## Resistance

The concept of vascular resistance is central in cardiovascular physiology and the formula for computing the total peripheral resistance is:

$$\Large TPR = \frac{MAP-RAP}{CO}$$

Where TPR is the total peripheral resistance, MAP is mean aortic pressure (average over time), RAP is mean right atrial pressure, and CO is the cardiac output.  The acronyms shown are common to medical texts.  For the discussions here, however, I'll use nomenclature for variable names and derived indices more typical of the bioengineering literature.

$$\Large R_{S}=\frac{\bar{p}_{Ao}-\bar{p}_{RA}}{\bar{q}_{Ao}}$$

Here we use $$R$$ to represent vascular resistance, $$p$$ for pressure, and $$q$$ for flow rate with overbars to indicate timeaveraged quantities.  Subscripts are employed to indicate the particular resistance, pressure, or flow. E.g. $$S$$ for systemic, $$Ao$$ for aortic, etc.

Now resistance is a concept borrowed from the study of electrical circuits where it has a very specific meaning,

$$\Large \Delta v = i R$$

$$\Large R = \frac{\Delta v}{i}$$

and the schematic symbol for a resistor is shown. Shown also is the characteristic equation for a resistor – one that characterizes its behavior.  In this setting, $$i$$ refers to electrical current ( Amperes ), $$v$$ is voltage, and $$R$$ is the resistance in Ohms.  The voltage is the difference in voltage, i.e. $$v_1-v_2$$ in the schematic.   Clearly there is a strong analogy between the electrical system and the fluid flow in a conduit.  We need only exchange symbols, $$q$$ (flow) for $$i$$ ( current ), and $$p$$ ( pressure ) for $$v$$ ( voltage ).

$$\Large \Delta p = q R$$

$$\Large R = \frac{\Delta p}{q}$$

In electrical circuits, resistance has to do specifically ( SPECIFICALLY ) with the conversion of energy into heat, i.e. a loss of energy manifest as a loss of voltage..  It is also applied this way in the study of physical hemodynamics and I will adhere to this representation.  However the study of flowing fluid embodies complexities that are not readily represented by electrical circuit elements.  Here for example is the Bernoulli equation which expresses the law of energy conservation on a fluid streamline:

$$\LARGE p_1+\frac{\rho}{2} v_1^2 + \rho g h_1 = p_2+\frac{\rho}{2} v_2^2 + \rho g h_2 + L_{\mu} + \rho \int_1^2 \frac{dv}{dt} ds$$

NOTE: In clinical cardiology, the simplified Bernoulli equation ( $$\Delta p = \rho v^2/2$$ ) is used to equate the pressure difference resulting from a change in kinetic energy where one of the $$v^2$$ terms in the above is "close enough" to zero.  All the other terms are assumed to be negligible under the specialized circumstances including the loss term $$L_{\mu}$$. The $$\rho v^2$$ terms don't have anything to do with energy loss -- they are NOT due to resistance!

I'll bring the Bernoulli equation to bear repeatedly as I believe many are familiar with it, particularly in the simplified form as commonly employed in common clinical practice. Shown is the pressure, gravitational potential ( involving $$\rho g h$$), a part due to kinetic energy (involving $$v^2$$ ), and what we will come to know soon as the inertial term ( involving an integration of $$dv/dt$$ along a streamline). Since this is a physical  equation, all the terms have the same physical units, apparently the same as pressure. The term that relates to resistance is shown all by itself ( $$L_{\mu}$$, i.e. loss due to viscosity through friction).  It has to do with fluid elements sliding against each other and the forces required to overcome friction as incurred through the viscosity of the fluid.  As this is only one equation, by itself we can use it to compute only one remaining term when all the others are known.  To figure out the loss term, we would have to know the quantitative values attributable to all the other terms OR we have to know exactly however the fluid elements are sliding against each other.  This is particular to the way the fluid flows so we have to know the entire velocity field and the relationship between shear stress and fluid relative motion.

#### Relationship between Fluid Relative Motion and Sheer Stress

Viscosity is the property of fluid that results in friction when fluid elements slide against each other.  The schematic below suggests how to devise one type of viscometer which is a device for measuring the viscous properties of a fluid.  There are a number of commonly employed devices to generate the conceptual arrangement shown.

