## Poiseuille Flow

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This video depicts a side view in a straight circular tube with steady ( time invariant ) flow occurring from left to right.  The parabolic velocity profile of Poiseuille flow is shown by the vectors and distortion of the fluid within the tube is suggested by the grid.  DOWNLOAD a larger avi file.

#### What's the point?

The Poiseuille flow relationship is included in basic physiology courses. Cardiologists are often familiar with the formula for the Poiseuille resistance. This article deals with the origins of this relationship and the assumptions and limitations inherent for Poiseuille flow.  The  Poiseuille relationship for unaccelerated (fully developed), steady (time invariant) flow in a straight circular tube provides an excellent opportunity for understanding some profound aspects of fluid dynamics.

#### Where does it come from?

The Poiseuille relationship comes not from experimental data but from a story problem or mathematical derivation. This is a purely intellectual endeavor derived from 3 basic assumptions: 1) Newton's fundamental law of motion, i.e. $$\bar{F} = m\bar{a}$$ ( force = mass times acceleration); 2) a geometric  assumption about the nature of the flow, i.e. a  circular cross-sectional geometry in which the fluid elements flow straight down the tube with no acceleration (steady or time invariant, fully developed flow); and 3) an assumed relationship between relative fluid motion and shear stress (a Newtonian fluid is assumed where shear stress is proportional to velocity gradient through the Newtonian viscosity).  While an understanding of the Poiseuille relationship is of great value, let us realize from the start that there is no location in the circulation where the last 2 assumptions are met.  However, the first assumption ( Newton's laws of motion ) is more reliable in this vicinity of the universe than any medical or physiological "fact" (and that includes the one about death).

The derivation of the Poiseuille relationship is simple enough to provide an inkling of the flavor of fluid dynamics and engineering science in general.  The derivation requires ruthless logic, scrupulous mathematical accounting of forces acting on fluid elements, and courage that we will be able to find our way to the result.

#### Getting the facts on to "paper"

To this end, consider a straight circular tube of radius $$r_0$$ and an annulus of fluid inside the tube.   The annulus consists of a thin rim of fluid with mean radius $$r$$ and thickness $$\Delta r$$ in the radial direction and length $$\Delta z$$ in the tube's axial direction,  $$z$$ .

We begin by actually assuming how the flow will occur, i.e. by moving in a straight line in the axial direction only, and at a constant speed (steady flow, not pulsatile or oscillatory). A moment's thought will surely convince you that this is also a statement that the acceleration of every fluid element is zero (acceleration is a change in velocity, a change in speed and/or direction of motion).  Newton's laws of motion tell us that an object having zero acceleration also must experience zero net force. Hence the sum of all forces acting on each and every fluid element ( each fluid annulus) must equal zero.

Our next step is to enumerate and account for the forces acting on the fluid annulus (each and every one).  Here we use a simple, but powerful conceptual tool called a free body diagram where the free body is the annulus of fluid.  The free body diagram allows us to isolate an arbitrary part of the fluid and define rigorously the forces acting on the fluid.  This process is equally valid for solid structures.

There are two basic kinds of forces when discussing mechanics of materials, fluid or solid: 1) Surface forces occur where the surface of the free body touches its surroundings, other fluid elements in this case. 2) Body forces are forces that seem to act from afar, to reach across space to affect an object.  The most physiologically relevant of these in cardiovascular applications is gravity (until we start accelerating people in various kinds of transportation gadgetry).  Even so, gravity is not very important to the discussion and cardiologists typically go out of their way to leave gravity out of the equation in a literal sense.  Consequently, we only have surface forces in this problem.

$$\Large p A = p|_{\small{z-\Delta z/2}} 2 \pi r \Delta r$$

In the following figures,  the surface under discussion is highlighted in color. The first surface force to consider is the pressure acting on the upstream surface of the annulus.   Pressure is an example of a stress which has physical units of force/area.  We calculate the value of this force by multiplying the pressure by the surface area it acts upon.  Because of the way the problem is formulated, pressure has a different value at each axial location in the tube ( each  value of  $$z$$ ).  For the annulus in question, we are assuming its position at arbitrary axial location $$z$$ with axial thickness $$\Delta z$$ , so that the upstream surface is at axial coordinate  $$z -\Delta z/2$$ .  Using common notation, we designate the pressure at this location as $$p|_{\small{z-\Delta z/2}}$$ .

