## The Gorlin Formula

**Straight to the Heart**

- For a given flow rate (\(q\)), the
velocity (\(\bar{u}\)) at a given location is**average axial**from the cross-sectional area there (\(A\)) as \(\bar{u} \equiv q/A\). This swings both ways, i.e. we can**defined**an area from a velocity and a flow rate, \(A \equiv q/u\). The clinical continuity equation makes use of this fact, \(q = \bar{u}_1 A_1 = \bar{u}_2 A_2\) where the subscripts refer to different locations in an outflow tract and we can compute an area by measuring 3 missing parts in the equation. NOTE: velocities computed from flow and area in this way do NOT include any**define**of the velocity that's perpendicular to the area. This continuity equation utilizes a long wavelength approximation which assumes that the flow rate doesn't vary with axial coordinate as if the conduit is incompliant.**component** - The so-called Gorlin formula (when correctly formulated) is the result of algebraic manipulation of appropriate terms of the Bernoulli equation and correctly represents the physical principles of stenosis flow. The next figure suggests the average velocity (vectors) varying with axial location; \(\bar{u}\) is inversely related to cross sectional area at each location and the average velocity at the physical stenosis is \(q/A_2\) where \(A_2\) is the cross-sectional area there.

I hasten to add that the physics implied by the Gorlin formula is commonly misrepresented in medical cardiology texts and papers.

- Now if we determine the
fluid acceleration, we'd find that the velocity at the vena contracta is**ACTUAL**(**greater**) than suggested by the area change alone. The fluid doesn't flow straight through the orifice but has a velocity component that is perpendicular to the flow orifice; the flow stream**ALWAYS**towards the centerline as it goes through the orifice. So the**contracts**velocity at the vena contracta is greater than suggested by the flow rate and \(A_2\). It's a velocity associated with a**ACTUAL**of \(A_2\).**fraction**

\(\Large A = C_C A_2\)

\(C_C\) is a thing called the * contraction coefficient*. It's a pure number less than \(1.0\) indicating that the fluid acceleration is more than suggested by the physical area change. Somewhat simplistically, \(\large A_{vc} \sim q/u_{vc}\) where \(\large u_{vc}\) is the spatial maximum at the vena contracta. This means that the contraction coefficient is approximated by the velocity ratio, \(\large C_C \sim \bar{u}_2/u_{vc}\) where \(\large \bar{u}_2 = q/A_2\)

- Now if we actually measure the pressure change across the orifice (the "gradient") as we do when employing the Gorlin formula, we'd find that
\(\large A_{vc}\). That's because not all the pressure went in to accelerating the fluid; some was**the gradient is even more than required to squirt the fluid through this last area,**to friction. To account for this loss we equate it to an additional change in kinetic energy, an amount required to force the fluid through an even smaller orifice.**lost**

\(\Large A_e = C_L C_C A_2 = C_D A_2\)

\(\large C_L\) is a thing called the loss coefficient, another pure number less than \(1.0\) so that the area \(\large A_e\) (effective area calculated by Gorlin) is smaller still. We can combine \(\large C_C\) and \(\large C_L\) (multiplying together as suggested by the equation) into a single number, the discharge coefficient, \(\large C_D = C_C C_L\). So when a pressure gradient is measured at a given flow rate across a stenosis, it's due NOT to the change in physical area alone but includes the additional pressure change resulting from the actual acceleration and from energy loss.

- The Gorlin formula uses the measured pressure gradient so it does NOT compute the physical stenosis area. It computes an effective stenosis area, \(\large A_e = C_D A_2\) which is a fraction (\(\large C_D\)) of the physical area. \(\large C_D\) depends on specific 3 dimensional details of the stenosis geometry as well as the Reynolds number. You don't know it when you perform a clinical cardiac cath and so you can't get the physical area using Gorlin. The implication that the Gorlin formula gives the physical stenosis area is simply incorrect. The Gorlins hoped that \(C_D\) could be approximated as a single number and suggested a value (but no such luck).
- The clinical Gorlin formula as it appears in the medical literature also contains some goofs. The \(44.3\) you see in your text book is incorrect for pressures measured in mmHg and the number cannot have physical units ascribed due to sloppy math. The pulsatile aspect of the flow throws an additional widget into the equation that is not accounted for clinically.

In this article we'll derive and explore the physical interpretation of the Gorlin formula. I think that this is a topic where misinterpretations have propagated through the clinical literature and I will freely disagree with other sources. I'll do my best not to contribute to the misunderstandings but implore you to read and understand what is represented by the equations and figures. This is also an opportunity to get a solid handle on stenosis flow physics; a qualitative description of stenosis flow was given in the PREVIOUS article.

The Gorlin formula is used to compute an area - ostensibly the physical cross sectional area of the stenosis. However that's NOT what it does. It computes an effective area that depends on the actual area and many other things. The area is computed from measurements of pressure and blood flow across the valve; given these inputs, it doesn't take much thought to recognize that the computed area will relate to the physics of how the flow occurs. This section is all about understanding exactly what is being computed by the Gorlin formula.

Before we even get started with the Gorlin equation, here's a quick recap of a concept we'll need. We can always associate an average (spatial) velocity with the flow rate through a conduit. For * steady* (non-pulsatile) flow through a

*(incompliant) conduit with no "tributaries", the average*

**rigid***velocity is:*

**axial**\(\Large \bar{u}_x(x) = \frac{q}{A(x)} \)

We're using \( u\) to represent velocity, as is commonly done in the engineering literature, to avoid confusing it with volume or voltage. \(q\) is the * flow rate* and this

*with axial location given the stated restrictions (rigid tube, no on- or off-ramp).*

**does not change**The above depicts a radial geometry with an axis of symmetry on the lower edge of the figure; the figure is ** revolved **around this lower edge to give the geometry suggested in the first figure. \(A(x)\) is the cross-sectional area of the conduit and the \((x)\) in parentheses on both \(A(x)\) and \(\bar{u}_x(x)\) is to indicate that the value can change with axial position. The \(x\) subscript on \(\bar{u}_x(x)\) refers to the fact that we're talking about the

*of velocity in the \(x\) direction only (axial direction in the figure above). The figure depicts a*

