Fluid Dynamics 2 Tynea: Unsolvable Pipe Network Problem?

[tynea]

Homework Statement

A professor gave a pipe network problem to be solved by the Hardy-Cross method. However when I attempted to create a scale drawing of the figure to get a better idea of the problem I realized it is physically impossible to create his design. Now i am wondering if because of this if the problem is unsolvable? If anyone interested needs to see the network with dimensions I could take a pic of the problem statement and post it if that would be the easiest way. I am completely new to this site and this is my second post. Thank you any and all for any guidance.

Tynea

Homework Equations

The Attempt at a Solution

 

[SteamKing]

Why is the pipe network physically impossible to create? Remember, you are basically given a schematic layout which just shows how the network is connected. The pipes connecting the nodes do not necessarily have to be straight runs.

In any event, you have not shown any work setting up your calculation.

 

[tynea]

well, if you notice the diagonals are 2000' and the top run is 6000' thus it is impossible for those two pipes to be connected to each other at a single juncture. So if the diagram is physically impossible than how can any method solve this problem? I don't like wasting my time on an unsolvable problem given by lazy professor who doesn't look thoroughly into the problems they assign. This is why I posed my question to you professionals.
Furthermore, I tried to contact the professor who assigned this problem over spring break and he put his email to auto-reply and is not available to answer questions. I find this VERY unprofessional, lazy and self serving but he's tenured thus he can do no wrong.

 

[peterpiper]

Are you taking into account that each of the junctures are at different elevations? It makes the drawing a little easier to visualize.

As far as lazy professors, if this prof isn't a brand new one, he's probably assigned this before. That would imply that it has a solution. My Statics and Dynamics class had some nasty looking problems that we were convinced were impossible but just needed a new frame of reference. Have some faith in your professors, they more than likely know what they're doing.

 

[tynea]

Dear peterpiper,
I know I sound bitter, but I've had this prof for three classes now and know his style. He often answers question with a simple, "I don't know," and moves on. He also routinely teaches for 2 hr in what is supposed to be a 3 hour class. I'm also not a 19 yr. old, I'm 46 and went back to school and I expect to get my monies worth not, I don't know. And, lastly node D is at the same elevation and this is were I am saying those pipes CANNOT physically be connected. If the Hardy- Cross method worked on a fictitious design then what good would it be in the real world?

Dear Steamking,
I could post my spreadsheet if you like, but the numbers are way out of whack and I checked my formulas and they are correct. It is so far from converging its seems futile. this is why I posed the question concerning a reality check for this design.

 

[tynea]

Ok SteamKing,

here is my excel spreadsheet and my reworked diagram with the
-pipe area in blue,
-distance in green,
-Friction factor in red,
-loop and direction of positive in purple
-and the rectangular boxes in pencil contain my first assumptions of the flow.

Believe me I do appreciate any and all help. I am not looking for a handout, I'll have to do this again on the final and I want to understand it.

 

[SteamKing]

I appreciate that you are attempting to work through your difficulties with this assignment.

One thing you must realize is that hydraulic networks are non-linear by nature, and thus are more difficult to solve than say, an electrical network.

Often times, for non-linear problems, an iterative solution is the only practical method. The Hardy-Cross method that was recommended by your instructor is one such technique.

Your spreadsheet has the start of a solution. I recommend that you apply Hardy-Cross to it to balance the flows in the network.

 

[tynea]

Dear SteamKing,
I am going to proceed as you suggest. However, I would be grateful if someone could give me an analogy or an explanation as to how it is possible to solve an un-buildable network? If the Hardy -Cross method can solve this than doesn't that then reflect poorly on the method itself and its validity. I'm sorry if I seem thick headed about this, but I cannot seem to wrap my head around this fact... My background is carpentry, land surveyor and field engineer all my experience is with practical physical structures and esoteric imaginative problems are lost on me. Thank you anyone...

 

[SteamKing]

I can't help you with what you think about this problem. You have convinced yourself, honestly I think, that because the network cannot be built, the practice which you gain from applying the Hardy-Cross method to its solution will result in wasted time and effort.

As I mentioned in an earlier post, I think you are mistaken in believing that the length of the pipes in the network represent straight runs. If you have ever seen a large chemical plant or oil refinery, you know that in addition to straight runs of pipe, there are many twists and turns in the piping. I think the sketch attached to the problem represents a schematic layout which shows only how the different branches are connected. A lot of problems which are given to student engineers to solve are highly simplified and may contain elements which are idealized and don't necessarily represent what actual engineers are capable of building.

 

[tynea]

SteamKing,
It was converging to the 5th iteration then it started increasing again. do you think I should rethink my initial flow pattern. the summation started at 43, 24, 9.7, 3.99, 3.36, then it started back up again at 4.41, 5.87and now at 7.47... what the #$%&!
You seem like the only person willing to offer any help. Now I am not embellishing any thing i have spent more than 8 hours on this problem I am posting my actual spreadsheet. Please someone HELP!

 

[tynea]

spreadsheet from tynea hydrology prob

 

[SteamKing]

I looked at your spread sheet calculations.

It appears that when you calculate K = fL/D, you divide fL/D byan additional factor of 39.7*D. Why is this?

I may be wrong, but it doesn't appear you account for the difference in elevation of the various node points.

 

[tynea]

Dear StreamKing,
I took a photo of the page out of the textbook we use it shows the formula's and how they are to be applied based on what units you are given in the problem statement. In some previous problems we were given the C value. Then we have to use the Hazen- Williams formula. in this problem we were given the f (friction factor) thus the Darcy-Weisbach is called for also given the units a more specific variation of the formula is used. I contacted a friend from class and he did the elevation portion of the problem after he got the proper flow per pipe and this is the way the books prescribes it be done. However, I find this particular textbook a bit lacking especially in the example dept. The professor also expects 0.0001 accuracy before he considers the problem solved as far as flow (Q) is concerned. I did a new spresdsheet assigning new initial flow values and it blew up on me as well after the 5th iteration. Do you also think there might be a problem with the way I add the negative values to represent the flow direction in each new iteration. Thank you again SteamKing, by the way are you guys out west the time on this web page is certainly not EST?

Tynea

 

[tynea]

hydro prob 2

This is the 2nd spreadsheet i created late last night the new iterations are on "sheet 2"

 

[SteamKing]

I'm not sure, but I think the times are GMT.

 

[SteamKing]

The attached file shows a worked out example of a flow network using the Hardy Cross method. It should help you check your iteration method from your spreadsheets.