Velocity equation for a cyclindrical tank and long pipe

Click For Summary
SUMMARY

The discussion focuses on deriving the velocity equation for water discharged from a cylindrical tank through a long pipe, as presented in "Numerical Methods for Engineers" by Steven C. Chapra and Raymond P. Canale. The equation is expressed as v = √(2gH)·tanh(√(2gH)/(2L)t), which describes transient fluid motion. Participants highlighted the relevance of the Unsteady Bernoulli Equation and the Hagen–Poiseuille Equation in understanding this derivation. The conversation emphasizes the importance of considering transient effects and the conditions under which the equation applies, such as the size of the tank and pipe.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically Bernoulli’s Equation.
  • Familiarity with transient fluid motion concepts.
  • Knowledge of the Hagen–Poiseuille Equation for viscous flow.
  • Basic mathematical skills for manipulating hyperbolic functions.
NEXT STEPS
  • Study the Unsteady Bernoulli Equation to grasp transient fluid behavior.
  • Explore the derivation of the Hagen–Poiseuille Equation for practical applications.
  • Review advanced fluid mechanics resources, such as MIT's course materials on inviscid flow.
  • Investigate the effects of tank and pipe dimensions on fluid discharge rates.
USEFUL FOR

Engineering students, fluid mechanics researchers, and professionals involved in hydraulic system design will benefit from this discussion, particularly those interested in transient fluid dynamics and practical applications of fluid equations.

Taylor_1989
Messages
400
Reaction score
14
TL;DR
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.
 
Last edited:
Engineering news on Phys.org
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.
 
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.
 
  • Like
Likes   Reactions: 256bits
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
 

Similar threads

  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 20 ·
Replies
20
Views
11K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
995
  • · Replies 17 ·
Replies
17
Views
3K
Replies
8
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 4 ·
Replies
4
Views
3K