Hazen Williams problem issues

[mjoneswvu]

I need some help solving a problem and appreciate any insight. Here goes,,,The problem has a reservoir with 20ft WSEL and has a 1ft dia. pipe exiting the bottom of the reservoir and the pipe drops 30ft over an unknown distance. Given: S=0.01, C=120, kinematic viscosity=1.217x10^-5, f=0.02; Find velocity using hazen williams equation and neglect discharge.

It appears that I could use the standard hazen williams equation: v=1.318*C*R^0.63*S*0.54 (assuming pipe full) but I'm not getting a correct answer and there has to be another method. It seems that the problem almost offers enough information to use Darcy as well.

THOUGHTS PLEASE?

 

[IlPadrino]

I need some help solving a problem and appreciate any insight. Here goes,,,The problem has a reservoir with 20ft WSEL and has a 1ft dia. pipe exiting the bottom of the reservoir and the pipe drops 30ft over an unknown distance. Given: S=0.01, C=120, kinematic viscosity=1.217x10^-5, f=0.02; Find velocity using hazen williams equation and neglect discharge.
It appears that I could use the standard hazen williams equation: v=1.318*C*R^0.63*S*0.54 (assuming pipe full) but I'm not getting a correct answer and there has to be another method. It seems that the problem almost offers enough information to use Darcy as well.

THOUGHTS PLEASE?

It would help if you had the exact problem to post... I'm not sure where you're looking to find the velocity and what you mean by "neglect discharge". Considering you're not given the length of pipe, you're essentially ignoring friction loss - which should greatly simplify the problem. Why can't you use the Bernoulli Equation to solve for v? I'm imagining the problem where the pressure terms are zero, the initial velocity term is zero, the initial potential term is 50 ft, and the final potential term is 0: v=sqrt(50 x 2g)

Edit: I missed the fact that you're given slope and vertical drop... so length is easily calculated and you can calculate friction loss.

 

[mjoneswvu]

Attached is the problem.

 

[JoeysVee]

Attached is the problem.

I don't think ti is attached.

 

[mjoneswvu]

Opps, attached is the problem. Pls. adviseproblem.doc

 

[IlPadrino]

Looks like a homemade problem... where did it come from? I think maybe the statement to use Hazen-Williams is misleading because I would approach as a simple Bernoulli problem as I mentioned before. You can add a term for friction head loss and use the more general energy equation:

View attachment 2706

where:

P/γ = pressue head (ft)

v2/2g = kinetic head (ft)

Z = potential head (ft)

hp = head of the pump (ft)

hL = head loss (ft)

It's possible to calculate the hL term using Hazen-Williams, but it's a little complicated:



where

hf = friction head loss (ft)

L = length of pipe (ft)

Q = discharge (ft3/sec)

C = Hazen-Williams Coefficient

D = diameter of pipe (ft)

You can calculate L using the vertical drop and slope but you need to know the discharge... that seems recursive if the goal is to find the velocity!

It's tempting to use the Hazen-Williams formula for circular pipes:



but this makes no account for the potential head of the reservoir and would give you the wrong answer.

I'll think about this some more when I'm at home... but for now, I'd suggest using the straight Bernoulli (ignoring friction head) and if that's not close enough to one of the answers given, use the velocity from Bernoulli to calculate a discharge that's used with Hazen-Williams to estimate friction head - and feed into the Energy Equation.

Make sense? What's the answer given?

 

[IlPadrino]

One more quick thought... why not use Darcy-Weisbach to calculate friction head:



Now you can combine the kinetic (velocity) head and the friction loss as they're both in terms of V2. A quick run through gives me an answer of 7.26 fps.

 

[Chucktown PE]

Here is what I came up with:Scan001.PDF

 

[IlPadrino]

Here is what I came up with:Scan001.PDF

Chuck... Nice solution... and it's the same answer I got, but help me understand: you think the Hazen-Williams equation for friction head also includes the velocity head? I haven't used Hazen-Williams much for friction loss, so maybe I misunderstood it's application.

 

[Chucktown PE]

Chuck... Nice solution... and it's the same answer I got, but help me understand: you think the Hazen-Williams equation for friction head also includes the velocity head? I haven't used Hazen-Williams much for friction loss, so maybe I misunderstood it's application.

I'm not sure what you're saying. Hazen-Williams is an empirical equation to approximate head loss due to friction. The variables are length, roughness, diameter, and flow/velocity. It is not dimensionally homogeneous.

 

[IlPadrino]

dimensionally homogeneous

Those are some big words!

What I'm asking is this: the Energy Equation says the energy between two points may change form (pressure, kinetic, potential, pump, and losses) but the total is always conserved. For this problem, there's no pressure or pump energy at the surface of the reservoir or the outlet of the pipe. That leaves kinetic (only at the outlet pipe because the water isn't moving at the surface of the reservoir), potential (only at the surface of the reservoir because we take the outlet of the pipe to be elevation zero), and loss (friction only because I assume the problem says to ignore the minor loss at the exit of the pipe).

In your equation, you say potential energy (50 ft) is all converted to the head specified by the Hazen-Williams equation. That means the Hazen-Williams equation you used includes both the kinetic (velocity) and the friction loss.

Or am I missing something obvious?

 

[Chucktown PE]

Those are some big words!
What I'm asking is this: the Energy Equation says the energy between two points may change form (pressure, kinetic, potential, pump, and losses) but the total is always conserved. For this problem, there's no pressure or pump energy at the surface of the reservoir or the outlet of the pipe. That leaves kinetic (only at the outlet pipe because the water isn't moving at the surface of the reservoir), potential (only at the surface of the reservoir because we take the outlet of the pipe to be elevation zero), and loss (friction only because I assume the problem says to ignore the minor loss at the exit of the pipe).

