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Crane 410 fittings 12
- Thread starter wfn217
- Start date
pleckner,
Great article. I appreciate your logical presentation. I am curious, however, about your representation that in 1979 Crane "...discussed and used the two-friction factor method for calculating the total pressure drop in a piping system...(f for straight pipe and ft for valves and fittings)." It is my understanding from reading the TP 410 Foreword (4th paragraph, quoted below) that Cranes's intent was to have the pipe friction factor, f, apply to the K factor calculation and that the K factor was intended to apply to the full range of flow regimes from laminar to full turbulence. Can you comment on this?
Great article. I appreciate your logical presentation. I am curious, however, about your representation that in 1979 Crane "...discussed and used the two-friction factor method for calculating the total pressure drop in a piping system...(f for straight pipe and ft for valves and fittings)." It is my understanding from reading the TP 410 Foreword (4th paragraph, quoted below) that Cranes's intent was to have the pipe friction factor, f, apply to the K factor calculation and that the K factor was intended to apply to the full range of flow regimes from laminar to full turbulence. Can you comment on this?
TP 410 FOREWORD said:
LeSabre,
The example in pleckner's article shows how the "old" (i.e. pre-1976) L/D method over-estimates the pressure drop in fittings for low Reynolds numbers. All my calcs agree with and confirm pleckner's result. Crane were quite correct to make this claim in the 1976 foreword.
It is true that using the "new" K values avoids the overstatement of the pressure drop in the laminar regime, but my example (See 5 Dec 5:15) shows how the Crane K values now understate the pressure drop by 90% at a Reynolds number of 100. If you were designing a new pipeline would you rather estimate the pressure drop as 40% too high or 90% too low?
Your query, and the latest post by wfn217, are typical examples of the confusion that is caused by the Crane method, and which lead to my rant near the start of this thread.
I have probably stretched everyone's patience to the limit by going on and on about this problem so I will leave it at this now.
Katmar Software
Engineering & Risk Analysis Software
The example in pleckner's article shows how the "old" (i.e. pre-1976) L/D method over-estimates the pressure drop in fittings for low Reynolds numbers. All my calcs agree with and confirm pleckner's result. Crane were quite correct to make this claim in the 1976 foreword.
It is true that using the "new" K values avoids the overstatement of the pressure drop in the laminar regime, but my example (See 5 Dec 5:15) shows how the Crane K values now understate the pressure drop by 90% at a Reynolds number of 100. If you were designing a new pipeline would you rather estimate the pressure drop as 40% too high or 90% too low?
Your query, and the latest post by wfn217, are typical examples of the confusion that is caused by the Crane method, and which lead to my rant near the start of this thread.
I have probably stretched everyone's patience to the limit by going on and on about this problem so I will leave it at this now.
Katmar Software
Engineering & Risk Analysis Software
To LaSabre:
Thanks.
To me that statement in the Foreword just means that they made a change on how they relate the new K value to pipe friction factor. The statment obviously doesn't give any specifics as to how to apply the two, that is left to the rest of the document. Indeed, everywhere K for a valve or fitting is calculated within TP410 they are very careful to note the friction factor as fT and NOT f.
CRANE notes that the friction factors (f and fT) will essentially be the same in the zone of complete turbulence.
But you are correct in that it appears CRANE was intent on having K applied throughout all flow zones and that is where they got it wrong as we've been pointing out in these Postings.
Thanks.
To me that statement in the Foreword just means that they made a change on how they relate the new K value to pipe friction factor. The statment obviously doesn't give any specifics as to how to apply the two, that is left to the rest of the document. Indeed, everywhere K for a valve or fitting is calculated within TP410 they are very careful to note the friction factor as fT and NOT f.
CRANE notes that the friction factors (f and fT) will essentially be the same in the zone of complete turbulence.
But you are correct in that it appears CRANE was intent on having K applied throughout all flow zones and that is where they got it wrong as we've been pointing out in these Postings.
katmar,
Thanks for your comments and their effects on this discussion.
BigInch,
You have pointed toward one of my pet peeves, the dreadful misunderstandings brought about by the ability of computers and calculators to thrash around 10 digit numbers. If the accuracy to which most parameters are known is only 1%, 5%, 10%, and sometimes even worse, most of those many digits presented by the mighty computer or calculator are really just so much drivel. Unfortunately, there is never a shortage of people who want to believe that all of those digits are significant (and are willing to make expensive/dangerous judgements and decisions as though they were valid).
Thanks for your comments and their effects on this discussion.
