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 π 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 = π r 4 Δ P 8 η L So if I were to plug in the continuity equations definition of flow rate into Poiseuilles Law: v A = v π r 2 = π r 4 Δ P 8 η L Therefore: v = r 2 Δ P 8 η 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.

When you write down Q = v A, it's implicit that the velocity profile is uniform over the cross-section A (and purely perpendicular to it).In general, Q = ∫ v ⋅ d A This no longer implies that v ∝ 1 R 2 If we assume v has purely radial dependence and is aligned with d A as in Poiseuille flow, then we have: Q = 2 π ∫ 0 R v ( r ) r d r

Solved: "Is there a contradiction between the continuity equation..."

Blaine Stein

Answered question

2022-05-15

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 π 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{π r^{4} Δ P}{8 η L}

So if I were to plug in the continuity equations definition of flow rate into Poiseuilles Law:

v A = v π r^{2} = \frac{π r^{4} Δ P}{8 η L}

Therefore:

v = \frac{r^{2} Δ P}{8 η 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.

Answer & Explanation

hodowlanyb1rq2

Beginner2022-05-16Added 12 answers

When you write down

Q = v A

, it's implicit that the velocity profile is uniform over the cross-section A (and purely perpendicular to it).
In general,

Q = \int v \cdot d A

This no longer implies that

v \propto \frac{1}{R^{2}}

If we assume

v

has purely radial dependence and is aligned with

d A

as in Poiseuille flow, then we have:

Q = 2 π \int_{0}^{R} v (r) r d r

Daphne Haney

Beginner2022-05-17Added 4 answers

For a fixed volumetric throughput rate Q,

Δ P

decreases as

r^{4}

, so v decreases as

r^{2}

, exactly what you would expect from the continuity equation.

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