Pushing against a fixed obstruction; or, the turbulent hemodynamics of aortic stenosis

More than once I’ve heard the phrase “pushing against a fixed obstruction” when referring to patients with aortic stenosis. This has been used to describe multiple different concepts:

The diameter of the LVOT in moderate to severe AS is unlikely to change despite your most desperate hemodynamic manipulations, and in this sense is “fixed” unlike, for example hypertrophic cardiomyopathy. This makes sense to me.
The left ventricle is unable to increase cardiac output via contractility, and this is why patients with aortic stenosis are particularly preload dependent. When I first heard this, this very much did not make sense to me.

This is the explanation I was given: in chronic moderate to severe aortic stenosis, the left ventricle is often seeing ridiculously high pressures as a result of the stenosis. This leads to concentric hypertrophic remodeling and loss of compliance, meaning that it’s not necessarily trivial to just flood the ventricle with more fluid to increase inotropy.

The question I had was: why can’t the left ventricle simply contract harder against a stenotic valve? Why can’t it just… try harder? In moderate to severe aortic stenosis, the pressure generated by blood making it past the resistance of the valve is generally somewhere around 40 mmHg – the transvalvular gradient. But systemic vascular resistance is often much higher than this number, which is why it is, generally, desirable for the SBP to remain 3 digits long. An additional 40 mmHg doesn’t seem like that much for a heart to compensate for.

Ultimately the answer to this seems to lie in the fact that the relationship between pressure and volumetric flow in ~~fixed spherical frictionless cows in a perfect vacuum~~ normal hearts is different than the relationship between pressure and flow in aortic stenosis.

Slow is smooth and smooth is fast

Suppose we cut out someone’s aortic valve, so there’s no resistance to LV contraction at the outflow tract. In this case, the pressure-radius-flow dynamics are generally modeled/explained in various physiology textbooks and so on with the Hagen-Poiseuille equation, which is most famous for simultaneously terrorizing a) the non-French with regards to its spelling, b) the French with regards to it being persistently misspelled, and c) medical students in that it is an equation with math and exponents.

Normally, this equation is brought up to make the point that, for a fixed rate $Q$ of volumetric flow, through a tube with radius $R$ and length $L$ , making the radius narrower increases the pressure $P$ required to maintain that flow with an exponent of 4. That probably didn’t make sense, but whatever, here’s your equation:

Δ P = \frac{8 μL Q}{π R ^{4}}

But, what is also demonstrated in this equation is that for a fixed radius, there is a linear relationship between pressure and flow – doubling the pressure would (roughly, within a scalar constant, big theta notation be damned) double your flow.

This may be a good time to point out that while the systemic vasculature isn’t actually a single cylindrical pipe, flow in the vasculature is probably reasonably well-modeled using this approach, as the Reynolds number appears to be well below 2000 as soon as the abdominal aorta.

But where did the Hagen-Poiseuille equation come from? Why is the exponent 4?

Skip this section if you don’t care

Disclaimer: I have always been bad at physics.

Suppose we have a Newtonian fluid with density $ρ$ and viscosity $μ$ flowing through a long smooth cylindrical pipe with length $L$ and radius $R$ . In this scenario, the pressure gradient throughout the section of pipe we’ve decided to care about will be constant. We assume the flow is symmetric about the axis of the pipe, with no radial component (no twisting, turbulence, or other shenanigans that might corrupt the youth).

We can describe the flow velocity vector $u$ in cylindrical coordinates $(r, θ, z)$ with time component $t$ :

u (r, θ, z, t) = u_{r} (r, θ, z, t) \cdot e_{r} + u_{θ} (r, θ, z, t) \cdot e_{θ} + u_{z} (r, θ, z, t) \cdot e_{z}

where we have basis vectors

e_{r}, e_{θ}, e_{z}

and the various components of

u

We assume, thanks to the afore-referenced spherical cows, that:

$\partial (___) / \partial t = 0$ (fully developed flow)
$u_{r} = u_{θ} = 0$ (no radial or axial movement)

Individuals much smarter than me have demonstrated that, under these conditions, the full Navier-Stokes equations reduce to

\frac{d P}{d z} = μ (\frac{1}{r} \frac{d}{d r} (r \frac{d u}{d r}))

where $u$ is the axial velocity and we’re working at some radial distance $r \leq R$ from the center of the pipe.

Since our pressure gradient is constant, it will decrease linearly as we go along the pipe. If the total pressure gradient across the entire pipe is $Δ P$ , then we have:

\frac{d P}{d z} = \frac{Δ P}{L}

Well, friends, it’s time for the only part of math class that I was ever good at… the ol’ plug n chug. We have:

\frac{Δ P}{L} = μ (\frac{1}{r} \frac{d}{d r} (r \frac{d u}{d r}))

which gives us

r \frac{Δ P}{μL} = \frac{d}{d r} (r \frac{d u}{d r})

Thus, we can integrate with respect to

r

r \frac{Δ P}{μL} = \frac{d}{d r} (r \frac{d u}{d r})

r^{2} \frac{Δ P}{2 μL} + C_{1} = r \frac{d u}{d r}

for some constant

C_{1}

. If you divide both sides by

r

you get

d u / d r \propto C_{1} / r + r

; but, at

r = 0

we know

d u / d r

is finite, which means that

C_{1}

has to be 0. Great, we now have

r \frac{d u}{d r} = r^{2} \frac{Δ P}{2 μL}

Divide both sides by

r

, and we have

\frac{d u}{d r} = r \frac{Δ P}{2 μL}

So we can integrate with respect to $r$ one more bloody time:

u (r) = r^{2} \frac{Δ P}{4 μL} + C_{2}

So, now we have flow

u

as a function of distance from the axis

r

. Because friction at the wall is annoying, we’ll assume that it doesn’t exist and that

u (R) = 0

, giving us

0 = u (R) = R^{2} \frac{Δ P}{4 μL} + C_{2}

and thus

C_{2} = - R^{2} \frac{Δ P}{4 μL}

which, when plugged back into the ol’ monstrosity, gives us

u (r) = r^{2} \frac{Δ P}{4 μL} - R^{2} \frac{Δ P}{4 μL} = \frac{Δ P}{4 μL} (r^{2} - R^{2})

