Velocity equation for a cyclindrical tank and long pipe

In summary, the equation given in the question is a model for the transient motion of a fluid. The equation is based on the Bernoulli Principle, which states that the pressure inside a fluid container is constant. When a packet of fluid enters the container, the pressure inside the container increases, and the speed of the packet is proportional to the pressure.
  • #1
Taylor_1989
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TL;DR Summary
Trying to understand the origin of a velocity equation by deriving it.
I was not really sure as to whether to ask this question here or the h/w forum as its something I am personally trying to understand. However, I shall let the admin decided.

The question comes from a book called :
Numerical Methods for Engineers by Steven C.Chapra and Raymond P.Canale
. The specific question I have an issue with is not the question itself but the equation given within the actual question. For those of have access to the book, it is on page 142, prob 5.15.

Here is the actual question along with the given diagram:

As depicted in Fig. P5.15, the velocity of water, v(m/s)v(m/s), discharged from a cylindrical tank through a long pipe can be computed as:


##v=\sqrt{\left(2gH\right)}\cdot tanh\left(\frac{\sqrt{\left(2gH\right)}}{2L}t\right)##​
Diagram

1581432337306.png


What I have been trying to figure is where this equation comes from. Now granted my fluids knowledge is not fantastic, which is why I am trying to figure out how to derive this equation as a learning opportunity.

So in my effort to try and derive the equation I have been using an old text I used in the past along with a lab report I found online, but I am still drawing blanks in how this equation been derived.

The lab report I found online is given by the following link: https://www.ias.ac.in/article/fulltext/reso/023/01/0069-0081

Now reading through this report section 3.
Modification of (9) Using the Hagen–Poiseuille Equation

seems to have some relevance maybe on what I should be looking into but I can't seem to take the next steps and form the hyperbolic part of the equation.

I do have a working knowledge of fluids of Bernoulli’s Equation etc and the book I normally use a resource to anything fluids is
Fluid Mechanics: Fundamentals and Applications by Yunus A.Cegal and John M.Cimbla

Which actually have and exercise where you have to calculate the exit velocity but only uses Bernoulli’s Equation to do so.

If anyone could give me some idea on an approach to deriving this equation or where I should be looking to learn how to derive it, I would be much grateful.
 
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  • #2
Taylor_1989 said:
Summary:: Trying to understand the origin of a velocity equation by deriving it.

As depicted in Fig. P5.15, the velocity of water, v(m/s)v(m/s), discharged from a cylindrical tank through a long pipe can be computed as:


v=√(2gH)⋅tanh(√(2gH)2Lt)v=(2gH)⋅tanh((2gH)2Lt)v=\sqrt{\left(2gH\right)}\cdot tanh\left(\frac{\sqrt{\left(2gH\right)}}{2L}t\right)​
Hi
You do realize that the equation is about the transient motion of the fluid from time t-0 to steady state.
A mass acted upon by a force from initial velocity zero takes some time to reach a particular final velocity.

Here you have packets of dm of fluid entering the tube, being accelerated by the force from the pressure.
If you take time intervals of Δt for each packet to enter the tube, the force accelerates the first packet, then another packet enters after 2Δt and both packets are accelerated by the force, then after 3Δt another packet enters, and so on.
 
  • #3
256bits said:
Hi
You do realize that the equation is about the transient motion of the fluid from time t-0 to steady-state.
A mass acted upon by a force from initial velocity zero takes some time to reach a particular final velocity.

Here you have packets of dm of fluid entering the tube, being accelerated by the force from the pressure.
If you take time intervals of Δt for each packet to enter the tube, the force accelerates the first packet, then another packet enters after 2Δt and both packets are accelerated by the force, then after 3Δt, another packet enters, and so on.

I didn't at the time it was about transient motion, but after reading your comment I did a little bit more research and came across the Unsteady Bernoulli Equation which has to do with transient motion, I am not sure if this will result in the given equation that I posted above but it does seem hopefull the more I read.

I honestly I have never really gone into fluids the deep before but, I am finding it a fansinating subject.
 
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  • #4
There is a good discussion from the guys at MIT regarding your equation,
https://ocw.mit.edu/courses/mechani...ow-and-bernoulli/MIT2_25F13_Unstea_Bernou.pdf
Note the stipulation"large tank", and long pipe. Large tank so that the head, h, does not vary considerably as the flow exits, and the long pipe so that the time constant becomes "reasonably realistic".

Compare that with this discussion where the head h does drop as the flow exits.
Note that they mention the a/A ( area of pipe/area of tank stipulation, which is your case is small.
For their discussion the length of pipe is short or non-existent.
I am assuming they are correct in their derivation.
https://physics.stackexchange.com/q...rect-to-use-bernoulli-and-continuity-equation
 

1. What is the velocity equation for a cylindrical tank and long pipe?

The velocity equation for a cylindrical tank and long pipe is given by the formula v = (Q/A), where v is the velocity, Q is the flow rate, and A is the cross-sectional area of the pipe.

2. How is the velocity affected by the flow rate in a cylindrical tank and long pipe?

The velocity is directly proportional to the flow rate in a cylindrical tank and long pipe. This means that as the flow rate increases, the velocity also increases, and vice versa.

3. What factors can affect the velocity in a cylindrical tank and long pipe?

The velocity in a cylindrical tank and long pipe can be affected by several factors, including the flow rate, the cross-sectional area of the pipe, the viscosity of the fluid, and any obstructions or bends in the pipe.

4. How does the cross-sectional area of the pipe impact the velocity in a cylindrical tank and long pipe?

The cross-sectional area of the pipe has an inverse relationship with the velocity in a cylindrical tank and long pipe. This means that as the cross-sectional area increases, the velocity decreases, and vice versa.

5. Can the velocity equation for a cylindrical tank and long pipe be used for all types of fluids?

Yes, the velocity equation for a cylindrical tank and long pipe can be used for all types of fluids, as long as the fluid is incompressible and the flow is laminar. However, for turbulent flow, a more complex equation is needed to accurately calculate the velocity.

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