In your equation, you say potential energy (50 ft) is all converted to the head specified by the Hazen-Williams equation. That means the Hazen-Williams equation you used includes both the kinetic (velocity) and the friction loss.

Or am I missing something obvious?

I think you're trying to look at this as derived equation, not an empirical equation. The only thing Hazen-Williams does is approximate headloss due to friction. Flow is in the numerator, therefore the higher the flow the higher the friction loss. C represents the roughness of the pipe. The the smoother the pipe, the higher the C. You can't balance this equation like you can the energy equation.

 

[IlPadrino]

I think you're trying to look at this as derived equation, not an empirical equation. The only thing Hazen-Williams does is approximate headloss due to friction. Flow is in the numerator, therefore the higher the flow the higher the friction loss. C represents the roughness of the pipe. The the smoother the pipe, the higher the C. You can't balance this equation like you can the energy equation.

Chuck,

I understand the nature of the Hazen-Williams equation, how it's derived from Chezy, and its empirical relationship that is only valid for a certain range of conditions. My question was why you set the result of the Hazen-Williams "head loss" equation equal to the loss in potential head (50 ft) without regard to the increase in kinetic head.

 

[Chucktown PE]

Because you start with 50 feet of head and end with 0 feet of head after the exit of the pipe. At the pipe outfall there is no velocity either. Think of it as a reservoir with the water surface at the invert of the pipe. Therefore you have 50 feet of headloss. Besides, the increase in velocity head is rather insignificant, 0.8 ft. Look at the hydraulic grade line for a two reservoir problem.

 

[IlPadrino]

Because you start with 50 feet of head and end with 0 feet of head after the exit of the pipe. At the pipe outfall there is no velocity either. Think of it as a reservoir with the water surface at the invert of the pipe. Therefore you have 50 feet of headloss. Besides, the increase in velocity head is rather insignificant, 0.8 ft. Look at the hydraulic grade line for a two reservoir problem.

OK... now we're getting somewhere. You chose to ignore the velocity head and I think that's a mistake in general, even if not a mistake here specifically. When you wrote "at the pipe outfall there is no velocity", that seems clearly at odds with the problem statement which says to find the velocity at the pipe outfall - agreed?. This isn't a two reservoir problem - why look at the HGL there?

For the purposes of the PE Exam, you can often choose to simplify the problem by ignoring some components... though it's sometimes hard to know where. I agree there's not much difference in the answer here, but with practice problems it's helpful to be clear on what assumptions and simplifications you're making - they won't always apply.

Again, thanks for the discussion - I hope it's useful for the OP.

 

[Chucktown PE]

OK... now we're getting somewhere. You chose to ignore the velocity head and I think that's a mistake in general, even if not a mistake here specifically. When you wrote "at the pipe outfall there is no velocity", that seems clearly at odds with the problem statement which says to find the velocity at the pipe outfall - agreed?.

I think you know what I mean. Bottom line is that from one end of the pipe to the other the water loses 50 feet of energy.

This isn't a two reservoir problem - why look at the HGL there?

Okay, leave the velocity head in, but then the problem becomes an iterative solution. Please provide a solution doing it your way.

 

[mjoneswvu]

Thank you guys. The answer was most nearly v=8FPS so you got it. Much appreciated. The extra information in the problem threw me off and I wasn't accounting for the head correctly.

Matt

 

[IlPadrino]

I think you know what I mean. Bottom line is that from one end of the pipe to the other the water loses 50 feet of energy.

I don't know what you mean... because it's only true that from one end of the pipe to the other, the water loses 50 feet of POTENTIAL energy. The point to all this discussion isn't to argue what's the right answer, but to explain what the right approach is. I suggest it's bad advice to discount the KINETIC energy out of hand, even if it doesn't make much of a difference here. I recommend always starting with the Energy Equation and then simplify to Bernoulli... canceling the terms that you want to discount given certain assumptions.

Okay, leave the velocity head in, but then the problem becomes an iterative solution. Please provide a solution doing it your way.

Well, yeah... that's what I said in one of my first posts! Then I suggested using Darcy-Weisbach which gives the friction loss in terms of V2 which can easily be combined algebraically with the velocity head. I don't think you really need me to work that out being as I already gave the answer. Using your approach, you could easily calculate the velocity and friction head for various velocities, including the one you get assuming the velocity head is negligible.

And yeah, I know the problem said to use Hazen-Williams, but you shouldn't get a radically difference answer seeing as they are both derived from de Chezy.

Again, the discussion is about helping the OP understand how to approach these problems. If you think something with my recommended approach is wrong, please speak up! I'm certainly not shy about speaking up where I think a part of your approach is wrong - EVEN if I agree with your answer (and the implicit assumption that velocity head is negligible compared to the friction head).

 

[IlPadrino]

Thank you guys. The answer was most nearly v=8FPS so you got it. Much appreciated. The extra information in the problem threw me off and I wasn't accounting for the head correctly.
Matt

Great! I think you'll find there are often some distractors in the problem statement.

 

[Chucktown PE]

Again, the discussion is about helping the OP understand how to approach these problems. If you think something with my recommended approach is wrong, please speak up! I'm certainly not shy about speaking up where I think a part of your approach is wrong - EVEN if I agree with your answer (and the implicit assumption that velocity head is negligible compared to the friction head).

Okay, well when someone specifically tells you to use Hazen and you do something different you aren't following directions. If this were an exam or if your employer asked you to do this and you disregarded the directions then you fail. The guy didn't ask what you thought about the approach to the problem, he asked for help with the problem, which specifically says find the velocity using Hazen.