BigInch,
You have pointed toward one of my pet peeves, the dreadful misunderstandings brought about by the ability of computers and calculators to thrash around 10 digit numbers. If the accuracy to which most parameters are known is only 1%, 5%, 10%, and sometimes even worse, most of those many digits presented by the mighty computer or calculator are really just so much drivel. Unfortunately, there is never a shortage of people who want to believe that all of those digits are significant (and are willing to make expensive/dangerous judgements and decisions as though they were valid).
The discussion has been very informative, but there is a certain logic for Crane to just continuing to publish equivalent lengths with conservative roughness at turbulent flows... its conservative and that's exactly how a lookup table intended for general application to all hydraulic systems should be.
Somewhere in Tips, I recall seeing a comment I liked questioning how many digits were actually needed before the answer became believable, or something to that effect. Guess we should have stayed with 8 bit computers. Even if all variables were known to 1%, it seems to me that most systems would spend an excessive amount of time operating outside the range where those accuracies were valid.
Lastly, risking repeating myself, sooner or later any given system reaches capacity at one extreme end of operational range or another, so over the long term is using Eq.Len at max turbulence really a bad thing?
BigInch![[worm]](/data/assets/smilies/worm.gif "[worm] [worm]")
-born in the trenches.
Somewhere in Tips, I recall seeing a comment I liked questioning how many digits were actually needed before the answer became believable, or something to that effect. Guess we should have stayed with 8 bit computers. Even if all variables were known to 1%, it seems to me that most systems would spend an excessive amount of time operating outside the range where those accuracies were valid.
Lastly, risking repeating myself, sooner or later any given system reaches capacity at one extreme end of operational range or another, so over the long term is using Eq.Len at max turbulence really a bad thing?
BigInch
-born in the trenches.katmar,
Thanks for directing my attention to your “rant”. I agree that the ft factor depends on geometry. Your j-factor theory, however, leads to the unlikely result that two different pipe I.D.s, say, 4” sch 40 (I.D. = 4.0260”) and 4” sch 160 (I.D. = 3.4380”) could have the same K-factor. How do you reconcile this situation? Also, how would you get the K-factor for a 36" XS L.R. Ell.?
Thanks.
Thanks for directing my attention to your “rant”. I agree that the ft factor depends on geometry. Your j-factor theory, however, leads to the unlikely result that two different pipe I.D.s, say, 4” sch 40 (I.D. = 4.0260”) and 4” sch 160 (I.D. = 3.4380”) could have the same K-factor. How do you reconcile this situation? Also, how would you get the K-factor for a 36" XS L.R. Ell.?
Thanks.
sailoday28
Mechanical
And the L/d or K for a run or branch of a "T" is?
Regards
Regards
You could have a look at these that Weldbend gives for fittings, including straight tees and they even give a method for assemblies as well,
and click on <next> for the following page
BigInch-born in the trenches.
and click on <next> for the following page
BigInch
-born in the trenches.Let's remember that CRANE TP410 doesn't publish equivalent lenghts for valves and fittings, it provides K values or the means to calculate K values from which we can then calculate the equivalent length using the appropriate friction factor at fully developed turbulent flow.
Let us also not forget that we can obtain various fT values from the graph on Page A-23. This graph allows us to choose the fT for the type of pipe being used (and this translates into a constant absolute roughness for that type of pipe) and the diameter of that pipe. So I don't agree that CRANE TP410 necessarily publishes conservative values of roughness. It all depends on how you want to manipulate the piping material you are using, e.g. For pharmaceutical grade highly polished SS pipe I might choose to use the roughness for Drawn Tubing (an absolute roughness of 0.000005 feet) rather than clean commercial steel pipe (0.00015 feet). Note that with this table, I can also try to interpolate and obtain fT for various pipe diameters!
Let us also not forget that the published fT values given in the table at the top of Page A-26 are strictly for CLEAN COMMERCIAL STEEL PIPE and for the schedule of pipe listed on Page 2-10. If one desires the the K value for another fitting of a different schedule pipe, they should adjust it using Equation 2-5 on that same page or go to the Graph on Page A-23.
One last thing for now, one can still use published equivalent lenghts as per the reference given by BigInch but just remember to calculate the pressure loss through that fitting using fT for that fitting rather than the pipe friction factor as I've shown in my paper and my post above.
Let us also not forget that we can obtain various fT values from the graph on Page A-23. This graph allows us to choose the fT for the type of pipe being used (and this translates into a constant absolute roughness for that type of pipe) and the diameter of that pipe. So I don't agree that CRANE TP410 necessarily publishes conservative values of roughness. It all depends on how you want to manipulate the piping material you are using, e.g. For pharmaceutical grade highly polished SS pipe I might choose to use the roughness for Drawn Tubing (an absolute roughness of 0.000005 feet) rather than clean commercial steel pipe (0.00015 feet). Note that with this table, I can also try to interpolate and obtain fT for various pipe diameters!
