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, for the flow through a drinking straw or through a hypodermic needle. It was experimentally derived independently by Gotthilf Heinrich Ludwig Hagen in 1839 and Jean Léonard Marie Poiseuille in 1838, and published by Poiseuille in 1840 and 1846.

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

Standard fluid dynamics notation

In standard fluid dynamics notation:[1][2]

$\Delta P = \frac{8 \mu L Q}{ \pi r^4}$

or

$\Delta P = \frac{128 \mu L Q}{ \pi d^4}$

where:

$\Delta P$ is the pressure loss
$L$ is the length of pipe
$\mu$ is the dynamic viscosity
$Q$ is the volumetric flow rate
$r$ is the radius
$d$ is the diameter
$\pi$ is the mathematical constant Pi

Physics notation

$\Phi = \frac{dV}{dt} = v \pi R^{2} = \frac{\pi R^{4}}{8 \eta} \left( \frac{- \Delta P}{\Delta x}\right) = \frac{\pi R^{4}}{8 \eta} \frac{ |\Delta P|}{L}$

where in compatible units (e.g., S.I.):

$\Phi$ is the volumetric flow rate (denoted as $Q$ above)
$V(t)$ is the volume of the liquid transferred as a function of time, $t$
$v$ is mean fluid velocity along the length of the tube
$x$ is distance in direction of flow
$R$ is the internal radius of the tube
$\Delta P$ is the pressure difference between the two ends
$\eta$ is the dynamic fluid viscosity (S.I. unit: pascal-second (Pa·s)),
$L$ is the length of the tube

The equation does not hold close to the pipe entrance.[3]: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 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

$\Phi_{max} = \pi R^2 \sqrt{2 \Delta P / \rho}$.

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 phenomenological 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:

$\Lambda = {64\over {\it \mathrm{Re}}} \; , \quad\quad \mathrm{Re} = {2\rho v r\over \eta} \; ,$

where Re is the Reynolds number and ρ fluid density. v is the mean flow velocity, which is half the maximum 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 maximum 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 specially in hemorheology and hemodynamics, both fields of physiology.[4]

The Poiseuilles' 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. Although more lengthy than directly using the Navier–Stokes equations, an alternative method of deriving the Hagen–Poiseuille equation is as follows.

Liquid flow through a pipe

a) A tube showing the imaginary lamina. b) A cross section of the tube shows the lamina moving at different speeds. Those closest to the edge of the tube are moving slowly while those near the center are moving quickly.

Assume the liquid exhibits laminar 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 the no-slip condition).

To figure out the motion of the liquid, all forces acting on each lamina must be known:

1. The pressure force pushing the liquid through the tube is the change in pressure multiplied by the area: $F = -\Delta P A$. This force is in the direction of the motion of the liquid. The negative sign comes from the conventional way we define $\Delta P = P_\text{end}-P_\text{top} < 0$.
2. Viscosity effects will pull from the faster lamina immediately closer to the center of the tube.
3. Viscosity effects will drag from the slower lamina immediately closer to the walls of the tube.

Viscosity

Two fluids moving past each other in the x direction. 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 a shear force between them. This force is proportional to the area of contact A, the velocity gradient in the direction of flow ${\Delta v_x}/{\Delta y}$, and a proportionality constant η (viscosity) and is given by

$F_{\text{viscosity, top}} = - \eta A \frac{\Delta v_x}{\Delta y}.$

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). By Newton'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 as Newtonian fluids.

Faster lamina

Assume that we are figuring out the force on the lamina with radius $s$. From the equation above, we need to know the area of contact and the velocity gradient. Think of the lamina as a ring of radius $r$, thickness $dr$, and length Δx. The area of contact between the lamina and the faster one is simply the area of the inside of the cylinder: $A = 2 \pi r \, \Delta x$ . 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 the change of the velocity with respect to the change in the radius at the intersection of these two laminae. That intersection is at a radius of $r$. 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

$F_{\text{viscosity, fast}} = - \eta 2 \pi r \, \Delta x \left . \frac{dv}{dr} \right|_r$

where the vertical bar and subscript r following the derivative indicates that it should be taken at a radius of $r$.