With the fluid of interest between the 2 plates, dragging the upper plate at a fixed velocity results in a well-defined, constant velocity gradient.  Because of the No Slip Condition, fluid at the wall of each solid boundary moves at the same velocity as the boundary.  Hence the velocity gradient in the above is $$v/h$$.  We would find that a force, $$F$$ is required to drag the upper plate at a constant velocity.  We would have to compute the sheer stress also ($$\tau$$) which is the measured force divided by the surface area of interaction between the plates ($$\tau = F/A$$).  The Newtonian viscosity is the stress divided by rate of strain which also suggests the physical dimensions of the quantity.

$$\Large \mu = \frac{\tau}{v/h}$$

For a Newtonian fluid this value is a constant, i.e. independent of the rate of strain although it can vary significantly with temperature and other aspects of the environment.  Blood is a non-Newtonian fluid.  Using the apparatus suggested above, we would that we do not obtain a constant number for $$\mu$$ but one that varies with the rate of shear stress.  The apparent viscosity is also highly dependent on hematocrit as we all know.  Furthermore blood has some of the characteristics of a solid; it is a viscoelastic fluid.  The study of the relationship between stress and the rate of fluid deformation is called rheology and hemorheology when the fluid of interest is blood.

It's tempting to consider viscosity and density as related fluid properties.  Certainly gases tend to be rather low viscosity fluids and also clearly have low density.  Consider oil and water as a counterexample to this relationship.  Oil is much more viscous than water but is less dense.  Another counterexample is blood which has nearly the same density as hematocrit increases, but significantly greater (apparent) viscosity.

Now we're all familiar with the idea of total peripheral resistance, but the resistance concept applies to individual vascular beds ( with varying applicability and success), and also to individual blood vessels.  Even if the resistance concept doesn't work very well for a given situation, there is always a rigorous way to define the amount of energy loss that uses a concept similar to the Bernoulli equation.  While the Bernoulli equation is applied to a streamline, an energy conservation law can be applied to a stream tube and encompass every conceivable flow situation.  (A streamline is just a very skinny stream tube.)

For flow in a circular conduit, there is only one situation where the electrical analogy is strictly applicable and that is the Poiseuille parabolic velocity profile for a Newtonian fluid resulting in the Poiseuiile resistance:

$$\Large R = \frac{8\mu L}{\pi r_0^4}$$

As was shown tediously in a previous article, there is no other steady fluid flow situation (no other velocity profile) for a straight circular tube where there are no fluid accelerations involved.  If we don't have a parabolic velocity profile, the fluid will accelerate  until the parabolic profile is achieved.  Those accelerations involve nonlinear terms that cause the apparent value of $$R$$ to change with flow rate. Pressure loss per unit of flow will change with flow rate and there will be a change in pressure not due to loss but due to acceleration. There is no place in the circulation where the parabolic profile exists because there is no section in the arterial tree that is long enough and straight enough for this profile to develop.  ( Pulsatility isn't the problem.)  Furthermore blood is a non-Newtonian fluid whose apparent viscosity changes with shear rate (flow rate).

The above image depicts fluid velocity vectors near the entrance region of a straight rectangular ( square ) conduit with flow entering from the left and proceeding to the right ( computational fluid dynamics solution ). Color represents velocity. The velocity profile at the entrance is uniform – all fluid elements moving at the same velocity.  As the flow proceeds through the conduit, fluid elements near the center of the conduit speed up; elements near the wall slow down due to the drag exerted by the nearby wall.  Whether speeding up or slowing down, they accelerate.  We would measure a pressure difference from upstream to downstream in this conduit that is not due to friction (not a loss), but due the acceleration of fluid elements.

Next we'll get an idea for how the velocity profile affects the friction and pressure loss.  As already noted, the Poiseuille profile is the only one for a conduit of  circular cross section that will not result in fluid accelerations as suggested by the above figure.  However we can still calculate frictional losses due to other profiles.  A commonly employed test case is this one:

$$\Large v = k \left[1 - \frac{r^n}{r_0^n} \right]$$

This represents an entire range of velocity profiles for a circular conduit depending on the parameter $$n$$. If $$n=2$$, the Poiseuille profile is obtained.  For increasing values of $$n$$, the profile flattens out but the velocity gradient becomes steeper at the wall. Profiles are shown below for several values of $$n$$. The nondimensional radial coordinate is shown on the x-axis, relative velocity on the y-axis ( actual velocity divided by the mean velocity over the cross section ).  The flow through the conduit is determined by integrating the above velocity function over the circular cross section.

$$\Large q = \int_0^{2\pi} \int_0^{r_0} v(r) \,r\, dr\, d\theta = 2\pi \int_0^{r_0} v(r)\, r\, dr =2\pi r_0^2 \frac{kn}{4+2n}$$

$$k$$ is chosen so that the velocity profile can be specified in terms of a flow rate, $$q$$:

$$\Large k = \frac{q(n+2)}{n\pi r_0^2}$$

$$\Large v =\frac{q}{\pi r_0^2}\frac{n+2}{n} \left[1-\frac{r^n}{r_0^n}\right]$$

In this form, the velocity function is normalized so that the velocity is specified in terms of the flow rate.  The figure below shows the velocity distributions for the various profiles with average velocity 1.0 and radius 1.0.  The velocity profile becomes flatter and flatter with increasing $$n$$, approaching  a uniform velocity profile in the limit as $$n \rightarrow \infty$$.