The surface area that this pressure acts upon is equal to a difference in areas between two circles; the outer one has radius  $$r + \Delta r/2$$ and the inner one has radius $$r - \Delta r/2$$ .  This comes out to $$2 \pi r \Delta r$$ . Without doing the math, you can think of this circular rim of surface as a rectangle that is been warped into a circle; the width of the rectangle is $$\Delta r$$ and the length is $$2 \pi r$$ , the perimeter length of the circle.

The above pressure force is opposed by a similar one acting on the downstream rim of the annulus.  The area of action is identical to the last calculation, but the pressure is different, occurring at a location $$\Delta z$$ downstream of the previous pressure determination and represented by the notation $$p|_{\small{z+\Delta z/2}}$$ .

$$\Large p A = p|_{\small{z+\Delta z/2}} 2 \pi r \Delta r$$

The first above pressure acts in the $$z$$ direction whereas the second acts in the $$-z$$ direction; these forces will have the opposite sign in the final accounting of forces.

Now we turn to the determination of shear stresses acting on the annulus.  These are designated as $$\tau$$ in general but we are specifically interested in shear stresses that translate to forces in the axial direction.  Mechanical stress is a second order tensor.  Vectors are first order tensors and have 3 components in a three-dimensional space, e.g. the $$x$$ , $$y$$ , and $$z$$ directions.  A second-order tensor has 9 components.  The notation $$\tau_{rz}$$ can be read as the "shear stress on the $$r$$ surface in the $$z$$ direction"; you can readily imagine how we would obtain 9 components in a three-dimensional space by having a stress on each of the 3 face orientations and in each of the 3 directions.

For the inner surface of the annulus, we must form an expression for  the force $$\tau_{rz}A$$ ,  i.e. the shear stress multiplied by the area it acts upon.  We can designate the shear stress at the location $$(r-\Delta r/2)$$ as  $$\tau_{rz}|_{\small{r-\Delta r/2}}$$ in keeping with previous notation.  For a Newtonian fluid, shear stress is equal to the viscosity multiplied by the appropriate velocity gradient and this translates to $$\mu \Large\frac{\partial u}{\partial r}|_{\small{r-\Delta r/2}}$$ for the problem at hand.  We are using $$u$$ to represent axial velocity and $$\Large\frac{\partial u}{\partial r}$$ is the velocity gradient, mathematically the derivative of $$u$$ with respect to the radial coordinate, $$r$$ .  $$\mu$$ is the Newtonian viscosity.

$$\Large \tau_{rz}A = \tau_{rz}|_{\small{r-\Delta r/2}} 2 \pi r|_{\small{r-\Delta r/2}} \Delta z=\mu \Large\frac{\partial u}{\partial r}|_{\small{r-\Delta r/2}} 2 \pi r|_{\small{r-\Delta r/2}} \Delta z$$

The surface area that the shear stress acts upon at this location is slightly tricky to express because the area changes with radius, just as the stress itself does.  Nevertheless, the area expression is simply that of a rectangle with thickness $$\Delta z$$ and length $$2 \pi (r - \Delta r/2)$$ so that the area expression is $$A|_{r-\Delta r/2} = 2 \pi \Delta z (r - \Delta r/2)$$ or $$2 \pi \Delta z r|_{\small{r - \Delta r/2}}$$ .  The entire expression for the force is shown adjacent to the figure above.  The shear stress acting on the outer rim of the annulus is derived similarly (below).

$$\Large \tau_{rz}A = \tau_{rz}|_{\small{r+\Delta r/2}} 2 \pi r|_{\small{r+\Delta r/2}} \Delta z=\mu \Large\frac{\partial u}{\partial r}|_{\small{r+\Delta r/2}} 2 \pi r|_{\small{r+\Delta r/2}} \Delta z$$

