## Fluid Statics

Statics -- sounds like a boring topic eh? So let's start with an exam question, one that actually appeared in the undergraduate cardiology curriculum at a veterinary school. I think there will be some who are surprised by the "correct" answer.

Which of the following exhibits the greatest average (time average) pressure in an 80kg, standing adult human?

- Superior vena cava
- Left ventricular chamber
- Pedal artery (in the foot)
- Ascending aorta
- Pulmonary capillaries

#### Fluid Statics

**F**and

**a**iindicate that these quantities, force and acceleration, are vectors having both magnitude and direction. Sigma (\( \Sigma\) ) represents a summation so the equation indicates that the sum of forces acting on the material is equal to 0 when there are no accelerations. The concept could be restated in terms of conservation of momentum but the equation says it all.

^{2}so that it's weight (a force which is also a vector) \( W=\rho g\Delta x\Delta y\Delta z\) For purposes of discussion we will assume that gravity acts in the negative \( z\) direction, i.e. "downward" on the figure shown below.

*stress*. Consequently a force due to pressure is the result of the pressure multiplied by the cross-sectional area it acts upon. In this case the cross-sectional area is \( {\Delta y\Delta z}\) . The pressures at these locations actually correspond to some kind of average at each location. Technically, we should choose \( p|_{x+\Delta x/2}\) so that \( p|_{x+\Delta x/2}{\Delta y\Delta z}\) is exactly equal to the force acting on that surface. However we will see shortly that this doesn't matter much as we're going to let the size of the surface approach zero so that the pressure will become the pressure at a point and the average issue goes away.

*p*in the

*x*direction:

*partial derivative*because the pressure could change with more than one variable; we've still got the \( y\) and \( z\) directions. I hope it will be obvious that the analysis of the equation of motion for the \( y\) direction proceeds identically as the \( x\) direction derivation and so I'll just leave it out.

*equation of motion*looks like this:

*Sum*of the forces acting on the fluid cube is zero. The minus sign in front of the body force is due to the fact that we've agreed that these act in the negative \( z\) direction. Dividing through by \( {\Delta x\Delta y\Delta z}\) :

*static*part of the Bernoulli equation. It's missing the parts having to do with acceleration, friction, etc. because all those things have to do with motion. You may be used to seeing \( h\) (as in

*height*) in place of the \( z\) shown here, but I've oriented the coordinate system so that \( z\) is exactly the same thing as \( h\) .

*density*of the air also changes ( decreases ) as we ascend. Nevertheless we can still agree arbitrarily that the pressure here in the room 0.

^{3}. In part A of the figure below, the opening of the catheter is at the same height as the manometer which reads ... well zero! (your choice of physical pressure units) That's what we agreed to in the last paragraph. The course of the catheter is completely immaterial as long as the fluid in the catheter is static. All straight?

_{2}0 which happens to correspond to 0.735 h mmHg. On this planet we're all familiar with substituting the height of a column of fluid with known density (e.g. mercury or water) in place of the actual pressure. We'll revisit this because it's a source of confusion and error in the medical literature. However the point of this last thought experiment is that

*So here it is, the obvious fact that just never got pointed out in your cardiology textbook:*

**the manometer does not read 0 despite the fact that the catheter tip is still at 0 pressure!!**The static pressure indicated by aproperly calibratedfluid filled catheter-manometer system is NOT THE ACTUAL STATIC PRESSURE. The indicated pressure is the actual static pressure,\( p\) , plus the quantity \( \rho g h\) where \( h\) is the height of the catheter tip opening above the manometer and \( \rho\) is the density of the fluid in the catheter. This fact is your "go to" for understanding many aspects relating to how these instruments behave!

Let's look at another situation. In figure A below, the catheter tip is poised to enter a reservoir of fluid that has the same density as the fluid in the catheter. What is the indicated pressure on the manometer? Well, this is no different than figure B above; the manometer indicates a pressure of \( \rho g h\) where the fluid density is that of the fluid in the catheter. Simple enough. And now what does it indicate in figure B below?