**component***, \(\bar{u}_x(x)\), at each axial location but varying with axial location and the cross-sectional area; no radial velocities appear.*

**uniform average velocity***Since the average axial velocity is defined in terms of the flow rate and cross-sectional area, an abrupt change in the cross-sectional area implies a reciprocal (abrupt) change in average axial velocity. \(\bar{u}_x(x)\) is*

**The figure suggests a discontinuity in the cross-sectional area and average velocity.***measured, it's*

**NOT***. This isn't what you measure with a Doppler gadget.*

**defined mathematically****In what follows, we'll also define certain cross-sectional areas*** mathematically*.

**These are not measured using a micrometer or ultrasound, they're***from the flow rate and a velocity.*

**computed**\(\Large A = \frac{q}{|\textbf{u}|} \)

Please don't let this bother you. Now we saw in the last article that the previous figure is a poor representation of the way the flow actually occurs. Here are some of the vectors computed from the computational fluid dynamics solution (CFD) for flow at a Reynolds number of 1000 in the parent tube:

The next figure is a contour plot showing the distribution of \(|\textbf{u}|\) within the region shown. Think of a topographic map that depicts where the mountains (high velocity,red) and valleys are (low velocity, blue).

The figure shows the location of the highest velocity at the vena contracta, a short distance downstream of the physical stenosis. That should orient you for the following; check out the previous article for a more detailed description.

### Gorlin I

For this first run through the Gorlin formula, let's assume that we have a steady flow situation (no pulsatility) so that there are no terms in the equation involving time derivatives. We are all familiar with some version of the Bernoulli equation for conservation of energy along a streamline in the flow. Here are some of the most important terms in the equation:

\(\Large p_1 + \frac{\rho}{2} |\textbf{u}_1|^2 = p_2 + \frac{\rho}{2} |\textbf{u}_2|^2 +L_{\mu}\)

The Bernoulli equation is about energy and \(\rho |\textbf{u}|^2 /2\) is kinetic energy per unit of volume. Technically we must use the absolute value of the velocity in this equation, i.e. \(|\textbf{u}|\); the vector quantity \(\textbf{u}\) can't be squared per se.

Here is a reasonable representation of streamlines for this particular model which has a parent tube diameter of \(1.0\) and a stenosis diameter of \(0.5\); the physical cross-sectional area of the stenosis is 1/4 the area of the parent tube. Fluid elements enter from the left and accelerate through the stenosis depicted about midway in the figure. For the above equation, the streamline of interest is at the lower edge of the figure (the centerline of the conduit) and locations corresponding to subscripts \(1\) and \(2\) of the above equation are labeled.

We're letting subscript 1 refer to the inlet and 2 for the location of the maximal velocity, i.e. at the vena contracta. **IF the upstream kinetic energy at the inlet (\(\rho\large u_1^2/2\)) and the viscous friction terms (\(L_{\mu}\)) are negligible with respect to the others, then the simplified Bernoulli equation (SBE) is obtained:**

\( \Large \Delta p = \frac{\rho \left|\textbf{u}_{2}\right|^2}{2} \)

where \(\Delta p = p_1-p_2\). The notation \( {\rho}/{2}\) is retained here because it helps us keep the physics in mind. The number "4" that everyone remembers for the clinical equation is a conversion factor (to get mmHg from squared m/sec); we'll figure out where the 4 comes from ELSEWHERE. ** **** For the rest of this section, we'll use the SBE approximation which allows us to proceed as if \(\Delta p\) and \(\rho \large |\textbf{u}_2|^2 /2\) are interchangable.** We explored some of the justification for the SBE in the last article also, including the negligible \(L_{\mu}\) approximation. To recap, here is another contour plot for this very same flow, this time depicting the highs and lows for the "total energy" (or "total pressure" or "stagnation pressure") defined as \(p+\rho |\textbf{u}|^2/2\).

This allows us to get a handle on whether the approximation of negligible \(L_{\mu}\) is justified. What this is showing is that the * total energy from the inlet of the tube to the vena contracta remains constant with negligible energy loss (more or less) along the center streamline*, at least for this particular flow example. This is a requirement for use of the SBE and you can also see that

*to any practical extent. The next figure shows the streamlines issuing from the inlet. You can see that*

**this requirement doesn't extend downstream of the vena contracta***the streamlines retain their energy to the location of the vena contracta (2).*

**not all***because energy losses aren't negligible. Remember however that the pressure is essentially constant at each axial location downstream of the vena contracta.*

**The Simplified Bernoulli approximation doesn't apply to those streamlines**

**With pressure and \(\Delta p\) essentially the same on each streamline, the difference between the streamlines that originate at the origin comes down to the fact that kinetic energy (and velocity) has been lost to friction (\(L_{\mu}\)).****We'll use this fact again shortly.**

To derive the Gorlin equation, all we have to do is replace the velocity with the flow rate, \( q\) , and an area. I've made an executive decision and decided to call this area \( A_u\), a * velocity area*, to imply that it's derived from a velocity.

\( \Large A_u \equiv \frac{q}{|\textbf{u}_2|} \)

As promised we're * defining* \(A_u\) in terms of other measured or known quantities, the velocity at the vena contracta and the flow rate. Simply stated, \(A_u\)

*Substituting this definition back into the SBE:*

**is the thing we divide the flow rate by to produce the actual velocity at the vena contracta.**\( \Large \Delta p = \frac{\rho |\textbf{u}_2|^2}{2} = \frac{\rho}{2} \left[\frac{q}{A_u}\right]^2\)

Rearranging algebraically:

\(\Large A_u = q \sqrt{\frac{\rho}{2\Delta p}} \)

This is a starting point for a steady flow version of the Gorlin formula with the above noted approximations in it and without the confusing stuff involving \(g\) and \(h\). Even in this first run through the problem, it's clear that this version of the Gorlin formula estimates an area (\(A_u\)) that relates to the vena contracta velocity and the flow physics, * not the physical area of the stenosis*.