In case it has not yet occurred to you,  it's (utterly) important to recognize that equations here are to express physical concepts.  "Equations" used in cardiology are sometimes the result of statistical regressions, or they may express a concept or approximation in a general way.  We can admit no conceptual or mathematical slack for the endevour at hand! We're deriving an equation whose physical units (thus far) are force.  Pressure must be expressed in terms of force/area, e.g. dyne/cm2 (a dyne is a gram-cm/sec2 which is a mass multiplied by an acceleration). If you can only think of pressure in terms of mmHg, you've been spending too much time in the cath lab.  When we multiply all of these things together, they have to have comparable physical units or we've got nothing but muck! The Newtonian viscosity, for example, is a physical quantity that multiplies the velocity gradient and yields a shear stress.  If your shear stress has units of dyne/cm2 and your velocity gradient is in cm/sec/cm (i.e. sec-1), then your viscosity had better have units of dyne-sec/cm2 (or gram-cm-1-sec-1) so that the physical units are correct!  It just so happens that the physical unit of viscosity just named is called the Poise after the gentleman under discussion.

#### Putting it all together

We are now in a position to begin combining (mathematical ) terms into a coherent expression (have courage!). As noted previously, the upstream and downstream pressure forces oppose each other:

$$\Large 2\pi r\Delta r[p|_{\small{z-\Delta z/2}}-p|_{\small{z+\Delta z/2}}]$$

Similarly, shear stresses at the inner and outer surfaces of the annulus oppose each other:

$$\Large 2\pi \Delta z \mu[(r \Large\frac{\partial u}{\partial r})|_{\small{r+\Delta r/2}}-(r \Large\frac{\partial u}{\partial r})|_{\small{r-\Delta r/2}}]$$

The net sum of all the forces acting on the annulus is equal to zero since the annulus is not accelerating:

$$\Large 2\pi r\Delta r[p|_{\small{z-\Delta z/2}}-p|_{\small{z+\Delta z/2}}]+2\pi \Delta z \mu[(r \Large\frac{\partial u}{\partial r})|_{\small{r+\Delta r/2}}-(r\Large\frac{\partial u}{\partial r})|_{\small{r-\Delta r/2}}]=0$$

We divide the entire equation by $$2\pi \Delta r \Delta z$$ and simplify:

$$\Large r\Large\frac{p|_{\small{z-\Delta z/2}}-p|_{\small{z+\Delta z/2}}}{\Delta z}+\mu \Large\frac{(r \Large\frac{\partial u}{\partial r})|_{\small{r+\Delta r/2}}-(r \Large\frac{\partial u}{\partial r})|_{\small{r-\Delta r/2}}}{\Delta r}=0$$

(The equation just acquired new physical units : force/cm2) The next step is allow $$\Delta r$$ and  $$\Delta z$$ approach zero and to recognize the definition of the derivative as defined in calculus:

$$\Large\frac{df}{dx} \equiv \Large\frac{f|_{x+\Delta x} - f|_{x}}{\Delta x}_{lim\Delta x \rightarrow 0}$$

This allows us to express the work thus far as a differential equation:

$$\Large -r\frac{\partial p}{\partial z}+\mu \Large\frac{\partial }{\partial r} (r \Large\frac{\partial u}{\partial r})=0$$

or

$$-\Large\frac{\partial p}{\partial z}+\Large\frac{\mu}{r}\Large\frac{\partial}{\partial r}(r \Large\frac{\partial u}{\partial r})=0$$

(The equation just acquired new physical units : force/cm3) It may not seem like it, but this is tremendous progress.  This is a linear, ordinary differential equation whose solution will tell us how the velocity changes as a function of radius.  It may look like there are 2 unknown functions in the equation, $$p(z)$$ and  $$u(r)$$ , but there's really only one. $$-\partial p/\partial z$$ is a given in this problem, a constant that you supply.

$$-\Large\frac{\partial p}{\partial z} = \Large\frac{p_1-p_2}{z_2-z_1}=\Large\frac{\Delta p}{L}$$

By convention, $$\Delta p$$ here is the upstream pressure minus the downstream and has the opposite sign of $$\Large\frac{\partial p}{\partial z}$$ . (Note: I'm playing a little bit fast and loose with notation.  $$\partial/\partial r$$ is more appropriately used to denote the partial derivative of a function that depends on more than one variable,  e.g. $$\partial u(r,z)/\partial r$$ .  We will soon tackle this very situation.)