We can calculate the forces due to friction from the assumed velocity profile.  The velocity gradient

$$\Large \frac{dv}{dr} = \frac{q}{\pi r_0^2}(n+2) \frac{r^{n-1}}{r_0^n}$$

is evaluated at the wall of the tube:

$$\Large \frac{dv}{dr}|_{r=r_0} = (n+2) \frac{q}{\pi r_0^3}$$

The shear stress due to friction is the velocity gradient multiplied by the Newtonian viscosity:

$$\Large \tau = \mu \frac{dv}{dr}|_{r=r_0}$$

To obtain the total force due to sheer stress at the wall, we multiply the sheer stress by the surface area it acts on ($$A=2\pi r_0 L$$ where $$L$$ is the length of the tube segment under consideration). The frictional force exerted on the wall is the same force that the wall exerts on the fluid.

$$\Large F_{\mu} = \mu \frac{dv}{dr}|_{r=r_0} 2\pi r_0 L = 2\pi \mu r_0 L \frac{q}{\pi r_0^3}(n+2) = \pi r_0^2 \Delta p$$

In the last step of the previous equation, the force due to sheer stress was equated with the pressure force driving the flow through the tube, simply the difference in pressure ( upstream minus downstream ) multiplied by the cross-sectional area of the tube.  The final step is simply to rearrange the equation to solve for the apparent resistance, $$R = p/q$$.

$$\Large \frac{\Delta p}{q} \equiv R = 2(n+2) \frac{\mu L}{\pi r_0^4}$$

Replacement of $$n$$ with the value 2 gives the Poiseuille resistance.  Higher values of $$n$$ show how the resistance increases due to the velocity profile and steeper velocity gradient that the wall.  Note that this velocity profile is NOT the same as we will find for pulsatile flow although there are qualitative similarities.

There are additional restrictions on the concept of resistance in the circulation.   If we're calculating resistance from measured pressure and flow, we have to be talking about time-averaged quantities, e.g. over an integer number of cardiac cycles.  Secondly, we are talking about a flow network where ALL the flow goes in at a single location ( single conduit ) and exits at a single conduit also.  While the network can be exceedingly complicated between the inlet and outlet, we are going to have quite a bit of mathematical difficulty if we don't have the same amount of flow going into the system as exiting and it's also going to be a big problem if the flow is exiting the system at multiple locations with potentially different pressures.

If we were to determine total peripheral vascular resistance, we would employee appropriate gadgetry to measure the pressure at the aorta and right atrium, the flow rate through the system, and find the time averaged values of each.  Then we would employ the formula above which implies the slope of the line shown in the next figure. "X" marks the spot of our pressure and flow measurements.

The expectation is that this single data point has told you something about how the circulation will behave, i.e. that we've determined a relationship, $$\Delta p/q = R$$, that would allow us to predict what will happen under other circumstaces. But consider what might happen if we extend this experiment and make multiple measurements of pressure and flow over a range of the variables.  Suppose then that we obtain the red line in the next figure:

Without this information, we were assuming that the single data point could be used to represent something intrinsic about the way the system behaves when in fact it does not.  It's simply a single  data point on the red curve, a curve that we didn't have until performing the more extensive experiment.   Calculating vascular resistance using the usual formula doesn't necessarily represent anything important about the system.  In fact, if we only reported the resistance in this situation ( slope of the blue line ) we've actually thrown information away – the single data point on the actual characteristic (red) line.  If we tried to characterize this system by calculating the resistance at multiple flow rates ( using the resistance formula ), we'd find that resistance depends on flow rate (or pressure) and might very well miss the fact of the actual curve that represents the pressure-flow relationship

Now this figure is not from data and I've exaggerated it to make a point but I'll enumerate some of the reasons we might see this or variations of it, i.e. various causes of apparently nonlinear vascular resistance.

