# Fluid Mechanics - Pipe Flow

1. Mar 29, 2005

### theoriginalh

Hi there, I'm struggling with a question that ive 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

#### Attached Files:

• ###### Pipe.jpg
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2. Mar 29, 2005

### minger

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. Mar 29, 2005

### FredGarvin

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. Mar 29, 2005

### minger

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. Apr 1, 2005

### theoriginalh

Thanks very much for the help! Ive got the problem figured out now.

6. Apr 2, 2005

### theoriginalh

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.

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7. Apr 2, 2005

### FredGarvin

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. Apr 6, 2005

many thanx

9. Apr 7, 2005

### minger

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. Apr 7, 2005

### FredGarvin

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:

$$h_f = \Sigma K (\frac{V^2}{2g}) = f \frac{L}{D} \frac{V^2}{2g}$$

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. Apr 8, 2005

### abodaood

hi all

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

12. Apr 8, 2005