#### Solving the Equation of Motion

The F=maequation in a physics problem is often denoted the equation of motion and is typically a differential equation like the one above. You may not be familiar with such equations, but the solution of this one is relatively simple and can be looked up in a book; there is also software that would solve the equation for you if you know how to pose the question correctly.  The full solution is as follows:

$$\Large u(r)=-\Large\frac{r^2 \Delta p}{4 \mu L}+C_1+C_2\ln(r)$$

$$C_1$$ and $$C_2$$ in this case are known as arbitrary constants of integration.  We would find that the original differential equation is satisfied no matter what values we choose for these constants.  Consequently we are free to choose any values we like; more specifically we are free to choose the values that satisfy the boundary conditions of the problem.  The boundary condition at $$r = r_0$$ is that  the velocity goes to zero.  This is called the no slip condition, that the fluid element at the wall moves at the same velocity as the wall (zero).  This may seem like an arbitrary and unjustified choice, but it is well verified experimentally.  Furthermore, it can be thought of as a microcosm of a much broader property of fluids – that they are continuous. (We will explore this issue in much greater detail elsewhere.)  The boundary condition at $$r=0$$ is that the velocity remains finite. You will notice that we cannot even evaluate $$\ln(r)$$ at $$r=0$$ because it approaches $$-\infty$$ as $$r$$ approaches zero ;hence the value of  $$C_2$$ is zero and we are free to choose $$C_1$$ to fulfill the first boundary condition.

$$\Large C_1=\Large\frac{r_0^2 \Delta p}{4 \mu L}$$

$$\Large u(r)=-\Large\frac{r^2 \Delta p}{4 \mu L}+\Large\frac{r_0^2 \Delta p}{4 \mu L}$$

$$\Large u(r)=\Large\frac{(r_0^2-r^2) \Delta p}{4 \mu L}$$

This equation for the axial velocity $$u(r)$$ describes a parabola such that the maximal velocity occurs at the tube centerline and tapers off parabolically (in proportion to $$r^2$$ ) with radius to a value of zero at the wall where $$r=r_0$$ .  It will turn out that the centerline velocity is twice the average velocity across the tube.  A schematic/graphic of the velocity profile for Poiseuille flow is shown:

#### What do we mean by the "solution"?

The solution to a differential equation like the one above is most typically a function, not a specific number.  $$u(r)$$ defines the parabolic profile that occurs for Poiseuille flow.  The velocity profile is parabolic regardless of the flow rate, tube radius, viscosity, or tube length.  The axial velocity is maximal at the tube centerline and equal to $$r_0^2 \Delta p/ {4 \mu L}$$ (obtained by substituting $$r=0$$ into the solution; the velocity at the wall is zero by design where $$r_0^2-r^2=0$$ .

We can verify that this is the correct solution by plugging the function back in to the differential equation.

$$\Large\frac{\mu}{r}\Large\frac{\partial}{\partial r}(r \Large\frac{\partial u}{\partial r})=-\Large\frac{\Delta p}{L}$$

The equation of motion can be expanded as follows for better understanding.  Recall from calculus that the derivative of the  product of two functions involves finding the derivative of each as follows:

$$\Large\frac{\partial}{\partial r}[f(r) g(r)]=f(r)\Large\frac{\partial g}{\partial r}+\Large\frac{\partial f}{\partial r}g(r)$$

In our problem, $$f(r)$$ and $$g(r)$$ are $$r$$ and $$\Large\frac{\partial u}{\partial r}$$ and their derivatives are $$1$$ and  $$\Large\frac{\partial^2 u}{\partial r^2}$$ respectively.  The expanded equation of motion:

$$\Large\frac{\mu}{r}(\Large\frac{\partial u}{\partial r} + r \Large\frac{\partial^2 u}{\partial r^2})=-\Large\frac{\Delta p}{L}$$

$$\Large \mu (\frac{1}{r}\Large\frac{\partial u}{\partial r} + \Large\frac{\partial^2 u}{\partial r^2})=-\Large\frac{\Delta p}{L}$$

