- Hagen-Poiseuille equation
The Hagen-Poiseuille equation is a
physical lawthat describes slow viscous incompressible flowthrough a constant circular cross-section. It is also known as the Hagen-Poiseuille law, Poiseuille law and Poiseuille equation.
tandard fluid dynamics notation
In standard fluid dynamics notation:: or :
Where: :"V" is a volume of the liquid poured (cubic meters):"t" is the time (seconds):"v" is mean fluid
velocityalong the length of the tube (meters/second):"x" is a distance in direction of flow (meters):"R" is the internal radius of the tube (meters):"ΔP" is the pressure difference between the two ends (pascals):"η" is the dynamic fluid viscosity(pascal- second(Pa·s)), :"L" is the total length of the tube in the "x" direction (meters).
Relation to Darcy-Weisbach
This result is also a solution to the phenomenological
Darcy-Weisbach equationin the field of hydraulics, given a relationship for the friction factor in terms of the Reynolds number:
where "Re" is the
Reynolds numberand "ρ" fluid density. 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 1856and 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
hemorheologyand hemodynamics, both fields of physiology. [ [http://www.cvphysiology.com/Hemodynamics/H003.htm Determinants of blood vessel resistance ] ]
The Poiseuilles' law was later in
1891extended to turbulent flowby L. R. Wilberforce, based on Hagenbach's work.
The Hagen Poiseuille equation can be derived from the Navier-Stokes equations.
The derivation of Poiseuille's Law is surprisingly simple, but it requires an understanding of
Viscosity. When two layers of liquid in contact with each other move at different speeds, there will be a forcebetween them. This force is proportional to the areaof contact "A", the velocity difference in the direction of flow "Δvx"/Δ"y", and a proportionality constant "η" 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
Liquid flow through a pipe
In a tube we make a basic assumption: the liquid in the center is moving fastest while the liquid touching the walls of the tube is stationary (due to
friction). To simplify the situation, let's assume 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.
To figure out the motion of the liquid, we need to know all forces acting on each lamina:
# The force pushing the liquid through the tube is the change in pressure multiplied by the area: "F" = -"ΔPA". This force is in the direction of the motion of the liquid - the negative sign comes from the conventional way we define .
# The pull from the faster lamina immediately closer to the center of the tube
# The drag from the slower lamina immediately closer to the walls of the tube
The first of these forces comes from the definition of
pressure. The other two forces require us to modify the equations above that we have for viscosity. In fact, we are not modifying the equations, instead merely plugging in values specific to our problem. Let's focus on the pull from the faster lamina (#2) first.
Assume that we are figuring out the force on the lamina with
radius"s". From the equation above, we need to know the areaof contact and the velocity gradient. Think of the lamina as a cylinder of radius "s" and thickness "ds". The area of contact between the lamina and the faster one is simply the area of the inside of the cylinder: "A" = "2πsΔx". 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 "s". 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 "s" following the
derivativeindicates that it should be taken at a radius of "s".
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 "s"+"ds" instead of "s". 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 liquid through a tube, we need to make one last assumption. There is no
accelerationof 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
Before we move further, we need to simplify this ugly equation. First, to get everything happening at the same point, we need to do a Taylor series expansion of the velocity gradient, keeping only the
linearand quadratic terms (a standard mathematical trick).
Let's use this relation in our equation. Also, let's use "r" instead of "s" since the lamina we chose was arbitrary and we want our expression to be valid for all laminae. Grouping like terms and dropping the vertical bar since all derivatives are assumed to be at radius "r",
Finally, let's get this in the form of a
differential equation, moving some terms around to make it easier to solve later, and neglecting the term quadratic in "dr" since this will be really small compared to the rest (another standard mathematical trick).
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 of the center of the tube). The equation may be re-arranged to:
This differential equation is subject to the following boundary conditions: :v(r) = 0 at r = R -- "No-slip" Boundary Condition at the Wall : 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 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:
: at r = 0
A solution possible only if A = 0. Next the no-slip boundary condition is applied to the remaining equation:
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.
Poiseuille's equation for compressible fluids
For a compressible fluid in a tube the
volumetric flow rateand the linear velocityis 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 and cooled. This meaning that the flow rate is dependant upon heat transfer to and from the fluid. For an ideal gasin the isothermalcase, 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:: inlet pressure: outlet pressure: is the length of tube: is the
viscosity: is the radius: is the volumeof the fluid at outlet pressure: is the velocityof 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 expressing the average pressure relative to the outlet pressure.
= Electrical Circuits analogy =
Electricity was originally understood to be a kind of fluid. This
hydraulic analogyis still conceptually useful.
Poiseuille's law corresponds to
Ohm's lawfor electrical circuits ("V" = "IR"), where the pressure drop Δ"P " is analogous to the voltage"V" and voluminal flow rate Φ is analogous to the current "I". Then the resistance :This concept is useful because the effective resistance in a tube is inversely proportional to the fourth power of the radius. This means that halfing the size of the tube increases the resistance to fluid movement by 16 times.
Both Ohm's law and Poiseuille's law illustrate
It was developed independently by
Gotthilf Heinrich Ludwig Hagen( 1797- 1884) and Jean Louis Marie Poiseuille.
Poiseuille's law was experimentally derived in
1838and formulated and published in 1840and 1846by Jean Louis Marie Poiseuille( 1797- 1869). Hagen did his experiments in 1839.
*S. P. Sutera, R. Skalak, "The history of Poiseuille's law," "Annual Review of Fluid Mechanics", Vol. 25, 1993, pp. 1-19
title=Poiseuille and his law.
* [http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid802.html Poiseuille's law for power-law non-Newtonian fluid]
* [http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid203.html Poiseuille's law in a slightly tapered tube]
* [http://www.calctool.org/CALC/eng/fluid/hagen-poiseuille Web-based calculator of the Hagen-Poiseuille equation]
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