Hagen-Poiseuille's Equation：Pressure drop formula for laminar flow

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Outline

The equation relating pressure drop and flow velocity in laminar pipe flow is called the Hagen-Poiseuille's equation.

It is expressed by Eq. (1):

Δ p = 32 μ L u ¯ d 2 ・ ・ ・ (1)

where Δp is the pressure drop, μ is the fluid viscosity, L is the pipe length, uー is the cross-sectional average velocity, d is the pipe diameter.

It is a fundamental equation in fluid mechanics, and I believe it is often tested in university exams and certifications.

This article explains the derivation of the Hagen-Poiseuille equation.

Derivation of Hagen-Poiseuille's equation

Consider the piping flow shown in the figure above.

The equation is formulated in terms of the equilibrium of forces in a small cylindrical region in the pipe.

P r e s s u r e a c t i n g o n t h e i n f l o w s u r f a c e = p ・ π r 2

P r e s s u r e a c t i n g o n t h e o u t f l o w s u r f a c e = (p + d p d x d x) ・ π r 2

S h e a r s t r e s s a c t i n g o n t h e c y l i n d r i c a l s i d e = τ ・ 2 π r d x

Since these three forces are balanced in the developed flow, Eq. (2) holds.

p ・ π r 2 - (p + d p d x d x) ・ π r 2 - τ ・ 2 π r d x = 0 ・ ・ ・ (2)

Eq. (2) can be rearranged into Eq. (3).

τ = - d p d x r 2 ・ ・ ・ (3)

Here, Eq. (4) follows from Newton's viscosity law.

τ = - μ d u d r ・ ・ ・ (4)

Substituting Eq. (4) into Eq. (3) yields Eq. (5).

- d p d x r 2 = - μ d u d r

d u d r = d p d x r 2 μ ・ ・ ・ (5)

Integrate equation (5) from r=R to r. When r=R, u=0 since it is a wall surface, and when r=r, u=u.

\int u 0 d u = \int r R d p d x r 2 μ d r

u = - 1 4 μ d p d x (R 2 - r 2)

u = R 2 4 μ (- d p d x) (1 - r 2 R 2) ・ ・ ・ (6)

Eq. (6) is the expression for the velocity distribution in a laminar pipe flow.

Incidentally, since the maximum flow velocity u_max is at the center of the pipe, it is expressed by Eq. (7) when r = 0 in Eq. (6).

u m a x = R 2 4 μ (- d p d x) ・ ・ ・ (7)

An equation is then formulated for the volumetric flow rate Q of the fluid flowing through the pipe.

The volume flow rate can be calculated by integrating the flow velocity u at any position r in Eq. (6) over the entire pipe cross section.

Q = \int R 0 u ・ 2 π r d r = π R 2 2 μ (- d p d x) \int R 0 r (1 - r 2 R 2) d r = π R 2 2 μ (- d p d x) [r 2 2 - r 4 4 R 2] R 0 = π R 4 8 μ (- d p d x) ・ ・ ・ (8)

Dividing the volumetric flow rate Q calculated by Eq. (8) by the cross-sectional area of the pipe, Eq. (9), which represents the cross-sectional average flow velocity, can be calculated.

u ¯ = Q π R 2 = R 2 8 μ (- d p d x) ・ ・ ・ (9)

Incidentally, comparing the maximum flow velocity in Eq. (7) with Eq. (9), Eq. (10) holds.

u ¯ = 1 2 u m a x ・ ・ ・ (10)

Keep in mind that it is well known that the cross-sectional average velocity is 1/2 of the maximum velocity in a laminar pipe flow.

Next, for the pressure gradient in Eq. (9), if pressure loss is applied by Δp in the section of pipe length L, Eq. (11) is obtained.

- d p d x = Δ p L ・ ・ ・ (11)

Substituting equation (11) into equation (9), we obtain the following equation.

u ¯ = R 2 Δ p 8 μ L

Δ p = 8 μ L u ¯ R 2

Rewrite the radius R of the pipe as a diameter d (R=d/2).

Δ p = 32 μ L u ¯ d 2 ・ ・ ・ (1)

We were able to derive the Eq. (1) of this article.

In conclusion

Derived the Hagen-Poiseuille's equation for use in laminar pipe flow.

Understanding the derivation process makes it easier to remember the equation. Let's derive it at least once.