Now we need to compute derivatives of $$u(r)$$ and plug them into the equation:

$$\Large\frac{\partial u}{\partial r} = \Large\frac{\partial}{\partial r} [\Large\frac{(r_0^2-r^2) \Delta p}{4 \mu L}] = \Large\frac{- r \Delta p}{2 \mu L}$$

$$\Large\frac{\partial^2 u}{\partial r^2} = \Large\frac{\partial^2}{\partial r^2} [\Large\frac{(r_0^2-r^2) \Delta p}{4 \mu L}] = \Large\frac{-\Delta p}{2 \mu L}$$

You will find that substituting these results back into the equation of motion yields the intended $$\Large\frac{\Delta p}{L}$$ .

#### The Poiseuille Resistance

The final component in the problem is to determine the formula for the Poiseuille resistance which first involves calculating the flow produced by the pressure gradient.  To accomplish this, we must integrate the axial velocity over the cross-sectional area of the tube.

$$\Large \int_0^{2\pi}\int_0^{r_0} u(r) r dr d\theta = \int_0^{2\pi}\int_0^{r_0} \Large\frac{(r_0^2-r^2) \Delta p}{4 \mu L} r dr d\theta=\Large\frac{\Delta p}{4 \mu L}\int_0^{2\pi}\int_0^{r_0}(r_0^2-r^2)r dr d\theta$$

$$\Large r dr d\theta$$ is the differential area element  in the polar coordinate system we are dealing with.  The integration with respect to $$\theta$$ :

$$\Large\frac{\Delta p}{4 \mu L}\int_0^{2\pi}\left[\int_0^{r_0}(r_0^2-r^2)r dr\right] d\theta=\Large\frac{\Delta p}{4 \mu L}\theta\left[\int_0^{r_0}(r_0^2-r^2)r dr\right]^{2\pi}_0=\Large\frac{\pi \Delta p}{2 \mu L}\int_0^{r_0}(r_0^2 r-r^3) dr$$

And, with respect to $$r$$ :

$$\Large\frac{\pi \Delta p}{2 \mu L}\int_0^{r_0}(r_0^2 r-r^3) dr = \Large\frac{\pi \Delta p}{2 \mu L}\left[r_0^2 \Large\frac{r^2}{2}-\Large\frac{r^4}{4}\right]^{r_0}_0 = \Large\frac{\pi \Delta p}{2 \mu L}\left[\Large\frac{r_0^4}{4}\right]$$

$$\Large q=\int_0^{2\pi}\int_0^{r_0} u(r) r dr d\theta = \Large\frac{\pi r_0^4 \Delta p}{8 \mu L}$$

This is the expression for the volumetric flow rate, $$q$$ , in the tube.  To obtain the Poiseuille resistance, we divide $$\Delta p$$ by $$q$$ (by definition).

$$\Large\frac{\Delta p}{q} \equiv R = \Delta p \Large\frac{8 \mu L}{\pi r_0^4 \Delta p}=\Large\frac{8\mu L}{\pi r_0^4}$$

#### Shear Stress

Now that we know a lot more about this flow, it turns out that we can calculate the frictional force on the fluid without having to do any more calculus.  We know that the net force on the fluid is zero. This applies to each and every bit of the fluid, but it also applies to the fluid in whole.  Pressure exerts a force on the fluid that can be calculated by multiplying the pressure by the cross-sectional area it acts upon.  If we do this for the curved walls of the tube, the forces simply cancel out.  However the pressures are different at the upstream and downstream ends of the tube and the force acting on the fluid due to these pressures can be readily determined from the resistance formula:

$$\Large \Delta p=q \Large\frac{8 \mu L}{\pi r_0^4 \Delta p}=\Large\frac{8q\mu L}{\pi r_0^4}$$

$$\Large F_p=\Delta p A=\Large\frac{8q\mu L}{\pi r_0^4}A=\Large\frac{8q\mu L}{\pi r_0^4}\pi r_0^2=\Large\frac{8q\mu L}{r_0^2}$$

$$\Large F_p$$ is the force due to pressure acting on the fluid where $$A$$ is the ccross-sectional area equal to $$\pi r_0^2$$ .  This force must be balanced by the frictional force due to shear stress, $$\tau$$ , acting at the wall.  The area that this shear stress acts on is the surface area of the tube, $$2\pi r_0 L$$ .