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 at r+dr instead of r. 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).

$F_{\text{viscosity, slow}} = \eta 2 \pi (r+dr) \Delta x \left . \frac{dv}{dr} \right \vert_{r+dr}$

Putting it all together

To find the solution for the flow of liquid through a tube, we need to make one last assumption. There is no acceleration of liquid in the pipe, and by Newton'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

$0 = F_{\text{pressure}} + F_{\text{viscosity, fast}} + F_{\text{viscosity, slow}}$

or

$0 = \Delta P 2 \pi rdr - \eta 2 \pi r \Delta x \left . \frac{dv}{dr} \right \vert_r + \eta 2 \pi (r+dr) \Delta x \left . \frac{dv}{dr} \right \vert_{r+dr}.$

First, to get everything happening at the same point, use the first two terms of a Taylor series expansion of the velocity gradient:

$\left . \frac{dv}{dr} \right \vert_{r+dr} = \left . \frac{dv}{dr} \right \vert_r + \left . \frac{d^2 v}{dr^2} \right \vert_r dr .$

The expression is valid for all laminae. Grouping like terms and dropping the vertical bar since all derivatives are assumed to be at radius r,

$0 = \Delta P 2 \pi r \, dr + \eta 2 \pi \, dr \, \Delta x \frac{dv}{dr} + \eta 2 \pi r \, dr \, \Delta x \frac{d^2 v}{dr^2} + \eta 2 \pi (dr)^2 \, \Delta x \frac{d^2 v}{dr^2}.$

Finally, put this expression in the form of a differential equation, dropping the term quadratic in dr.

$- \frac{1}{\eta} \frac{\Delta P}{\Delta x} = \frac{d^2 v}{dr^2} + \frac{1}{r} \frac{dv}{dr}$

It can be seen that both sides of the equations are negative: there is a drop of pressure along the tube (left side) and both first and second derivatives of the velocity are negative (velocity has a maximum value at the center of the tube, where r = 0). Using the product rule, the equation may be re-arranged to:

$- \frac{1}{\eta} \frac{\Delta P}{\Delta x} = \frac{1}{r} \frac{d}{dr} \left(r \frac{dv} {dr} \right).$

This differential equation is subject to the following boundary conditions:

$v(r) = 0$ at $r = R$ -- "no-slip" boundary condition at the wall
$\frac{dv} {dr} = 0$ at $r = 0$ -- axial symmetry.

Axial symmetry means that the velocity v(r) is maximum at the center of the tube, therefore the first derivative $\frac{dv}{dr}$ is zero at r = 0.

The differential equation can be integrated to:

$v(r) = - \frac{1}{4 \eta}r^2\frac{\Delta P}{\Delta x} + A\ln(r) + B.$

To find A and B, we use the boundary conditions.

First, the symmetry boundary condition indicates:

$\frac{dv}{dr} = - \frac{1}{2 \eta} r \frac{\Delta P}{\Delta x} + A \frac{1}{r} = 0$ at r = 0.

A solution possible only if A = 0. Next the no-slip boundary condition is applied to the remaining equation:

$v(R) = - \frac{1}{4 \eta} R^2 \frac{\Delta P}{\Delta x} + B = 0$

so therefore

$B = \frac{1}{4 \eta} R^2 \frac{\Delta P}{\Delta x}.$

Now we have a formula for the velocity of liquid moving through the tube as a function of the distance from the center of the tube

$v = \frac{1}{4 \eta} \frac{\Delta P}{\Delta x} (R^2 - r^2)$

or, at the center of the tube where the liquid is moving fastest (r = 0) with R being the radius of the tube,

$v_{max} = \frac{1}{4 \eta} \frac{\Delta P}{\Delta x}R^2.$

Poiseuille's law

To get the total volume that flows through the tube, we need to add up the contributions from each lamina. To calculate the flow through each lamina, we multiply the velocity (from above) and the area of the lamina.