Let us also not forget that the published fT values given in the table at the top of Page A-26 are strictly for CLEAN COMMERCIAL STEEL PIPE and for the schedule of pipe listed on Page 2-10. If one desires the the K value for another fitting of a different schedule pipe, they should adjust it using Equation 2-5 on that same page or go to the Graph on Page A-23.
One last thing for now, one can still use published equivalent lenghts as per the reference given by BigInch but just remember to calculate the pressure loss through that fitting using fT for that fitting rather than the pipe friction factor as I've shown in my paper and my post above.
on the <next> page, they give a method for dP, from which you can backcalculate Ks, right?
BigInch-born in the trenches.
BigInch
-born in the trenches.@LeSabre,
Yes, indeed (assuming we are still talking LR bends) the Sch40 and Sch160 fittings would have the same K value using my J factor, or using the Crane method. The table at the top of page A-26 has a note "K is based on use of schedule pipe as listed on page 2-10". And page 2-10 seems to say that the K values apply to Schedules 40 to 160, but that the velocity that is used to calculate the velocity head must be based on the actual ID of the fitting. Makes sense to me. If you take a look at Figure 2-16 on page 2-13 of Crane 410 you will doubt every calculation you have ever made. This figure shows the variability in the experimental data on which the K values we use are based. A great deal of license has been used in getting to the "averages" we accept as gospel.
This question again highlights the reason I disagree with the Crane fT method. People want to start fiddling with the fT for their particular pipe, whereas Crane's intention was that if you have a 4" fitting you use an fT of 0.017 irrespective of the schedule or actual roughness of that pipe. That is why I suggested that we call it "J" and eliminate the false link to the friction factor that is confusing everybody. All the experimental work shows that the K value has almost no dependency on the roughness, and if you look at Fig 2-16 again you will see that there is no room for hair splitting here.
Using the Darby 3-K method to calculate K values for the Sch40 and Sch160 bends gives values on 0.28 and 0.29 respectively (at Re = 300,000). If I was doing a calculation that involved these fittings I would use these two different values for the two different schedules, but only for the sake of computational consistency and to allow anyone to later check my calcs using the same methods. In my heart I would know that in fact they are for all intents and purposes the same. Of course for the same flowrate the pressure drop is higher through the Sch160 bend because of the higher velocity, but not because of any real change in K value. To try to calculate the actual K value from the Crane method by interpolating fT between Sch40 and Sch160 is IMO like measuring the length of a football pitch with a micrometer.
For the 36" bend I would use 3-K and get a K value of about 0.18. The rate of decrease with size gets less as the sizes get bigger. You could use Crane fT of 0.0105 (interpolating from the figure on page A-24) because there is a substantial increment to 36" from the data on page A-26, but the fact that Crane neglected to give values on page A-26 for 36" pipe does not detract from my argument that their procedure is confusing.
@Sailoday28,
Crane have neglected to give K values for welded or flanged Tee's. I have no idea why. As always I would use 3-K. A good reference for this type of data is the classic article by Larry Simpson and Martin Weirick (Chemical Engineering, April 3, 1978).
@pleckner,
I am confused over what you are saying Phil. If we apply your example of polished SS with a roughness of 0.000005 inch to our 4" LR bend we would have to have a Reynolds number of over 100 million to get to full turbulence (See Crane A-24). Under these conditions you would get an fT of about 0.007. This compares with an fT of 0.017 for a 4" fitting given on page A-26. Using the value of 0.007 would make a 4" highly polished LR bend have a K value of 17x0.007 = 0.119 and not the 0.28 that I would calculate from Darby's 3-K. Is this what you are saying, or have I got the wrong end of the stick?
Katmar Software
Engineering & Risk Analysis Software
Yes, indeed (assuming we are still talking LR bends) the Sch40 and Sch160 fittings would have the same K value using my J factor, or using the Crane method. The table at the top of page A-26 has a note "K is based on use of schedule pipe as listed on page 2-10". And page 2-10 seems to say that the K values apply to Schedules 40 to 160, but that the velocity that is used to calculate the velocity head must be based on the actual ID of the fitting. Makes sense to me. If you take a look at Figure 2-16 on page 2-13 of Crane 410 you will doubt every calculation you have ever made. This figure shows the variability in the experimental data on which the K values we use are based. A great deal of license has been used in getting to the "averages" we accept as gospel.
