# I have a hydrodynamic loss head question

• Hauzen
Hauzen
Hi!
I have a loss head question.
There is a difficulty in understanding the hydrodynamic loss head concept.
Assuming that there is a one-sided tube with incompressibility, visibility, tube friction coefficient and cross-sectional area A..
There is friction in the loss head formula, but is the flow rate and flow rate of the Inlet and Oulets the same?
I know that continuous equations are useful when there is no friction.
I don't understand that there is friction, but the flow rate of the Oulet does not decrease. And in the loss head, it is stated that v is the average velocity, but is it the v of the inlet? Is it the v of the outlet?
I don't understand the concept exactly.

Last edited by a moderator:
Hauzen said:
There is friction in the loss head formula, but is the flow rate and flow rate of the Inlet and Oulets the same?
It has to be. Where would the fluid go otherwise?

Hauzen said:
I don't understand that there is friction, but the flow rate of the Oulet does not decrease.
But the pressure is not the same.

Hauzen said:
it is stated that v is the average velocity, but is it the v of the inlet? Is it the v of the outlet?
The average velocity remains the same but the velocity profile changes:

https://i.stack.imgur.com/fTEdG.png

Last edited by a moderator:
berkeman, Hauzen, Chestermiller and 1 other person
Welcome, @Hauzen !

The problem is that the internal energy of the fluid, not the volumetric flow, is degraded by friction.

That internal energy is what pushes the vertical columns of fluid up; therefore, those column's heights are proportional to the static pressure at each cross-section, or number of molecules collisions/time-area, in the direction that is perpendicular to the main flow.

The walls of the tube at the cross-section where the vertical tube is located "feel" the same static pressure than the lowest section of the vertical column, but no velocity.

Hauzen said:
Hi!
I have a loss head question.
There is a difficulty in understanding the hydrodynamic loss head concept.
Assuming that there is a one-sided tube with incompressibility, visibility, tube friction coefficient and cross-sectional area A..
There is friction in the loss head formula, but is the flow rate and flow rate of the Inlet and Oulets the same?
I know that continuous equations are useful when there is no friction.
I don't understand that there is friction, but the flow rate of the Oulet does not decrease. And in the loss head, it is stated that v is the average velocity, but is it the v of the inlet? Is it the v of the outlet?
I don't understand the concept exactly.View attachment 341555
In fully developed, steady, incompressible, viscous flow the velocity profile is not changing between 1 and 2. However, the internal energy of the flow is being lost to heat, warming the flow, and (usually) escaping to the environment. The pressure is decreasing along the pipe because of this heat generation. The bulk kinetic energy flow is unchanged between 1 and 2 in the setup shown i.e. no change in velocity.

The equation that most basic engineering application utilize is to perfom the energy acounting between 1 and 2 is:

$$\frac{P_1}{\rho g} + z_1 + \frac{V_1^2}{2g} = \frac{P_2}{\rho g} + z_2 + \frac{V_2^2}{2g} + \overbrace{\sum_{1 \to 2} K_i \frac{V_i}{2g}}^{>0}$$

Where ##K_i## is a variety of functions that may (or may not) be dependent on several factors like, pipe roughness, flow speed, pipe geometry, etc...

Last edited:
Hauzen and Lnewqban
jack action said:
It has to be. Where would the fluid go otherwise?

But the pressure is not the same.

The average velocity remains the same but the velocity profile changes:

Thank you for your detailed explanation. :)
After studying the concepts of static pressure and dynamic pressure, I came to understand them.
Then, because the length of L is infinitely long, does the dynamic pressure energy decrease from the moment the static pressure energy becomes zero?

And if the axis is set up on the y-axis rather than the x-axis, will it be the same?

erobz said:
In fully developed, steady, incompressible, viscous flow the velocity profile is not changing between 1 and 2. However, the internal energy of the flow is being lost to heat, warming the flow, and (usually) escaping to the environment. The pressure is decreasing along the pipe because of this heat generation. The bulk kinetic energy flow is unchanged between 1 and 2 in the setup shown i.e. no change in velocity.

The equation that most basic engineering application utilize is to perfom the energy acounting between 1 and 2 is:

$$\frac{P_1}{\rho g} + z_1 + \frac{V_1^2}{2g} = \frac{P_2}{\rho g} + z_2 + \frac{V_2^2}{2g} + \overbrace{\sum_{1 \to 2} K_i \frac{V_i}{2g}}^{>0}$$

Where ##K_i## is a variety of functions that may (or may not) be dependent on several factors like, pipe roughness, flow speed, pipe geometry, etc...
Hi.

If it is assumed that it is normal air, not incompressible, can the Inlet and Outlet flow rates be different if the pressure drop is large?

Hauzen said:
Hi.

If it is assumed that it is normal air, not incompressible, can the Inlet and Outlet flow rates be different if the pressure drop is large?
Mass flow rate. No. Not in steady compressible flow. What happens is the velocity and and density change along the length of the pipe.

Mass is still conserved.

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