In light of the given definitions (for \(A_u\)), the Gorlin formula results from an algebraic rearrangement of the Bernoulli equation(subject to certain physical approximations when appropriate). It cannot be used, by itself, to compute the physical area of a stenosis. While we're not measuring a velocity when employing the Gorlin formula,applied specifically to the streamline that traverses the vena contracta\(u = \sqrt{2\Delta p/\rho}\); this corresponds tothe formula implies that the measured pressure "gradient" results in a fluid acceleration from \(0.0\) to the the value\(A_e = q/u\). These last 2 statements are truethe Gorlin effective areaof the approximations made and apply to the rest of the article too. I mean it!regardless

That's as far as you get in a clinical situation, i.e. where you don't know what the physical stenosis area is! However we can gain further insights as to how the Gorlin area relates to the physical stenosis area by * defining* something called the

*which is a ratio; the average velocity at the physical stenosis divided by the vena contracta velocity:*

**contraction coefficient**\(\Large C_C = \frac{q}{A_p} \frac{1}{|\textbf{u}_2|} \)

Also, rearranging this fact for later use:

\(\Large \frac{q}{|\textbf{u}_2|} = C_C A_p \)

\(A_p\) is the physical stenosis area and \(q/A_p\) is the average axial velocity at the stenosis. The * contraction coefficient* implies that the flow "contracts" downstream of the physical area to achieve the higher velocity at the vena contracta. \(C_C\) is a number less than 1.0; the average axial velocity at the stenosis is the vena contracta velocity muliplied by \(C_C\). Starting with the above definition of the velocity area, we can see how the contraction coefficient and physical area are interrelated:

\(\Large A_u \equiv \frac{q}{|\textbf{u}_2|} = C_C A_p \)

\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{C_C^2 A_p^2} \)

\(A_u\) is the product of the contraction coefficient and the physical stenosis area. * When the losses and and upstream velocity are negligible*:

\(\Large C_C A_p = A_u = q \sqrt{\frac{\rho}{2\Delta p}} \)

and

\(\Large A_p = \frac{q}{C_C} \sqrt{\frac{\rho}{2\Delta p}} \)

Getting ahead of myself for a moment, we'll talk about something called a loss coefficient (\(C_L\)) and a discharge coefficient (\(C_D\)) in a subsequent section. These are related mathematically to the contraction coefficient as follows:

\(\Large C_D = C_L C_C\)

With the Simplified Bernoulli assumption, energy loss on the center streamline is negligible and this corresponds to a loss coefficient of \(1.0\); it turns out that this corresponds to NO energy loss as a result of the way in which \(C_L\) is defined. Consequently \(C_D\) and \(C_C\) are numerically equal under Simplified Bernoulli conditions and \(C_D\) is interchangeable with \(C_C\) in the Gorlin formula and Simplified Bernoulli equation.

\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{C_D^2 A_p^2} \)

\(\Large A_p = \frac{q}{C_D} \sqrt{\frac{\rho}{2 \Delta p}} \)

To summarize, the maximal velocity at the vena contracta (\(|\textbf{u}_2|\)) is * ALWAYS* greater than the average velocity (\(\bar{u}_x\)) at the physical stenosis. The contraction coefficient \(C_C\) is a number less than 1.0 (

*); so \(C_C A_p\) is an area smaller than the physical area and \(C_C\) is the*

**ALWAYS***of the physical area that results in the actual velocity at the vena contracta. This particular example has an input velocity of 1.0 and the velocity color scale on the right side of the plot shows the full velocity range so that vena contracta velocity can be read right off the color bar (5.865). The average velocity at the physical stenosis is 4.0, because of the 4:1 physical contraction, so the velocity at vena contracta is greater by a factor of 1.466; \(C_C\) is the reciprocal of that, i.e. 0.682.*

**fraction**If we've done everything correctly, the physical units have to be correct also. For example: \( \rho\) is in *gm-cm ^{-3}*, \( q\) in

*cm*, and \( p\) in

^{3}-sec^{-1}*dynes-cm*where a

^{-2}*dyne*is a

*gm-cm-sec*so pressure works out to units of

^{-2}*gm-sec*. Consequently the stuff under the square root sign has units of

^{-2}-cm^{-1}*sec*and we have no problem taking the square root of it (

^{2}-cm^{-2}*sec-cm*) because the physical units are a perfect square. Multiplying this quantity by the flow rate gives us an area, (

^{-1}*sec-cm*)(

^{-1}*cm*) =

^{3}-sec^{-1}*cm*; yup, that's an area.

^{2}### Misconceptions

\(\Large A_e = q \sqrt{\frac{\rho}{2\Delta p}} \)

I know, this doesn't really look like the Gorlin formula to you but we're going to fill in the missing pieces a little at a time. The clinical formula as it appears in cardiology texts is derived next as applied to data obtained at cardiac catheterization. We measure the pressure gradient and flow, specifically over the cardiac ejection, and perform some kind of averaging process. Exactly what that averaging process is turns out to be a matter of concern that we'll address in Gorlin III. Since we're clinicians, we are going to express the pressure in units of mmHg and this is one important source of misunderstanding. The Bernoulli equation tells us that pressure can be expressed as the height of a column of static liquid:

\( \Large p=\rho gh\)

In this equation, \(\rho\) * refers to the density of the fluid in a manometer that you're using for a pressure measurements, not the density of the blood*. For water, density is 1.0 gm/cm

^{3}(g = 980 cm/sec

^{2}). In the following, density is subscripted with

**or**

*b***to indicate density of the blood (\(\rho_b\)) or manometer fluid (\(\rho_m\)) respectively.**

*m*\( \Large A_e = q \sqrt{\frac{\rho_b }{2\rho_m g \Delta h}}\)

* The Gorlin paper is explicit in that pressure was determined in centimeters of a water column, NOT millimeters of mercury*. Substituting those quantities into the equation (assuming a blood density of 1.0 gm/cm

^{3}also which is not perfectly accurate) and proceeding with an illegal mathematical operation (to be discussed):

\( \Large A_e = q \sqrt{\frac{1.0}{2 \times 1.0 \times 980 \Delta h}} = \frac{q}{44.3} \sqrt{\frac{1}{\Delta h}}\)

You'll now recognize the magic number in the denominator as the so-called Gorlin constant. I've intentionally left \(h\) in the equation to remind you that this is a ** height **that's been measured and entered into the formula. \( p=\rho g h\) was the pressure in centimeters of water and 2 of the terms involved in the pressure calculation have been pulled outside of the square root. What's illegal about this operation is that we have taken the square root of a physical quantity that wasn't a perfect square -- you can't take the square root of something that has units of 1/cm! You may think I'm being nitpicky here, but this little flaw is at the root of some common misconceptions! We're trying to calculate an area in physical units e.g. cm

^{2}. That's exactly what occurs if we use the correct physical units inside the square root sign before performing the operation. Without doing it that way, the physical interpretation has been lost and an incorrect value for the Gorlin constant has been propagated through the literature. Mercury has a density of ~13.6 gm/cm

^{3}or 1.36 gm/cm

^{2}-mm. The correct value of the Gorlin constant, for pressure expressed in mm of a mercury column is \( \small{\sqrt{{2}\times{1.36}\times{980}}}\) = 51.6 (and you will see this value used in physical hemodynamic texts and engineering publications).