$$\Large F_p=F_\tau=\Large\frac{8q\mu L}{r_0^2}$$

$$\Large \tau=F_\tau /2\pi r_0 L=\Large\frac{8q\mu L}{r_0^2}/[2\pi r_0 L]=\Large\frac{4q\mu}{\pi r_0^3}$$

Apparently shear stress is linearly proportional to the flow rate and the viscosity, but inversely proportional to the cube of the tube radius.

#### Why does it do that?

In the foregoing the Poisseuille flow profile and resistance were derived mathematically.  It was found that the velocity profile for a Newtonian fluid in a straight circular tube approaches a parabola with maximal velocity at the center of the tube and a velocity of zero at the wall; the centerline velocity is twice the average velocity.

Why does it do that?!  To look at that question more closely we'll use some figures that have to do with flow in a straight circular tube where the flow velocity profile is not a parabola.  These figures are derived from computational fluid dynamics solutions where the centerline is at the lower part of the computational figure and the wall of the tube is at the upper part.  To orient you to these figures, consider the following figure which is from a computation of stenosis flow where flow is from left to right through an orifice plate – a circular ridge or ring within the tube.

Hopefully this orients you well enough for the subsequent figures where we'll look at a simple straight tube without a stenosis.

The following figure shows the development of the velocity profile in a straight tube where the flow starts on the left with a uniform velocity; i.e. fluid elements have the same velocity all the way across the inside of the tube.  The figure shows the development of velocity proceeding left to right.

This solution was computed a relatively low Reynolds number ( discussed elsewhere) which has the effect of allowing the velocity profile to develop into the parabola over a small number of tube diameters.   Hence you can see the development of Poisseuille flow in a relatively small figure.  Color depicts the absolute value of the velocity which is also depicted by the vectors of course.  Fluid elements near the wall ( upper edge of the figure ) decelerate whereas those near the center line ( lower edge of the figure ) accelerate.

The changing of velocity as fluid elements flow through the tube is synonymous with acceleration.  Acceleration is a vector so the fact that elements near the wall are slowing down simply means that acceleration is towards the left-hand side of the figure.  Why do they accelerate?  Well, we know that all of the fluid elements obey Newton's laws of motion.  Fluid elements accelerate because there is a net force acting upon them.  In a nutshell:

Flow of a Newtonian fluid in a straight circular tube achieves a parabolic profile at a sufficient distance downstream in the tube because that is the only velocity profile ( for that geometry ) where net forces on all the fluid elements sum to zero. If the profile is not a parabola, then there will be a net force on fluid elements.  Fluid element velocities will change ( accelerate ) until the parabolic profile is achieved.

Fluid at the wall necessarily moves at the same velocity as the wall itself, i.e. 0 - the no slip condition.  Consequently / subsequently the slow moving fluid near the wall exerts a "drag" force (shear stress, exerted through viscosity and relative motion of the fluid elements) on neighboring fluid elements which leads to the parabolic profile.  Slow moving fluid near the wall requires that fluid near the centerline moves faster than the average.  The slow moving fluid near the wall causes a decrease of the effective cross sectional area of the tube.

The Poisseuille resistance derives from the situation where a pressure force ( higher pressure upstream and downstream ) is exactly balanced by frictional forces of fluid elements ( fluid layers or laminae )  sliding against each other.  The  Poisseuille resistance was derived above:

$$\Large\frac{\Delta p}{q} \equiv R =\Large\frac{8\mu L}{\pi r_0^4}$$

This value of this resistance is entirely dependentinextricably related to the parabolic profile itself.  The parabola leads to the lowest possible resistance and the lowest pressure gradient to push the fluid through the tube.  Since there is no place in the circulation where the parabolic profile exists, resistance is always greater than the Poisseuiile value.