• The vasculature is made up of conduits that are compliant and change their dimensions with distending pressure and therefore flow rate.  Therefore vascular channels dilate with increasing flow and channels that are closed may actually open to allow additional avenues of flow ( distention and recruitment in the literature ).
• Blood is non-Newtonian and its apparent viscosity decreases with increasing shear rate.
• There is no place in the circulation where Poiseuille flow occurs and therefore the blood velocity profiles are not constant in shape.  Velocity gradients will not be proportional to flow rate.
• Certain vascular beds ( e.g. pulmonary ) undergo rather dramatic physical distortion which changes the physical characteristics of the vascular bed and resistance in consequence.  Distention of vessels within the pulmonary circulation in particular are dependent upon the gravitational term of the Bernoulli equation.  Intravascular pressure varies in the direction of gravity ( as it does in all the blood vessels ), but the external pressure is essentially the same as alveolar pressure ( a constant more or less ).  So vascular distending pressure increases in the direction of the dependent portions of the lung.
• In muscular vascular beds ( e.g. the coronary circulation ), stresses within the tissue also affect the distending forces of the vessel.  The apparent apparent coronary resistance changes dramatically over the course of the cardiac cycle and the strict definition of resistance is no longer applicable.

To make matters quite a bit worse, data depicted by the red line plot above can't even be collected reliably in the circulation.  Autonomic, neurohumoral, and intrinsic control systems act over short time intervals to adjust the resistance through vasomotion as soon as there is any significant change in the flow rate through the system. Thus vascular resistance changes as soon as any attempt is made to collect pressure and flow data over a range of inputs.

#### Resistors in Series and in Parallel

Back to things we can handle, i.e. linear resistors.  Even though we've just discussed the fact that it's a bit of a stretch to consider any part of the circulation in these terms, I'll include now the usual discussion of "equivalent" resistance (we're going to need this for the discussion of impedance also).  I'll also use the electrical analog unabashedly to refer to fluidic systems; a tube or tube network with flowing fluid can be a "resistance" and $$\Delta p/q$$ (flow and pressure, not current and voltage) is how we calculate the value.  Our task is to replace the 2 resistors in series (top of figure) with a single equivalent resistor ($$R_e$$ ) that exhibits the same characteristics, i.e. the same trans-resistance pressure (voltage) for a given flow rate (current).  For flow rate $$q$$, the pressure across $$R_1$$  is $$\Delta p_1 = q R_1$$; also $$\Delta p_2 = q R_2$$.  So the total $$\Delta p$$ is $$\Delta p_1 + \Delta p_2 = q (R_1+R_2)$$.  Hence the total resistance can be replaced by a single resistor, $$R_e = R_1+R_2$$. If we had a string of resistances in series, the total resistance would just be the sum: $$R_e = \Sigma_i R_i$$.  $$R_e$$ for a series combination is always greater than  the single resistors involved  (it's the sum).

For the parallel combination at a given trans-resistance pressure (voltage) we have that $$q_1 = p/R_1$$, and $$q_2 = p/R_2$$.  So the total flow is $$q_1+q_2 = p(1/R_1+1/R_2)$$.  Hence $$R_e = p/q= p/(q_1+q_2) = 1/( 1/R_1+1/R_2 )$$ and $$R_e = 1/( 1/R_1+1/R_2 ) = R_1 R_2/(R_1+R_2)$$.  While it may not be obvious from looking at the formula,  $$R_e = R_1 R_2/(R_1+R_2)$$ is less than either $$R_1$$ or $$R_2$$.

Arranging 2 resistors in parallel results in a division of average flow that's determined by the resistances.

The total resistance of this system is $$R_1 R_2/(R_1+R_2)$$  as we just determined.  The pressure at the upstream junction is $$p = q R_e = q R_1 R_2/(R_1+R_2)$$.  The flow through $$R_1$$ is $$p/R_1 = q R_2/(R_1+R_2)$$ and hence the fraction of flow through $$R_1$$ is $$R_2/(R_1+R_2)$$; similarly the fractional flow through  $$R_2$$ is $$R_1/(R_1+R_2)$$ and the sum of these 2 fractions is 1.0 of course.  Folks are fond of saying that the flow takes the "path of least resistance" and you will see by analyzing the expressions for the flow division that a greater fraction of the flow takes the lower resistance pathway.  But of course it doesn't ALL take the path of least resistance.

