Timmy the EE's WR Question

[Timmy!]

Six-Minute Solutions, 3rd Edition, Question 26:

Given: Concrete rectangular channel [assume n=0.013], 8m wide by 1.5m deep, velocity = 2.5 m/s at 0.002m/m slope. Transition 50m beneath a highway [DISTRACTOR!!!] to a circular 9 FOOT diameter concrete concrete culvert of same slope and "n" value, flowing half-full.

Find: Number of 9' diameter culverts required.

The answer is "four", but that is beside the point.

My initial inclination was to use Manning's equation to find Qchannel upstream, then find Qcircular [9-foot diameter half-full] for a single pipe downstream, and then divide and round up to the nearest whole number to determine the number of 9-foot diameter pipes required.

NOT!!! Intuitively wrong again. I spent two hours working on this to gain conceptual understanding, and the best I can figure is that the velocity changes downstream [to 2.67 m/s] which negates this concept.

Does this make sense? The solution offered is to find V2 downstream, determine A2 for a single circular 9' pipe, and use the continuity equation. Is it true that the change in downstream velocity is the reason why my initial inclination is wrong?

I wait with bated breath for the wise and sagacious CEs to enlighten me...

 

[Guest]

Timmy,

I would like to start out my response by stating this is a !@#$%&*%$# problem. The question, structure, and the need to convert from metric to english units is beyond absurd for the purposes of a 'six-minute' problem. I wanted to put that out there to start with because I did want to relieve some of the panic you must have at seeing problems like this one and thinking 'impossible'. I will soften the blow by also saying that these problems are very good to work through because they help you understand inter-related concepts ---> that is very important.

Okay, on to the problem. I have a copy of my Six-Min Solns at my house, but not my refereneces

. I will have to update my post tomorrow for my complete response.

However, I would like to address your initial approach to the problem. You noted that there were some distractors in the problem. That's good - you will need to be able to filter necessary information and parameters from extraneous ones quickly when you are taking the exam.

You indicated, "My initial inclination was to use Manning's equation to find Qchannel upstream ..."

If you look at the information you were given about the rectangular channel, it states length, width, and velocity. Because the velocity is given in the problem statement, you know what the conveyance of the channel will be based on Q = vA ---> you don't need to using the Manning form of the equation because the slope and roughness of the channel are not being brought into the fold based on the problem statement.

So, as far as the channel is concerned, the conveyance (flow) = 30 m3/s. So, by using the continuity equation we can equate the conveyance for the channel to the conveyance of the culvert pipe system. I will need my references to complete the explanation, but wanted to give you an initial point of view so you can work with it.

*** The Rest of the Story ***

I took the oppurtunity to look at the Errata, and I noted that "the other board" corrected the answers for Prob #26 to show that all of the pipes have a 9-ft diameter. This helps to simplify the problem and make it more rationale. I have attached a pdf of the "the other board" Errata for WR to this post.

So working with what we know, the conveyance (flow) of the system = 30 m3/s which = 1060 ft3/s converted to english units. In terms of ease, I would approach the remainder of the problem in this way:

I know I am going to approach the pipe flow by using the Manning Equation. The general equation for the flow is Qfull = 1.49/n*{*AR2/3S1/2}. However, I know that I am dealing with a half full pipe. Instead of trying to calculate the changes in factors - I use Appendix 19.C, Page A-37 (Circular Channel Ratios).

---> Tab this one !! It will save you a lot of time if you learn how to use this chart.

Okay ... we know that d/D = 0.5 (right-hand y-axis) ---> reading from the curve ----> Q/Qfull = 0.5 (coincidence).

Solve for Qfull

n = 0.013

D = 9-ft

A = pi()*D/4

R = D/4

S = 0.002

Don't forget to multiply by 1.49 (English Units)

Qfull = 559 ft3/s. If we wanted to know Qhalf-full, we multiply by our factor from Appendix 19.C (0.5) to get Qhalf-full = 280 .

So, in other words, that is how much flow (conveyance) that one (1) 9-ft diameter pipe flowing half-full can accomodate.

Now, let's set up the equivalency. Equivalency is based on continuity.

