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The aorta

  1. Jul 9, 2006 #1
    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? :)
     
  2. jcsd
  3. Jul 9, 2006 #2
    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...
     
  4. Jul 9, 2006 #3

    Hootenanny

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    Perhaps you could share with us your thoughts?
     
  5. Jul 9, 2006 #4
    The Hagen-Poiseuille law is the key...
     
  6. Jul 9, 2006 #5

    Hootenanny

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    Poiseuille's law requires lamina flow; blood flow is decidedly turbulent.
     
  7. Jul 9, 2006 #6
    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...:)
     
  8. Jul 9, 2006 #7

    Hootenanny

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    Okay, if we assume lamina flow then the 'resistance' of a circular tube would be given by;

    [tex]\Re = \frac{8L\eta}{\pi r^{4}}[/tex]

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

    [tex]\Re = \frac{8L\eta}{Ar^{2}}[/tex]

    You can work the rest out for yourself...:wink:
     
  9. Jul 9, 2006 #8
    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?
     
  10. Jul 9, 2006 #9

    Hootenanny

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    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.
     
  11. Jul 9, 2006 #10
    8.89 *10^-8? :)
     
  12. Jul 9, 2006 #11

    Hootenanny

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    Indeed [itex]\frac{800}{9\times 10^9}[/itex]. 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.
     
  13. Jul 9, 2006 #12
    The radius is: sqrt (A/pi), isn't it?

    (thanks again for your help...really appreciate it. :))
     
  14. Jul 9, 2006 #13

    Hootenanny

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    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;

    [tex]\Re = \frac{8L\eta}{\pi r^{4}}[/tex]

    [tex]\Re = \frac{8L\eta}{\frac{A^{2}}{\pi}}[/tex]
    Its my pleasure :smile:
     
  15. Jul 9, 2006 #14
    I don't fully understand though...what about the 8.89 *10^-8?



    Edit: My bad...but what's the final result?
     
    Last edited: Jul 9, 2006
  16. Jul 9, 2006 #15

    Hootenanny

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    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?
     
  17. Jul 9, 2006 #16
    Yes...so the resistance per unit length of a single capillary is to be raised to the power of 9 billion?
     
  18. Jul 9, 2006 #17

    Hootenanny

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    Why would you wish to do this?
     
  19. Jul 9, 2006 #18
    Edit: My bad again.


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

    right?
     
    Last edited: Jul 9, 2006
  20. Jul 9, 2006 #19

    Hootenanny

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    Spot on :smile:
     
  21. Jul 9, 2006 #20
    Okay...but there is R1 - R9billion...I can't possibly do that!


    Ah...so: R1 * 9billion?
     
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