Fluid Mechanics Problems, "Flow in Pipes"

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Yes I understand what you saying, but the thing is i'm doing a self study, and it not that I need answers but I need to discuss with someone who have a better understanding in fluid, that not an assignment or homework it just a question I came across with...

My mistake. Sounds a ton like a homework question to me.

Probably has something to do with balanced flow or something.

Maybe somebody will take the time to help you. I have trouble balancing my checkbook.

It cool,I thought as much, hey thanks anyway this site is helping, I just went through it thanks alot, well i'll patiently wait for that person...

He said he was trying to understand and came to us and this site for help. I get your point but in a way this IS doing his homework. He wants to Understand. He is not looking for is the answer A or B or C or D. He wants to know. I see postings like yours often. I do not know if the guy did well in school or not. But I would like to know the answer and will look for it and one way is to see if anyone else here may know. Now. When I was a teen I was learning not only to drive Large trucks but also I wanted to be a mechanic. And I ended up being a mechanic and I never went to school for it. I learned from those that knew. Eventualy I went to a school ONLY to get the degree and I figured I would learn something. But guess what. In many cases I knew more than the instructor did. We did not have internet back then or even BBSes. I learned from ANYONE that would talk to me that may have the info I needed until it was clear in my mind. After that I was eventually in charge of Maintainance of a fleet of 42 Wheeler trucks. We had some small 18 Wheelers as well but most were much larger. Now. Are you stating that I am a cheat?? I could take down and overhaul and engine as I understood it. That is what this kid is asking. HE is Asking to UNDERSTAND. Not what is the correct answer for a test. Cesare

The height of the water behind the dam is the relevant height for the potential energy calculation, even if a given parcel of water is collected at the bottom of the dam. This is because the pressure of the water above provides the motive force. In the absence of turbines or other restrictions, the water would emerge from the penstock at a velocity of v = sqrt(2gh) so that a flow, F, would require an area A = F/v. For example, Hoover Dam, at 222 m high (in the days when Lake Mead was full!) would eject water at a stunning 67 m/s (150 m.p.h.) if a big hole opened up in the bottom. At the nominal flow rate of 1000 m³/s, this corresponds to a hole about 4 m in diameter. adjust your numbers accordingly

These might help....

http://www.engineeringtoolbox.com/hazen-williams-water-d_797.html

http://www.esha.be/fileadmin/esha_files/documents/publications/GUIDES/GUIDE_SHP/GUIDE_SHP_EN.pdf

I'm not sure of the specific meanings being used here, but will guess that the gross head of 260 m equals the available head plus the losses. Since the loss is 1/3 the available, this works out to 195 m available and 65 m frictional loss.

From the 6 MW and 195 m available head, you can work out the flow needed to deliver the power. (Possibly including a factor for turbine efficiency.) From 1/3 of the flow per pipe, the 65 m of friction loss, and 3000 m pipe length, you can work out the pipe diameter.

Hazen-Williams is one commonly used formula; Darcy-Weisbach is another; possibly even Colebrook.

I do not know the answer yet but am thinking about it and watching to see if there are any good answers. And what I want to say is sometimes someone in some industry writes a statement and no one knows what it means for sure due to the wording. In this case it most likely can be figured out. In fact I am certain of it as all of the data is there. But my point is to the jerk that says "Do your Own homework" and to the other Jerk that said it sounds like homework Well You are not asking for the answer that would give you a grade. You made it clear you want to understand. As long as the goal is to understand and also in collage one thing you learn is to learn how to find answers where ever you can get them. So even if you were in school I see no problem as long as the goal is to understand it. Considering what you are doing I would say you already know a great deal. The wording of that statement is from someone that has a strange way of communicating. But that is part of communication. And communication is part of learning. Cesare

Let him talk for himself....we don't need somebody to interpret....

We don't spend all this time on here to not help people, that's pretty much all we do...It sure sounds like a test question to me, but we still have no problem helping the OP to understand the problem and find the answer...We get a lot of students looking for shortcuts on here, we're happy to help, but like the man said, this isn't a cheat site....You seem to want us to believe you can read the OP's mind,,, is that so....

Oh sure I forgot you are running the show here. And your are conflicting yourself. It does not mater where he gets the info even if he is in school and this is part of a test as long as he gets And most of All understands it. That is in fact part of what the University system does. Not all answers are in the books. You point him to other sites. What is wrong with this one. Anyway you think your way and I will think mine. Also and most important. I am right and you are wrong. Cesare

By all means, point HIM in the right direction. So far, all you've done is spew bile.

Which personality are you this time? Never mind.

I'll leave you to your righteous indignation.

<unsubscribe>

Based on your Jackass posts I can say I do not like you and I do not like anyone that does.

You seem very arrogant . In fact you flat are.

But you also are closed minded and rude. Never did like folks like you. Never will.