We’ll then do the same thing yet again, and integrate over

r

to get volumetric flow

Q

Q Q = r = 0 \int R u (r) 2 π r d r = r = 0 \int R \frac{Δ P}{4 μL} (r^{2} - R^{2}) 2 π r d r = r = 0 \int R \frac{Δ P}{2 μL} (r^{2} - R^{2}) π r d r = \frac{π Δ P}{2 μL} r = 0 \int R r (r^{2} - R^{2}) d r = \frac{π Δ P}{2 μL} r = 0 \int R (r^{3} - r R^{2}) d r = \frac{π Δ P}{2 μL} (\int_{r = 0}^{R} r^{3} d r - R^{2} \int_{r = 0}^{R} r d r) = \frac{π Δ P}{2 μL} (\frac{R ^{4}}{4} - R^{2} \cdot \frac{R ^{2}}{2}) = \frac{π Δ P}{2 μL} (\frac{R ^{4}}{4} - \frac{R ^{4}}{2}) = \frac{π Δ P}{2 μL} \cdot \frac{- R ^{4}}{4} = - \frac{π Δ P R ^{4}}{8 μL}

Which, up to a sign, is Hagen-Poiseuille. That sign inconsistency might be my mistake, I’m not sure.

Now, the problem with all this is that I can’t derive Navier-Stokes from scratch, meaning that on some level there are parts of this I cannot prove, which does undermine the value of “deriving it yourself” to a nontrivial degree.

Anyways, back to our regularly scheduled programming.

Turbulent flow and inertia at the aortic valve

Okay, great, we did some approximation of deriving the Hagen-Poiseuille equation, which makes the argument that in free-flowing nice pipes (notably not the case with a stenosed aortic valve), pressure and flow are linearly related – double the pressure, (kind of) double the flow.

But if we’re pushing liquid through a narrow orifice that rapidly expands immediately after – like through a stenosed valve – the losses caused by the viscosity (that $μ$ in Hagen-Poiseuille) are a lot less than losses caused by inertia. You push a bunch of liquid into a narrow jet, that then gets into a much bigger chamber – there it decelerates, spreads, rolls around, engages in various affronts to God and Country, and loses a bunch of kinetic energy to fluid separation, vortices, and turbulence.

The energy cost here appears to be not so much managing the viscosity-induced friction along the length of the pipe, but rather the kinematic energy needed to overcome the inertia of the fluid. This relationship isn’t really what Hagen-Poiseuille is talking about, but is what’s at play with the Bernoulli equation, which in simplified form here can be written as:

Δ P = \frac{1}{2} ρ v^{2}

where

ρ

is the density of the fluid and

v

is the velocity of the fluid. For a pipe with cross-sectional area

A = π R^{2}

, we have

v = Q / A = Q / (π R^{2})

, giving us

Δ P = \frac{1}{2} ρ \frac{Q ^{2}}{π ^{2} R ^{4}}

and if you squint at this with some intensity, you’ll note that instead of

Δ P

being linearly varying with

Q

, it’s now quadratic in Q, so to double

Q

you need 4 times the amount of pressure.

Various abuses of notation

If you remember the earlier setup, we decomposed the overall resistance that the heart sees into resistance caused by systemic vasculature ( $Δ P \propto Q$ ) and resistance caused by the stenotic valve ( $Δ P \propto Q^{2}$ ). We can add these together with broad and unrelenting abuses of mathematical notation and quantitative rigor:

Δ P \propto a Q + b Q^{2}

where

a Q

is the Hagen-Poiseuille-like laminar component from the systemic vasculature, and

b Q^{2}

is the turbulent component at the aortic valve. Of course,

a

and

b

are not real physical parameters, but catch-all terms that allows us to say the following:

At relatively low-turbulence regimes, when the valve is not stenotic, $a Q ≫ b Q^{2}$ , meaning $Δ P \propto Q$ and we can pretty easily increase the flow $Q$ by squeezing harder and increasing $Δ P$ . However, in high-turbulence regimes, like when the valve is stenotic, $b Q^{2} ≫ a Q$ , so $Δ P \propto Q^{2}$ , and it becomes way, way, way harder to increase the flow even with heroic efforts of pressure generation.

So,

The reason the left ventricle is unable to augment cardiac output via inotropic augmentation in aortic stenosis is that changes in pressure produce relatively modest changes in volumetric flow due to the energetic cost of turbulence
Induced distributive shock is not a viable therapeutic strategy for patients with moderate to severe aortic stenosis.

Some fun papers to read in all your spare time

All that said, the mental model above is more a cognitive shorthand to understand what’s happening. The reality is a messier and doesn’t provide a clear dichotomy between linear and quadratic pressure-flow regimes. A couple of sources:

Takeda, Rimington, and Chambers 1999 – plotted transaortic pressure differences against flow and found a nice linear relationship in many cases (though all at rest)

Johnson et al 2018 – studied pressure-flow relationships during several TAVIs. They make the claim that the observed data fit neither linear nor quadratic models of pressure-flow dynamics, but I’m pretty sure I’ve seen worse fits paraded around with much pomp and ceremony, so, who’s to say.

Oh, also, everything here might be wrong. What do I know?