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Discussion in 'Step I' started by HelicopterBen, Aug 14, 2009.
Generally, it should be the same for all areas, as viscosity is an inherent physical property of a fluid.
However, it is possible in the specific case with increased vessel hydrostatic or interstitial colloidal pressure, viscosity will be the highest in the capillaries. This is due to plasma filtering out of capillaries into the interstitium, leading to a higher cell to plasma ratio.
Actually viscosity is lowest in capillaries and at the same time the velocity is the highest in them!
not true. velocity is lowest in the capillaries. total cross sectional are is highest in the capillaries, leading to the lowest velocity there. why would you want high velocity in the area of exchange?
woah... are you gents talking velocity or viscosity?
If I recall from my biotransport phenomena class, hematocrit is highest in capillaries due to the rouleau effect hence viscosity is also highest at capillaries.
Gozar's absolutely right about the inverse relation between velocity and total surface area.
In this link , they explain a little bit more about visscosity:
They dont mention an area specifically , and I cant find that in BRS neither , so I think as Somedoc says , that if should be the same.
Oops! Thanks for the reminder Gozar!
Sorry fellows I've made a mistake!
Viscosity is inversely related to velocity. Since, capillaries have the highest velocity they have the lowest viscosity!
I hope the case is closed now!
I guess the answer is inversely related to what you just said then..
Excuse me if I bring up this thread but... this is confusing.
It's true that flow velocity is LOWEST in capillaries - and that mostly occurs because capillaries DO have the highest total cross-section area.
But you are implying that viscosity increases in capillaries which it doesn't seem right to me.
What about plasma skimming?
I guess there's a game of apparent and effective viscosity here.
well I think that the link above is pretty conclusive...viscosity is lower in small diameter vessels.