Fluids and pressures (artery question)

[Cypripedium]

Homework Statement

The diameter of a certain artery has decreased by 22% due to arteriosclerosis.

(a) If the same amount of blood flows through it per unit time as when it was unobstructed, by what percentage has the blood pressure difference between its ends increased?

(b) If, instead, the pressure drop across the artery stays the same, by what factor does the blood flow rate through it decrease? (In reality we are likely to see a combination of some pressure increase with some reduction in flow.)

Homework Equations

flow rate = (pi/8) * deltaP/(viscosity*L) * r^4. Fill in for point A (before the blockage) and fill in for point B (inside the blockage) and then set equal because flow rate doesn't change (for part (a)).

The Attempt at a Solution

For (a), I know that everything is constant except for deltaP and r, so the equation can be simplified down to deltaP of B/delta P of A = (r of A)^4/(r of B)^4 which is, in turn, = to (1/.78)^4, or at least I think it is. That gives a percentage of 2.7% which is incorrect.

For (b), I am similarly able to reduce the equations down to (.78/1)^4. That gives a factor of .37 drop in blood flow rate. Once again, I'm incorrect.I'm clearly not doing something correctly here. Can anyone shed a bit of light? Thank you.

 

[chaoseverlasting]

Going by your equation, rate, $$r=k P D^4$$ where $$k=\frac{2\pi}{L\nu}$$.

Differentiating this equation, $$dr=kP(4D^3)d(D)$$, now we know d(D)=.22 or 22%. Substituting this, you get your change in rate. Similarly for constant pressure.

 

[Cypripedium]

Going by your equation, rate, $$r=k P D^4$$ where $$k=\frac{2\pi}{L\nu}$$.

Differentiating this equation, $$dr=kP(4D^3)d(D)$$, now we know d(D)=.22 or 22%. Substituting this, you get your change in rate. Similarly for constant pressure.

chaoseverlasting, thank you for the response. However, to be honest, I don't really have any idea what those equations mean.

For part (a), I am able to get the equation down to deltaP of B / deltaP of A = (r of A / r of B)^4. That comes out to (1/.78)^4 = 2.7%. I believe my calculations are correct but I'm unable to frame them into exactly what the questions ask for.

I have the same issue with part (b). (.78/1)^4 = .37, but I don't really know how to frame that into what the question is asking for.

 

[Cypripedium]

Ok, I figured out my issue.

For part (a), I took my number, subtracted 1 from it, and then multiplied by 100 to get the %.

Similarly, for part (b), I subtracted my number from 1. Makes sense now.

 

[88elephants]

Alright, I am working a similar problem, but I don't understand what you said in the last section. I understand this part:

For part (a), I am able to get the equation down to deltaP of B / deltaP of A = (r of A / r of B)^4. That comes out to (1/.78)^4 = 2.7%. I believe my calculations are correct but I'm unable to frame them into exactly what the questions ask for.

I have the same issue with part (b). (.78/1)^4 = .37, but I don't really know how to frame that into what the question is asking for.

but after that, I simply get lost. How is the percentage for the blood pressure difference ACTUALLY calcuated from here? Sorry for unearthing this post, but any help would be appreciated!