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Diffstat (limited to 'source/know/concept/hagen-poiseuille-equation')
| -rw-r--r-- | source/know/concept/hagen-poiseuille-equation/index.md | 45 |
1 files changed, 21 insertions, 24 deletions
diff --git a/source/know/concept/hagen-poiseuille-equation/index.md b/source/know/concept/hagen-poiseuille-equation/index.md index 6484631..52d3ce8 100644 --- a/source/know/concept/hagen-poiseuille-equation/index.md +++ b/source/know/concept/hagen-poiseuille-equation/index.md @@ -11,9 +11,8 @@ 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. -Due to its viscosity, the fluid clings to the sides, -limiting the amount that can pass through, for a pipe with radius $$R$$. +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 $$\rho$$. @@ -27,13 +26,12 @@ $$\begin{aligned} \nabla \cdot \va{v} = 0 \end{aligned}$$ -Into this, we insert the ansatz $$\va{v} = \vu{e}_z \: v_z(r)$$, -where $$\vu{e}_z$$ is the $$z$$-axis' unit vector. -In other words, we assume that the flow velocity depends only on $$r$$; -not on $$\phi$$ or $$z$$. -Plugging this into the Navier-Stokes equations, -$$\nabla \cdot \va{v}$$ is trivially zero, -and in the other equation we multiply out $$\rho$$, yielding this, +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: $$\begin{aligned} @@ -56,7 +54,7 @@ $$\begin{aligned} = - G \end{aligned}$$ -The former equation, for $$p(z)$$, is easy to solve. +The former equation for $$p(z)$$ is easy to solve. We get an integration constant $$p(0)$$: $$\begin{aligned} @@ -64,13 +62,12 @@ $$\begin{aligned} = p(0) - G z \end{aligned}$$ -This gives meaning to the **pressure gradient** $$G$$: -for a pipe of length $$L$$, -it describes the pressure difference $$\Delta p = p(0) - p(L)$$ -that is driving the fluid, -i.e. $$G = \Delta p / L$$ +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 $$v_z(r)$$, +As for the latter equation for $$v_z(r)$$, we start by integrating it once, introducing a constant $$A$$: $$\begin{aligned} @@ -148,8 +145,8 @@ $$\begin{aligned} = \pi R^2 L G \end{aligned}$$ -We would like to get rid of $$G$$ for being impractical, -so we substitute $$R^2 G = 8 \eta \Expval{v_z}$$, yielding: +$$G$$ is an inconvenient quantity here, so we remove it +by substituting $$R^2 G = 8 \eta \Expval{v_z}$$: $$\begin{aligned} \boxed{ @@ -159,8 +156,8 @@ $$\begin{aligned} \end{aligned}$$ 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: +does work on the fluid at a rate $$P$$. +Since power equals force (i.e. pressure times area) times velocity: $$\begin{aligned} P @@ -179,14 +176,14 @@ $$\begin{aligned} = D \Expval{v_z} \end{aligned}$$ -In conclusion, the power $$P$$, -needed to drive a fluid through the pipe at a rate $$Q$$, -is given by: +In conclusion, the power $$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}$$ |