Determining Velocity of Water From a Pipe

Click For Summary

Homework Help Overview

The discussion revolves around determining the velocity of water exiting a pipe under pressure. The problem involves a pipe partially filled with water and air, with specific dimensions and pressure conditions provided. Participants are exploring the application of fluid dynamics principles, particularly Bernoulli's equation and linear momentum concepts, to analyze the situation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss using Bernoulli's equation, questioning the orientation of the pipe and the applicability of the equation given the changing mass and pressure conditions. There are attempts to relate pressure and force to velocity, and some participants suggest considering linear momentum methods as an alternative approach.

Discussion Status

The discussion is active, with various interpretations being explored. Some participants express uncertainty about the applicability of Bernoulli's equation due to the dynamic nature of the problem, while others suggest that momentum methods may provide a clearer path forward. Guidance has been offered regarding the relationship between pressure and velocity, but no consensus has been reached on the best approach.

Contextual Notes

Participants note the absence of a diagram, which may hinder understanding. There is also discussion about the pressure conditions at different points in the system, with some uncertainty about whether the pressure remains constant or changes over time.

alibond07
Messages
10
Reaction score
0

Homework Statement


I have a pipe with a capacity of 1.5 liters. Half of the pipe is filled with water and the other half air. The pressure in the pipe will be pumped to a certain value then the valve will be released. I want to determine the speed of the water as it leaves the pipe.

Variables known:

Length of Pipe:1.3m
Radius of cross-section:0.0075m
Pressure inside of pipe: 60psi
Nozzle diameter: 3mm

(Will disregard the nozzle if its makes determining the velocity too hard)


Homework Equations


Pressure*Area=Force
pΔV=Work Done

The Attempt at a Solution



Firstly I though I could approach the problem by thinking of the air in the pipe doing work against the 'piston' of water when the valve is released. However I couldn't determine the change in volume of the gas. Is there a way to do this knowing the dimensions of the pipe?

Next I tried to use Pressure*Area=Force. Then simply doing (413685*cross section)=73N

I know F=ma but I can't seem to link this to velocity.

Are my previous assumptions correct?
 
Physics news on Phys.org
my first guess was to use Bernoulli's eqn
bu for that we need to know if pipe is horizontal or vertical ...

isnt there any figure?
 
cupid.callin said:
my first guess was to use Bernoulli's eqn
bu for that we need to know if pipe is horizontal or vertical ...

isnt there any figure?

The pipe is horizontal. I had a look at Bernoulli's equation and I had 0% of an idea how to apply to my situation.
 
can you attach the figure because i can't really understand some things about the question
 
cupid.callin said:
can you attach the figure because i can't really understand some things about the question

What do you mean by figure?
 
figure i mean the diagram of pipe if given with the question ...
 
cupid.callin said:
figure i mean the diagram of pipe if given with the question ...

Sorry this is an experiment I carried out. I'd just like to be able to relate the pressure and the distance the water goes from the pipe. Is there information missing that you need?
 
after increasing the pressure when you open the valve ... so at nozzle pressure will become = atmospheric pressure ...

what about the piston end ... is 60psi maintained ... or it drops slowly (which will be very hard to work with) or it is also = atmospheric pressure?
 
cupid.callin said:
after increasing the pressure when you open the valve ... so at nozzle pressure will become = atmospheric pressure ...

what about the piston end ... is 60psi maintained ... or it drops slowly (which will be very hard to work with) or it is also = atmospheric pressure?

I'm afraid it drops slowly. The valve releases the pressure. The nozzle after the valve release is atmospheric pressure.

I'm willing to just say that 60psi is maintained if it is easier. I will simply add it too my analysis in my lab report.
 
  • #10
alibond07 said:
The pipe is horizontal. I had a look at Bernoulli's equation and I had 0% of an idea how to apply to my situation.

If you look at the first form of Bernoulli's equation as linked, you see that the LHS is constant along streamlines. That means that the expression is the same along the flow at a beginning point (take it to be the air-water surface in the pipe, where the pressure is known) and an end point (e.g., outside the nozzle, where the flow has fully accelerated and the pressure has dropped to ambient).
 
