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Continuum mechanics  

Laws


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 assumptions of the equation are that the fluid is incompressible and Newtonian; the flow is laminar through a pipe of constant circular crosssection 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.
In standard fluiddynamics notation:^{[3]}^{[4]}^{[5]}
where:
The equation does not hold close to the pipe entrance.^{[6]}^{: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
Normally, HagenPoiseuille 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 socalled 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, 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.^{[7]}
Poiseuille's law was later in 1891 extended to turbulent flow by L. R. Wilberforce, based on Hagenbach's work.
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.
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 noslip condition).
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 a shear force between them. This force is proportional to the area of contact A, the velocity gradient in the direction of flow Δv_{x}/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). 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.
Assume that we are figuring out the force on the lamina with radius r. 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 = 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 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
where the vertical bar and subscript r following the derivative indicates that it should be taken at a radius of r.
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).
To find the solution for the flow of a laminar layer 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
or
First, to get everything happening at the same point, use the first two terms of a Taylor 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 radius r,
Finally, put this expression in the form of a differential equation, dropping the term quadratic in 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 rearranged to:
The righthand side is the radial term of the Laplace operator ∇^{2}, so this differential equation is a special case of the Poisson equation. It is subject to the following boundary conditions:
— "noslip" boundary condition at the wall  
— axial symmetry. 
Axial symmetry means that the velocity v(r) is maximum at the center of the tube, therefore the first derivative dv/dr is zero at r = 0.
The differential equation can be integrated to:
To find A and B, we use the boundary conditions.
First, the symmetry boundary condition indicates:
A solution possible only if A = 0. Next the noslip boundary condition is applied to the remaining equation:
so therefore
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
or, at the center of the tube where the liquid is moving fastest (r = 0) with R being the radius of the tube,
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.
Finally, we integrate over all lamina via the radius variable r.
When a constant pressure gradient 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 NavierStokes equations reduce to
with initial and boundary conditions,
The velocity distribution is given by
where is the Bessel function of the first kind of order zero and are the positive roots of this function and is the Bessel function of the first kind of order one. As , Poiseuille solution is recovered.
If is the inner cylinder radii and is 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.
Plane Poiseuille flow is flow created between two infinitely long parallel plates, separated by a distance with a constant pressure gradient is applied in the direction of flow. The flow is essentially unidirectional because of infinite length. The NavierStokes equations reduce to
with noslip condition on both walls
Therefore, the velocity distribution and the volume flow rate per unit length are
Boussinesq^{[11]} derived the velocity profile and volume flow rate in 1868 for rectangular channel and tubes of equilateral triangular crosssection and for elliptical crosssection. Proudman^{[12]} derived the same for isosceles triangles in 1914. Let 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 and width are
The velocity and the volume flow rate of tube with equilateral triangular crosssection of side length are
The velocity and the volume flow rate in the rightangled isosceles triangle are
The velocity distribution for tubes of elliptical crosssection with semiaxis and is^{[8]}
Here, when , Poiseuille flow for circular pipe is recovered and when , plane Poiseuille flow is recovered.
For a compressible fluid in a tube the volumetric flow rate and the linear 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 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
where:
This equation can be seen as Poiseuille's law with an extra correction factor P_{i} + P_{o}/2P_{o} expressing the average pressure relative to the outlet pressure.
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 fluidmechanical 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
where S = πr^{2}, i.e. ΔF = πr^{2} ΔP, 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 = ×10^{−19} C 1.6.)
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
But ΔF = Eq, where q is the total charge in the volume of the tube. The volume of the tube is equal to πr^{2}L, so the number of charged particles in this volume is equal to nπr^{2}L, and their total charge is
Now,
Since the voltage V = EL, we get
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 = πr^{2} 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.^{[13]}
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.
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).