# Relative roughness of a pipe?



## Aurora09 (Feb 20, 2019)

When determining the roughness, e, of a pipe - which value do we use from the range provided in the table? For example, the values of e for cast iron range from 0.0006 - 0.003 (im looking at the table on the moody diagram from the PE handbook). Do I take the average of the highest and lowest value? Just wondering how I should go about this on the actual exam lol


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## Slay the P.E. (Feb 20, 2019)

Aurora09 said:


> When determining the roughness, e, of a pipe - which value do we use from the range provided in the table? For example, the values of e for cast iron range from 0.0006 - 0.003 (im looking at the table on the moody diagram from the PE handbook). Do I take the average of the highest and lowest value? Just wondering how I should go about this on the actual exam lol


Yes, tables usually give a range and also a "design" value (See Appendix 17.A in page A-48 of MERM13. I'm not sure what the "PE Handbook" that you're referencing is, but in MERM13, the Moody Diagram of page 17-7 doesn't have a table).

But anyway, you're right. Values are typically listed as ranges. So, what to do? Let's look at the problems in the Thermal and Fluid Systems practice exam from NCEES that use a friction factor or a friction head loss:


Problem 126: They say "smooth pipe" but they also just give you the friction head, so *no need to calculate the Darcy friction factor*.

Problem 133: You're asked to calculate the friction head loss, but you don't have enough info to use the Darcy equation. You find hf from the definition of NPSH. So *no need to calculate the Darcy friction factor*.

Problem 505: The friction factor is given in the problem statement, so there's *no need to calculate the Darcy friction factor*.

Problem 515: The friction loss is given, so there's *no need to calculate the Darcy friction factor*.

Problem 519: The friction factor is given in the problem statement, so there's *no need to calculate the Darcy friction factor*.

I notice a pattern. 

I suspect that most likely you won't need to calculate a friction factor with a Moody chart in the test. HOWEVER, if you do, I strongly suspect they will tell you exactly which roughness _e_ to use because given the ambiguity in the ranges, you will not land on the correct multiple-choice answer unless you use their same roughness. Finally, if push comes to shove and you find yourself having to calculate a friction factor and they are not explicitly giving you a roughness in the problem statement then the safest bet is to use the "design" value from Appendix 17.A


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## Audi Driver P.E. (Feb 21, 2019)

Aurora09 said:


> When determining the roughness, e, of a pipe - which value do we use from the range provided in the table? For example, the values of e for cast iron range from 0.0006 - 0.003 (im looking at the table on the moody diagram from the PE handbook). Do I take the average of the highest and lowest value? Just wondering how I should go about this on the actual exam lol


Look at the range, and how it is used in the problem, if you select the mid range value, how far off will your answer be if the correct value to use was at either extreme?


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## Aurora09 (Feb 23, 2019)

Thanks for your responses! @Audi driver, P.E. @Slay the P.E.


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## Aurora09 (Feb 23, 2019)

I don't want to create too many topics so I hope its okay to post my additional questions here lol. but anyways, can someone assist me on pipes in parallel (please see question #7 in the following link: http://bu.edu.eg/portal/uploads/Engineering, Shoubra/Civil Engineering/3015/crs-14332/Files/FLUIDS - III.pdf)

I used bernoullis eqn between the two reservoirs to find V1 in which I then solved for Q1 = 0.43 m^3/sec.

Since the two branches are the same size and have the same friction factor, then Q2 = Q3 = 21.5 m^3/sec.

I tried a different method using the darcy weisbach equation Hf = f (L/D) V^2 /2g. I substituted all the variables for the main pipe to solve for V and then multiplied by the Area of the main pipe; I still end up getting Q1 = 0.43 m^3/sec. 

I really feel like I am solving this correctly but this is driving me crazy


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## Slay the P.E. (Feb 23, 2019)

Aurora09 said:


> I don't want to create too many topics so I hope its okay to post my additional questions here lol. but anyways, can someone assist me on pipes in parallel (please see question #7 in the following link: http://bu.edu.eg/portal/uploads/Engineering, Shoubra/Civil Engineering/3015/crs-14332/Files/FLUIDS - III.pdf)
> 
> I used bernoullis eqn between the two reservoirs to find V1 in which I then solved for Q1 = 0.43 m^3/sec.
> 
> Since the two branches are the same size and have the same friction factor, then Q2 = Q3 = 21.5 m^3/sec.


Can you show your work?

The approach you're describing here is exactly what they did in the link to solve the problem.


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## Slay the P.E. (Feb 23, 2019)

Aurora09 said:


> I tried a different method using the darcy weisbach equation Hf = f (L/D) V^2 /2g. I substituted all the variables for the main pipe to solve for V and then multiplied by the Area of the main pipe; I still end up getting Q1 = 0.43 m^3/sec.


How did you solve for V1 if you don't have hf for branch 1?


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## Aurora09 (Feb 24, 2019)

Slay the P.E. said:


> Can you show your work?
> 
> The approach you're describing here is exactly what they did in the link to solve the problem.


@Slay the P.E. Here is a picture of my work below:

(Note: Top reservoir is labeled A and the lower one is labeled B). 

I wasn't sure if Vb should be 0 so I ended up combing it with the same V in the Darcy head loss equation. My assumption might be wrong here. Also, should I also be adding another hf in the Bernoulli equation to account for one of the branches?


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## Slay the P.E. (Feb 24, 2019)

If "B" is a reservoir, then VB = 0, so that's the first thing you're doing wrong.

Also, when you write the Bernoulli equation from A to B, then hf has to be from A to B. From what you wrote above, your hf is only from A to the end of branch 1. So, yes, you need to add the friction loss across either one of the other branches. This is what they did in the link you provided.


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## Aurora09 (Feb 24, 2019)

Thank you - I went ahead and resolved the equation! My answer is still different than the link provided. However, I did notice a mistake in their calculation for K2. Even when I tried correcting their equations, the answer was still different than the way I set it up 

Anyways, do you think my work below is close enough? If anyone can check my math, that would be great. I've been stuck on this question all weekend.Thank you!


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