The answer is \( \rho g h\) ; the manometer indicates the same pressure that was indicated before the catheter was plunged into the liquid. It happens to be the actual pressure also, but in fact the manometer indicates the same value for pressure no matter what level the catheter tip is immersed in the liquid (but only if the liquid in the catheter and in the reservoir have the same density!) In figure B, the catheter tip is at the same level as the manometer. According to the law given above, that means that the manometer will read the actual static pressure. The manometer always indicates \( p+\rho g h\) -- and \( h=0\) in that situation so it indicates \( p\) . Furthermore we know that the **actual** pressure at that location is \( \rho g h\) since the catheter is submerged to depth \( h\) in the reservoir; \( p=\rho g h\) at that depth.

This situation is more readily understood if you can grasp the fact that the reservoir is really just a part of ... an extension of the catheter. Once the catheter is submerged in the reservoir, we have a fixed situation where the top of the fluid surface of the reservoir is really where the tip of the catheter is. We can no longer affect the reading by repositioning the catheter in the reservoir. The fluid in the catheter is continuous with the fluid in the reservoir – it's all just one body of fluid!

Given a fluid-filled catheter whose tip is submerged in a fluid that has the same density as the fluid in the catheter, the manometer no longer registers pressure changes due to gravitational potential.

The situation above can be expressed analytically as follows: The pressure indicated by the manometer is

\(\Large p_i= p_a+\rho gz\)

The actual pressure in the reservoir is

\(\Large p_a= \rho g(h-z)\)

where \( h\) is the height of the reservoir surface above the manometer and \( z\) is the vertical coordinate relative to the manometer. We're letting \( p_i\) stand for the pressure *indicated* by the manometer and \( p_a\) is the *actual* pressure You can see that when \( z=h\) we are talking about the surface of the reservoir where the actual pressure should be atmospheric, i.e. zero (and \( h-z=0\) ). At the level of the manometer, we have that \( z=0\) and the pressure function gives \( p_a= \rho g(h-0)=\rho gh\) . Since that's the actual pressure, let's see what the manometer reads:

\(\Large p_i=p_a+\rho gz=\rho g(h-z)+\rho gz=\rho gh\)

i.e. a constant value. So a fluid filled catheter-manometer system essentially deletes the effects of gravitational potential when immersed in a fluid that has density equal to the fluid within the catheter. In many situations where pressures in the cardiovascular system are being measured, this is actually the desired result. However you have to be clear that this is happening and make sure that this IS the desired result. A micro-manometer tipped catheter, for example, DOES NOT BEHAVE THIS WAY. When properly calibrated, it indicates the actual pressure** regardless.**

And now we are ready for the ..

#### Answer to the Exam Question

The answer to the exam question is #3, the pressure is highest in the pedal artery. I voted to throw out the exam question where the correct answer had been listed as the ascending aorta. If that's the answer you picked, I'm sure you're not alone. Cardiologists and many physiologists are preconditioned to the concept that blood flow always proceeds from high pressure to low pressure. If you looked at the intervening explanations for fluid statics, then you've reviewed the fact that this concept is incorrect. In cardiology/physiology we often conceptualize hemodynamic function in terms of the pressure obtained from a measurement with a fluid filled catheter. As you saw above, a correctly calibrated fluid filled catheter does not indicate the actual pressure! I think a question of this type is acceptable (e.g. for cardiology board exams) with the addition of a clause stating what kind of a pressure measuring device is being used, but should also ask "what does the device read (or 'indicate')?" ( as opposed to what is the actual pressure ).

This was not a completely "static" hemodynamic situation and you needed to understand or approximate pressure differences due to gravity versus losses due to resistance, i.e. friction related to flow and blood viscosity in the aorta. In rough terms, the dorsal pedal artery is about 100 cm below the level of the heart/ascending aorta in a standing adult human. That pressure translates to a meter (100 cm) H_{2}O which corresponds to ~74 mmHg; the time averaged pressure is A LOT higher in the foot than in the aorta due to the difference in gravitational potential (and don't get me started on giraffes or King Kong!). In contrast, the average pressure loss due to viscous (pressure) losses in the aorta is only a few mmHg.

