Solving Water Pressure Problem with 8" Pipe and 90 Degree Turn

[meggs521]

Hey guys, I've been reading threads on here for a little bit and now I have a question I need some help with. I tried doing a search, but I didn't really know what I was looking for so it wasn't too successful. But here it is:

My boss gave me this question to figure out and I'm not really sure where to start. I'm interning with an engineering firm and the following question arose.

If we have an 8" Steam pipe that has steam at 80 psia and some of the steam condenses within the pipe forming a "plug of condensate" about 10' long, how much force will be applied on the end of the pipe if it goes into a 90 degree turn with a short radius?

The velocities I'm supposed to check this for are 4000 fpm, 6000 fpm, and 10000 fpm. I'm currently a sophomore in engineering, so I don't know too much about how to do this. Of course F=ma and all that, but without the acceleration I'm kind of lost.

I also know that in an 8" pipe the mass of water is approximately 21 lbs/ ft, so at 10' there's roughly 210 lbs of water. What sort of equation or method should I be using?

Any help would be very helpful as I'm pretty lost. Here's a picture of the pipe in case my description wasn't good enough.

Thanks again!

~Meg

 

[meggs521]

I also have two alternatives to the 90 degree bend. One is a 45 degree slant and the other simply has the pipe end with a blind flange. I've attached pictures of each of these as well.

Would I solve these two the same way as the above problem?

 

[minger]

Well his boss wants an answer, not an alternative...and I think we all know how bosses are. Anyways, this problem is basically a Converstation of Momentum problem. I have attached a diagram showing how I would go about solving the problem. If you would like to be more accurate, we can calculate a pressure drop across the short radius bend.

Anyways, basically, you have the momentum of the water being transferred to a load on your support. Mass flow is (kg/s), so multiplied by velocity (m/s) equals (kg m/s²), or a unit of force. You also need not to forget about the actual pressure in the pipe F= pA, and the actual weight of the water + pipe which will add to the component of the force in the y direction acting on your support.

edit: I have assumed that as the condensate flows through the elbow, that the entire elbow will be full of water, and thus can be treated for that particular instant as constant flow.

 

[meggs521]

Thanks!

For some reason I never think to use conservation of momentum. Maybe I should tatoo it on my hand or something.

 

[meggs521]

Quick clarification- if I'm wanting to know the force on the pipe itself, then I would just have the total weight be the weight of the water? I would only use the weight of the pipe itself if I'm wanting to find the force on the support, correct? As I understand it, all I want is the force on the pipe wall when the water hits it as it comes into the 90 degree bend.

Also, when I posted the "alternatives" my boss had asked me to determine those. Am I correct in assuming that the force in the x direction will be the same with the 90 degree bend as it would be with a blunt end (the flange)? Or am I'm oversimplifying the problem?

 

[minger]

Yes, I guess you're correct about the first part. Well...on the pipe itself. I'm not so sure if there is any real force from moving liquid on the pipe... Um..there will be some force, but because it's like every dx distance, the pipe curves a differential amount, and so there is a differential amount of momentum change.

You really shouldn't have to worry about too much 'force' on the pipe. Rather you should worry about the pipe overpressurizing and bursting. Other than that, you should look to make sure the pipe supports can withstand the momentum.

Anyways, about the alternatives, with the blunt end, from what it looks like, are you just considering like a cap on the end of the pipe? If so, there won't be any flow, and thus no momentum, so the only force you will have will be the pressure times the area.