NCEES water sample prob 501

The solution for the afternoon water sample problem #501 does not make sense to me. HGL is defined as the pressure head plus the elevation head, so why does the change in HGL - pipe friction = pressure at that point? if you assume the elevation is constant for the whole run, then the change in HGL could give you the change in pressure head, but how does it equal the actual pressure at that point?

It appears they may be calculating total head at the end of the pipe and using the HGL as elevation, but I'm not sure if this is the case. Can anyone explain?

For anyone who may have looked at this, I can get the problem to work out if I take the change in hydraulic gradeline to be only a change in elevation (which does not fit the definition of hydraulic gradeline), and assume the initial pressure to be atmospheric - taken to be zero. Then I can solve for a gauge pressure at the end of the line. However, this seems to have to include an incorrect definition for hydraulic gradeline and an assumption that may not necessarily be true (the initial zero pressure). It would seem there is still something incorrect about this problem, despite them issuing changes to it in the errata.

Can you post the problem?

The problem states:

A 15,000 ft long, 8-in diameter PVC line (C=140) serves 500 connections at the end of the line. The elevation of the hydraulic grade line is 495 ft at the beginning and 365 ft at the end. The pressure (PSI) at the end of the line during peak conditions (1gpm/connection) is most nearly:

A. 12

B. 28

C. 56

D. 158

There's a note in the errata to change the reading of the question, but it seems to state exactly what's in the book already, so I'm not sure what they intended to change. Like I said, the only way I can get this to work out is to take the hydraulic grade line to mean elevation/potential energy change only, not including the change in pressure, and to take the initial pressure as zero.

jamie said:

The problem states:A 15,000 ft long, 8-in diameter PVC line (C=140) serves 500 connections at the end of the line. The elevation of the hydraulic grade line is 495 ft at the beginning and 365 ft at the end. The pressure (PSI) at the end of the line during peak conditions (1gpm/connection) is most nearly:

A. 12

B. 28

C. 56

D. 158

There's a note in the errata to change the reading of the question, but it seems to state exactly what's in the book already, so I'm not sure what they intended to change. Like I said, the only way I can get this to work out is to take the hydraulic grade line to mean elevation/potential energy change only, not including the change in pressure, and to take the initial pressure as zero.

Click to expand...

1. Calculate head loss since your given Q, C, D and L

2. You know elevation loss, therefore the pressure at the end of the pipe in psi = (495-365-Head loss) / 62.4 X 144

Depending on which head loss equation you're using, the difference can range in 10-20 psi. In this case, I believe the answer is D just based on estimate calcs.

jamie said:

There's a note in the errata to change the reading of the question, but it seems to state exactly what's in the book already, so I'm not sure what they intended to change. Like I said, the only way I can get this to work out is to take the hydraulic grade line to mean elevation/potential energy change only, not including the change in pressure, and to take the initial pressure as zero.

Click to expand...

OK... some basic definition we can agree on: The hydraulic grade line (HGL) is the graph of the sum of the pressure and gravitational heads. The elevation of the HGL is the pressure head plus the gravitational head.

For a problem like this, I'd assume that at the "beginning" there is still fluid at a free surface. That means there's no velocity, the EGL = HGL, there's no pressure, and the elevation of the HGL is equal to the gravitational head. This means that Zbeginning = 495ft.

Unfortunately, I don't see any way to conclude what Zend is. The piping network could end at any elevation and this would obviously affect the pressure head. I'd probably assume Zend = 0 and then this turns into a trivial problem: The elevation of the HGL at the end (365tf) is all pressure head... 158psi.

What's the answer given?

The given answer is B, 27.6 psi.

My issue was exactly as you stated above. you can assume the initial pressure is zero, but if all they're telling you is the change in hydraulic gradeline, you have no way of quantifying the end pressure, since you don't know how much of that change is due to elevation loss and how much due to pressure change.

the way they solve the problem it's as though they intended it to say change in elevation only. they tried to change something from that statement in the errata, but maybe that was a typo as well, as it didn't change the reading of it at all.

jamie said:

the way they solve the problem it's as though they intended it to say change in elevation only. they tried to change something from that statement in the errata, but maybe that was a typo as well, as it didn't change the reading of it at all.

