Resistor Change from Aorta to Capillaries

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The discussion centers on the change in resistance from the aorta to capillary vessels, highlighting that the aorta branches into approximately 9 billion capillaries, significantly increasing the total cross-sectional area. The Hagen-Poiseuille law is referenced to calculate resistance, assuming laminar flow despite blood flow being turbulent. Participants clarify that the cross-sectional area of an individual capillary is much smaller than that of the aorta, specifically 8.89 x 10^-8 times smaller. The total resistance of the capillaries is derived from the individual resistance, leading to a formula that incorporates the number of capillaries. The conversation concludes with expressions of gratitude and encouragement for future medical studies.
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The aorta branches into up to 30-40 billion capillary vessels, 8-10 billion of which are used effectively. As a consequence the cross-sectional area increases by a factor of 800. If we start from the assumption that there are 9 billion capillary vessels (all of which are equally thick) – how does the resistor change from the aorta to the (parallel) capillary vessels?

Anyone please? :)
 
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Come on guys...you use to talk about things like pulling the Moon towards Earth...now, here's a real question and you don't know what to say anymore? You can't be serious on that one...
 
Perhaps you could share with us your thoughts?
 
The Hagen-Poiseuille law is the key...
 
Poiseuille's law requires lamina flow; blood flow is decidedly turbulent.
 
Ya, you're right. But I have to write down something, so let's just pretend that we've got lamina flow here.

Thank you by the way...:)
 
Okay, if we assume lamina flow then the 'resistance' of a circular tube would be given by;

\Re = \frac{8L\eta}{\pi r^{4}}

We can express this in terms of cross-sectional area thus;

\Re = \frac{8L\eta}{Ar^{2}}

You can work the rest out for yourself...:wink:
 
Ah, no...I can't really.

We have the "resistance" of one circular tube now, right? So we have to raise it to the power of 9 billion?
 
Alcyon said:
Ah, no...I can't really.

We have the "resistance" of one circular tube now, right? So we have to raise it to the power of 9 billion?
Not quite, if the total cross-sectional area (the sum of the cross-sectional areas of the individual capillaries) increases by a factor of eight hundred and there are nine billion capillaries, each capillary will have a cross-sectional area of how many times than that of the aorta. HINT: You should obtain a number <<1 since a capillary is many times narrower than the aorta. We are of course assuming, as the question states, that each capillary is of the same dimensions.
 
  • #10
8.89 *10^-8? :)
 
  • #11
Alcyon said:
8.89 *10^-8? :)
Indeed \frac{800}{9\times 10^9}. So the cross-sectional area of an individual capillary is 8.89x10-8 times that of the aorta. Now what happens to the radius of each capillary?

Just for your information, after some research it appears that blood flow from the aorta behaves as lamina flow; I stand corrected.
 
  • #12
The radius is: sqrt (A/pi), isn't it?

(thanks again for your help...really appreciate it. :))
 
  • #13
Alcyon said:
The radius is: sqrt (A/pi), isn't it?
Indeed, however, you may wish to leave it in terms of r2 (it just takes a further step out of the calculation. So using that information we can form as equation in terms of area exclusively;

\Re = \frac{8L\eta}{\pi r^{4}}

\Re = \frac{8L\eta}{\frac{A^{2}}{\pi}}
Alcyon said:
(thanks again for your help...really appreciate it. :))
Its my pleasure :smile:
 
  • #14
I don't fully understand though...what about the 8.89 *10^-8?



Edit: My bad...but what's the final result?
 
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  • #15
Alcyon said:
I don't fully understand though...what about the 8.89 *10^-8?
That is how many times larger the cross-sectional area of a single capillary compared to the aorta. If you substitute that into the last equation for A that will give you the resistance per unit length of a single capillary when compared to the aorta. Do you follow?
 
  • #16
Yes...so the resistance per unit length of a single capillary is to be raised to the power of 9 billion?
 
  • #17
Alcyon said:
Yes...so the resistance per unit length of a single capillary is to be raised to the power of 9 billion?
Why would you wish to do this?
 
  • #18
Edit: My bad again.


1:R_total = 1/R1 + 1/R2 ...

right?
 
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  • #19
Alcyon said:
Edit: My bad again.


1:R_total = 1/R1 + 1/R2 ...

right?
Spot on :smile:
 
  • #20
Okay...but there is R1 - R9billion...I can't possibly do that!


Ah...so: R1 * 9billion?
 
  • #21
Alcyon said:
Okay...but there is R1 - R9billion...I can't possibly do that!


Ah...so: R1 * 9billion?
Yes, as each capillary is identical, the resistance of each will also be identical;

\frac{1}{\Re_{total}} = 9\times10^{9}\cdot\frac{1}{\Re}
 
  • #22
Okay, so R_total = R * 9*10^9 ?
 
  • #23
Alcyon said:
Okay, so R_total = R * 9*10^9 ?
Not quite, you may want to recheck your math there;

\frac{1}{\Re_{total}} = 9\times10^{9}\cdot\frac{1}{\Re} = \frac{9\times10^{9}}{\Re}

\Re_{total} = \frac{\Re}{9\times10^{9}}
 
  • #24
Of course...:D

I can't believe I'm going to be a doctor a few years from now.
Thank you very much, you have been of great help.

Greetings from Holland! :)
 
  • #25
Alcyon said:
Of course...:D

I can't believe I'm going to be a doctor a few years from now.
Thank you very much, you have been of great help.

Greetings from Holland! :)
Good luck with your studies, I imagined this was some sort of biophysics course. It was my pleasure to help you :smile:, an interesting problem; greetings from England.
 
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