1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
|
---
title: "Hagen-Poiseuille equation"
sort_title: "Hagen-Poiseuille equation"
date: 2021-04-13
categories:
- Physics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---
The **Hagen-Poiseuille equation**, or simply the **Poiseuille equation**,
describes the flow of a fluid with nonzero [viscosity](/know/concept/viscosity/)
through a cylindrical pipe: the fluid clings to the sides,
limiting the amount that can pass through per unit time.
Consider the [Navier-Stokes equations](/know/concept/navier-stokes-equations/)
of an incompressible fluid with spatially uniform density $$\rho$$.
Assuming that the flow is steady $$\ipdv{\va{v}}{t} = 0$$,
and that gravity is negligible $$\va{g} = 0$$, we get:
$$\begin{aligned}
(\va{v} \cdot \nabla) \va{v}
= - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
\qquad \quad
\nabla \cdot \va{v} = 0
\end{aligned}$$
Let the pipe have radius $$R$$, and be infinitely long and parallel to the $$z$$-axis.
We insert the ansatz $$\va{v} = \vu{e}_z \: v_z(r)$$,
where $$\vu{e}_z$$ is the $$z$$-axis' unit vector,
and we are assuming that the flow depends only on $$r$$, not on $$\phi$$ or $$z$$.
With this, $$\nabla \cdot \va{v}$$ trivially vanishes,
and in the main equation multiplying out $$\rho$$ yields this,
where $$\eta = \rho \nu$$ is the dynamic viscosity:
$$\begin{aligned}
\nabla p
= \vu{e}_z \: \eta \nabla^2 v_z
\end{aligned}$$
Because only $$\vu{e}_z$$ appears on the right-hand side,
only the $$z$$-component of $$\nabla p$$ can be nonzero.
However, $$v_z(r)$$ is a function of $$r$$, not $$z$$!
The left thus only depends on $$z$$, and the right only on $$r$$,
meaning that both sides must equal a constant,
which we call $$-G$$:
$$\begin{aligned}
\dv{p}{z}
= -G
\qquad \quad
\eta \frac{1}{r} \dv{}{r}\Big( r \dv{v_z}{r} \Big)
= - G
\end{aligned}$$
The former equation for $$p(z)$$ is easy to solve.
We get an integration constant $$p(0)$$:
$$\begin{aligned}
p(z)
= p(0) - G z
\end{aligned}$$
This gives meaning to $$G$$: it is the **pressure gradient**,
which for a pipe of length $$L$$
describes the pressure difference $$\Delta p = p(0) - p(L)$$
that is driving the fluid, i.e. $$G = \Delta p / L$$.
As for the latter equation for $$v_z(r)$$,
we start by integrating it once, introducing a constant $$A$$:
$$\begin{aligned}
\dv{}{r}\Big( r \dv{v_z}{r} \Big)
= - \frac{G}{\eta} r
\quad \implies \quad
\dv{v_z}{r}
= - \frac{G}{2 \eta} r + \frac{A}{r}
\end{aligned}$$
Integrating this one more time,
thereby introducing another constant $$B$$,
we arrive at:
$$\begin{aligned}
v_z
= - \frac{G}{4 \eta} r^2 + A \ln{r} + B
\end{aligned}$$
The velocity must be finite at $$r = 0$$, so we set $$A = 0$$.
Furthermore, the Navier-Stokes equation's *no-slip* condition
demands that $$v_z = 0$$ at the boundary $$r = R$$,
so $$B = G R^2 / (4 \eta)$$.
This brings us to the **Poiseuille solution** for $$v_z(r)$$:
$$\begin{aligned}
\boxed{
v_z(r)
= \frac{G}{4 \eta} (R^2 - r^2)
}
\end{aligned}$$
How much fluid can pass through the pipe per unit time?
This is denoted by the **volumetric flow rate** $$Q$$,
which is the integral of $$v_z$$ over the circular cross-section:
$$\begin{aligned}
Q
= 2 \pi \int_0^R v_z(r) \: r \dd{r}
= \frac{\pi G}{2 \eta} \int_0^R R^2 r - r^3 \dd{r}
= \frac{\pi G}{2 \eta} \bigg[ \frac{R^2 r^2}{2} - \frac{r^4}{4} \bigg]_0^R
\end{aligned}$$
We thus arrive at the main Hagen-Poiseuille equation,
which predicts $$Q$$ for a given setup:
$$\begin{aligned}
\boxed{
Q
= \frac{\pi G R^4}{8 \eta}
}
\end{aligned}$$
Consequently, the average flow velocity $$\Expval{v_z}$$
is simply $$Q$$ divided by the cross-sectional area:
$$\begin{aligned}
\Expval{v_z}
= \frac{Q}{\pi R^2}
= \frac{G R^2}{8 \eta}
\end{aligned}$$
The fluid's viscous stickiness means it exerts a drag force $$D$$
on the pipe as it flows. For a pipe of length $$L$$ and radius $$R$$,
we calculate $$D$$ by multiplying the internal area $$2 \pi R L$$
by the [shear stress](/know/concept/cauchy-stress-tensor/)
$$-\sigma_{zr}$$ on the wall
(i.e. the wall applies $$\sigma_{zr}$$, the fluid responds with $$- \sigma_{zr}$$):
$$\begin{aligned}
D
= - 2 \pi R L \: \sigma_{zr} \big|_{r = R}
= - 2 \pi R L \eta \dv{v_z}{r}\Big|_{r = R}
= 2 \pi R L \eta \frac{G R}{2 \eta}
= \pi R^2 L G
\end{aligned}$$
$$G$$ is an inconvenient quantity here, so we remove it
by substituting $$R^2 G = 8 \eta \Expval{v_z}$$:
$$\begin{aligned}
\boxed{
D
= 8 \pi \eta L \Expval{v_z}
}
\end{aligned}$$
Due to this drag, the pressure difference $$\Delta p = p(0) - p(L)$$
does work on the fluid at a rate $$P$$.
Since power equals force (i.e. pressure times area) times velocity:
$$\begin{aligned}
P
= 2 \pi \int_0^R \Delta p \: v_z(r) \: r \dd{r}
\end{aligned}$$
Because $$\Delta p$$ is independent of $$r$$,
we get the same integral we used to calculate $$Q$$.
Then, thanks to the fact that $$\Delta p = G L$$
and $$Q = \pi R^2 \Expval{v_z}$$, it follows that:
$$\begin{aligned}
P
= \Delta p \: Q
= G L \pi R^2 \Expval{v_z}
= D \Expval{v_z}
\end{aligned}$$
In conclusion, the power $$P$$ needed to drive a fluid
through the pipe at a rate $$Q$$ is given by:
$$\begin{aligned}
\boxed{
P
= 8 \pi \eta L \Expval{v_z}^2
= \frac{8 \eta L}{\pi R^4} Q^2
}
\end{aligned}$$
## References
1. B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
CRC Press.
|