_{2}O is sometimes seen also. It is sometimes important to remember that the physical units of these measurements are literally

**length**, centimeters or millimeters. The density of mercury is about 13.6 gm/cm

^{3}, i.e. 13.6 times as dense as water, but mercury pressure columns are commonly reported in millimeters, water pressure in centimeters. To convert either column height into an actual pressure we need to multiply by \( \rho g\) ; the thing does not have physical units of pressure until you do that. \( g\) is the acceleration of gravity and is equal to about 980 cm/s

^{2 }which can also be expressed as 9.8 m/s

^{2}or about 32 ft/s

^{2}. If we try directly multiply a column of mercury by \( \rho g\) , we end up with ( millimeters ) x ( gm/cm

^{3}) x (cm/s

^{2}). This has to be a pressure, but it's not in any recognizable, commonly employed units. Consequently we have to use a funny value for the density of mercury : 1.36 gm/(cm

^{2}-mm). Taking that approach, millimeters in the numerator and denominator cancel and we end up with units of gm/cm-s

^{2 }which is also known as a dyne/cm

^{2}which is a true unit of pressure. (A dyne is a unit of force equal to a gm-cm/s

^{2}.)

^{2}is a pretty small unit of pressure for clinical purposes and there are about 1333 of them per mmHg taking the latter as a measure of pressure which also corresponds to about 980 of them per cm H

_{2}O. Now here's an interesting number: the square root of (2 x 980) is ~44.3 happens to be the "Gorlin constant". This number shows up in the denominator of the Gorlin equation where we are taking the square root of a pressure; the 2 is in there because fluid kinetic energy density s calculated as \( \rho v^2/2\) and somewhere along the line they just bundled the 2 in there with the pressure. This has caused a lot of misunderstanding. First off, the Gorlin constant has nothing to do with any experiment done by those researchers; it's simply a physical units conversion constant that derives from the laws of physics. The literature suggests ERRONEOUSLY that this constant varies with flow rate. The thing that (potentially) varies with flow rate in stenosis flow is called the

**discharge coefficient**and I think this is sufficiently clear in the paper. Secondly, the original paper is absolutely clear that pressures were being measured in centimeters of water, NOT millimeters of Mercury. However this constant (44.3) has been used throughout the medical literature in conjunction with pressures measured in mmHg. (The number that actually appears in the paper is 44.5 due to a minor miscalculation, maybe done with a slide rule?) The correct "Gorlin constant" for pressures reported in mmHg is the square root of (2 x 1333) = 51.6. A relatively minor point that I will discuss elsewhere is due to the fact that we can't take the square root of a pressure at all; the physical units of pressure do not permit it. This is one of those curious failings in the medical literature that has propagated without correction for decades.

^{2}to obtain a pressure in mmHg? Well first of all we have to recognize that kinetic energy density has the same physical units as pressure. As alluded to above, kinetic energy density is calculated as \( \rho_b v^2/2\) where \( \rho_b\) is the density of the blood. Our task is to convert this pressure into mmHg.

^{3}x \( \Large\frac{v^2}{2}\) m/s = \( \rho_b\) gm/cm

^{3}x \( \Large\frac{v^2}{2}\) m/s x (100 cm/m)

^{2}= \( \rho_{Hg}g h\)

^{2}

^{2}so the constant comes out to 1.09 x 100

^{2}/ (2 x 1333) ~ 4.08; the physical units of the constant are pretty messy due to mixing of nonstandard units for pressure and velocity. So anyway, the 4.0 that we use in the simplified Bernoulli equation is pretty close to an appropriate number. The purpose of this was to 1) verify the number and 2) think about the fact that we have 2 fluid densities to deal with – the blood or fluid of interest and (separately) the density of the pressure reference fluid (mercury in this case).