The equivalent resistor arising from multiple resistors in parallel is also readily determined.

$$\Large q = \Sigma_{i=1}^n q_i = p \Sigma_{i=1}^n \frac{1}{R_i}$$

$$\Large \frac{p}{q} = R_e = \frac{1}{\Sigma_{i=1}^n \frac{1}{R_i}}$$

The fraction of flow through $$R_i$$ is just $$R_e/R_i$$.

#### The Vascular "Steal"

You may run across discussions pertaining to one vascular bed "stealing" blood flow from another.  It's worth considering how this might occur from the information above, at least as an exercise in manipulating the concepts. I hasten to preface the discussion with a reminder that this thievery typically refers to the coronary circulation in conjunction with an upstream stenosis that is nonlinear and coronary resistance is difficult to conceptualize in the first place since contraction of the heart changes the vasculature dramatically.  Linear resistors will help us conceptualize the situation but do not apply for quantitative analysis.

$$R_u$$ here is used to represent an "upstream" resistor. From our recent discussion, an equivalent resistor for the 2 in parallel is $$R_1 R_2/(R_1+R_2)$$.  For the entire system:

$$\Large R_e = R_u + \frac{R_1 R_2}{R_1+R_2} = \frac{R_u(R_1+R_2)+R_1 R_2}{R_1+R_2}$$

If the perfusion pressure of this system is $$p$$, then the flow through the system is:

$$\Large q = p/R_e = p \frac{R_1+R_2}{R_u(R_1+R_2)+R_1 R_2}$$

The fraction through $$R_2$$:

$$\Large \frac{q_2}{q} = \frac{R_1}{R_1+R_2}$$

And the actual flow through $$R_2$$:

$$\Large q_2 = p \frac{R_1}{R_u R_2 + R_1(R_2+R_u)}$$

At this time we will say that $$R_2$$ corresponds to a vascular bed at risk for a steal by $$R_1$$.  We'll say that perfusion pressure for the entire system is adequate at value $$p$$ and perfusion of $$R_2$$ at pressure $$p$$  corresponds to a relative flow of 1.0.  Then the following plot summarizes the problem for the steal for various ratios of the resistances.

The plot at the top shows a constant normal flow (1.0) through $$R_2$$ when the value of $$R_u$$ is zero.  $$R_1$$ cannot influence the flow through the vascular bed at risk when there is no upstream resistor.  This is due partly to the way the problem has been stated; we've assumed that there is a "pressure source" perfusing the system so that changing $$R_1$$ does not result in a perfusion pressure change to $$R_2$$ if it's connected right to the source. Considering the heart as a pressure source would be good approximation for this model as long as the system shown constitutes a relatively small component of the total system.  Since we are typically talking about the coronary circulation for this model, it's a good approximation with mean aortic pressure a constant at the coronary ostia.

With increasing resistance for the upstream resistor, we shift to lower and lower curves on the graph.  At constant perfusion pressure, increasing the upstream resistor means there will be an ever decreasing perfusion pressure for both $$R_2$$ and $$R_1$$.  This is obvious enough, but it's not what we mean by a "steal", i.e. how does the parallel vascular bed ($$R_1$$) influence flow through the bed at risk ($$R_2$$)?  That's what's depicted by the x-axis variable, the ratio of resistance of the two vascular beds in parallel.  We see that for $$R_1 >> R_2$$ ( to the right side of the plot and continuing to the far right ) the curves are relatively flat.  This is for the circumstance where the vascular bed at risk is much larger ( lower resistance ) than the bed doing the stealing.  If $$R_1$$ is the larger vascular bed, however ($$R_1 << R_2$$, left side of plot), then it begins to have a much greater influence on flow through $$R_2$$.  Basically the flow through $$R_2$$ is at risk when it is in parallel with a much larger vascular bed ( with lower resistance ) and there is an upstream resistance.

The situation is most concerning (for $$R_2$$ ) for smaller values of both $$R_u$$ and $$R_1$$.  That circumstance puts us on one of the upper curves, but also towards the far left of the plot.  In this region of the parameter space, a change in $$R_1$$ has the most dramatic effect on flow through $$R_2$$ and the effect of a "steal" is best visualized.  With $$R_u$$ fixed, vasodilation of $$R_1$$ ( relative to $$R_2$$ ) can result in a more drastic reduction in flow through $$R_2$$.  Even though larger values of $$R_u$$ clearly decrease $$R_2$$ perfusion, we would say that flow through $$R_2$$ is most "sensitive" to changes in $$R_1$$ at low values of both $$R_u$$ and $$R_1$$.  The system is most sensitive at low values of $$R_u$$ also indicated by the fact that the curves are farther apart for the smaller values of $$R_u$$.