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Influid dynamics, theHagen–Poiseuille equation, also known as theHagen–Poiseuille law,Poiseuille law orPoiseuille equation, is aphysical law that gives thepressure drop in anincompressible andNewtonian fluid inlaminar flow flowing through a long cylindrical pipe of constant cross section. It can be successfully applied to air flow inlungalveoli, or the flow through a drinking straw or through ahypodermic needle. It was experimentally derived independently byJean Léonard Marie Poiseuille in 1838[1] andGotthilf Heinrich Ludwig Hagen,[2] and published by Hagen in 1839[1] and then by Poiseuille in 1840–41 and 1846.[1] The theoretical justification of the Poiseuille law was given byGeorge Stokes in 1845.[3]
The assumptions of the equation are that the fluid isincompressible andNewtonian; theflow is laminar through a pipe of constant circular cross-section that is substantially longer than its diameter; and there is noacceleration of fluid in the pipe. For velocities and pipe diameters above a threshold, actual fluid flow is not laminar butturbulent, leading to larger pressure drops than calculated by the Hagen–Poiseuille equation.
Poiseuille's equation describes the pressure dropdue to the viscosity of the fluid; other types of pressure drops may still occur in a fluid (see a demonstration here).[4] For example, the pressure needed to drive a viscous fluid up against gravity would contain both that as needed in Poiseuille's lawplus that as needed inBernoulli's equation, such that any point in the flow would have a pressure greater than zero (otherwise no flow would happen).
Another example is when blood flows into a narrowerconstriction, its speed will be greater than in a larger diameter (due tocontinuity ofvolumetric flow rate), and its pressure will be lower than in a larger diameter[4] (due to Bernoulli's equation). However, the viscosity of blood will causeadditional pressure drop along the direction of flow, which is proportional to length traveled[4] (as per Poiseuille's law). Both effects contribute to theactual pressure drop.
In standard fluid-kinetics notation:[5][6][7]
where
The equation does not hold close to the pipe entrance.[8]: 3
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 theDarcy–Weisbach equation. The ratio of length to radius of a pipe should be greater than 1/48 of theReynolds number for the Hagen–Poiseuille law to be valid.[9] If the pipe is too short, the Hagen–Poiseuille equation may result in unphysically high flow rates; the flow is bounded byBernoulli's principle, under less restrictive conditions, by
because it is impossible to have negative (absolute) pressure (not to be confused withgauge pressure) in an incompressible flow.
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 theDarcy–Weisbach equation in the field ofhydraulics, given a relationship for the friction factor in terms of the Reynolds number. In the case of laminar flow, for a circular cross section:
whereRe is theReynolds number,ρ is the fluid density, andv 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 theDarcy friction factor, theenergy (head) loss factor,friction loss factor orDarcy (friction) factorΛ 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 Poiseuille's law.
The law is also very important inhemorheology andhemodynamics, both fields ofphysiology.[10]
Poiseuille's law was later in 1891 extended toturbulent flow by L. R. Wilberforce, based on Hagenbach's work.
The Hagen–Poiseuille equation can be derived from theNavier–Stokes equations. Thelaminar flow through 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 coordinates(r,θ,x) by making the following set of assumptions:
Then the angular equation in the momentum equations and thecontinuity equation are identically satisfied. The radial momentum equation reduces to∂p/∂r = 0, i.e., thepressurep is a function of the axial coordinatex only. For brevity, useu instead of. The axial momentum equation reduces to
whereμ is the dynamic viscosity of the fluid. In the above equation, the left-hand side is only a function ofr and the right-hand side term is only a function ofx, implying that both terms must be the same constant. Evaluating this constant is straightforward. If we take the length of the pipe to beL and denote the pressure difference between the two ends of the pipe byΔp (high pressure minus low pressure), then the constant is simply
defined such thatG is positive. The solution is
Sinceu needs to be finite atr = 0,c1 = 0. The no slipboundary condition at the pipe wall requires thatu = 0 atr =R (radius of the pipe), which yieldsc2 =GR2/4μ. Thus we have finally the followingparabolicvelocity profile:
The maximum velocity occurs at the pipe centerline (r = 0),umax =GR2/4μ. The average velocity can be obtained by integrating over the pipecross section,
The easily measurable quantity in experiments is the volumetric flow rateQ = πR2uavg. Rearrangement of this gives the Hagen–Poiseuille equation
Although more lengthy than directly using theNavier–Stokes equations, an alternative method of deriving the Hagen–Poiseuille equation is as follows.
To figure out the motion of the liquid, all forces acting on each lamina must be known:
When two layers of liquid in contact with each other move at different speeds, there will be ashear force between them. This force isproportional to thearea of contactA, the velocity gradient perpendicular to the direction of flowΔvx/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.
Assume that we are figuring out the force on the lamina withradiusr. From the equation above, we need to know thearea of 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 surface area 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 radius at 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 subscriptr following thederivative indicates that it should be taken at a radius ofr.
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 + dr instead 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).
To find the solution for the flow of a laminar layer through a tube, we need to make one last assumption. There is noacceleration of 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 expansion of 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.