Click to expand...

Was there a drawing given? If so, please post as I can't seem to find my WR/ENV sample problems. Questions shouldn't require you to assume anything about the problem construct, even if reasonable assumptions can be used to get you to a quick answer.

jamie said:

they tried to change something from that statement in the errata, but maybe that was a typo as well, as it didn't change the reading of it at all.

Click to expand...

Given the solution in the errata (where it says delta H = 495-365 = 130ft), there's something odd here. Following the solution, they're working from the top (495ft), subtracting off the head loss (66ft), the velocity head (negligent), and the gravitation head (365ft) to leave you with the pressure head (64ft = 28psi). But as we've both said, there's no way to know how much of the HGL elevation at the end is pressure or gravitation head.

I agree the solution reads like they mean the HGL elevations to be actual elevations. And you have to assume at the "beginning" there is still fluid at a free surface.

no drawing was given. the problem statement i posted was all the info.

the solution in the errata greatly expanded the solution in the book, but still does not make clear how they are getting that. I think there must be a typo.

I have a question to this problem: The CERM uses a different formula for the Hazen Williams Friction Factor,

hf=(3.022(v^1.85)(L))/ ((C^1.85)(D^1.17))

When using the equation above by converting Q = 500gpm to Velocity. I do not get the same result. Anyone else tried this? Thanks.

jamie said:

the solution in the errata greatly expanded the solution in the book, but still does not make clear how they are getting that. I think there must be a typo.

Click to expand...

It's a "simple" Bernoulli's Theorem problem if you just replace "HGL" with "pipe". Agreed? The errata solution even reads "where ΔH is total drop in hydraulic grade line" and we know this isn't necessarily true as there are other things that can change the HGL elevation

But that's got me to thinking... if you assume that the change in pressure is only due to elevation change then the problem, as written, does work to the same solution.

maximus808 said:

I have a question to this problem: The CERM uses a different formula for the Hazen Williams Friction Factor,
hf=(3.022(v^1.85)(L))/ ((C^1.85)(D^1.17))

When using the equation above by converting Q = 500gpm to Velocity. I do not get the same result. Anyone else tried this? Thanks.

Click to expand...

I get the same answer as the errata using your formula above and V=3.19ft/s. My preferred HW equation for calculating friction head loss is



where

hf = friction head loss (ft)

L = length of pipe (ft)

Q = discharge (ft3/sec)

C = Hazen-Williams Coefficient

D = diameter of pipe (ft)

jamie said:

The solution for the afternoon water sample problem #501 does not make sense to me. HGL is defined as the pressure head plus the elevation head, so why does the change in HGL - pipe friction = pressure at that point? if you assume the elevation is constant for the whole run, then the change in HGL could give you the change in pressure head, but how does it equal the actual pressure at that point?
It appears they may be calculating total head at the end of the pipe and using the HGL as elevation, but I'm not sure if this is the case. Can anyone explain?

Click to expand...

That solution has been changes, see the errata.

In the question, the sentence 2 has changes as:

The elevation of the hydraulic grade line is 495 ft at the beginning of the pipe and 365 ft at the end of the pipe.

All the solution has been change too.

The problem is solved using the change in energy from one end to the other. 495-365-65=65ft, 65 ft X 62.4 lb/cf=4056 lb/sf, 4056 lb/sf X 144 sq in/sf=28.16 lb/sq in. 65 ft is the head loss calculated using the Hazen equation. The change in energy from one end to the other is only due to the potential energy and friction loss.

mattbail said:

The problem is solved using the change in energy from one end to the other. 495-365-65=65ft, 65 ft X 62.4 lb/cf=4056 lb/sf, 4056 lb/sf X 144 sq in/sf=28.16 lb/sq in. 65 ft is the head loss calculated using the Hazen equation. The change in energy from one end to the other is only due to the potential energy and friction loss.

Click to expand...

Correct,

Remembering that 2.31ft of head corresponds to 1psi of pressure simplifies this problem further. This quick conversion saves alot of time.

65ft/(2.31ft/psi) = 28.14psi