The so-called Gorlin constant has nothing to do with any experiment or measurement performed by those authors. The number 44.3 is strictly a units conversion factor that stems from measuring the stenosis pressure gradient in terms of a column of water measured in centimeters.

The number is not "flow dependent" in any way but depends on the density of the blood, the density of the fluid in your imaginary/virtual manometer (e.g. 44.3 for pressure in cm HPhysical units cannot be ascribed to these constants because mathematical protocol has not been adhered to._{2}0, 51.6 for pressure in mmHg), and what planet you're from (forg).

### Gorlin II

For a second look at the formula, we'll take a step back and pick up some of the stuff that was ignored the first time around. To proceed, we need to be very comfortable with the notion that a velocity area can be defined in terms of the flow rate and a velocity, \(A = q/|\textbf{u}|\), and that an average velocity is defined by \(q/A\). Gorlin I made the Simplified (clinical) Bernoulli approximations. Consequently this section is also a discussion of whether / when those approximations are justified. For reference, here is the Bernoulli equation again but including the terms omitted in Gorlin I:

\(\Large p_1 + \frac{\rho}{2} |\textbf{u}_1|^2 = p_2 + \frac{\rho}{2} |\textbf{u}_2|^2 +L_{\mu}\)

What if the velocity \(|\textbf{u}_1|\) is NOT negligible with respect to \(u_2\) or other terms? To analyze this situation we define "velocity areas" as before that encapsulate the actual upstream and downstream velocities in terms of the flow rate:

\(\Large A_{u1} \equiv \frac{q}{|\textbf{u}_1|} \)

\(\Large A_{u2} \equiv \frac{q}{|\textbf{u}_2|} \)

With these definitions, the following is an * exact* rendition of the above version of the Bernoulli equation:

\(\Large \Delta p = \frac{\rho}{2} \left[ \frac{q^2}{A^2_{u2}} - \frac{q^2}{A^2_{u1}}\right] + L_{\mu} = \frac{\rho}{2} q^2 \left[ \frac{1}{A^2_{u2}} - \frac{1}{A^2_{u1}}\right] + L_{\mu} = \frac{\rho}{2} q^2 \left[ \frac{A_{u1}^2-A_{u2}^2}{A_{u1}^2 A_{u2}^2} \right] + L_{\mu} \)

Now ALL the stuff inside the square brackets has the same physical units, reciprocal squared area. We'll define a new area, \(A_{uE}\) (velocity equivalent area), based on the math inside the brackets:

\(\Large A_{uE} \equiv \left[ \frac{A^2_{u1}A^2_{u2}}{A^2_{u1}-A^2_{u2}} \right]^\frac{1}{2}\)

\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{A_{uE}^2} + L_{\mu}\)

This last shows us something worth taking a moment to internalize. The Simplified Bernoulli approximation involves accelerating a fluid element from \(0\) to the vena contracta velocity, the latter being equivalent to a flow rate divided by a velocity area. If the initial velocity weren't \(0\), the pressure change due to the acceleration can still be related to a flow rate and a velocity area that was just defined from the math: \(A_{uE}\). Consequently the following is * exactly* true for the center streamline because of the way \(A_{uE}\) is defined:

\(\Large \frac{q^2}{A_{uE}^2} = |\textbf{u}_2|^2 - |\textbf{u}_1|^2 \)

On the off chance it's easier to follow, \(\large \rho/2 \left[|\textbf{u}_2|^2 - |\textbf{u}_1|^2\right]\) corresponds to the pressure required to cause the acceleration from \(\large |\textbf{u}_1|\) to \(\large |\textbf{u}_2|\). This is the same pressure that's required to accelerate the fluid from \(0\) to \(\large u_E\) (velocity equivalent) where:

\(\Large u_E \equiv \sqrt{|\textbf{u}_2|^2-|\textbf{u}_1|^2} \)

The area associated with this velocity is \(A_{uE}\):

\(\Large A_{uE} = \frac{q}{u_E} = \frac{q}{\sqrt{|\textbf{u}_2|^2-|\textbf{u}_1|^2} }\)

The 2 expressions for \(A_{uE}\) are identical mathematically. With that in mind, perhaps it will be easier to appreciate that a mild stenosis with non-negligible \(|\textbf{u}_1|\) can contribute to overestimation of the stenosis area. The fluid acceleration is less and pressure gradient is less than would be required to get from \(0\) to \(|\textbf{u}_2|\) so the computed area will be greater.

Now for our particular example of a rather severe stenosis, there's not much point in going the extra mile. With the Simplified Bernoulli equation we have \(A_u = q/|\textbf{u}_2| = \pi [\frac{1}{2}]^2 / 5.865 = 0.134 \); the area of the physical stenosis is \(\pi [\frac{1}{2}]^2 = 0.196\) so the contraction coefficient is \(0.134/0.196 = 0.682\). With the correction for the upstream velocity we have \(A_{uE} = q/\sqrt{|\textbf{u}_2|^2-|\textbf{u}_1|^2} = \pi [\frac{1}{2}]^2 / \sqrt{5.865^2 - 1.0^2} = 0.136\). This difference is clinically meaningless for this situation of course. However a mild stenosis where the upstream and vena contracta velocities are more similar presents more of an issue. We can investigate further the effect of the upstream area by defining a parameter:

\(\Large \beta \equiv \sqrt{\frac{A_{u2}}{A_{u1}} } \)