$\Phi (r) \, dr = \frac{1}{4 \eta} \frac{|\Delta P|}{\Delta x} (R^2 - r^2) 2 \pi r \, dr = \frac{\pi}{2 \eta} \frac{|\Delta P|}{\Delta x} (rR^2 - r^3) \, dr$

Finally, we integrate over all lamina via the radius variable r.

$\Phi = \frac{\pi}{2 \eta} \frac{|\Delta P|}{\Delta x} \int_0^R (rR^2 - r^3)\, dr = \frac{|\Delta P| \pi R^4}{8 \eta \Delta x}$

Poiseuille's equation for compressible fluids

For a compressible fluid in a tube the volumetric flow rate and the linear velocity is 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 an ideal gas in the isothermal case, where the temperature of the fluid is permitted to equilibrate with its surroundings, and when the pressure difference between ends of the pipe is small, the volumetric flow rate at the pipe outlet is given by

$\Phi = \frac{dV}{dt} = v \pi R^{2} = \frac{\pi R^{4} \left( P_{i}-P_{o} \right)}{8 \eta L} \times \frac{ P_{i}+P_{o}}{2 P_{o}} = \frac{\pi R^{4}}{16 \eta L} \left( \frac{ P_{i}^{2}-P_{o}^{2}}{P_{o}} \right)$

where:

$P_{i}$ inlet pressure
$P_{o}$ outlet pressure
$L$ is the length of tube
$\eta$ is the viscosity
$R$ is the radius
$V$ is the volume of the fluid at outlet pressure
$v$ is the velocity of the fluid at outlet pressure

This is usually a good approximation when the flow velocity is less than mach 0.3

This equation can be seen as Poiseuille's law with an extra correction factor $\frac{P_{i}+P_{o}}{2} \times \frac{1}{P_{o}}$ 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

$\Delta F = S \Delta P$ ,    where    $S = \pi r^2$ ,    i.e.    $\Delta F = \pi r^2 \Delta P$ ,

then from Poiseuille's law    $\Delta P = \frac{8 \mu LQ}{\pi r^4}$

it follows that    $\Delta F = \frac{8 \mu LQ}{r^2}$ .

For electrical circuits,
let   $n$   be the concentration of free charged particles,   $[n] = m^{-3}$ ;
let   $q^{*}$   be the charge of each particle,   $[q^{*}] = C$ .

(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,    $Q = \frac{I}{nq^{*}}$ ,   and
$\Delta F = \frac{8 \mu LI}{nr^2 q^{*}}$ .

But    $\Delta F = Eq$ ,    where   $q$   is the total charge in the volume of the tube.
The volume of the tube is equal to   $\pi r^2 L$ ,
so the number of charged particles in this volume is equal to   $n\pi r^2 L$ ,
and their total charge is   $q = n\pi r^2 Lq^{*}$ .

Now,   $E = \frac{\Delta F}{q} = \frac{8\mu I}{n^2\pi r^4 (q^{*})^2}$ .   Since the voltage    $V = EL$ ,

we get    $V = \frac{8\mu LI}{n^2\pi r^4 (q^{*})^2}$ .   This is exactly Ohm's law,

where the resistance    $R=\frac{V}{I}$   is described by the formula
$R = \frac{8\mu L}{n^2\pi r^4 (q^{*})^2}$ .

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  $S=\pi r^2$  of the resistor, which is wrong!

The correct relation is    $R=\frac{\rho L}{S}$ ,    where   $\rho$   is the specific resistance;
i.e. the resistance $R$  is inversely proportional to the first power of the cross section area  $S$  of the resistor.

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 currency. 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.