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?

  1. Superior vena cava
  2. Left ventricular chamber
  3. Pedal artery (in the foot)
  4. Ascending aorta
  5. Pulmonary capillaries
If you can't stand the suspense, click HERE to jump to the answer. Otherwise let's talk about some fluid statics first.

Fluid Statics

Although studying something that is static doesn't sound like it would be very revealing, there are some very interesting things that derive from this consideration.  Solid materials ( including the heart)  typically can be treated as if they are static – not moving.  More specifically, materials can be considered static when accelerations can be considered negligible for one reason or another.  Although the heart obviously moves, the accelerations typically are minimal in comparison with the forces or stresses within the material.  When accelerations are minimal, we can apply an unimpeachable law of physics:
\(\large \Sigma \bar{F}=m \bar{a} = 0\)
Force equals mass times acceleration. Overbars on 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.
This equation applies to every piece of matter, in part or in whole.  This allows us to consider ANY piece of material, fluid or solid, and apply the equation to analyze forces acting on the material.  Engineers use an incredibly simple, devastatingly powerful tool for this purpose called a free body diagram. The FBD is simply a sketch or illustration of an object, or any part of the object, demonstrating all of the forces acting on the object. From the FBD, one or more equations are derived showing how the forces and accelerations are related. The latter equations are called the equations of motion ( even if there's no motion).
Let's now take a look at the free body diagram of an element of fluid within a larger body of fluid.  For simplicity, consider a small cube of the fluid; this turns out to be arbitrary enough.
The fluid cube is centered at arbitrary point \( (x,y,z)\) with cube dimensions \( \Delta x\) , \( \Delta y\) , \( \Delta z\) . Consequently the mass of the cube is \( \rho\Delta x\Delta y\Delta z\) where \( \rho\) is the fluid density (mass per unit of volume).  Assuming we're still on planet Earth, the force acting on the fluid cube due to gravity is its mass times acceleration of gravity g = 980 cm/s2 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.
The force of gravity is an example of a "body" force, one that seems to reach out with an invisible hand to pull (or push) on an object.  Other examples include magnetic or electrostatic forces.  However your frame of reference can make it seem like there is a body force due to acceleration.  We have all experienced this force, e.g. riding on a "merry-go-round" ( in which case you are accelerating) and there seems to be a force pushing you to the outside of the circle.
In the next figure, pressure forces acting on the fluid cube in the \( x\) direction are shown. With the cube centered at position \( (x,y,z)\) and with a cube width in the \( x\) direction of \( \Delta x\) , one of the \( x\) faces is located at position \( x-\Delta x/2\) and the other at \( x+\Delta x/2\) .  These two faces of the cube are highlighted below.
The pressures at these locations are designated \( p|_{x-\Delta x/2}\) and \( p|_{x+\Delta x/2}\) respectively.  Pressure at ANY location is said to act equally in all directions, but when we're talking about a surface the pressure force acts perpendicular to the surface.  Hence the pressure at location \( {x-\Delta x/2}\) acts to push the cube in the \( x\) direction and the pressure at \( {x+\Delta x/2}\) acts to push it in the \( -x\) direction.
Pressure has physical units of force per unit area; it's a 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.
We are now ready to write an equation of force balance for the \( x\) direction.
\(\large  p|_{x-\Delta x/2}{\Delta y\Delta z}-p|_{x+\Delta x/2}{\Delta y\Delta z}=0\)
These forces sum to 0 specifically because 1) the fluid cube is not accelerating (in fact we've assumed it is stationary) and 2) these are the only forces acting in the \( x\) direction.  The next step is to divide the equation by \( {\Delta x\Delta y\Delta z}\) which yields:
\( \large\frac{p|_{x-\Delta x/2}-p|_{x+\Delta x/2}}{\Delta x}=0\)
And then take the limit of this expression as \(\Delta x\rightarrow 0\) which yields none other than the definition of the (negative) derivative of p in the x direction:
\( -\large\frac{\partial p}{\partial x}=0\)    or      \( \large\frac{\partial p}{\partial x}=0\)
As noted above, the pressure is now the value at the point \( (x,y,z)\) so the choice of how to define the average pressure has gone away.  This was nothing other than a long-winded, (rigorous, analytical) way of saying that the pressure does not change in the \( x\)  direction, i.e. perpendicular to the body force ( gravity ).  We are not saying that the pressure itself is zero; just that the pressure doesn't change in that direction.  The expression contains a 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.
The analysis for the \( z\)  direction isn't quite as simple ...
The forces on the 2 \( z\)  faces due to pressure look pretty similar to the above development for the \( x\)  direction:
\( \large p|_{z-\Delta z/2}{\Delta x\Delta y}\)   and   \( -p|_{z+\Delta z/2}{\Delta x\Delta y}\)
However there is a third force acting in the \( z\) direction, the body force due to gravity.  The force balance equation or equation of motion looks like this:
\( \large (p|_{z-\Delta z/2}{\Delta x\Delta y})-(p|_{z+\Delta z/2}{\Delta x\Delta y})-(\rho g\Delta x\Delta y\Delta z)=0\)
This equation says that the 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 \( {\Delta x\Delta y\Delta z}\) :
\(\large \frac{p|_{z-\Delta z/2}-p|_{z+\Delta z/2}}{\Delta z}-\rho g=0\)
and allowing   \( \Delta z\) to approach 0:
\(-\large \frac{\partial p}{\partial z}=\rho g\)  or \( \large\frac{\partial p}{\partial z}=-\rho g\)
This is a very simple differential equation that's easily integrated:
\( \large \int dp=-\int \rho g dz\)
\( \large p|_{1}^{2}=-\rho g z|_{1}^{2}\)
\( \large p_{2}-p_{1}=-\rho g (z_{2}-z_{1})\)
\( \large p_{1}+\rho g z_{1}=p_{2}+\rho g z_{2}\)
Recognize it? This is the 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 \( h\) .
The consequences of this development are far-reaching and it's worth considering them closely.  We've just observed that gravity ( or any other body force ) induces a stress distribution, a pressure gradient, in a static piece of material.  We would also see exactly the same thing for a piece of solid material ( as opposed to fluid ).  Under typical circumstances materials deform under the stress imposed by gravity. This doesn't always make much of a difference, but it makes a heck of a difference when you are building something here on earth that will eventually have the effects of gravity removed ( for example ).  I don't know how much this had to do with the original goof on the Hubble space telescope whose mirror had to be shaped to incredibly close tolerances.  Imagine having to understand exactly how much the mirror was deformed on earth and exactly what shape it would assume after the stress of gravity was removed!!
I would also point out that all the fluid cubes in the above are at rest despite the fact that the pressure is higher on the underside of the cube than the top. Clinicians sometimes have it ingrained in their thinking that this necessarily results in motion of the fluid, from high pressure to low pressure.  This is the place to divest yourself of this misconception.  Fluid elements are just pieces of matter. They sometimes obey the resistance equation.  They always obey the \( \mathbf{F}=m\mathbf{a} \)equation.

