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Homework Help: Pressure difference to maintain blood flow

  1. Sep 16, 2008 #1
    hey guys.. i need a bit of help on this question...

    For a non-smoker, with blood viscosity of 2.5x10-3Pa's, normal blood flow requires a pressure difference of 8.0 mm of Hg between the two ends of an artery. If this person were to smoke regularly, his blood viscosity would increase to 2.7x10-3 Pa's, and the arterial diameter would constrict to 90% of its normal value. What pressure difference would be needed to maintain the same blood flow?


    any help here would be greatly appreciated!!

    cheers :smile:
    KC
     
  2. jcsd
  3. Sep 16, 2008 #2

    tiny-tim

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    Hi KC! :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Sep 17, 2008 #3
    ok, this is what i had so far... not sure if i am on the right track or not though...

    non smoker:2.5 x 10^-3 Pa.s
    smoker: 2.7 x 10^-3 Pa.s

    density of blood @ 1 atm: 1060 kg/m3

    normal blood flow: [tex]\Delta[/tex] 8.0 mmHg

    pressure difference:
    smoker - nonsmoker = 2 x 10^-4

    pressure difference: ________________________________

    that's where i am not sure about what to do.

    i tried: 8.0 mmHg / 2 x 10^-4 = 40,000 mmHg or 4.0 x 10^-3......
    but i know that that's not the answer coz i havent taken into account for the 10% constriction of the artery.:bugeye:

    any clues or nudges in the right direction would be GREAT!! :biggrin:
     
  5. Sep 17, 2008 #4

    tiny-tim

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    Hi twiztdlogik! :smile:

    I have to confess I only know about non-viscous flow. :redface:

    But this looks to me like a dimensions question … y'know, of the "if three men take two days to paint a cube, how long do seven men take to paint a cube with twice the radius and with brushes half as small and paint twice as thin" sort. :wink:

    So what equation do you know relating pressure to viscosity and diameter and flow? :smile:
     
  6. Sep 17, 2008 #5
    erm.. are you thinking of the poiseuilles equation...?
    i think that would work if i can transpose it to get the right formula for this situation

    thanks any ways!
     
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