Q1 = Q2 * (# of pipes)

---> 1060 ft3/s = 280 ft3/s * (# of pipes)

Therfore, (# of pipes) = 3.78 (round up to 4 pipes)

That's how I would do it. If you have specific questions about my approach, feel free to post follow-up questions. This is an important concept.

JR

 

[Timmy!]

I'm getting a bit confused as to determining "Q".

I know that Q=vA and that I can also obtain "Q" via Manning. For this problem, the upstream flow rate is Q=vA=2.5 *8=12 cubic meters per second.

But if I use Manning to find a value for Q upstream, I come up with something like 65.7 cubic meters per second.

I note that Manning's equation does not require a velocity value, so I'm guessing that water velocity is incorporated into the slope, hydraulic radius and "n" coefficient [?]

 

[Road Guy]

those Q's should give you the same value (if the areas and material are the same)

also mannings is different for metric & english (that used to bug me all the time)

& yes mannings incorporates the velocity through the terms you have mentioned.

have you worked all the hydraulics sections of the CERM? I think the water was one of the few sections that lindeburg did a good job with.

it had been a while for me, but I worked all those **** example problems to get back into the hydraulics mindset.

 

[Guest]

Timmy,

I think a few guided responses might help out here:

I know that Q=vA and that I can also obtain "Q" via Manning.

That's true. Those are two ways of finding Q. Notice in the first way that you have velocity, so all you need to know is the cross-sectional area. This is the same as saying we are taking the velocity for granted, so what would the flow be through a particular geometry. The second is by using Manning - which as you noted incorporates slope, geometry, and 'roughness' into the velocity term through an emperically derived equation.

For this problem, the upstream flow rate is Q=vA=2.5 *8=12 cubic meters per second.

You are correct in so much that Q=vA. However, the terms of the expression are (2.5 m/s)*(8 m)*(1.5 M) = 30 m3/s.

But if I use Manning to find a value for Q upstream, I come up with something like 65.7 cubic meters per second.

You will not use Manning to find Q upstream in this case because you are given v - if you know v, you do not need to use manning. You will use Manning for the 'downstream' section since you know the conveyance of water is dependent on the cross-sectional geometry, depth of water, slope of pipe (channel), and roughness of the pipe (channel).

I note that Manning's equation does not require a velocity value, so I'm guessing that water velocity is incorporated into the slope, hydraulic radius and "n" coefficient [?]

This is true based on my previous statements above.

I suggest moving on from this problem to another to try and work through the mechanics. This way, you can try applying these principles to see if you understand the applications and limitations. Or, in the alternative, I can try to come up with a few problems from resources and references I have to see how you do. In fact, I think I will try to do that - I will post something in the next day or so.

JR

 

[Guest]

I found a problem that is similar to this one that will test the same concepts. It comes from 101 Solved Civil Engineering Problems, 4th Edition.

Problem Statement:

An exisiting 22-inch diameter PVC pipe (mannings n = 0.01) is installed on a 2% slope. The flow rate that needs to be carried between two points separated by 1,000 feet is 35 ft3/sec. However, upon inspection, the pipe does not appear to have such capacity.

Questions:

1. What is the existing capacity of the 22-inch pipe?

2. What diameter pipe should be installed parallel to the first pipe to carry the excess?

3. What will the diameter of the 2nd pipe be if it is intalled over a different route with an average slope of 0.025?

4. If the 2nd pipe starts at elevation 4,257-ft, what will be the elevation of the 2nd pipe's invert at the end of the run? (Assume the two pipes have the same length)

Try working those and I will check back and post solutions.

JR

 

[Guest]

Solutions:

1. What is the existing capacity of the 22-inch pipe?

33 ft3/s

2. What diameter pipe should be installed parallel to the first pipe to carry the excess?

0.64-ft (7.7-in) ---> Select 8-inch pipe

3. What will the diameter of the 2nd pipe be if it is intalled over a different route with an average slope of 0.025?

0.61-ft (7.3-in) ---> Select 8-inch pipe

4. If the 2nd pipe starts at elevation 4,257-ft, what will be the elevation of the 2nd pipe's invert at the end of the run? (Assume the two pipes have the same length)

4,232-ft

Let me know if you have any questions/problems.

