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# Hagen–Poiseuille equation

In nonideal fluid dynamics, the Hagen–Poiseuille equation, also known as the Hagen–Poiseuille law, Poiseuille law or Poiseuille equation, is a physical law that gives the pressure drop in an incompressible and Newtonian fluid in laminar flow flowing through a long cylindrical pipe of constant cross section. It can be successfully applied to air flow in lung alveoli, or the flow through a drinking straw or through a hypodermic needle. It was experimentally derived independently by Jean Léonard Marie Poiseuille in 1838[1] and Gotthilf Heinrich Ludwig Hagen,[2] and published by Poiseuille in 1840–41 and 1846.[1] The theoretical justification of the Poiseuille law was given by George Stokes in 1845.[3]

The assumptions of the equation are that the fluid is incompressible and Newtonian; the flow is laminar through a pipe of constant circular cross-section that is substantially longer than its diameter; and there is no acceleration of fluid in the pipe. For velocities and pipe diameters above a threshold, actual fluid flow is not laminar but turbulent, leading to larger pressure drops than calculated by the Hagen–Poiseuille equation.

## Equation

In standard fluid-kinetics notation:[4][5][6]

where:

Δpis the pressure difference between the two ends,Lis the length of pipe,μis thedynamic viscosity,Qis thevolumetric flow rate,Ris the piperadius.

The equation does not hold close to the pipe entrance.[7] []

The equation fails in the limit of low viscosity, wide and/or short pipe. Low viscosity or a wide pipe may result in turbulent flow, making it necessary to use more complex models, such as Darcy–Weisbach equation. If the pipe is too short, the Hagen–Poiseuille equation may result in unphysically high flow rates; the flow is bounded by Bernoulli's principle, under less restrictive conditions, by

## Relation to Darcy–Weisbach

Normally, Hagen–Poiseuille flow implies not just the relation for the pressure drop, above, but also the full solution for the laminar flow profile, which is parabolic. However, the result for the pressure drop can be extended to turbulent flow by inferring an effective turbulent viscosity in the case of turbulent flow, even though the flow profile in turbulent flow is strictly speaking not actually parabolic. In both cases, laminar or turbulent, the pressure drop is related to the stress at the wall, which determines the so-called friction factor. The wall stress can be determined phenomenologically by the Darcy–Weisbach equation in the field of hydraulics, given a relationship for the friction factor in terms of the Reynolds number. In the case of laminar flow, for a circular cross section:

where Re is the Reynolds number, ρ is the fluid density, and v is the mean flow velocity, which is half the maximal flow velocity in the case of laminar flow. It proves more useful to define the Reynolds number in terms of the mean flow velocity because this quantity remains well defined even in the case of turbulent flow, whereas the maximal flow velocity may not be, or in any case, it may be difficult to infer. In this form the law approximates the Darcy friction factor, the energy (head) loss factor, friction loss factor or Darcy (friction) factor Λ in the laminar flow at very low velocities in cylindrical tube. The theoretical derivation of a slightly different form of the law was made independently by Wiedman in 1856 and Neumann and E. Hagenbach in 1858 (1859, 1860). Hagenbach was the first who called this law the Poiseuille's law.

The law is also very important in hemorheology and hemodynamics, both fields of physiology.[8]

Poiseuille's law was later in 1891 extended to turbulent flow by L. R. Wilberforce, based on Hagenbach's work.

## Derivation

The Hagen–Poiseuille equation can be derived from theNavier–Stokes equations. Thelaminar flowthrough a pipe of uniform (circular) cross-section is known as Hagen–Poiseuille flow. The equations governing the Hagen–Poiseuille flow can be derived directly from theNavier–Stokes momentum equations in 3D cylindrical coordinatesby making the following set of assumptions:
1. The flow is steady ( ).

2. The radial and azimuthal components of the fluid velocity are zero ( ).

3. The flow is axisymmetric ( ).

4. The flow is fully developed ( ). Here However, this can be proved via mass conservation, and the above assumptions.

Then the angular equation in the momentum equations and thecontinuity equationare identically satisfied. The radial momentum equation reduces to, i.e., thepressureis a function of the axial coordinateonly. For brevity, useinstead of. The axial momentum equation reduces to
whereis the dynamic viscosity of the fluid. In the above equation, the left-hand side is only a function ofand the right-hand side term is only a function of, implying that both terms must be the same constant. Evaluating this constant is straightforward. If we take the length of the pipe to beand denote the pressure difference between the two ends of the pipe by(high pressure minus low pressure), then the constant is simplydefined such thatis positive. The solution is
Sinceneeds to be finite at,. The no slipboundary conditionat the pipe wall requires thatat(radius of the pipe), which yieldsThus we have finally the followingparabolicvelocityprofile:
The maximum velocity occurs at the pipe centerline (),. The average velocity can be obtained by integrating over the pipecross section,
The easily measurable quantity in experiments is the volumetric flow rate. Rearrangement of this gives the Hagen–Poiseuille equation
Elaborate derivation starting directly from first principles
Although more lengthy than directly using theNavier–Stokes equations, an alternative method of deriving the Hagen–Poiseuille equation is as follows.

#### Liquid flow through a pipe

Assume the liquid exhibitslaminar flow. Laminar flow in a round pipe prescribes that there are a bunch of circular layers (lamina) of liquid, each having a velocity determined only by their radial distance from the center of the tube. Also assume the center is moving fastest while the liquid touching the walls of the tube is stationary (due to theno-slip condition). To figure out the motion of the liquid, all forces acting on each lamina must be known:
1. Thepressureforce pushing the liquid through the tube is the change in pressure multiplied by the area:F = −A Δp. This force is in the direction of the motion of the liquid. The negative sign comes from the conventional way we defineΔp = pendptop< 0.
2. Viscosityeffects will pull from the faster lamina immediately closer to the center of the tube.
3. Viscosityeffects will drag from the slower lamina immediately closer to the walls of the tube.

#### Viscosity

Two fluids moving past each other in thexdirection. The liquid on top is moving faster and will be pulled in the negative direction by the bottom liquid while the bottom liquid will be pulled in the positive direction by the top liquid.
When two layers of liquid in contact with each other move at different speeds, there will be ashear forcebetween them. This force isproportionalto theareaof contactA, the velocity gradient perpendicular to the direction of flowΔ*vx
/Δy
, and a proportionality constant (viscosity) and is given by
The negative sign is in there because we are concerned with the faster moving liquid (top in figure), which is being slowed by the slower liquid (bottom in figure). ByNewton's third law of motion, the force on the slower liquid is equal and opposite (no negative sign) to the force on the faster liquid. This equation assumes that the area of contact is so large that we can ignore any effects from the edges and that the fluids behave asNewtonian fluids.

#### Faster lamina

Assume that we are figuring out the force on the lamina withradiusr. From the equation above, we need to know theareaof contact and the velocitygradient. Think of the lamina as a ring of radiusr, thicknessdr, and lengthΔx. The area of contact between the lamina and the faster one is simply the area of the inside of the cylinder:A = 2πr Δx. We don't know the exact form for the velocity of the liquid within the tube yet, but we do know (from our assumption above) that it is dependent on the radius. Therefore, the velocity gradient is thechange of the velocity with respect to the change in the radiusat the intersection of these two laminae. That intersection is at a radius ofr. So, considering that this force will be positive with respect to the movement of the liquid (but the derivative of the velocity is negative), the final form of the equation becomes
where the vertical bar and subscriptrfollowing thederivativeindicates that it should be taken at a radius ofr.

#### Slower lamina

Next let's find the force of drag from the slower lamina. We need to calculate the same values that we did for the force from the faster lamina. In this case, the area of contact is atr + drinstead ofr. Also, we need to remember that this force opposes the direction of movement of the liquid and will therefore be negative (and that the derivative of the velocity is negative).

#### Putting it all together

To find the solution for the flow of a laminar layer through a tube, we need to make one last assumption. There is noaccelerationof liquid in the pipe, and byNewton's first law, there is no net force. If there is no net force then we can add all of the forces together to get zero
or
First, to get everything happening at the same point, use the first two terms of aTaylor series expansionof the velocity gradient:
The expression is valid for all laminae. Grouping like terms and dropping the vertical bar since all derivatives are assumed to be at radiusr,
Finally, put this expression in the form of adifferential equation, dropping the term quadratic indr.
The above equation is the same as the one obtained from the Navier-Stokes equations and the derivation from here on follows as before.

### Startup of Poiseuille flow in a pipe[9]

When a constant pressure gradientis applied between two ends of a long pipe, the flow will not immediately obtain Poiseuille profile, rather it develops through time and reaches the Poiseuille profile at steady state. TheNavier-Stokes equationsreduce to

with initial and boundary conditions,

The velocity distribution is given by

whereis theBessel function of the first kindof order zero andare the positive roots of this function andis theBessel function of the first kindof order one. As, Poiseuille solution is recovered.

## Poiseuille flow in annular section[10]

Poiseuille flow in annular section

Ifis the inner cylinder radii andis the outer cylinder radii, with applied pressure gradient between the two ends, the velocity distribution and the volume flux through the annular pipe are
When, the original problem is recovered.

## Poiseuille flow in a pipe with oscillating pressure gradient

Flow through pipes with oscillating pressure gradient finds applications in blood flow through large arteries[11][12][13][14]. The imposed pressure gradient is given by

where,andare constants andis the frequency. The velocity field is given by

where

whereandare theKelvin functionsand.

## Plane Poiseuille flow

Plane Poiseuille flow

Plane Poiseuille flow is flow created between two infinitely long parallel plates, separated by a distancewith a constant pressure gradientis applied in the direction of flow. The flow is essentially unidirectional because of infinite length. TheNavier-Stokes equationsreduce to

with no-slip condition on both walls

Therefore, the velocity distribution and the volume flow rate per unit length are

## Poiseuille flow through some non-circular cross-sections[15]

Joseph Boussinesq[16] derived the velocity profile and volume flow rate in 1868 for rectangular channel and tubes of equilateral triangular cross-section and for elliptical cross-section.Joseph Proudman[17] derived the same for isosceles triangles in 1914. Letbe the constant pressure gradient acting in direction parallel to the motion.
The velocity and the volume flow rate in a rectangular channel of heightand widthare
The velocity and the volume flow rate of tube with equilateral triangular cross-section of side lengthare
The velocity and the volume flow rate in the right-angled isosceles triangleare
The velocity distribution for tubes of elliptical cross-section with semi-axisandis[9]
Here, when, Poiseuille flow for circular pipe is recovered and when, plane Poiseuille flow is recovered. More explicit solutions with cross-sections such as snail-shaped sections, sections having the shape of a notch circle following a semicircle, annular sections between homofocal ellipses, annular sections between non-concentric circles are also available, as reviewed byRatip Berker[18].

## Poiseuille flow through arbitrary cross-section

The flow through arbitrary cross-sectionsatisfies the condition thaton the walls. The governing equation reduces to[19]

If we introduce a new dependent variable as

then it is easy to see that the problem reduces to that integrating a Laplace equation

satisfying the condition

on the wall.

