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
Contents 1 Equation 2 Relation to Darcy–Weisbach 3 Derivation 3.1
4 Poiseuille flow in annular section[9] 5 Plane Poiseuille flow 6 Poiseuille flow in non-circular cross-sections[10] 7 Poiseuille's equation for compressible fluids 8 Electrical circuits analogy 9 Medical applications – intravenous access and fluid delivery 10 See also 11 Notes 12 References 13 External links Equation[edit] In standard fluid-dynamics notation:[3][4][5] Δ P = 8 μ L Q π R 4 , displaystyle Delta P= frac 8mu LQ pi R^ 4 , where: ΔP is the pressure difference between the two ends, L is the length of pipe, μ is the dynamic viscosity, Q is the volumetric flow rate, R is the pipe radius. 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
Q max = π R 2 2 Δ P ρ . displaystyle Q_ max =pi R^ 2 sqrt frac 2Delta P rho . Relation to Darcy–Weisbach[edit]
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
Λ = 64 R e , R e = ρ v d μ , displaystyle Lambda = frac 64 mathrm Re ,quad mathrm Re = frac rho vd mu , 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.
Derivation[edit]
Main article: Hagen–Poiseuille flow from the Navier–Stokes
equations
The
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.
The pressure force pushing the liquid through the tube is the change
in pressure multiplied by the area: F = −A ΔP. This force is in the
direction of the motion of the liquid. The negative sign comes from
the conventional way we define ΔP = Pend − Ptop < 0.
Viscosity[edit] 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 Δvx/Δy, and a proportionality constant (viscosity) and is given by F viscosity, top = − μ A Δ v x Δ y . displaystyle F_ text viscosity, top =-mu 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[edit] 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 F viscosity, fast = − 2 π r μ Δ x d v d r
r displaystyle F_ text viscosity, fast =-2pi rmu ,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[edit] 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 viscosity, slow = 2 π ( r + d r ) μ Δ x d v d r
r + d r displaystyle F_ text viscosity, slow =2pi (r+dr)mu ,Delta xleft. frac dv dr right_ r+dr Putting it all together[edit] 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 0 = F pressure + F viscosity, fast + F viscosity, slow displaystyle 0=F_ text pressure +F_ text viscosity, fast +F_ text viscosity, slow or 0 = Δ P 2 π r d r − 2 π r μ Δ x d v d r
r + 2 π ( r + d r ) μ Δ x d v d r
r + d r . displaystyle 0=Delta P2pi r,dr-2pi rmu ,Delta xleft. frac dv dr right_ r +2pi (r+dr)mu ,Delta x,left. frac dv dr rightvert _ 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: d v d r
r + d r = d v d r
r + d 2 v d r 2
r d r . displaystyle left. frac dv dr right_ r+dr =left. frac dv dr right_ r +left. frac d^ 2 v dr^ 2 right_ 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 = Δ P 2 π r d r + 2 π μ d r Δ x d v d r + 2 π r μ d r Δ x d 2 v d r 2 + 2 π μ ( d r ) 2 Δ x d 2 v d r 2 . displaystyle 0=Delta P2pi r,dr+2pi mu ,dr,Delta x frac dv dr +2pi rmu ,dr,Delta x frac d^ 2 v dr^ 2 +2pi mu (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. − 1 μ Δ P Δ x = d 2 v d r 2 + 1 r d v d r displaystyle - frac 1 mu 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 rearranged to: − 1 μ Δ P Δ x = 1 r d d r ( r d v d r ) . displaystyle - frac 1 mu frac Delta P Delta x = frac 1 r frac d dr left(r frac dv dr right). The right-hand 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: v ( r ) = 0 at r = R displaystyle v(r)=0quad mbox at r=R — "no-slip" boundary condition at the wall d v d r = 0 at r = 0 displaystyle frac dv dr =0quad mbox 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 dv/dr is zero at r = 0. The differential equation can be integrated to: v ( r ) = − 1 4 μ r 2 Δ P Δ x + A ln ( r ) + B . displaystyle v(r)=- frac 1 4mu r^ 2 frac Delta P Delta x +Aln(r)+B. To find A and B, we use the boundary conditions. First, the symmetry boundary condition indicates: d v d r = − 1 2 μ r Δ P Δ x + A r = 0 at r = 0. displaystyle frac dv dr =- frac 1 2mu r frac Delta P Delta x + frac A r =0quad mbox at r=0. A solution possible only if A = 0. Next the no-slip boundary condition is applied to the remaining equation: v ( R ) = − 1 4 μ R 2 Δ P Δ x + B = 0 displaystyle v(R)=- frac 1 4mu R^ 2 frac Delta P Delta x +B=0 so therefore B = 1 4 μ R 2 Δ P Δ x . displaystyle B= frac 1 4mu 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 = 1 4 μ Δ P Δ x ( R 2 − r 2 ) displaystyle v= frac 1 4mu 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 m a x = 1 4 μ Δ P Δ x R 2 . displaystyle v_ max = frac 1 4mu frac Delta P Delta x R^ 2 . Poiseuille's law[edit] 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. Q ( r ) d r = 1 4 μ
Δ P
Δ x ( R 2 − r 2 ) 2 π r d r = π 2 μ
Δ P
Δ x ( r R 2 − r 3 ) d r displaystyle Q(r),dr= frac 1 4mu frac Delta P Delta x (R^ 2 -r^ 2 )2pi r,dr= frac pi 2mu frac Delta P Delta x (rR^ 2 -r^ 3 ),dr Finally, we integrate over all lamina via the radius variable r. Q = π 2 μ
Δ P
Δ x ∫ 0 R ( r R 2 − r 3 ) d r =
Δ P
π R 4 8 μ Δ x displaystyle Q= frac pi 2mu frac Delta P Delta x int _ 0 ^ R (rR^ 2 -r^ 3 ),dr= frac Delta Ppi R^ 4 8mu Delta x Startup of Poiseuille flow in a pipe[8][edit] When a constant pressure gradient G = − d p / d x = constant displaystyle G=-dp/dx= text constant 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 ∂ u ∂ t = G ρ + ν ( ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r ) displaystyle frac partial u partial t = frac G rho +nu left( frac partial ^ 2 u partial r^ 2 + frac 1 r frac partial u partial r right) with initial and boundary conditions, u ( r , 0 ) = 0 , u ( R , t ) = 0. displaystyle u(r,0)=0,quad u(R,t)=0. The velocity distribution is given by u ( r , t ) = G 4 μ ( R 2 − r 2 ) − 2 G R 2 μ ∑ n = 1 ∞ 1 λ n 3 J o ( λ n r / R ) J 1 ( λ n ) e − λ n 2 ν t R 2 , J o ( λ n ) = 0 displaystyle u(r,t)= frac G 4mu (R^ 2 -r^ 2 )- frac 2GR^ 2 mu sum _ n=1 ^ infty frac 1 lambda _ n ^ 3 frac J_ o (lambda _ n r/R) J_ 1 (lambda _ n ) e^ -lambda _ n ^ 2 frac nu t R^ 2 ,quad J_ o (lambda _ n )=0 where J o ( λ n r / R ) displaystyle J_ o (lambda _ n r/R) is the
λ n displaystyle lambda _ n are the positive roots of this function and J 1 ( λ n ) displaystyle J_ 1 (lambda _ n ) is the
