# Hagen–Poiseuille equation vs Bernoulli's equation Conflict

1. Jun 13, 2009

### tiredryan

Hey,

Here's something that confuses me with the Hagen–Poiseuille equation and the Bernoulli's equation. From the Hagen–Poiseuille equation and in general, fluid will tend to flow from high to low pressure. From Bernoulli's equation pressure is a function of cross-sectional area and velocity. So for the figure attached I would expect P1 to be high pressure, P2 to be low pressure so fluid would flow from the high to low pressure. This makes sense.

At P2 the pressure is low, but the pressure increases in the region after P2, lets call it P3. Then how does fluid from from P2 to P3 if the pressure goes from low to high again?

Thanks.

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2. Jun 15, 2009

### minger

What happens when you expand your control volume such that the left boundary lies in P1 and the right boundary lies in P2?

3. Jun 15, 2009

### tiredryan

Hey,

Thanks for your response. I am little confused about the question. From the textbooks I have come across they have drawn the figures similar to bernouli3.jpg. From Bernoulli's equation it seems to be that P1=P3>P2 and V1=V3<V2. From Hagen–Poiseuille equation I would expect that fluid would flow from high to low pressure, from P1 to P2 and from P3 to P2.

Something that makes more sense would be something similar to figure bernouli2.jpg where P1>P3 and V1<V3. Maybe I am missing something very trivial.

Thanks.

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• ###### bernouli2.jpg
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4. Jun 16, 2009

### ank_gl

I don't remember it flowing from high to low pressure, it flows from high to low potential(or energy, as you might want to call it).
In post#3, you are correct in saying that P1>P3, as the pressure drops due to pressure losses(viscous drag & pressure drag), but the flow you assumed is partially correct, flow does expand, but does not recover fully to its initial state(this is called incomplete pressure recovery), pressure is higher at 3 than 2, but isn't equal to 1.

5. Jun 16, 2009

### tiredryan

Thanks ank_gl.

So I tried to draw out what you said in a diagram (bernouli4.jpg). So it flows from high to low potential as seen in the top graph. Also by Bernouli's equation the constricted area has a lower pressure.

Now here's where it gets confusing. I isolated the region of straight tube before the constriction and showed it in another figure (bernouli5.jpg). How would I apply Hagen-Poiseuille equation if dP/dx = zero in this region. The Hagen–Poiseuille equation is:
Q = ((Pi*r^4)/(8u))*(dP/dx)
So the flow rate is zero if the pressure is constant and the gradient is zero.

Am I missing a key assumption or something trivial? Thanks.

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6. Jun 16, 2009

### minger

What I was referring to is a key concept of fluid dynamics, that of a control volume. A control volume is kind of like putting a black box around something. Once this box is in place, you don't really care about what happens inside. There could be loop-d-loops and barrel rolls, but its inside the box. All that matters is what comes in vs. what goes out.

If you draw your control volume such as I mentioned, then the nozzle effectively doesn't exist (yes yes, there will be losses due to friction.

Now, in your image, you are failing the principle of continuity, which states that mass flow in equals mass flow out. Recall that:
$$\dot{m} = \rho A V$$
edit: tex still jacked up
m_dot = p A V
Assuming that density stays constant, if A1 and A2 are the same, how can V2 be greater than V1?

7. Jun 16, 2009

### ank_gl

Yes, you are. But that isn't any assumption, its a bit of theory.

Pressure again in a flow is made of two separate parts, static pressure & dynamic pressure. It is the dynamic pressure which remains constant, static pressure is what keeps the flow flowing. The P in term dP/dx is the gradient of static pressure, which balances the fluid friction. (Remember the flow is steady, therefore net force in X direction is 0, so what is balancing fluid friction??)

I missed the part where you mentioned V3<V1
As minger pointed out the principle of continuity, V1 = V3, all the energy loss across the control volume occurs as the pressure loss, velocity remain constant, if A1 = A3

8. Jun 16, 2009

### tiredryan

[edited; corrected figures and updated to reflect reading]

Hey,

Thanks. So I've read more and here's what I'm getting.

So is it correct to say that in Bernoulli's equation, P+(1/2)pmv^2+pgh = constant, that P equals static pressure and the (1/2)pmv^2 equals dynamic pressure?

And is it correct to say that in the Hagen–Poiseuille equation, the dP/dx refers to the static pressure+dynamic pressure but the dynamic pressure is constant so gradient is cause only by the static pressure?

The difference in the applications of these two equations is that Bernoulli's equation assumes a frictionless environment whereas the Hagen–Poiseuille equation requires a friction loss which is compensated by a gradient in static pressure. Is that correct?

So the first attached figure (benouli6.jpg) shows the pressures assuming there is friction so that there requires a gradient in static pressure to keep the flow constant. On the other hand, the second attached figure (benouli7.jpg) shows the pressures assuming there is no friction so that the total pressure is constant. Since the flow is steady and there is no friction, then there is no pressure gradient in the second figure. Is this correct?

Thanks for your help. I greatly appreciate it.

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9. Jun 16, 2009

### ank_gl

Correct. Though the equation has an extra m, it shouldn't be.

yes it is.
Correct
Correct.

10. Jun 16, 2009

### tiredryan

Hey,

Wonderful! Thanks ank_gl and minger for your responses!

Thank,
Ryan