\(\beta\) then has the interpretation of a diameter (or radius) ratio of the 2 velocity areas. A small value for \(\beta\) (approaching 0.0) means it's a severe stenosis; a value of 1.0 is no stenosis at all (no area change). A little bit of algebra yields:

\(\Large A_{uE} = A_{u2} \sqrt{\frac{1}{1-\beta^4}} \) or

\(\Large \frac{A_{uE}}{A_{u2} } = \sqrt{\frac{1}{1-\beta^4}} \)

A plot of the area ratio against \(\beta\) shows that we're in pretty good shape ignoring the upstream velocity until \(\beta\) gets to about 0.65 (meaning that the stenosis diameter is 65% of the parent diameter or more). For mild stenoses (e.g. \(\beta > 0.65\)), the error is greater than 10%; this aspect of the simplified Bernoulli equation starts to fail and the approximation is relatively inadequate. You can actually think of this as a plot showing how the stenosis area is overestimated for mild stenoses (\(\beta \rightarrow 1.0\)).

Now we go after the ** energy loss term** and start by taking a largish leap of faith; we assume that the losses are related to the squared velocity:

\(\Large L_{\mu} \sim \rho q^2\)

and specifically:

\(\Large L_{\mu} \sim \frac{\rho}{2} \left[\frac{q}{A_{uE}}\right] ^2\)

I'm not sure how justifiable this is as a function; i.e. I would not expect the losses due to viscosity to be proportional to \(q^2\) except as a result of turbulence which is minimal within the region we're considering. What I expect we * can* assume is that the ratio of \(\rho q^2/L_{\mu}\) is

*.*

**fixed at a specific Reynolds number***-- a ratio of inertial to viscous forces. Since the Reynolds number changes with flow rate (other components remaining unchanged),*

**That's what the Reynolds number is***. Changing the flow rate will affect the relationship between the losses and the acceleration so that the area computed by the formula will change. We'll explore flow dependence in greater detail ELSEWHERE.*

**this is also a likely source of flow dependence of the Gorlin formula**Proceeding with this assertion, \(\Delta p\) will be * greater* in the presence of energy loss than what would occur for the acceleration only. This result can be incorporated into the formula by

*a loss coefficient, \(C_L\), as follows:*

**defining**\(\Large L_{\mu} = \frac{\rho}{2} \left[\frac{q}{A_{uE}}\right] ^2 \left[ \frac{1 - C_L^2}{C_L^2} \right] = \frac{\rho}{2} \left[|\textbf{u}_2|^2-|\textbf{u}_1|^2 \right] \left[ \frac{1 - C_L^2}{C_L^2} \right]\)

\(\Large C_L = \sqrt{ \frac{\rho q^2}{\rho q^2+2 A_{uE}^2 L_{\mu}}} = \sqrt{\frac{ \rho \left[|\textbf{u}_2|^2-|\textbf{u}_1|^2\right] } {2L_{\mu} + \rho \left[|\textbf{u}_2|^2-|\textbf{u}_1|^2\right]} } \)

\(C_L\) is a pure number between \(0.0\) and \(1.0\) (UNLESS nuances of the flow result in positive shear work on the streamline). With this definition of the loss coefficient, the equation now appears as follows:

\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{C_L^2 A_{uE}^2}\)

\(C_L\) implies ** no losses** when the value is \(1.0\) and

*when the value is \(0.0\). I don't expect this aspect of the formula is very intuitive, but consider the result. When \(C_L = 1.0\) the area is the same as \(A_{uE}\), the equivalent area associated with the pressure gradient due to the velocity change alone.*

**total loss***. \(C_L\) has the effect in the formula of augmenting the pressure gradient*

**If there are energy losses, the pressure gradient is greater***the flow were forced through a smaller orifice, \(C_L A_{uE}\). \(C_L\) is a*

**as if***of the equivalent velocity area.*

**fraction**\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{A_e^2} \)

\(\Large A_e = q \sqrt{\frac{\rho}{2 \Delta p}} \)

\(C_L A_{uE}\) is an **area**** **called the * effective area of the stenosis*, \(A_e\).

*is the area determined from the steady flow Gorlin formula so we'd better understand what it is.*

**THIS**\(\Large A_e = C_L A_{uE} = \sqrt{\frac{\rho q^2}{2 L_{\mu} + \rho q^2}} \frac{q}{\sqrt{|\textbf{u}_2|^2-|\textbf{u}_1|^2}} \)

The area computed by the Gorlin formula (from measured pressure gradient and flow rate) is anthat encompasses the physics of flow. The calculated area includes a part that relates to the flow acceleration since the measured pressure corresponds to the increase velocity from the upstream value (wherever you measured the pressure) to the vena contracta velocity. There is (potentially) an additional part of the pressure gradient due to energy loss that decreases the calculated area still further. (A better explanation resides in the form of the equations shown.) The physical areas are involved in producing the flow acceleration and energy loss, but the actual flow and pressure gradient engender complexities not apparent from simplistic parameters. However these complexities can be encapsulated into a discharge coefficient.effective area

To put the finishing touches on this section, we want to go back and express \(A_{uE}\) in * terms of physical areas*. Here is the figure used above to illustrate

*associated with those areas.*

**average velocities**

The * average axial velocity *at the inlet is \( q/A_{p1}\) and at the stenosis it's \(q/A_{p2}\); the subscript \(p\) in \(A_p\) is to imply

*with numeral subscripts to indicate the 2 sites. According to the Bernoulli equation, the pressure change implied by this change in velocity (alone) is:*

**physical areas**\( \Large \Delta p = \frac{\rho}{2} \left[ \left(\frac{q}{A_{p2}}\right)^2 - \left(\frac{q}{A_{p1}} \right)^2 \right] = \frac{\rho}{2} q^2 \left[ \frac{1}{A_{p2}^2}-\frac{1}{A_{p1}^2}\right] \)

While * this is not the actual acceleration or resulting pressure*, it will behoove us to understand exactly how the physical areas are involved. As before, we see that the quantities inside the square brackets have the same physical units, reciprocal squared area, and the difference can be performed algebraically:

\(\Large \left[ \frac{1}{A_{p2}^2}-\frac{1}{A_{p1}^2}\right] = \frac{A_{p1}^2-A_{p2}^2}{A_{p1}^2A_{p2}^2} \)

As before we can define a * single* area mathematically to describe this acceleration; I'm calling it \(A_{puE}\) - an

*:*

**equivalent velocity area due to the physical area constriction**\(\Large A_{puE} \equiv \left[\frac{A_{p1}^2A_{p2}^2}{A_{p1}^2-A_{p2}^2} \right]^{\frac{1}{2}} \)

**In other words**, we're talking about accelerating the fluid from the ** average axial velocity** at the inlet (\( \bar{u}_{x1} = q/A_{p1}\)) to the

*at the physical stenosis (\(\bar{u}_{x2}= q/A_{p2} \)). This takes the same \(\Delta p\) as accelerating from \(0\) to an equivalent velocity, \(\bar{u}_{xE}\):*

**average axial velocity**\(\Large \bar{u}_{xE} \equiv \sqrt{\bar{u}_{x2}^2-\bar{u}_{x1}^2} \)

and

\(\Large A_{puE} \equiv \frac{q}{\bar{u}_{xE}} = \frac{q}{\sqrt{\bar{u}_{x2}^2-\bar{u}_{x1}^2}} \)

This expression for \(A_{puE}\) is * mathematically identical* to the one above. Consequently the pressure required for

*acceleration is:*

**this**\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{A_{puE}^2} \)

This time we'll define a * ratio of physical areas*, \(\beta\), as follows:

\(\Large \beta \equiv \sqrt{\frac{A_{p2}}{A_{p1}}}\)

Again, \(\beta\) is the ratio of the diameter of physical stenosis to that of the parent conduit. The physical area equivalent now appears as:

\( \Large A_{puE} = A_{p2} \sqrt{\frac{1}{1-\beta^4}} \)

\(A_{puE}\) is a "physical" velocity equivalent area that encompasses the fluid acceleration when the upstream velocity isn't negligible. So here is the pressure change due fluid acceleration (only) as if the fluid velocity increases from the average upstream velocity to the average at the physical stenosis:

\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{A_{p2}^2} (1-\beta^4) \)

Now we're going to * update* the contraction coefficient that was defined previously for the Simplified Bernoulli approximation to include the mild stenosis possibility. As before, \(C_C\) is the ratio of the actual velocity contraction to the contraction due only to the physical area change; it's a ratio of 2 areas:

\(\Large C_C \equiv \frac{A_{uE}}{A_{puE}} = \frac{\left[ \frac{A^2_{u1}A^2_{u2}}{A^2_{u1}-A^2_{u2}} \right]^\frac{1}{2}} {\left[\frac{A_{p1}^2A_{p2}^2}{A_{p1}^2-A_{p2}^2} \right]^{\frac{1}{2}}} = \frac{\left[ \frac{A^2_{u1}A^2_{u2}}{A^2_{u1}-A^2_{u2}} \right]^\frac{1}{2}} {A_{p2} \sqrt{\frac{1}{1-\beta^4}}} = \frac{\frac{q}{\sqrt{|\textbf{u}_2|^2-|\textbf{u}_1|^2}}}{A_{p2}\sqrt{\frac{1}{1-\beta^4}}} = \frac{\sqrt{1-\beta^4}}{A_{p2}}\frac{q}{\sqrt{|\textbf{u}_2|^2-|\textbf{u}_1|^2}}\)

(Yeah that's a mess, but we're in the home stretch now.) We can now summarize a previous rendition of the Bernoulli equation as:

\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{C_L^2 A_{uE}^2} = \frac{\rho}{2} \frac{q^2}{C_L^2 C_C^2 A_{p2}^2} (1-\beta^4) = \frac{\rho}{2} \frac{q^2}{C_D^2 A_{p2}^2} (1-\beta^4) \)

And finally we combine the loss coefficient and contraction coefficient into a single value, the discharge coefficient, \(C_D\):

\(\Large C_D = C_L C_C = \sqrt{ \frac{\rho q^2}{\rho q^2+2 A_{uE}^2 L_{\mu}}} \frac{A_{uE}}{A_{p2} \sqrt{\frac{1}{1-\beta^4}}} = \left[ \sqrt{\frac{ \rho \left[|\textbf{u}_2|^2-|\textbf{u}_1|^2\right] } {2L_{\mu} + \rho \left[|\textbf{u}_2|^2-|\textbf{u}_1|^2\right]}}\right] \left[ \frac{\sqrt{1-\beta^4}}{A_{p2}} \frac{q}{\sqrt{|\textbf{u}_2|^2-|\textbf{u}_1|^2}}\right] \)

The square brackets have been left surrounding each of the coefficients, \([C_L][C_C]\) so the formula for each can be identified. It's tempting to simplify these expressions further, but you can end up trying to take an impossible square root due to the physical units. (Be careful!) And now here is the Bernoulli equation, rewritten in a form that is less familiar where the meaning of each symbol has been defined above:

\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{C_D^2 A_{p2}^2} (1-\beta^4) \)

If you were to substitute the above expression for \(C_D\) into this equation, you would find that * this expression simplifies exactly to steady flow Bernoulli equation* given at the top of the section.

*by which I mean there has been no distortion of the physical meaning of the equation. What has been accomplished is a direct representation of \(C_L\), \(C_C\), and \(C_D\) in terms of the flow physics.*

**We've accomplished nothing**The last equation is rearranged algebraically into a final steady flow * engineering* version of the Gorlin formula:

\(\Large A_e = C_D A_{p2}= q \sqrt{\frac{\rho}{2 \Delta p}} \sqrt{1-\beta^4}\)

I've presented the formula in this form to show what is accomplished by performing the calculation using the measured quantities on the right hand side (the Gorlin formula). As before the Gorlin formula does * NOT* allow us to calculate the physical stenosis area per se, \(A_{p2}\), but an

*(\(A_e\)) that includes some rather complicated aspects of the flow and geometry. If you've never thought about it before, the clinical Gorlin formula loses its credibility for mild stenoses (\(\beta \rightarrow 1.0\)). As \(\Delta p\) gets small, the computed area starts to hare off for infinity. The formula isn't applicable at all for very mild stenoses without including \(\beta\) in the calculation as shown above (see plot of \(\beta\) above).*

**effective area**Now if you're able to swallow the math, the above demonstrates the * exact* physical interpretation of the discharge, contraction, and loss coefficients. But these aren't computed in a practical setting; they're determined from a bench top experiment (or a CFD solution) where the physical areas are known. In the cath lab we're stuck with trying to estimate an area from measured pressure and flow (and known density of blood).

