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#1




50% Decreased Flow in the Coronary Artery
A 49 YO woman with CAD undergoes coronary arteriography that shows a 50% decrease in the lumen of the LAD. For any arteriovenous pressure gradient, the flow thru this artery (compared with normal) wall decrease by a factor of which of the following?
a) 2 b) 4 c) 8 d) 16 e) 32 need explanation of equation please? 
#2




its very simple... lumen is reduced by half (50%) which mean that the radius is 1/2 the size, therefore
1/(0.5)^4 = 16 The flow thru LAD artery wall decrease by a factor of 16. 
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Mondoshawan (01272011) 
#3




I thought Poiseulle Law is talking about resistance not flow, Can we apply it to flow?



#4




Quote:
and Flow = ΔP/resistance so the factor of change in ΔP (1/r^4 in this case) is the same as that in Flow, by the transitivity of multiplication. 
#5




Quote:
The italicized part should say "Flow is Q in the equation, so the factor of change in Flow is the inverse of that in ΔP (1/r^4 in this case) = r^4 = 1/16. So, as aktorque said, the flow would decrease by a factor of 16." Hope that's clearer, and sorry for the bad explanation before! 
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aktorque (01272011) 
#6




This is really the gist, and all we really need to know about Poiseuille's equation. Flow is directly proportional to radius^4, and resistance is inversely proportional to radius^4.

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Jorabim (06282014) 
#7




Mondoshawan is right. this is another way to look at it;
Resistance = (8 x viscosity x length)/ radius^4 Resistance = Pressure change/Flow If radius reduces by 1/2, then resistance increases by 16. (inverse proportnl) If resistance increases by 16,then Flow reduces by 16.(inversely proportional) for questions like this, Goljan said, just remember '16', the boards wont go beyond that. hope this helps. 
The above post was thanked by:  
Jorabim (06282014), Mondoshawan (01272011) 
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