t → ∞ displaystyle trightarrow infty , Poiseuille solution is recovered. Poiseuille flow in annular section[9][edit] Poiseuille flow in annular section If R 1 displaystyle R_ 1 is the inner cylinder radii and R 2 displaystyle R_ 2 is the outer cylinder radii, with applied pressure gradient between the two ends G = − d p / d x = constant displaystyle G=-dp/dx= text constant , the velocity distribution and the volume flux through the annular pipe are u ( r ) = G 4 μ ( R 1 2 − r 2 ) + G 4 μ ( R 2 2 − R 1 2 ) ln ( r / R 1 ) ln ( R 2 / R 1 ) , Q = G π 8 μ [ R 2 4 − R 1 4 − ( R 2 2 − R 1 2 ) 2 ln R 2 / R 1 ] . displaystyle begin aligned u(r)&= frac G 4mu (R_ 1 ^ 2 -r^ 2 )+ frac G 4mu (R_ 2 ^ 2 -R_ 1 ^ 2 ) frac ln(r/R_ 1 ) ln(R_ 2 /R_ 1 ) ,\Q&= frac Gpi 8mu left[R_ 2 ^ 4 -R_ 1 ^ 4 - frac (R_ 2 ^ 2 -R_ 1 ^ 2 )^ 2 ln R_ 2 /R_ 1 right].end aligned When R 2 = R , R 1 = 0 displaystyle R_ 2 =R, R_ 1 =0 , the original problem is recovered. Plane Poiseuille flow[edit] Plane Poiseuille flow Plane Poiseuille flow is flow created between two infinitely long parallel plates, separated by a distance h displaystyle h with a constant pressure gradient G = − d p / d x = constant displaystyle G=-dp/dx= text constant is applied in the direction of flow. The flow is essentially unidirectional because of infinite length. The Navier-Stokes equations reduce to d 2 u d y 2 = − G μ displaystyle frac d^ 2 u dy^ 2 =- frac G mu with no-slip condition on both walls u ( 0 ) = 0 , u ( h ) = 0 displaystyle u(0)=0,quad u(h)=0 Therefore, the velocity distribution and the volume flow rate per unit length are u ( y ) = G 2 μ y ( h − y ) , Q = G h 3 12 μ . displaystyle u(y)= frac G 2mu y(h-y),quad Q= frac Gh^ 3 12mu . Poiseuille flow in non-circular cross-sections[10][edit] Boussinesq[11] 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. Proudman[12] derived the same for isosceles triangles in 1914. Let G = − d p / d x = constant displaystyle G=-dp/dx= text constant 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 displaystyle 0leq yleq h and width 0 ≤ z ≤ l displaystyle 0leq zleq l are u ( y , z ) = G 2 μ y ( h − y ) − 4 G h 2 μ π 3 ∑ n = 1 ∞ 1 ( 2 n − 1 ) 3 sinh ( β n z ) + sinh ( β n ( l − z ) ) sinh ( β n l ) sin ( β n y ) , β n = ( 2 n − 1 ) π h , Q = G h 3 l 12 μ − 16 G h 4 π 5 μ ∑ n = 1 ∞ 1 ( 2 n − 1 ) 5 cosh ( β n l ) − 1 sinh ( β n l ) . displaystyle begin aligned u(y,z)&= frac G 2mu y(h-y)- frac 4Gh^ 2 mu pi ^ 3 sum _ n=1 ^ infty frac 1 (2n-1)^ 3 frac sinh(beta _ n z)+sinh(beta _ n (l-z)) sinh(beta _ n l) sin(beta _ n y),quad beta _ n = frac (2n-1)pi h ,\Q&= frac Gh^ 3 l 12mu - frac 16Gh^ 4 pi ^ 5 mu sum _ n=1 ^ infty frac 1 (2n-1)^ 5 frac cosh(beta _ n l)-1 sinh(beta _ n l) .end aligned The velocity and the volume flow rate of tube with equilateral triangular cross-section of side length 2 h / 3 displaystyle 2h/ sqrt 3 are u ( y , z ) = − G 4 μ h ( y − h ) ( y 2 − 3 z 2 ) , Q = G h 4 60 3 μ . displaystyle begin aligned u(y,z)&=- frac G 4mu h (y-h)(y^ 2 -3z^ 2 ),\Q&= frac Gh^ 4 60 sqrt 3 mu .end aligned The velocity and the volume flow rate in the right-angled isosceles triangle y = π , y ± z = 0 displaystyle y=pi , ypm z=0 are u ( y , z ) = G 2 μ ( y + z ) ( π − y ) − G π μ ∑ n = 1 ∞ 1 β n 3 sinh ( 2 π β n ) sinh [ β n ( 2 π − y + z ) ] sin [ β n ( y + z ) ] − sinh [ β n ( y + z ) ] sin [ β n ( y − z ) ] , β n = n − 1 2 , Q = G π 4 12 μ − G 2 π μ ∑ n = 1 ∞ 1 β n 5 [ coth ( 2 π β n ) + csc ( 2 π β n ) ] . displaystyle begin aligned u(y,z)&= frac G 2mu (y+z)(pi -y)- frac G pi mu sum _ n=1 ^ infty frac 1 beta _ n ^ 3 sinh(2pi beta _ n ) sinh[beta _ n (2pi -y+z)]sin[beta _ n (y+z)]-sinh[beta _ n (y+z)]sin[beta _ n (y-z)] ,quad beta _ n =n- frac 1 2 ,\Q&= frac Gpi ^ 4 12mu - frac G 2pi mu