# TFS Lindeburg #33



## Kohos (Oct 20, 2018)

Condensed Problem statement: 40ft of 4in sch. 40 steel pipe containing five long radius 90deg. elbows + one gate valve + one sharp edge. V=16ft/s, f=0.019

Book Solution: K_entrance = 0.5 and all others included in the equivalent length as 40ft + 5(6.4ft) + 2.5ft = 65.5 and total head loss calculated as K*v^2/(2g) + f*(65.5)*v2/(2Dg) and answer = 16.7ft

My method: K_entrance + K_all others = 0.5+5(0.6)+0.19 = 3.69 and total head loss = v^2/(2g)*(3.69+f*(40)/D) = 23ft

I always thought the same answer is achieved whether we take K as a loss or included as an equivalent length, what am I missing here?

Thank you


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## Slay the P.E. (Oct 21, 2018)

Kohos said:


> Condensed Problem statement: 40ft of 4in sch. 40 steel pipe containing five long radius 90deg. elbows + one gate valve + one sharp edge. V=16ft/s, f=0.019
> 
> Book Solution: K_entrance = 0.5 and all others included in the equivalent length as 40ft + 5(6.4ft) + 2.5ft﻿ = 65.5 and total head loss calculated as K*v^2/(2g) + f*(65.5)*v2/(2Dg) and answer = 16.7ft
> 
> ...


Well, for starters 40ft + 5(6.4ft) + 2.5ft﻿ is actually 74.5 ft, not 65.5 -- but that's not all.

There also seems to be something going on with the "conversion" from K to L_eq. For example, according to your data, the elbows have an equivalent length L_eq = 6.4 feet, and a loss coefficient K = 0.6, correct?

Well, the relationship is L_eq = K*(D/f).  If we use K=0.6, f=0.019, and D=4.026 in, we get L_eq = 0.6*[(4.026/12 ft)/0.019] = 10.6 ft (i.e. not 6.4 feet). You're only going to get the same answer if the Ks and the L_eqs are related by L_eq = K*(D/f).

Also, what book is this?


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## MikeGlass1969 (Oct 21, 2018)

Slay,

I think 6.4 is a typo.  I looked up the Leq for the elbows and was only 4.6 ft each.

He is also using "typcial loss coefficients" from table 17.4 in the MERM.

I looked at the problem earlier.  I had some questions too.  Is the Kentrance given as .5?  Why could we not use a leq=9.5 ft from A-54 in MERM.  It is a square mouth inlet.


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## Slay the P.E. (Oct 22, 2018)

MikeGlass1969 said:


> Slay,
> 
> I think 6.4 is a typo.  I looked up the Leq for the elbows and was only 4.6 ft each.
> 
> ...


Ok. 4.6ft for the elbows makes more sense, but K is still not 0.6


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## Kohos (Oct 22, 2018)

Hello, this is the book: https://ppi2pass.com/pe-mechanical-thermal-and-fluids-systems-practice-exam.html?mkwid=&amp;amp;pcrid=282914701360&amp;amp;pmt=&amp;amp;pkw=&amp;amp;pdv=c&amp;amp;slid=&amp;amp;product=METSPE&amp;amp;gclid=Cj0KCQjw6rXeBRD3ARIsAD9ni9AEuPdVtA214H2xNDVF1Y_Ge-rC_l5N5Vigz4Lbbos5hTap5z21FGQaAtqaEALw_wcB


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## Kohos (Oct 22, 2018)

yes 6.4 was my typo, it should have been 4.6 but I am still not getting the solution values. 

I opted to use k values, so I have:

5-long radius 90 elbows: 5x0.9 = 3, 1 gate valve - 0.19, 1 sharp edge  - 0.5

Total k = 3.69

Total loss = kv2/2g = 14.68

The the 40ft length loss comes to: 9ft *so total for me = 26.68ft* whereas solution has *16.7 ft* following these steps:

Book solution:

K_entering = 0.5

Effective length with elbow and valve - 40+5*4.6+2.5 = 65.5

*Giving total loss of 16.7ft*

Any advice would be helpful, I was under the assumption that either k or L_eq could be used but now getting different answers, I am not so sure anymore....


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## Slay the P.E. (Oct 22, 2018)

Kohos said:


> yes 6.4 was my typo, it should have been 4.6 but I am still not getting the solution values.
> 
> I opted to use k values, so I have:
> 
> ...


I think their 17 feet and your 23 feet are within the uncertainty of these problems in the real world, given the approximate nature of these values for loss coefficients (especially for valves). The problem in the PPI book is a throwback from the olden days when a human being actually laid eyes on your hand-written essay solution. Those days are obviously long-gone since the scantron and 100% multiple choice format took over the exam. Back in the day, you would have received full credit for your solution (I think) when graded by a human.

In modern times, we cannot have this much wiggle room and NCEES has to craft these problems so that you land squarely on the one right answer.  How do they do that? Well, they make it quite obvious what values of K (or L_eq) you have to use. For examples, look at these in the NCEES official practice exam:


*Problem 516*. They are explicitly telling you the equivalent lengths of the elbows and the globe valve. You then use those values. DO NOT go all nuts trying to look up values of K for these fittings. Use what they're giving you in the problem statement.

*Problem 519.* They are explicitly telling you the values of K to use for the elbow and the entrance. Use that. Don't go trying to get equivalent lengths in some table.

Note that the problem you are posting about doesn't provide Ks or L_eqs in the problem statement; hence the ambiguity. You will not (or shouldn't) have this in the exam.


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## Kohos (Oct 22, 2018)

Thank you! and yes I did notice in the NCEES practice problem we are given all K or equivalent length values


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## Audi Driver P.E. (Oct 23, 2018)

Slay the P.E. said:


> I think their 17 feet and your 23 feet are within the uncertainty of these problems in the real world, given the approximate nature of these values for loss coefficients (especially for valves). The problem in the PPI book is a throwback from the olden days when a human being actually laid eyes on your hand-written essay solution. Those days are obviously long-gone since the scantron and 100% multiple choice format took over the exam. Back in the day, you would have received full credit for your solution (I think) when graded by a human.
> 
> In modern times, we cannot have this much wiggle room and NCEES has to craft these problems so that you land squarely on the one right answer.  How do they do that? Well, they make it quite obvious what values of K (or L_eq) you have to use. For examples, look at these in the NCEES official practice exam:
> 
> ...


This.  Well done.


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