*\(C_D\)*-- how much flow contraction occurs and how much energy is lost. \(C_D\)

**embodies everything specific about the way the flow occurs***Notice that you would need to know upstream and downstream velocities (\(|\textbf{u}_1|\) and \(|\textbf{u}_2|\)) on the center streamline as well as \(\Delta p\) to be able to compute \(C_L\) and \(C_C\), the loss and contraction coefficients. That would NOT be the usual situation in the cath lab.*

**is the thing that can be "flow dependent", NOT the "Gorlin constant".**Some final comments before Gorlin III. I suspect that no one reading this (neither of you) has ever seen the equation with the \(\beta\) in it, what I've called the "engineering Gorlin formula"; \(\beta\) never shows up in clinical use. * That's because we don't know what its value is!* Engineers have other purposes for the application of these concepts. When evaluating what happens to flow through a nozzle, they already know all about the geometry (they designed the thing!). We're using the Gorlin formula to try to figure out an area; if we knew what \(\beta\) was, we'd measure the upstream diameter and calculate the physical stenosis area using \(\beta\)! So like the Simplified Bernoulli equation, the clinical Gorlin formula is just plain inappropriate when the upstream velocity is not negligible. In a clinical setting we can

*use it when \(\beta\) is close to \(0\) (meaning the area computed by the clinical Gorlin formula gets wildly erroneous for very mild stenoses). Without simply throwing all the above work away, here's what happens to it when \(\beta \sim 0\) and \(|\textbf{u}_1| \sim 0\):*

**only**\(\Large C_D = C_L C_C = \left[ \sqrt{\frac{\rho |\textbf{u}_2|^2}{2 L_{\mu}+\rho |\textbf{u}_2|^2}} \right] \left[ \frac{1}{A_{p2}} \frac{q}{|\textbf{u}_2|} \right] \)

and

\(\Large \Delta p = \frac{\rho}{2} \frac{q^2}{C_D^2 A_{p2}^2} \)

\(\Large A_e = C_D A_{p2} = q \sqrt{\frac{\rho}{2 \Delta p}} \)

It's the same as Gorlin I but with \(C_D\) enhanced to include the possibility of losses (\(C_L \lt 1.0\)).

We've already seen that while each of the streamlines issuing through the stenosis orifice can have a different velocity, they all have essentially the same \(\Delta p\) once the fluid has reached the vena contracta. Consequently we would get (potentially) a different value for losses (\(C_L\)) and for contraction (\(C_C\)) for each streamline. However.we would obtain the essentially the same discharge coefficient (\(C_D\)) no matter which streamline we pickIn a subsequent article, we'll see how to define \(p\) and \(|\textbf{u}|\) from the entire flow so that \(C_C\) and \(C_L\) are not streamline dependent.The Gorlin formula corresponds to the fluid acceleration from \(0\) to \(|\textbf{u}|\) where\(|\textbf{u}| = \sqrt{2\Delta p /\rho}\) and \(\Delta p\) is the measured pressure gradient. The effective area, \(A_e = C_D A_{p2} = q/|\textbf{u}|\), doesn't depend on differences between the streamlines as long as \(\Delta p\) is the same between them (we've seen that \(\Delta p\) is the same).

### Gorlin III

For a final go, we'll try to explore how the pulsatile nature of cardiac flow affects the interpretation of the Gorlin formula. There's no problem incorporating time-varying flows into CFD solutions -- ALL of the solutions shown have to "march through time" until reaching the steady state that's presented. However there's an immense problem presenting the information and I don't have any decent way of displaying a time-varying solution for you. Furthermore you can conceive that each pulsatile flow solution would be a little bit different due to the particular time course of the flow. That adds an infinite dimension to the parameter space that we just can't afford to explore. However we can examine some very basic aspects of the clinical application that often escape attention. We'll start with the full time-varying Bernoulli equation:

\( \Large p_1(t)+\frac{\rho}{2} |\textbf{u}_1(t)|^2+\rho gh_1(t) = p_2(t)+\frac{\rho}{2} |\textbf{u}_2(t)|^2+\rho gh_2(t)+\int_1^2 \rho\frac{\partial u}{\partial t} ds + L_{\mu}(t) \)

All of the terms above now include \((t)\) to indicate explicitly that they can change with time (due to pusatility); however I'll immediately discontinue the notation in favor of brevity. \( h\) is the height above some arbitrary level of elevation, \( g\) is the acceleration of gravity (980 cm/sec^{2 }on this planet). The \( \partial u/\partial t\) term has to do with time varying velocity, i.e. pulsatility. Subscripts 1 and 2 refer to the upstream and vena contracta locations on the center streamline as before. The integration with respect to \(s\), ( \(\int ds\) ) refers to integration along a streamline. In the figure, streamlines are represented accurately for the simplified geometry and velocity magnitude is also represented by color with warmer colors suggesting higher velocity. I'll remind you again however that the streamlines are from * average velocities*; turbulence adds a seemingly random component to the trajectories (not depicted).

It turns out we can do away immediately with the gravitational terms. A fluid-filled catheter actually does this for you automatically if the fluid has the same density as blood (close enough). Explore further HERE.

\( \Large p_1+\frac{\rho}{2} |\textbf{u}_1|^2 = p_2+\frac{\rho}{2} |\textbf{u}_2|^2+\int_1^2 \rho\frac{\partial u}{\partial t}ds+L_{\mu}\)

The next step is to average the entire equation over the ejection cycle. The time average of a time varying function, \(f(t)\), is defined as follows:

\( \Large \frac{\int_{t_1}^{t_2}f(t) dt}{t_2-t_1} = \frac{\int_{\Delta t}f(t) dt}{\Delta t}\)

\(\int_{\Delta t}\) is to indicate integration over the ejection cycle.

\( \Large \int_{\Delta t} \Delta p \; dt= \frac{\rho}{2} \int_{\Delta t} \left[|\textbf{u}_2|^2 - |\textbf{u}_1|^2\right] \; dt +\int_{\Delta t} \left[\int_1^2 \rho\frac{\partial u}{\partial t}ds\right] \; dt+ \int_{\Delta t} L_{\mu} \; dt\)

We can exchange the order of integration of the \( \partial u/\partial t\) term:

\( \Large \int_{\Delta t} \Delta p \; dt= \frac{\rho}{2} \int_{\Delta t} \left[|\textbf{u}_2|^2-|\textbf{u}_1|^2\right] \; dt +\int_1^2 \left[ \int_{\Delta t} \rho\frac{\partial u}{\partial t}dt\right] \; ds+ \int_{\Delta t} L_{\mu} \; dt\)

and be rid of the thing by integrating over the ejection cycle:

\( \Large \int_1^2 \left[ \int_{\Delta t} \rho\frac{\partial u}{\partial t}dt\right] \; ds = 0 \)

In what follows, I've divided through by \(\Delta t\) and overbars are used to indicate the time-averaged values.

\( \Large \overline{\Delta p} = \frac{\rho}{2} \left[\overline{|\textbf{u}_2|^2}-\overline{|\textbf{u}_1|^2}\right] + \overline{L_{\mu}}\)

Considering that we learned in the last section -- that we NEED to confine the clinical Gorlin formula to situations where the upstream velocity is negligible -- here's a further simplification:

\( \Large \overline{\Delta p} = \frac{\rho}{2} \overline{|\textbf{u}_2|^2}+ \overline{L_{\mu}}\)

Now we're going to rewrite the Bernoulli equation in a Gorlin-like format and see how the pieces might fit together:

\(\Large \frac{\rho}{2} \frac{\overline{q^2}}{\overline{A_e}^2} = \frac{\rho}{2} \overline{|\textbf{u}_2|^2}+ \overline{L_{\mu}}\)

We could readily solve this for the effective area, \(A_e\) with a little algebra. However the average of the squared flow rate, \(\overline{q^2}\), is not available to us from a standard cardiac catheterization procedures. \(\bar{q}\) is determined in the cath lab from the cardiac output and the fraction of time spent in systole (ejection). \((\overline{q})^2\) is NOT the same as \(\overline{q^2}\). This can be illustrated with a stylized approximation of the aortic flow; suppose \(q(t)\) can be represented as a triangular function with a peak flow rate of \(q_p\) as suggested by the normal Doppler flow signal:

In this case the area under the signal is \(q_p \Delta t/2\) (1/2 the base multiplied by the height) so the average flow rate is \(q_p/2\) and the squared average is \(q_p^2/4\). * However the average of the squared flow rate* is \(q_p^2/3\) for this example. So it isn't the average of the flow (\(\bar{q}\)) that we want for the Gorlin formula, it's actually the root-mean-square value, \(q_{rms}\).

\(\Large q_{rms} = \sqrt{ \int_{\Delta t} q(t)^2 \; dt } = \sqrt{\overline{q^2}} \)

We can encapsulate this aspect of the discussion with the ratio \(q_{rms}/\bar{q}\) which is a consequence of the * shape of the flow signal* (think shape of the Doppler velocity envelope, \(\sqrt{4/3} \approx 1.15\) for the example).

\(\Large S_q \equiv \frac{q_{rms}}{\bar{q}}\)

This value of \(S_q\) must be \(\ge 1.0\) and will approach \(1.0\) for a constant flow rate throughout ejection. Now we rewrite the equation in terms of \(\overline{q}\):

\( \Large \overline{\Delta p} = \frac{\rho}{2} S_q^2 \frac{\overline{q}^2}{\overline{A_{e}}^2} = \frac{\rho}{2} \overline{|\textbf{u}_2|^2}+ \overline{L_{\mu}}\)

The time-averaged Gorlin formula then appears as follows:

\( \Large \overline{A_e} = \overline{q} S_q \sqrt{\frac{\rho}{2 \overline{\Delta p}}} \)

**As always,****the Gorlin formula computes an effective area due to the flow physics.**** **The clinical Gorlin formula never includes the correction term for the shape of the flow (\(S_q\)) and so this aspect of the formula actually underestimates the effective area somewhat (0-20%??).

The problem with the next step is that the discharge coefficient depends on the physical area; **what if the physical area changes during ejection??**** **Well that of course is one of the issues with formula. If we could determine how to define the average physical area, then we could proceed with the math. However I'm going to take the usual approach and assume that the physical stenosis area is constant. Proceeding with that assumption, and going through similar gyrations as shown above in Gorlin II, here are the the time-averaged discharge, loss, and contraction coefficients :

\(\Large \overline{C_D} = \overline{C_L} \; \overline{C_C} = \left[ \sqrt{\frac{\rho \overline{|\textbf{u}_2|^2}}{2 \overline{L_{\mu}}+\rho \overline{|\textbf{u}_2|^2}}} \right] \left[ \frac{1}{A_{p2}} \frac{S_q \overline{q}}{\sqrt{\overline{|\textbf{u}_2|^2}}} \right] \)

(Notice that the * rms* value for \(\large|\textbf{u}_2|\) shows up in the contraction coefficient.) So the effective area is again some complicated function that multiplies the physical stenosis area (\(A_2\)).

\(\Large \overline{A_e} = \overline{C_D} A_2\)

And here's the time-averaged Bernoulli equation in terms of the stenosis parameters:

\(\Large \overline{\Delta p} = \frac{\rho}{2} \frac{S_q^2 \overline{q}^2}{\overline{C_D}^2 A_2^2} \)

You would find that substitution of the mess for \(C_D\) into this equation regenerates the time-averaged Bernoulli equation (with the stated approximations). Note again that this is all by way of informed consent. When you apply the Gorlin formula, you calculate an area. I'm just giving you (myself) a rigorous, mathematical description of what that calculated area actually is. You don't have access to \(\large \overline{|\textbf{u}_2|^2} \), \(A_2\), \(S_q\), etc when you use the formula, but the calculated Gorlin area still depends on those things, like it or not.

It looks a lot worse if we let \(A_2\) vary in size over the ejection. I'll shut up now.