In the subsequent figures we're looking at the velocity profile ( from tube centerline to wall as before ) somewhere far enough downstream from the inlet so that the profile has obtained the Poiseuille parabola.

Here now is a figure that shows the forces acting on the fluid elements.  The red arrows show the force due to friction ( through fluid viscosity and relative velocity ); the blue arrows show the force due to the pressure gradient.

This shouldn't be much of a surprise to you, particularly if you slogged through the math given above.  We started this problem with an assumption that fluid elements were not accelerating and hence we should find that the net force on each element is exactly equal to zero.  This is what the parabolic profile accomplishes for a Newtonian fluid. It's the only profile in a circular tube that accomplishes this feat.

Next we're looking at the shear stresses on the fluid elements.  You can see by looking at the velocity profile that elements near the centerline are moving at nearly the same velocity as neighboring elements whereas velocity near the wall changes rapidly with the radial coordinate; the velocity gradient near the wall is greater than the velocity gradient at the centerline.  Shear stress is the force that fluid elements exert on one another as a result of relative motion; relative motion is expressed as a velocity gradient, i.e. the velocity changes with location.  Consequently there is greater shear stress on fluid elements near the wall than at the centerline;  shear stress at the centerline is actually zero.  The following vectors show the shear stress vectors on the fluid elements ( centerline at the bottom of the figure, tube wall at the top).

This shows us that the parabolic velocity profile results in a linear shear stress profile.  If you consider any one of the fluid elements, you'll realize that there is greater shear stress on one side ( the side closer to the wall ) than on the other ( the side closer to the centerline ).  The difference in shear stress constitutes a force.  This shear stress (when correctly multiplied by the surface area it acts upon ) exactly counterbalances the pressure force acting to push the fluid elements in the downstream direction.  That's how we get the figure above showing the balance of forces between friction and pressure.

#### Vorticity

If you take just a moment to consider, you'll recognize that individual fluid elements are spinning in Poiseuille flow! Each element has a greater velocity on its centerline side than on the wall side.  When fluid elements spin, they possess vorticity. This is not the same thing as a vortex where the gross fluid motion is circular. Vorticity is a vector, with magnitude and direction, that is determined from the velocity field of the fluid motion; in general it varies from point to point within the flow and also likely with time.  The sense of the vorticity vector is defined by the "right-hand rule". Using your right hand with your fingers curling in the direction of spin, your thumb points in the direction of the vorticity vector.  Of course there is a much more specific and quantitative definition of this aspect of the flow field.

$$\Large \omega=\Large\frac{1}{2}(\Large\frac{\partial w}{\partial y}-\Large\frac{\partial v}{\partial z}){i}+\Large\frac{1}{2}(\Large\frac{\partial u}{\partial z}-\Large\frac{\partial w}{\partial x}){j}+\Large\frac{1}{2}(\Large\frac{\partial v}{\partial x}-\Large\frac{\partial u}{\partial y}){k}$$

Here,   $$u$$ , $$v$$ , and $$w$$ are the $$x$$ , $$y$$ , and $$z$$ components of the velocity.  $$i$$ , $$j$$ , and $$k$$ are unit vectors that correspond to those directions.  Looking at this math may not mean much to you, but you can see that each of the components of the vorticity vector are due to velocity gradients ( derivatives of velocity with respect to location ).

The vorticity vector is a little difficult to show for Poiseuille flow; it points straight out of the screen at you for the figures shown above. In the figure above of the shear stress, vorticity has been represented by color ( warmer color, more vorticity ) so you can see that maximal vorticity occurs at the wall of the tube.  ( These figures were made with computational fluid dynamics viewing software that you can download from this site.  The software allows you to display vorticity vectors associated with various flow solutions.)  Vorticity usually originates where the fluid flows past a solid surface.  Obviously the tube wall is that solid surface in Poiseuille flow.  While vorticity is a kinematic feature of the flow ( having to do with the motion ), it behaves like a substance – almost as if the inner surface of the tube were exuding a dye that can be traced.  Vorticity flows along with the fluid and is transported just like everything else that is contained within the fluid.  It also diffuses through the fluid as if viscosity were the diffusion coefficient.