Hagen–Poiseuille equation

Relation to the Darcy–Weisbach equation

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:

${\displaystyle \mathrm{\Lambda }=\frac{64}{\mathrm{R}\mathrm{e}},\mathrm{R}\mathrm{e}=\frac{\rho vd}{\mu },}$

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 Poiseuille's law.

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

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 the Navier–Stokes equations. The laminar 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 the Navier–Stokes momentum equations in 3D cylindrical coordinates (r,θ,x) by making the following set of assumptions:

1. The flow is steady ( ⁠∂.../∂t⁠ = 0 ).
2. The radial and azimuthal components of the fluid velocity are zero ( u_r = u_θ = 0 ).
3. The flow is axisymmetric ( ⁠∂.../∂θ⁠ = 0 ).
4. The flow is fully developed ( ⁠∂u_x/∂x⁠ = 0 ). Here however, this can be proved via mass conservation, and the above assumptions.

Then the angular equation in the momentum equations and the continuity equation are identically satisfied. The radial momentum equation reduces to ⁠∂p/∂r⁠ = 0, i.e., the pressure p is a function of the axial coordinate x only. For brevity, use u instead of ${\displaystyle u_x}$. The axial momentum equation reduces to

${\displaystyle \frac{1}{r}\frac{\mathrm{\partial }}{\mathrm{\partial }r}\left(r\frac{\mathrm{\partial }u}{\mathrm{\partial }r}\right)=\frac{1}{\mu }\frac{\mathrm{d}p}{\mathrm{d}x}}$

where μ is the dynamic viscosity of the fluid. In the above equation, the left-hand side is only a function of r and the right-hand side term is only a function of x, implying that both terms must be the same constant. Evaluating this constant is straightforward. If we take the length of the pipe to be L and denote the pressure difference between the two ends of the pipe by Δp (high pressure minus low pressure), then the constant is simply

${\displaystyle -\frac{\mathrm{d}p}{\mathrm{d}x}=\frac{\mathrm{\Delta }p}{L}=G}$

defined such that G is positive. The solution is

${\displaystyle u=-\frac{G{r}^2}{4\mu }+c_1\ln r+c_2}$

Since u needs to be finite at r = 0, c₁ = 0. The no slip boundary condition at the pipe wall requires that u = 0 at r = R (radius of the pipe), which yields c₂ = ⁠GR²/4μ⁠. Thus we have finally the following parabolic velocity profile:

${\displaystyle u=\frac{G}{4\mu }\left(R^2-r^2\right).}$

The maximum velocity occurs at the pipe centerline (r = 0), u_max = ⁠GR²/4μ⁠. The average velocity can be obtained by integrating over the pipe cross section,

${\displaystyle u_{\mathrm{a}\mathrm{v}\mathrm{g}}=\frac{1}{\pi R^2}{\int }_0^{R}2\pi ru\mathrm{d}r=\frac{1}{2}{u}_{\mathrm{m}\mathrm{a}\mathrm{x}}.}$

The easily measurable quantity in experiments is the volumetric flow rate Q = πR² u_avg. Rearrangement of this gives the Hagen–Poiseuille equation

${\displaystyle \mathrm{\Delta }p=\frac{8\mu QL}{\pi R^4}.}$

Startup of Poiseuille flow in a pipe

When a constant pressure gradient G = −⁠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. The Navier–Stokes equations reduce to

${\displaystyle \frac{\mathrm{\partial }u}{\mathrm{\partial }t}=\frac{G}{\rho }+\nu \left(\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{r}^2}+\frac{1}{r}\frac{\mathrm{\partial }u}{\mathrm{\partial }r}\right)}$

with initial and boundary conditions,

${\displaystyle u(r,0)=0,u(R,t)=0.}$

The velocity distribution is given by

${\displaystyle u(r,t)=\frac{G}{4\mu }\left(R^2-r^2\right)-\frac{2G{R}^2}{\mu }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\lambda }_n^3}\frac{J_0({\lambda }_{n}r/R)}{J_1({\lambda }_n)}{e}^{-{\lambda }_n^2\nu t/R^2},J_0\left({\lambda }_n\right)=0}$

where J₀(⁠λ_nr/R⁠) is the Bessel function of the first kind of order zero and λ_n are the positive roots of this function and J₁(λ_n) is the Bessel function of the first kind of order one. As t → ∞, Poiseuille solution is recovered.