sum _ n=1 ^ infty frac 1 beta _ n ^ 5 [coth(2pi beta _ n )+csc(2pi beta _ n )].end aligned The velocity distribution for tubes of elliptical cross-section with semi-axis a displaystyle a and b displaystyle b is[8] u ( y , z ) = G 2 μ ( 1 a 2 + 1 b 2 ) ( 1 − y 2 a 2 − z 2 b 2 ) , Q = π G a 3 b 3 4 μ ( a 2 + b 2 ) . displaystyle begin aligned u(y,z)&= frac G 2mu 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 Ga^ 3 b^ 3 4mu (a^ 2 +b^ 2 ) .end aligned Here, when a = b displaystyle a=b , Poiseuille flow for circular pipe is recovered and when a → ∞ displaystyle arightarrow infty , plane Poiseuille flow is recovered. Poiseuille's equation for compressible fluids[edit] 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 Q = d V d t = v π R 2 = π R 4 ( P i − P o ) 8 μ L × P i + P o 2 P o = π R 4 16 μ L ( P i 2 − P o 2 P o ) displaystyle Q= frac dV dt =vpi R^ 2 = frac pi R^ 4 left(P_ mathrm i -P_ mathrm o right) 8mu L times frac P_ mathrm i +P_ mathrm o 2P_ mathrm o = frac pi R^ 4 16mu L left( frac P_ mathrm i ^ 2 -P_ mathrm o ^ 2 P_ mathrm o right) where: Pi is inlet pressure Po is outlet pressure L is the length of tube μ displaystyle mu 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 equation can be seen as Poiseuille's law with an extra correction factor Pi + Po/2Po expressing the average pressure relative to the outlet pressure. Electrical circuits analogy[edit] This section does not cite any sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (September 2016) (Learn how and when to remove this template message) 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
Δ F = S Δ P , displaystyle Delta F=SDelta P, where S = πr2, i.e. ΔF = πr2 ΔP, then from Poiseuille's law Δ P = 8 μ L Q π r 4 displaystyle Delta P= frac 8mu LQ pi r^ 4 it follows that Δ F = 8 μ L Q r 2 displaystyle Delta F= frac 8mu 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 = 6981160000000000000♠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 Δ F = 8 μ L I n r 2 q ∗ . displaystyle Delta F= frac 8mu LI nr^ 2 q^ * . 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 = n π r 2 L q ∗ . displaystyle q=npi r^ 2 Lq^ * . Now, E = Δ F q = 8 μ I n 2 π r 4 ( q ∗ ) 2 . displaystyle E= frac Delta F q = frac 8mu I n^ 2 pi r^ 4 (q^ * )^ 2 . Since the voltage V = EL, we get V = 8 μ L I n 2 π r 4 ( q ∗ ) 2 . displaystyle V= frac 8mu LI n^ 2 pi r^ 4 (q^ * )^ 2 . This is exactly Ohm's law, where the resistance R = V/I is described by the formula R = 8 μ L n 2 π r 4 ( q ∗ ) 2 displaystyle R= frac 8mu 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 = πr2 of the resistor, which is wrong according to the electrical analogy. The correct relation is R = ρ L S , displaystyle R= frac rho L S , 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
This section does not cite any sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (September 2016) (Learn how and when to remove this template message) The
Couette flow Darcy's law Pulse Wave Hydraulic circuit Notes[edit] ^ a b Sutera, Salvatore P.; Skalak, Richard (1993). "The History of
Poiseuille's Law". Annual Review of
References[edit] Sutera, S. P.; Skalak, R. (1993). "The history of Poiseuille's law".
Annual Review of
External links[edit] Poiseuille's law for power-law non-Newtonian fluid Poiseuille's law in a slightly tapered tube Hagen–Poiseuille e |