3.33 (cm^3 s^-1). Now we need to work out the volume of ONE hollow-fibre unit. V = cross-sectional area X length. CSA = pi*r^2 = pi*(200x10^-6 / 2)*2 = pi*0.0002. The reason I made the length x10^-6 was to put the unit in metres, so the equation works. Leave this in terms of pi for now. V = pi*0.0002*(30x10^-2) = 1.88x10^-4.

From here, you can calculate the entrance length.

Hope this helps.