Poiseuille flow in an annular section

If R₁ is the inner cylinder radii and R₂ is the outer cylinder radii, with constant applied pressure gradient between the two ends G = −⁠dp/dx⁠, the velocity distribution and the volume flux through the annular pipe are

${\displaystyle \begin{array}{cc}u(r)& =\frac{G}{4\mu }\left(R_1^2-r^2\right)+\frac{G}{4\mu }\left(R_2^2-R_1^2\right)\frac{\ln (r/R_1)}{\ln (R_2/R_1)},\\ Q& =\frac{G\pi }{8\mu }\left[R_2^4-R_1^4-\frac{{\left(R_2^2-R_1^2\right)}^2}{\ln (R_2/R_1)}\right].\end{array}}$

When R₂ = R and R₁ tends to zero, the original problem is recovered.

Poiseuille flow in a pipe with an oscillating pressure gradient

Flow through pipes with an oscillating pressure gradient finds applications in blood flow through large arteries. The imposed pressure gradient is given by

${\displaystyle \frac{\mathrm{\partial }p}{\mathrm{\partial }x}=-G-\alpha \cos \omega t-\beta \sin \omega t}$

where G, α and β are constants and ω is the frequency. The velocity field is given by

${\displaystyle u(r,t)=\frac{G}{4\mu }\left(R^2-r^2\right)+[\alpha F_2+\beta (F_1-1)]\frac{\cos \omega t}{\rho \omega }+[\beta F_2-\alpha (F_1-1)]\frac{\sin \omega t}{\rho \omega }}$

where

${\displaystyle \begin{array}{cc}{F}_1(kr)& =\frac{\mathrm{b}\mathrm{e}\mathrm{r}(kr)\mathrm{b}\mathrm{e}\mathrm{r}(kR)+\mathrm{b}\mathrm{e}\mathrm{i}(kr)\mathrm{b}\mathrm{e}\mathrm{i}(kR)}{{\mathrm{b}\mathrm{e}\mathrm{r}}^2(kR)+{\mathrm{b}\mathrm{e}\mathrm{i}}^2(kR)},\\ F_2(kr)& =\frac{\mathrm{b}\mathrm{e}\mathrm{r}(kr)\mathrm{b}\mathrm{e}\mathrm{i}(kR)-\mathrm{b}\mathrm{e}\mathrm{i}(kr)\mathrm{b}\mathrm{e}\mathrm{r}(kR)}{{\mathrm{b}\mathrm{e}\mathrm{r}}^2(kR)+{\mathrm{b}\mathrm{e}\mathrm{i}}^2(kR)},\end{array}}$

where ber and bei are the Kelvin functions and k² = ⁠ρω/μ⁠.

Plane Poiseuille flow

Plane Poiseuille flow is flow created between two infinitely long parallel plates, separated by a distance h with a constant pressure gradient G = −⁠dp/dx⁠ is applied in the direction of flow. The flow is essentially unidirectional because of infinite length. The Navier–Stokes equations reduce to

${\displaystyle \frac{{\mathrm{d}}^{2}u}{\mathrm{d}{y}^2}=-\frac{G}{\mu }}$

with no-slip condition on both walls

${\displaystyle u(0)=0,u(h)=0}$

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

${\displaystyle u(y)=\frac{G}{2\mu }y(h-y),Q=\frac{G{h}^3}{12\mu }.}$

Poiseuille flow through some non-circular cross-sections

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. Joseph Proudman derived the same for isosceles triangles in 1914. Let G = −⁠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 height 0 ≤ y ≤ h and width 0 ≤ z ≤ l are

${\displaystyle \begin{array}{cc}u(y,z)& =\frac{G}{2\mu }y(h-y)-\frac{4G{h}^2}{\mu {\pi }^3}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n-1{)}^3}\frac{\sinh ({\beta }_{n}z)+\sinh [{\beta }_n(l-z)]}{\sinh ({\beta }_{n}l)}\sin ({\beta }_{n}y),{\beta }_n=\frac{(2n-1)\pi }{h},\\ Q& =\frac{G{h}^{3}l}{12\mu }-\frac{16G{h}^4}{{\pi }^5\mu }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n-1{)}^5}\frac{\cosh ({\beta }_{n}l)-1}{\sinh ({\beta }_{n}l)}.\end{array}}$

The velocity and the volume flow rate of tube with equilateral triangular cross-section of side length ⁠2h/√3⁠ are

${\displaystyle \begin{array}{cc}u(y,z)& =-\frac{G}{4\mu h}(y-h)\left(y^2-3z^2\right),\\ Q& =\frac{G{h}^4}{60\sqrt{3}\mu }.\end{array}}$

The velocity and the volume flow rate in the right-angled isosceles triangle y = π, y ± z = 0 are

