The heart muscle is supplied with nutrients and oxygen through the coronary vessels. The blood flow through the coronary vessels by more than a factor of 15 through the deposition of Plaques on the vessel inner wall, occur in athletic stress complaints. For the sake of simplicity, we consider blood as Newtonian fluid, flowing through a laminar tube. The current is described by the law of Hagen and Poiseuille : $$V = \frac {d V} {d t} = \frac {π r 4 Δp} {8\dot(viscosity)l}$$ , Δ p is the pressure difference at the ends of the tube of length l . The Viscosity of blood is $$2.5 · 10^{-3} $$
a) By what factor is the diameter of the coronary vessels reduced when the Flow rate has fallen by a factor of 15! The pumping action of the heart is constant (same Δp).
This is not a homework Question, I already have the answer but I do not understand how they got to it and was hoping someone could explain it.
Answer: Since everything but r, V and Δp are constant we can use the following equation $$Δp=\frac V{r^{4}}$$ Hence, $$\frac {V_1}{r_1^{4}} = \frac {V_2}{r_2^{4}}$$ Comment:Why? How did they get this?
then $$V_2=\frac {V_1}{15}$$ $$15=\frac {V_1}{V_2}=\frac {r_1^{4}}{r_2^{4}}$$ Comment: Why is it not $$**\frac {r_2^{4}}{r_1^{4}}**$$ Therefore the answer is: $$\frac {r_1}{r_2}= \sqrt[4]15\approx2$$
There's a typo somewhere. I'm not sure if it's your post or your textbook. You divide both sides of the first equation by $r^4$ to get $$\frac{V}{r^4}=C\Delta p$$ for some constant $C,$ which is ignored. I'll leave it in though, so you can see how it drops out. Your post has $$\frac{V^4}{r}=C\Delta p$$ which is clearly a typo.
We are told that $\Delta p$ is constant, so $C\Delta p$ is constant, and therefore, $$\frac{V_1}{r_1^4} = C\Delta p = \frac{V_2}{r_2^4}.\tag 1$$ This is the same result that your text gets. Dividing both sides by $V_2$ and multiplying both sides by $r_1^4$ gives $$ \frac{V_1}{V_2}=\frac{r_1^4}{r_2^4}\tag 2.$$ Since we are told that $V_1/V_2=15,$ $(2)$ gives $$\frac{r_1^4}{r_2^4}=15.$$