Fluid Mechanics - Pipe Flow

In summary: The user "cristo" gives an answer for a 90 degree bend. You should be able to modify his solution for a 135 degree bend.In summary, the conversation discusses a question about the force on a bend in a pipe. The question asks for the upstream pressure and the magnitude and direction of the force on the bend, as well as the loss of head in the pipe bend. The solution involves using Bernoulli's equation and conservation of momentum. The conversation also mentions a similar problem involving a junction in a horizontal pipe, where the force on the junction must be calculated using the momentum and pressure components. The conversation also touches on the concept of minor losses in a pipe system.
  • #1
theoriginalh
5
0
Hi there, I'm struggling with a question that I've been given regarding the force on a bend in a pipe, and I wondered if any of you guys could help me out with a few clues. The question is as follows:

A contracting pipe bend turns water through 135° (In a horizontal plane) before discharging it to the atmosphere. The inlet diameter is 80mm and the outlet diameter is 60mm. The volumetric flow rate is 20 Ls

a) Neglecting energy losses, calculate the upstream pressure and the magnitude and direction of the force on the bend.

b) If the upstream pressure is actually 24kPa, calculate the loss of head in the pipe bend.

Diagram below


I really don't know where to get started with this problem, so any help would be welcome.

Thanks
 

Attachments

  • Pipe.jpg
    Pipe.jpg
    6.5 KB · Views: 1,279
Engineering news on Phys.org
  • #2
Well to calculate the upstream pressure, you will just equate Bernoulli's Equation. It should be simple enough to find your two velocities. I assume for part a) you are neglecting frictional head losses, so just solve for your other pressure.

To calculate the force on the bend, you will use conservation of momentum as I showed in the post below this one. It isn't a 90° though, so you will have a couple extra components. (i.e. the force in the x-direction on the bend will not only be the momentum of the incoming fluid, but also of the x-component of the exiting fluid, plus incoming pressure, minus exiting pressure in the x-direction).

For part b, you will simple use Bernoulli's again. You are now given intial and final pressures, and you have already calculated your velocities. The only thing now is to find your head loss.
 
  • #3
Won't the "p*a" components both be zero for the outlet (i.e. exiting to atm)? I think all you'll have is the momentum components in the X and Y at the exit along with the Y support force. Since they don't give a weight for the pipe, I wouldn't worry about the weight of the pipe and water.
 
  • #4
Oh, discharging to atmosphere. I need to learn to read. Yes, your pA term for the exit will be 0, but it will not for the inlet though. You should have some positive gage pressure at the inlet.
 
  • #5
Thanks very much for the help! I've got the problem figured out now.
 
  • #6
Hey sorry, one more question. I'm having another problem with a pipe flow question. I guess that its similar in ways to the last problem, using bernoulli and suchlike, but although I have a few ideas I'm not positive about how to go about solving it. A few hints would be appreciated.

A junction splits a horizontal pipe of diameter 80mm into two horizontal pipes of diameter 60mm and 50mm at 30 degrees to the original flow direction (as shown in the picture). The junction is arranged so as to split the flow in the ratio 2:1 between these two pipes. For water flow, if the upstream pressure and velocity are 10 kPa and 5 ms respectively, and the outlet pipes discharge to the atmosphere, calculate the force on the junction in the x-y coordinate system shown.

If anyone can help with this one, thanks.
 

Attachments

  • Y Junction.jpg
    Y Junction.jpg
    9.1 KB · Views: 4,711
  • #7
There's no realdifference between the two problems. You have to look at the momentum coming in and going out in the X and Y components and balance that against the pressure in all of the parts. It's the same as before, but with one more term in each part of the equations.
 
  • #8
minger said:
The only thing now is to find your head loss.

cAN u tell us please how to find the head loss.
many thanks :biggrin:
 
  • #9
Well I would assume that you can neglect frictional head loss in the pipe due to the fact that there's probably only a couple of meters there. So, the only head loss you will have is due to the tee. You will have to look up in a table to find a resistance coefficient or something of the like. Then you can plug that head loss into your Bernoulli's now to find the rest of the stuff.
 
  • #10
The frictional loss is a minor loss that is an additional term in the energy balance equation for flow. Every type of component in a system will have one of two things (depending on the reference you use) to describe it's effects with respect to friction. They will have either a "K" value or they will have an equivilent length value (L/D). They eventually eneter into this term:

[tex] h_f = \Sigma K (\frac{V^2}{2g}) = f \frac{L}{D} \frac{V^2}{2g}[/tex]

The equivilent length is a length of straight pipe that would produce the same loss as the fitting, valve, etc... that you are considering. It's much easier to find items in terms of the K value if you can, but that is not always possible.

Which example are you trying to find the minor losses for?
 
  • #11
hi all
Minger... thanks for ur reply...


FredGarvin, well ur explanation is very obvious.( am trying to do the first example part b). However, i did search for the K value of a 135 degrees bent pipe through various sources but i couldn't find it. At the same time we're not given the length of the pipe...

so any further procedures may solve this problem??! please

many thanx
 

1. What is the difference between laminar and turbulent flow in a pipe?

Laminar flow refers to a smooth, orderly flow of fluid in a pipe, where the fluid particles move in parallel layers with little to no mixing. Turbulent flow, on the other hand, is chaotic and unpredictable, with fluid particles moving in various directions and mixing together.

2. How does the diameter of a pipe affect fluid flow?

The diameter of a pipe has a significant impact on fluid flow. A larger diameter pipe allows for a higher flow rate and lower pressure drop, while a smaller diameter pipe will have a lower flow rate and higher pressure drop.

3. What is the equation for calculating pressure drop in a pipe?

The equation for pressure drop in a pipe is: ΔP = f*(L/D)*(ρ*v^2/2), where ΔP is the pressure drop, f is the friction factor, L is the length of the pipe, D is the diameter, ρ is the density of the fluid, and v is the average velocity of the fluid.

4. How do you determine the flow regime in a pipe?

The flow regime in a pipe can be determined by calculating the Reynolds number (Re), which is a dimensionless number that compares the inertial forces to the viscous forces in the fluid. A Re value below 2300 indicates laminar flow, while a value above 4000 indicates turbulent flow.

5. What is the difference between steady and unsteady flow in a pipe?

Steady flow refers to a condition where the fluid properties (velocity, pressure, etc.) at any given point in the pipe remain constant over time. Unsteady flow, on the other hand, describes a situation where the fluid properties are changing with time at a specific point in the pipe.

Similar threads

  • Mechanical Engineering
Replies
20
Views
7K
  • Mechanical Engineering
Replies
31
Views
2K
Replies
3
Views
215
  • Mechanical Engineering
Replies
4
Views
814
  • Mechanical Engineering
Replies
2
Views
857
Replies
8
Views
1K
  • Mechanical Engineering
Replies
8
Views
361
  • Mechanical Engineering
Replies
4
Views
2K
  • Mechanical Engineering
Replies
11
Views
1K
  • Mechanical Engineering
Replies
3
Views
1K
Back
Top