${\displaystyle \begin{array}{cc}u(y,z)& =\frac{G}{2\mu }(y+z)(\pi -y)-\frac{G}{\pi \mu }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\beta }_n^3\sinh (2\pi {\beta }_n)}\left\{\sinh [{\beta }_n(2\pi -y+z)]\sin [{\beta }_n(y+z)]-\sinh [{\beta }_n(y+z)]\sin [{\beta }_n(y-z)]\right\},{\beta }_n=n+\frac{1}{2},\\ Q& =\frac{G{\pi }^4}{12\mu }-\frac{G}{2\pi \mu }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\beta }_n^5}\left[\coth (2\pi {\beta }_n)+\csc (2\pi {\beta }_n)\right].\end{array}}$

The velocity distribution for tubes of elliptical cross-section with semiaxes a and b is

${\displaystyle \begin{array}{cc}u(y,z)& =\frac{G}{2\mu \left(\frac{1}{a^2}+\frac{1}{b^2}\right)}\left(1-\frac{y^2}{a^2}-\frac{z^2}{b^2}\right),\\ Q& =\frac{\pi G{a}^3{b}^3}{4\mu \left(a^2+b^2\right)}.\end{array}}$

Here, when a = b, Poiseuille flow for circular pipe is recovered and when a → ∞, 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 by Ratip Berker.

Poiseuille flow through arbitrary cross-section

The flow through arbitrary cross-section u(y,z) satisfies the condition that u = 0 on the walls. The governing equation reduces to

${\displaystyle \frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{y}^2}+\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{z}^2}=-\frac{G}{\mu }.}$

If we introduce a new dependent variable as

${\displaystyle U=u+\frac{G}{4\mu }\left(y^2+z^2\right),}$

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

${\displaystyle \frac{{\mathrm{\partial }}^{2}U}{\mathrm{\partial }{y}^2}+\frac{{\mathrm{\partial }}^{2}U}{\mathrm{\partial }{z}^2}=0}$

satisfying the condition

${\displaystyle U=\frac{G}{4\mu }\left(y^2+z^2\right)}$

on the wall.

Poiseuille's equation for an ideal isothermal gas

For a compressible fluid in a tube the volumetric flow rate Q(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 an ideal gas in the isothermal case, where the temperature of the fluid is permitted to equilibrate with its surroundings, an approximate relation for the pressure drop can be derived. Using ideal gas equation of state for constant temperature process (i.e., ${\displaystyle p/\rho }$ is constant) and the conservation of mass flow rate (i.e., ${\displaystyle \dot{m}=\rho Q}$ is constant), the relation Qp = 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,

${\displaystyle -\frac{\mathrm{d}p}{\mathrm{d}x}=\frac{8\mu Q}{\pi R^4}=\frac{8\mu Q_2{p}_2}{\pi p{R}^4}\Rightarrow -p\frac{\mathrm{d}p}{\mathrm{d}x}=\frac{8\mu Q_2{p}_2}{\pi R^4}.}$

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. Since μ is independent of pressure, the above equation can be integrated over the length L to give

${\displaystyle p_1^2-p_2^2=\frac{16\mu L{Q}_2{p}_2}{\pi R^4}.}$

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

${\displaystyle Q_2=\frac{\pi R^4}{16\mu L}\left(\frac{p_1^2-p_2^2}{p_2}\right)=\frac{\pi R^4\left(p_1-p_2\right)}{8\mu L}\frac{\left(p_1+p_2\right)}{2p_2}.}$

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. This hydraulic 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 a hydraulic circuit. Poiseuille's law corresponds to Ohm's law for electrical circuits, V = IR. Since the net force acting on the fluid is equal to ΔF = SΔp, where S = πr², i.e. ΔF = πr² ΔP, then from Poiseuille's law, it follows that

${\displaystyle \mathrm{\Delta }F=\frac{8\mu LQ}{r^2}}$.

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

${\displaystyle \mathrm{\Delta }F=\frac{8\mu LI}{n{r}^2{q}^{\ast }}.}$

But ΔF = Eq, where q is the total charge in the volume of the tube. The volume of the tube is equal to πr²L, so the number of charged particles in this volume is equal to nπr²L, and their total charge is q = nπr² Lq*. Since the voltage V = EL, it follows then

${\displaystyle V=\frac{8\mu LI}{n^2\pi r^4{\left(q^{\ast }\right)}^2}.}$

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

${\displaystyle R=\frac{8\mu L}{n^2\pi r^4{\left(q^{\ast }\right)}^2}}$.

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 different from the electrical formula. The electrical relation for the resistance is

${\displaystyle R=\frac{\rho L}{S},}$

where ρ is the resistivity; i.e. the resistance R is inversely proportional to the cross section area S of the resistor. The reason why Poiseuille's law leads to a different 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 vascular resistance and hence flow rate of intravenous (IV) 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, 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. in blood transfusion). Delivery of fluids such as antibiotics or analgesics by means of an elastomeric pump can also be understood in terms of a Poiseuille-flow model.

Cited references