JR

 

[owillis28]

A couple of questions:

Why did they use a mannings coefficient of 0.01 for both the existing and proposed pvc pipe. Wouldn't the 'n' value for the existing pipe be smaller than that for a new pipe.

Also, could you help me on the 4th question? I am thinking it is very easy but I have not been able to figure it out.

I did send you an email for the 2nd question.

I appreciate any help that you can offer.

Thank you,

owillis

 

[Guest]

owillis28 --

In response to your questions:

Why did they use a mannings coefficient of 0.01 for both the existing and proposed pvc pipe. Wouldn't the 'n' value for the existing pipe be smaller than that for a new pipe.

Good question - I guess it was just a conservative assumption. The ranges of 'n' for PVC are something like 0.009 - 0.013, so 0.01 was probably considered OK for design.

Also, could you help me on the 4th question? I am thinking it is very easy but I have not been able to figure it out.

Think rise over run y/x = 0.025. The drop in elevation over 1,000-ft = 0.025*1,000 ft = 25-ft. We subtract that from the elevation at the top of the pipe (4257-ft) ----> 4232-ft.

I did send you an email for the 2nd question

You are correct - Equation 19.16( b ) from the Quick Reference Manual is used. I reworked #2 and came out with the correct answer. You might check the order of operation for your calculation and that the power 3/8 is applied correctly. Let me know how it comes out when you rework it .. if it is still wrong, maybe you can post follow-up with #'s.

JR

 

[civilsid]

Abuse the force Luke, kill the Ewoks!

 

[bay]

2. What diameter pipe should be installed parallel to the first pipe to carry the excess?

beer:30 would come alot sooner if you were to just pipe-burst the existing 22-incher with a 24-incher.

 

[RIP - VTEnviro]

Why did they use a mannings coefficient of 0.01 for both the existing and proposed pvc pipe. Wouldn't the 'n' value for the existing pipe be smaller than that for a new pipe.
Good question - I guess it was just a conservative assumption. The ranges of 'n' for PVC are something like 0.009 - 0.013, so 0.01 was probably considered OK for design.

If anything, I'd think the value for older pipe would have a higher n value than a newer pipe, as it would get rougher over time.

 

[cantaloup]

Six-Minute Solutions, 3rd Edition, Question 26:

Given: Concrete rectangular channel [assume n=0.013], 8m wide by 1.5m deep, velocity = 2.5 m/s at 0.002m/m slope. Transition 50m beneath a highway [DISTRACTOR!!!] to a circular 9 FOOT diameter concrete concrete culvert of same slope and "n" value, flowing half-full.

Find: Number of 9' diameter culverts required.

The answer is "four", but that is beside the point.

I think the way the question wording is not clear. Look at this statement: "Concrete rectangular channel [assume n=0.013], 8m wide by 1.5m deep, velocity = 2.5 m/s at 0.002m/m slope" This means that the size of the CONCRETE CHANNEL is 8m wide and 1.5m deep, but the size of the flow is different. Because the slope is given right there, we have to find the flow depth Y (that means the size of the water body!!!), with width B, velocity V, slope S, n factor are given; using the HP-33s to get a quick answer I found Y = 2.4 ft that means the flow is not full. Then I found Q=A*V = 2.4ft* 8*3.28ft * 2*3.28 ft/s = 478.6 cfs .

If they ("the other board") wants us to use 8m wide X 1.5m deep as the crosssection of flow they had to lstate the slope S=0.002m/m at the end, so it should apply to the 9' pipes only.

THIS QUESTION STATEMENT IS UNCLEAR

 

[jfusilloPE]

You know VTE, I don't think so. I think that the flow would have a tendency to smooth the inner wall of a pipe out. Sort of like rocks in a stream, they are rounded and smooth because of the constant erosion from the water. I might be wrong about that, but that is my hunch.

That is a good assumption, however, pipe (concrete, ductile, cast iron) acts in a different way. Concrete "pits", as do ductile and cast iron, and also tuberculation due to corrosion occurs. Regardsless of the pipe material, corrosion will be evident and your pipe "C" value will decrease of the "n" value will increase.

A typical "C" value in industry for ductile iron waterline is 100, while a typical "n" value for pvc sanitary sewer is 0.013.

 

[RIP - VTEnviro]

^ What he said!! :th_rockon: