# Is there a contradiction between the continuity equation and Poiseuilles Law? The continuity equati

Is there a contradiction between the continuity equation and Poiseuilles Law?
The continuity equation states that flow rate should be conserved in different areas of a pipe:
$Q={v}_{1}{A}_{1}={v}_{2}{A}_{2}=v\pi {r}^{2}$
We can see from this equation that velocity and pipe radius are inversely proportional. If radius is doubled, velocity of flow is quartered.
Another way I was taught to describe flow rate is through Poiseuilles Law:
$Q=\frac{\pi {r}^{4}\mathrm{\Delta }P}{8\eta L}$
So if I were to plug in the continuity equations definition of flow rate into Poiseuilles Law:
$vA=v\pi {r}^{2}=\frac{\pi {r}^{4}\mathrm{\Delta }P}{8\eta L}$
Therefore:
$v=\frac{{r}^{2}\mathrm{\Delta }P}{8\eta L}$
Now in this case, the velocity is proportional to the radius of the pipe. If the radius is doubled, then velocity is qaudrupled.
What am I misunderstanding here? I would prefer a conceptual explanation because I feel that these equations are probably used with different assumptions/in different contexts.
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hodowlanyb1rq2
When you write down $Q=vA$, it's implicit that the velocity profile is uniform over the cross-section A (and purely perpendicular to it).
In general,
$Q=\int \mathbf{v}\cdot \mathrm{d}\mathbf{A}$
This no longer implies that $v\propto \frac{1}{{R}^{2}}$
If we assume $\mathbf{v}$ has purely radial dependence and is aligned with $\mathrm{d}\mathbf{A}$ as in Poiseuille flow, then we have:
$Q=2\pi {\int }_{0}^{R}v\left(r\right)r\mathrm{d}r$
###### Not exactly what you’re looking for?
Daphne Haney
For a fixed volumetric throughput rate Q, $\mathrm{\Delta }P$ decreases as ${r}^{4}$, so v decreases as ${r}^{2}$, exactly what you would expect from the continuity equation.