# How do physics notions of fluid dynamics relate to pressure gradients in circulation?

I'm having a hard time comprehending why sometimes physiology notions seem to contradict each other and contradict physics teachings. More specifically I don't understand why aortic coarctation causes the pressure gradient to increase. In this condition the radius of the blood vessel decreases which should mean that pressure decreases too. According to Bernoulli theorem ΔP+Δvp/2=o where v is velocity and p density. This theorem and the fact that velocity has to increase to maintain a stable flow through the coarctation should mean that the pressure gradient drops. Moreover in every textbook I've found it is said that tension on the aortic wall increases too but this seems even more counterintuitive, because if the radius decreases due to La Place T=Pr tension (T) should decrease too. Then the most puzzling thing to me is the pressure drop when there vasoconstriction of the arterioles, this is fine according to my reasoning so far (except it seems to contradict aortic coarctation) but resistance is supposed to increase too with vasoconstriction, and because of this equation: ΔP=Q*R (where Q is flow and R is resistance) the pressure gradient should go up. Now I think I understand that it increases because of the successive drop in pressure, correct? What am I missing with aortic coarctation? I'm sure it has to do with confusing the PTM in that specific place and the pressure gradient but I can't figure it out.

As far as we know so far, biology isn't able to break any laws of physics.

However, biology isn't passive: biological systems react to changing environments to maintain some level of homeostasis. Arguably this is one of the defining characteristics of life.

If you narrow the aorta, you will always get compensatory changes. Ultimately, the circulatory system is tasked with delivering sufficient blood to the body. It might be a bit of an oversimplification, but let's assume that blood flow stays constant, because flow is a good measure of the ability of circulatory system to deliver enough oxygen/nutrients to the periphery.

Blood flow is pulsatile, which also adds some challenges in understanding things simply, but let's simplify and think in terms of mean pressures. Again, this is wrong, but it's not terrible, kind of like when in a physics class you discuss a "uniform spherical cow" or something like that.

ΔP=Q*R

You write:

In this condition the radius of the blood vessel decreases which should mean that pressure decreases too

This is false. If we keep flow (Q) constant, but R increases due to a constriction, then ΔP, the change in pressure, must increase.

Consider:

ΔPaorta=Q*Raorta

If Q stays constant, but R (resistance) increases due to the decreased radius, then ΔPaorta must increase: there is a larger pressure drop. Perhaps you've confused the larger change in pressure with a "decrease in pressure" because there is this larger pressure drop?

Consider also that we can consider the rest of the parallel vasculature (all branches off the proximal aorta) in series with the aorta. If we think of the vasculature after the aorta as:

ΔPdistal=Q*Rdistal

then you can think of the whole system as:

(ΔPaorta + ΔPdistal) = Q*(Raorta + Rdistal)

If only the aorta is constricted and the distal vasculature keeps the same pressure and flow, then ΔPdistal must also not change. Therefore, the total pressure drop (equal to the difference in pressure between the aorta and great veins, the sum ΔPaorta + ΔPdistal) will be larger when ΔPaorta increases.

As far as Bernoulli:

This theorem and the fact that velocity has to increase to maintain a stable flow through the coarctation should mean that the pressure gradient drops. Moreover in every textbook I've found it is said that tension on the aortic wall increases too but this seems even more counterintuitive

Due to Bernoulli, if you go from large (A), to small (B), to large (C) diameter in some vessel, yes, you will have less outward pressure in the segment (B). You will also always have decreasing energy from A to B to C. However, think about what we were just talking about: if you want to keep the perfusion of the rest of the vasculature constant, that means you have to keep the energy at C (interchangeable with pressure in this context) constant, despite the constriction. You shouldn't be thinking about A B and C here when comparing a vessel with and without stenosis, you have to think about the non-stenosed case where you have large (X), large (Y), large (Z). In this case, you still have decreasing energy from X to Y to Z, but without the stenosis, you don't lose as much going through segment "Y". It is segment A that requires more energy/pressure/tension on the vascular wall compared to segment X.

There's some simplification here regarding vascular compliance, non-Newtonian fluids, and vascular permeability but these can be mostly ignored for these purposes. I think this website will help you a lot:

https://www.cvphysiology.com/Hemodynamics/H001

especially

https://www.cvphysiology.com/Hemodynamics/H012

for the part about Bernoulli's principle.

First, you are misapplying Bernoulli's law. At each point, Bernoulli's law says that $$\text{constant}=P+\frac{\rho v^2}{2}$$ But remember, that constant is not constant in time for unsteady flows; it is constant in space. To make use of Bernoulli's law, we should consider two different points in the fluid flow.

I will call the one closer to the aorta "upstream" and the one closer to the superior vena cava "downstream." Since the constant in Bernoulli's law is the same for these two points, it cancels if we subtract. Thus $$0=\Delta P+\frac{\rho}{2}\Delta(v^2)$$ (In fact, this is still not true, but it's a good place to start — see below.) If the fluid returning to the heart is faster than the the fluid pumped out, then $$\Delta(v^2)$$ will be positive, but the vena cava will have lower pressure than the aorta, so that $$\Delta P$$ will be negative.

So yes, if the change in velocity increases, then $$\Delta P$$ will "decrease", but decreasing a negative number gives it a larger magnitude.

Now for the technicality: it might seem peculiar that I assumed $$\Delta(v^2)$$ is nonzero. If fluid flows faster out of region than in, then either the density changes, or, before long, there will be no fluid left there! So if $$\Delta(v^2)$$ is nonzero, either the heart will soon exsanguinate, or the blood is somehow rarefacting and compressing. Is that the case? Of course not!

Bernoulli's law does not apply to blood flow, because Bernoulli's law assumes viscous forces are negligible. If viscous forces were negligible, you wouldn't need a heart; your blood would just flow on its own, a perpetual motion machine. Instead, each millimeter of blood vessel has a boundary layer of slow-moving fluid at its edges — slowed down, because it touches the capillary walls. Keeping this fluid moving absorbs some of the hydraulic head.

Since blood vessels are largely homogenous, the amount of head absorbed is proportional to distance traveled. This means a good model for blood flow is actually pumping against gravity: we can consider "downstream" points as "pretend higher" than upstream points.

Bernoulli's law in the presence of substantial altitude (or "altitude") changes has another term: $$0=\Delta P+g\Delta h+\frac{\rho}{2}\Delta(v^2)$$ To translate between our model (in which fluid is pumped up) and blood flow, $$h$$ describes the length traveled by a blood cell and $$g$$ describes the viscous drag per millimeter. Decreasing the arterial thickness increases the cross-sectional proportion of blood flow in which the viscous drag from the capillary walls is relevant. In short, it increases $$g$$. So capillary constriction has much the same effect as an increased $$\Delta(v^2)$$: it must decrease $$\Delta P$$ to compensate. But now the same analysis as before shows that we are decreasing a negative number, giving it a larger magnitude.

(I've assumed in my formulations of Bernoulli's law that our density $$\rho$$ is constant. This is not true, but a reasonable approximation, as Bryan Krause's answer says.)