So now we start relating this to practical cardiology. To start the discussion, we have to agree upon something that is simply not true -- that the pressure here in the Cath Lab is 0 (mmHg or whatever units you care to use).  As a matter of fact the pressure is actually about 14.7 pounds per square inch which corresponds to about 760.2 mmHg.  As we all know, that's because we are at the bottom of a sea of air.  As we've just learned, that pressure will decrease as we ascend a mountain or go up in an airplane.  As a matter of fact, airplane altimeters are based on this fact and are designed to convert decreasing pressure in the atmosphere to an altitude.  It's a little more complicated in the atmosphere than for the problem above because the density of the air also changes ( decreases ) as we ascend.  Nevertheless we can still agree arbitrarily that the pressure here in the room 0.
Now consider a fluid filled catheter, attached to a properly calibrated manometer, which contains water whose density is 1.0 gm/cm3.  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?
Now in Figure B, the tip of the catheter has been raised above the level of a manometer to a height, h.  What does the manometer read now?
Well, \( \rho g h\) of course or just \( h\)  cm H20 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 the manometer does not read 0 despite the fact that the catheter tip is still at 0 pressure!! So here it is, the obvious fact that just never got pointed out in your cardiology textbook:
The static pressure indicated by a properly calibrated fluid 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) H2O 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.


Constants and Conversions

A potential source of problems and confusion is due to the substitution of the height of a fluid column in place of a pressure.  Obviously this is done all the time in cardiology and clinical pressures are most commonly reported in mmHg (or Torr); cm H2O 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/cm3, 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/s2 which can also be expressed as 9.8 m/s2 or about 32 ft/s2.  If we try directly multiply a column of mercury by  \( \rho g\) , we end up with ( millimeters ) x ( gm/cm3 ) x (cm/s2 ).  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/(cm2-mm).  Taking that approach, millimeters in the numerator and denominator cancel and we end up with units of gm/cm-s2 which is also known as a dyne/cm2 which is a true unit of pressure.  (A dyne is a unit of force equal to a gm-cm/s2.)
A dyne/cm2 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 H2O.   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.
While we're at it, let's figure out the constant for the simplified Bernoulli equation.  What should you multiply a squared velocity in (m/s)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.
\(\large  \rho_b\) gm/cm3 x  \( \large\frac{v^2}{2}\) m/s =  \( \rho_b\) gm/cm3 x  \( \large\frac{v^2}{2}\) m/s x (100 cm/m)2 =  \( \rho_{Hg}g h\)
The density of blood is close to the density of water, but I'll go along with tweaking the equation since blood is actually a little denser than water at about 1.09 gm/cm3 (RBCs sink, remember?).
\(\large  h=\large\frac{\rho_b}{2\rho_{Hg}g}v^2\) x 1002
We already figured out that there are 1333 mmHg in a dyne/cm2 so the constant comes out to 1.09 x 1002 / (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).
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