Downward Turbulent Flow in a Vertical Tube: Pressure Change

[gamz95]

"For a particular flow rate, the pressure drop due to the downward flow of the fluid is
balanced by the pressure gain due to gravity—that is, at this flow rate, the static
pressure in the pipe is independent of the distance along the pipe."

Book says this. Can you explain how the pressure gained by gravity?

 

[256bits]

"For a particular flow rate, the pressure drop due to the downward flow of the fluid is
balanced by the pressure gain due to gravity—that is, at this flow rate, the static
pressure in the pipe is independent of the distance along the pipe."

Book says this. Can you explain how the pressure gained by gravity?

Here is a hint..
For a tank of fluid, say water, how do you find the pressure at the bottom of the tank?

 

[gamz95]

Pressure= rho*g*h; that is okay. But, rho doesn't change with the height, so how is it possible gaining by gravity?

 

[Chestermiller]

Pressure= rho*g*h; that is okay. But, rho doesn't change with the height, so how is it possible gaining by gravity?

Are you saying that the hydrostatic pressure in a tank does not vary with depth because rho is constant?

 

[gamz95]

No sir. I am okay with that pressure can vary with the height. But what I don't understand is how pressure only affected due to gravity. I had wanted to say gravity, but incidentally I wrote rho.

My problem is with :"by the pressure gain due to gravity"

.

 

[Chestermiller]

In your problem, the pressure is not varying at all. The pressure at the top of the tube is atmospheric and the pressure at the bottom of the tube is atmospheric. If you isolate the fluid within the tube as a free body, what are the forces acting on it?

Chet

 

[gamz95]

In that case, gravity forces(weights) and pressure forces acts.

Assume that fluid stands still at vertical position.Let's also say that the top of the tube has a pressure of P2, and the bottom has P1, and P1>P2 ; since it is deeper than top.
Then, assume now fluid starts to move downward; so what properties may change? I assume rho, and g wouldn't change. So how the pressure is gained by gravity?

 

[Chestermiller]

In that case, gravity forces(weights) and pressure forces acts.

Assume that fluid stands still at vertical position.Let's also say that the top of the tube has a pressure of P2, and the bottom has P1, and P1>P2 ; since it is deeper than top.
Then, assume now fluid starts to move downward; so what properties may change? I assume rho, and g wouldn't change. So how the pressure is gained by gravity?

I already told you that the pressures on the top and bottom P1 and P2 are equal, so their net force is zero. Besides the weight of the fluid, what other force is acting on it?

Chet

 

[gamz95]

There is friction forces also?

 

[Chestermiller]

There is friction forces also?

Yes. The viscous shear force at the wall (resulting from the turbulent flow) supports the weight of the fluid.

 

[gamz95]

Thank you for your understanding Sir. But why writer referred to the gravity, instead of saying the friction

 

[Chestermiller]

Thank you for your understanding Sir. But why writer referred to the gravity, instead of saying the friction

I have no idea what the writer is referring to. But gravity comes in because it determines the weight of the free body (downward force). The friction force is the upward force in the opposite direction which balances the weight.

The words "Can you explain how the pressure gained by gravity?" makes no sense either grammatically or physically.

So I can't account for what the writer wrote. All I can say is that I have confidence in what I said.

 

[mfig]

I have an idea of what the book is saying, but I am not positive.
Let's consider laminar flow since we have nice formulas. Turbulent flow will follow the same physical principles, but due to a different velocity profile the equations will be different. For laminar flow in a pipe inclined(+) or declined(-) at angle θ, the flow rate Q is given by:##Q = π⋅D^4⋅\frac{(Δp - ρ⋅g⋅L⋅sin(θ))}{(128⋅μ⋅L)} ##The second term in parenthesis has units of pressure, and so this is (probably) what your book is referring to as the pressure gained due to gravity. Note that if the angle is less than zero we have flow downward, and for a given Q the static pressure difference (Δp) required to maintain the same flow rate is smaller because of the effective pressure contribution due to the gravity term. Also note that if there is no pressure difference (Δp=0) then the L in the numerator cancels the L in the denominator and the flow rate is constant over any distance along the pipe, as your book describes.