  • #11
As you know atmospheric pressure is 14.7 psi

so applying Bernoulli eqn ... (let point near piston is A and at nozzle is B)

P_A + \rho hg + \frac{1}{2}\rho v_A^2 = P_B + \rho hg + \frac{1}{2}\rho v_B^2

as pipe is horizontal so h is same

PB = Patmospheric
PA = 60psi

and for relation b/w vA, vB use eqn of continuity

hope this helps :)
 
  • #12
I am not sure Bernoulli is helpful with this problem. The mass and pressure in the cylinder is changing with time and although the water is discharged to atmospheric pressure I think there is a pressure at "B". This problem resembles a water rocket problem that is fixed to a stand. These problems are typically solved using Linear Momentum equations.
 
  • #13
RTW69 said:
I am not sure Bernoulli is helpful with this problem. The mass and pressure in the cylinder is changing with time and although the water is discharged to atmospheric pressure I think there is a pressure at "B". This problem resembles a water rocket problem that is fixed to a stand. These problems are typically solved using Linear Momentum equations.

Mass is changing, so?
and according to OP, pressure is not charging ...

can you tell us how we can solve this problem with linear momentum concept?

and i'll look at the question again and see if the solution i gave is correct or not ... :smile:
 
  • #14
cupid.callin said:
Mass is changing, so?
and according to OP, pressure is not charging ...

can you tell us how we can solve this problem with linear momentum concept?

and i'll look at the question again and see if the solution i gave is correct or not ... :smile:

RTW69 said:
I am not sure Bernoulli is helpful with this problem. The mass and pressure in the cylinder is changing with time and although the water is discharged to atmospheric pressure I think there is a pressure at "B". This problem resembles a water rocket problem that is fixed to a stand. These problems are typically solved using Linear Momentum equations.

Do you mean the simple (m1)(u1)+(m2)(u2)=(m1)(v1)+(m2)(v2)?

Can you show me how this would apply? I could work out the mass of the water but not it's initial velocity.
 
  • #15
Do a Google search on "Science bits water rockets" they have a good explanation of using momentum methods for analyzing water rockets. Your problem looks similar except your pipe (rocket) isn't moving. It is not clear to me that the pressure at your nozzle is atmospheric pressure.
 
  • #16
RTW69 said:
Do a Google search on "Science bits water rockets" they have a good explanation of using momentum methods for analyzing water rockets. Your problem looks similar except your pipe (rocket) isn't moving. It is not clear to me that the pressure at your nozzle is atmospheric pressure.


It should be atmospheric pressure? The water does not go all the way to the nozzle, only to a valve just before the nozzle.

Thanks for pointing me in the direction of that website, looks like it's going to be a great help.
 
  • #17
RTW69 said:
I am not sure Bernoulli is helpful with this problem. The mass and pressure in the cylinder is changing with time and although the water is discharged to atmospheric pressure I think there is a pressure at "B". This problem resembles a water rocket problem that is fixed to a stand. These problems are typically solved using Linear Momentum equations.
RTW69 said:
Do a Google search on "Science bits water rockets" they have a good explanation of using momentum methods for analyzing water rockets. Your problem looks similar except your pipe (rocket) isn't moving. It is not clear to me that the pressure at your nozzle is atmospheric pressure.

The first page returned by that Google search gives an analysis for the rocket exit velocity that finds exactly the same expression as found using Bernoulli. I submit that this happens because Bernoulli is derived by integrating the momentum equations. :)
 
  • #18
cupid.callin said:
Mass is changing, so?
and according to OP, pressure is not charging ...

can you tell us how we can solve this problem with linear momentum concept?

and i'll look at the question again and see if the solution i gave is correct or not ... :smile:

As you can see in my attachment I've been directed to something I think is going to help me.

I have a question. What does the p_{w} stand for? Is it power?

Thanks
 

Attachments

  • Nozzle Velocity and Thrust.jpg
    Nozzle Velocity and Thrust.jpg
    30.8 KB · Views: 439

Similar threads

Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K