And I see where you go several that are on this post are on other sections posting with you. All Together. I find that curious. Yet you accuse the man that posted of this action.

Spot on Cesare, you understand me very well, people here don't get my last lines, I wanna understand the question actually the whole concept of Fluid concerning questions like this,the reason I wrote as an "Homework question" I want someone who is an expert of fluid mechanics to get the source of my question I mentioned in my last line of my topic, thanks Cesare for understanding

What is this a noob tagteam?

Their writing styles do seem oddly similar.

You've gone a bit quiet! Did any of the posts help? What answers have you calculated?

So far, you've wasted a lot of words spewing sour grapes and pounding your chest , while saying absolutely nothing of value to the OP.

Way to go! You two belong together.

I take it as a compliment to be called a jerk by such as you.

Have a nice smug life.

I contributed. And there was no chest pounding until now. As far as enjoying my smug life. I am. Your not. Shove it bitch. And you and Solar idiot should get married. Also I did go to school and got a PhD But I learned to be a mechanic and did that for a trade for years before school and I learned from folks that knew. Not from folks like you. Cesare

phD.....let me guess, pretty handsome dog..?

not everyone went to "collage" like yourself

That's prolly why he thought he didn't need to read the CR4 FAQ

14) Are there specific rules for posting cause he's an "old timer.

His join date is 07/26/2014. Wait, that's................................... What? Today.

He's been here for less than 24 hours????

Never mind, the OP's in good hands, now.

no one decides to build a power plant and begins with..."I have 3 pipes.......its a poorly worded homework question

I would approach it as you are - since there are 3 identical piping systems, just calculate for one. One pipe carries 1/3 of the water so treat it as supplying a 2MW turbine (6 * 1/3). But is the 6 MW the power input to the turbine or generated by the turbine?Without knowing the efficiency, only the input power can be assumed. And saying that the turbine makes more power when some of the head is lost to the piping sounds bogus, but so be it. OP knows the power, the head, and the head loss. Calculate the required flow rate to provide that much power. Then use the flow rate and head loss to calculate the pipe size.

I agree. But I do not think it was a question that was posed to the poster. I think it was a statement. But a poorly written statement. Maybe engish is a second language or maybe it is a manual that was translated to english. I have seen several from German to english that have statements that seem to make no sense. And english is my second language as well. But I have to agree that a full head is going to give you the most power not one third. But as you stated it may have meant one third after all parasite drag is considered. The way it is stated I do not get it. And it does NOT look like a textbook question as the arguments I was included in. AS a side note that is relevant. Grand coulee DAM in WA State. One cool DAM by the way. They had a maintenance man that worked on building the DAM. Then after it was finished due to his knowledge he had learned on his own during building the thing he got a job as a maintaince man. Then some Federal government official decided that since he did NOT have a degree he could no longer work there. Well. the men they hired that did have degrees caused a flood that almost destroyed the DAM. IT cam very close to total destruction. They ended up calling in the fellow they had fired that did Not have a degree but knew the dam from top to bottom and how this dam worked and how it was quite different from the others. Anyway he saved the DAM. If they had waited another hour it would have been all over for that remarkable piece of engineering and construction. It used to be one of the 10 wonders of the world when I was a kid. Have had a love of Hydro power since then. I have watched this thread for years but did not join until yesterday. Was too busy with other things.

I shall try to give a step by step explanation:The power of a turbine is the product between flow and pressure drop in your case P= Q[m^3/s]*Ht[m]*γ[N/m^3] in which γ[N/m^3]=ρ[kg/m^3]*g[m/s^2]You know it is water so that you know either the specific mass or the density and can compute the flow since the power is given by the problem text.The head loss is in the pipes due to the flow you computed. You know that the 3 pipes are in parallel so that the pressure loss will be the same and in each pipe flows 1/3 of the total flow you computed.Now you have to make sure which equation you use since there are 2 equations for the pressure drop in pipes.You have the flow, the pressure drop, the friction coefficient and the pipe length so that with the well chosen equation you can compute the diameter.You can verify if all is correct computing back the pressure drop with your values for flow, velocity, viscosity,aso.Three pipes with the same diameter and length deliver water from a reservoir to a turbine. Each pipe is 3000 long. The turbine develops 6 MW against a gross head of 260 m. Assume each pipe delivers a third of the water and the maximum power will be achieved when the head loss is one-third of the available head Neglect all losses except pipe friction and calculate the diameter and the velocity of water in the pipes when f = 0,005 My question is what do they mean if they say "...each pipe delivers a third of the water and the maximum power will be achieved when the head loss is one-third of the available head" Please make it clear for me on this, i'm trying to understand but I fail to do so

I don't know why you bother . I asked OP for an update on 27 July, nothing back and I doubt we'll hear from him again.

I was not on line for a long time so that I did not notice all details.

Thank you for the note.

You are right too many times we think we can help and do our best but without any return at least by politeness. We should be less involved.