When aconstant pressure gradientG = −dp/dx is 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 equations reduce to
with initial and boundary conditions,
The velocity distribution is given by
whereJ0(λnr/R) is theBessel function of the first kind of order zero andλn are the positive roots of this function andJ1(λn) is theBessel function of the first kind of order one. Ast → ∞, Poiseuille solution is recovered.[11]
IfR1 is the inner cylinder radii andR2 is the outer cylinder radii, withconstant applied pressure gradient between the two endsG = −dp/dx, the velocity distribution and the volume flux through the annular pipe are
WhenR2 = R andR1 tends to zero, the original problem is recovered.[12]
Flow through pipes with an oscillating pressure gradient finds applications in blood flow through large arteries.[13][14][15][16] The imposed pressure gradient is given by
whereG,α andβ are constants andω is the frequency. The velocity field is given by
where
whereber andbei are theKelvin functions andk2 =ρω/μ.
Plane Poiseuille flow is flow created between two infinitely long parallel plates, separated by a distanceh with a constant pressure gradientG = −dp/dx is applied in the direction of flow. The flow is essentially unidirectional because of infinite length. TheNavier–Stokes equations reduce to
withno-slip condition on both walls
Therefore, the velocity distribution and the volume flow rate per unit length are
Joseph Boussinesq 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.[17]Joseph Proudman derived the same for isosceles triangles in 1914.[18] LetG = −dp/dx be the constant pressure gradient acting in direction parallel to the motion.
The velocity and the volume flow rate in a rectangular channel of height0 ≤y ≤h and width0 ≤z ≤l are
The velocity and the volume flow rate of tube with equilateral triangular cross-section of side length2h/√3 are
The velocity and the volume flow rate in the right-angled isosceles triangley = π,y ±z = 0 are
The velocity distribution for tubes of elliptical cross-section with semiaxesa andb is[11]
Here, whena =b, Poiseuille flow for circular pipe is recovered and whena → ∞,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 [tr;de].[19][20]
The flow through arbitrary cross-sectionu(y,z) satisfies the condition thatu = 0 on the walls. The governing equation reduces to[21]
If we introduce a new dependent variable as
then it is easy to see that the problem reduces to that integrating aLaplace equation
satisfying the condition
on the wall.
For a compressible fluid in a tube thevolumetric flow rateQ(x) and the axial velocity are not constant along the tube; but the mass flow rate is constant along the tube length. The volumetric flow rate is usually expressed at the outlet pressure. As fluid is compressed or expanded, 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 gas in theisothermal case, where the temperature of the fluid is permitted to equilibrate with its surroundings, an approximate relation for the pressure drop can be derived.[22] Using ideal gas equation of state for constant temperature process (i.e., is constant) and the conservation of mass flow rate (i.e., is constant), the relationQp =Q1p1 =Q2p2 can 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 compressibility effects. Though locally we ignored the effects of pressure variation due to density variation, over long distances these effects are taken into account. Sinceμ is independent of pressure, the above equation can be integrated over the lengthL to 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 factorp1 +p2/2p2 expressing the average pressure relative to the outlet pressure.
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Electricity was originally understood to be a kind of fluid. Thishydraulic analogy is 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 law for electrical circuits,V =IR. Since the net force acting on the fluid is equal toΔF =SΔp, whereS = πr2, i.e.ΔF = πr2 ΔP, then from Poiseuille's law, it follows that
For electrical circuits, letn be the concentration of free charged particles (in m−3) and letq* be the charge of each particle (incoulombs). (For electrons,q* =e =1.6×10−19 C.) ThennQ is the number of particles in the volumeQ, andnQq* is their total charge. This is the charge that flows through the cross section per unit time, i.e. thecurrentI. Therefore,I =nQq*. Consequently,Q =I/nq*, and
ButΔF =Eq, whereq is 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 isq =nπr2Lq*. Since thevoltageV =EL, it follows then
This is exactly Ohm's law, where theresistanceR =V/I is described by the formula
It follows that the resistanceR is proportional to the lengthL of the resistor, which is true. However, it also follows that the resistanceR is inversely proportional to the fourth power of the radiusr, i.e. the resistanceR is inversely proportional to the second power of the cross section areaS = πr2 of the resistor, which is different from the electrical formula. The electrical relation for the resistance is
whereρ is the resistivity; i.e. the resistanceR is inversely proportional to the cross section areaS of the resistor.[23] The reason why Poiseuille's law leads to a different formula for the resistanceR is the difference between the fluid flow and the electric current.Electron gas isinviscid, 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 thehydraulic analogy are useful only within certain limits when applied to electricity. Both Ohm's law and Poiseuille's law illustratetransport phenomena.
The Hagen–Poiseuille equation is useful in determining thevascular resistance and hence flow rate ofintravenous (IV) fluids that may be achieved using various sizes of peripheral and centralcannulas. 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, 26G. As an example, assuming cannula lengths are equal, the flow of a 14G cannula is 1.73 times that of a 16G cannula, and 4.16 times that of a 20G cannula. 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, an increased change in pressure (∆p) — such as by pressurizing the bag of fluid, squeezing the bag, or hanging the bag higher (relative to the level of the cannula) — can be used to speed up flow rate. It is also useful to understand that viscous fluids will flow slower (e.g. inblood transfusion).Delivery of fluids such asantibiotics oranalgesics by means of anelastomeric pump, such as antibiotics, can also be understood in terms of a Poiseuille-flow model.[citation needed]