Understanding Rigid Column Theory in Fluid Mechanics

In summary, the conversation discusses the concept of rigid column theory and how it relates to pressure and fluid flow in a closed or opened valve. The question asks about the relationship between the head difference, ΔH, and the head loss, Hf. The conversation also touches on the concept of pressure surge and the calculation of ΔH in different scenarios. Ultimately, the main focus is on understanding the dynamics of fluid flow in a closed or opened valve and the role of ΔH in this process.
  • #1
axe34
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0
Hello, please see attached text below - this concerns ''rigid column theory''. When a valve is closed, the pressure rises at the valve; the pressure rises above the reservoir pressure by ΔP with a height increase ΔH above H. When the valve is opened, this right hand pressure will disappear and the fluid will start to flow through the open valve.

Why is ΔH = H - Hf?
upload_2017-1-2_9-52-28.png


upload_2017-1-2_9-52-43.png


upload_2017-1-2_9-52-59.png
upload_2017-1-2_9-53-11.png
 
Last edited:
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  • #2
So the driving force / ##\left ( {\pi\over 4} d^2 \rho g \right ) \ \ ## ΔH is the head H at the entrance of the pipe minus the head loss in the pipe.
 
  • #3
I've drawn so many diagrams and simply cannot show that delta H = H - Hf!
 
  • #4
That a question ?
 
  • #5
The explanations given in the attached document (the first post in this thread) are poor in my opinion.
Here is how I interpret it: when the valve closes, it seems to imply that a piezometer at the valve with show height H + ΔH
upload_2017-1-5_13-45-15.png


Thus, if we define positive direction to the right,

∑forces on the fluid in horizontal pipe = m.a

(ρgH + ρ0).A - (ρg(H+ΔH) + ρ0).A = ρAL. dv/dt where: ρAL is the mass of fluid in the horizontal pipe and ρ0 is atmospheric pressure (A is pipe cross section)

This gives ΔH = - L/g dv/dt which implies that when delta H is positive, then there will be a deceleration of flow. Similarly, upon valve opening, we will see:
upload_2017-1-5_13-45-44.png

If positive direction still to the right,
∑forces on the fluid in horizontal pipe = m.a

(ρgH + ρ0)A - (ρg(H-ΔH) + ρ0)A = ρAL. dv/dt

thus ΔH = L/g dv/dt which implies that when delta H is as shown on the diagram, then there will be a deceleration of flow. Now, I always though that head loss was as shown in the following diagram:
upload_2017-1-5_13-54-5.png


I've drawn the energy grade line (EGL) and hydraulic grade line (HGL) as the piezometer. ΔH, as defined in diagram 2 on this post is shown in white. Delta H is not equal to H - Hf (Hf also called HL).

upload_2017-1-5_13-49-58.png


So, where have I gone wrong?
 

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  • #6
To be honest, I so far only reacted to the part of your question I could recognize (##H_f##). For the pressure surge I have to read up too.
Text says they close the valve gradually, so the stuff keeps flowing. Instead of assuming a linear deceleration as in the link, the book stops with (2.2a), basically ## F = ma ##. So I agree with you, up to and including
axe34 said:
This gives ΔH = - L/g dv/dt which implies that when delta H is positive, then there will be a deceleration of flow.
Note that in fact so far we haven't said anything about the magnitude of ##\Delta H##.
Rememer that ##\Delta H## is a difference.

Upon valve opening, the inital head at he valve is H and the valve has to counter it with a force ##\rho g H A##. Any form of opening the valve lowers the force the valve can exercise on the liquid (the head at the valve), so there is a negative ##\Delta H## available to accelerate the cylinder of liquid.

So imho the baseline for ##\Delta H## in your second picture is the top line and ##\Delta H < 0##.

As such, I see no wrong in the treatment in the text: there are no excessive claims, just manipulation of a few equations.

Your third diagram depicts a steady state: ##{du\over dt}=0## so there is no ##\Delta H##.
 

1. What is a rigid column in fluid mechanics?

A rigid column in fluid mechanics refers to a hypothetical structure that is assumed to be completely rigid, meaning it does not deform or bend under the influence of external forces. This simplification is often used in fluid mechanics analysis to make calculations and models easier.

2. How is a rigid column different from a flexible column?

A flexible column is one that can bend or deform under the influence of external forces, while a rigid column does not. In fluid mechanics, the assumption of a rigid column is often used to simplify calculations, while a flexible column may require more complex analysis.

3. What are the applications of rigid column fluid mechanics?

Rigid column fluid mechanics is commonly used in the analysis of structures such as dams, bridges, and buildings that are subjected to fluid forces. It can also be applied in the study of fluid flow in pipes, channels, and other engineering systems.

4. How is the behaviour of a rigid column in fluid mechanics represented?

A rigid column in fluid mechanics is typically represented using mathematical equations and principles, such as Newton's laws of motion and Bernoulli's principle. These equations are used to analyze the forces and pressures acting on the column in a given fluid environment.

5. What are the limitations of using the rigid column assumption in fluid mechanics?

The rigid column assumption is a simplification that may not accurately represent real-world scenarios. In some cases, it may lead to errors in calculations and analysis. Additionally, the assumption may not hold true in situations where the column is subjected to extremely high or dynamic forces.

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