## Poiseuille's equation for compressible fluids

For a compressible fluid in a tube thevolumetric flow rate(but not the mass flow rate) and the axial velocity are not constant along the tube. The flow is usually expressed at outlet pressure. As fluid is compressed or expands, work is done and the fluid is heated or cooled. This means that the flow rate depends on the heat transfer to and from the fluid. For anideal gasin theisothermalcase, where the temperature of the fluid is permitted to equilibrate with its surroundings, an approximate relation for the pressure drop can be derived[20]. Using ideal gas equation of state for constant temperature process, the relationcan be obtained. Over a short section of the pipe, the gas flowing through the pipe can be assumed to be incompressible so that Poiseuille law can be used locally,
Here we assumed the local pressure gradient is not too great to have any compressiblity effects. Though locally we ignored the effects of pressure variation due to density variation, over long distances these effects are taken into account. Sinceis independent of pressure, the above equation can be integrated over the lengthto give

Hence the volumetric flow rate at the pipe outlet is given by

This equation can be seen as Poiseuille's law with an extra correction factor p1 + p2/2p2 expressing the average pressure relative to the outlet pressure.

## Electrical circuits analogy

Electricity was originally understood to be a kind of fluid. Thishydraulic analogyis still conceptually useful for understanding circuits. This analogy is also used to study the frequency response of fluid-mechanical networks using circuit tools, in which case the fluid network is termed ahydraulic circuit. Poiseuille's law corresponds toOhm's lawfor electrical circuits,V = IR. Since the net force acting on the fluid is equal to, whereS = πr2, i.e.ΔF = πr2ΔP, then from Poiseuille's law, it follows that
.

For electrical circuits, let n be the concentration of free charged particles (in m−3) and let q* be the charge of each particle (in coulombs). (For electrons, q* = e = 1.6×10−19 C.) Then nQ is the number of particles in the volume Q, and nQq* is their total charge. This is the charge that flows through the cross section per unit time, i.e. the current I. Therefore, I = nQq*. Consequently, Q = I/nq*, and

ButΔF = Eq, whereqis the total charge in the volume of the tube. The volume of the tube is equal toπr2L, so the number of charged particles in this volume is equal tonπr2L, and their total charge isSince thevoltageV = EL, it follows then

This is exactly Ohm's law, where the resistance R = V/I is described by the formula

.

It follows that the resistance R is proportional to the length L of the resistor, which is true. However, it also follows that the resistance R is inversely proportional to the fourth power of the radius r, i.e. the resistance R is inversely proportional to the second power of the cross section area S = πr2 of the resistor, which is wrong according to the electrical analogy. The correct relation is

where ρ is the specific resistance; i.e. the resistance R is inversely proportional to the cross section area S of the resistor.[21] The reason why Poiseuille's law leads to a wrong formula for the resistance R is the difference between the fluid flow and the electric current. Electron gas is inviscid, so its velocity does not depend on the distance to the walls of the conductor. The resistance is due to the interaction between the flowing electrons and the atoms of the conductor. Therefore, Poiseuille's law and the hydraulic analogy are useful only within certain limits when applied to electricity. Both Ohm's law and Poiseuille's law illustrate transport phenomena.

## Medical applications – intravenous access and fluid delivery

The Hagen–Poiseuille equation is useful in determining the flow rate of intravenous fluids that may be achieved using various sizes of peripheral and central cannulas. The equation states that flow rate is proportional to the radius to the fourth power, meaning that a small increase in the internal diameter of the cannula yields a significant increase in flow rate of IV fluids. The radius of IV cannulas is typically measured in "gauge", which is inversely proportional to the radius. Peripheral IV cannulas are typically available as (from large to small) 14G, 16G, 18G, 20G, 22G. As an example, the flow of a 14G cannula is typically twice that of a 16G, and ten times that of a 20G. It also states that flow is inversely proportional to length, meaning that longer lines have lower flow rates. This is important to remember as in an emergency, many clinicians favor shorter, larger catheters compared to longer, narrower catheters. While of less clinical importance, the change in pressure can be used to speed up flow rate by pressurizing the bag of fluid, squeezing the bag, or hanging the bag higher from the level of the cannula. It is also useful to understand that viscous fluids will flow slower (e.g. in blood transfusion).

• Couette flow

• Darcy's law

• Pulse

• Wave

• Hydraulic circuit

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