This question again highlights the reason I disagree with the Crane fT method. People want to start fiddling with the fT for their particular pipe, whereas Crane's intention was that if you have a 4" fitting you use an fT of 0.017 irrespective of the schedule or actual roughness of that pipe. That is why I suggested that we call it "J" and eliminate the false link to the friction factor that is confusing everybody. All the experimental work shows that the K value has almost no dependency on the roughness, and if you look at Fig 2-16 again you will see that there is no room for hair splitting here.
Using the Darby 3-K method to calculate K values for the Sch40 and Sch160 bends gives values on 0.28 and 0.29 respectively (at Re = 300,000). If I was doing a calculation that involved these fittings I would use these two different values for the two different schedules, but only for the sake of computational consistency and to allow anyone to later check my calcs using the same methods. In my heart I would know that in fact they are for all intents and purposes the same. Of course for the same flowrate the pressure drop is higher through the Sch160 bend because of the higher velocity, but not because of any real change in K value. To try to calculate the actual K value from the Crane method by interpolating fT between Sch40 and Sch160 is IMO like measuring the length of a football pitch with a micrometer.
For the 36" bend I would use 3-K and get a K value of about 0.18. The rate of decrease with size gets less as the sizes get bigger. You could use Crane fT of 0.0105 (interpolating from the figure on page A-24) because there is a substantial increment to 36" from the data on page A-26, but the fact that Crane neglected to give values on page A-26 for 36" pipe does not detract from my argument that their procedure is confusing.
@Sailoday28,
Crane have neglected to give K values for welded or flanged Tee's. I have no idea why. As always I would use 3-K. A good reference for this type of data is the classic article by Larry Simpson and Martin Weirick (Chemical Engineering, April 3, 1978).
@pleckner,
I am confused over what you are saying Phil. If we apply your example of polished SS with a roughness of 0.000005 inch to our 4" LR bend we would have to have a Reynolds number of over 100 million to get to full turbulence (See Crane A-24). Under these conditions you would get an fT of about 0.007. This compares with an fT of 0.017 for a 4" fitting given on page A-26. Using the value of 0.007 would make a 4" highly polished LR bend have a K value of 17x0.007 = 0.119 and not the 0.28 that I would calculate from Darby's 3-K. Is this what you are saying, or have I got the wrong end of the stick?
Katmar Software
Engineering & Risk Analysis Software
@pleckner,
I got confused between inches and feet for your roughness in my working of your example with the polished SS. The fT is more like 0.010, making the K value 0.17. The numbers are different from my earlier calc but the principle is the same. My question is, are you proposing that we use the calculated fT value of about 0.010 rather than Crane's 0.017 to calculate the K value of a highly polished LR bend?
Sorry for the calculation error - its nearly midnight here!
Harvey
Katmar Software
Engineering & Risk Analysis Software
I got confused between inches and feet for your roughness in my working of your example with the polished SS. The fT is more like 0.010, making the K value 0.17. The numbers are different from my earlier calc but the principle is the same. My question is, are you proposing that we use the calculated fT value of about 0.010 rather than Crane's 0.017 to calculate the K value of a highly polished LR bend?
Sorry for the calculation error - its nearly midnight here!
Harvey
Katmar Software
Engineering & Risk Analysis Software
Have any of you looked at the “Handbook of Hydraulic Resistance” by I. E. Idelchik? It was my impression that this book was the ultimate reference for calculating flow resistance through fittings. Perhaps someone who studied this text might have some comments on how Idelchik handles these issues.
Yes, CRANE TP410, page 2-10 says that the K values apply to Schedules 40 to 160 but they are not constant at all these schedules. Look again and you will see that Schedule 40 pipe only applies to Class 300 and less; Schedule 80 pipe for Class 400 and 600 and so on. The way I interpret this is that you would have to adjust the K value using equation 2-5 for the different internal diameter if you were using a schedule 80 pipe in a Class 300 or less application. For example, 2" pipe, schedule 80 has an ID of 1.939" but that same pipe in schedule 40 has an ID of 2.067". If you were using schedule 80 pipe in a Class 150 application, the K value would have to be adjusted accordingly because the K values are published for the schedule 40 Class 150 pipe, not the schedule 80 Class 150 pipe; different velocities.
@katmar: My example is to show how one can manipulate, or try to manipulate absolute roughness to suite the type of piping material they have rather than just blindly following the table on A-26 that most people do. And yes, for this instance, I don't see anything wrong with using a lower K value for polished stainless steel pipe (actually tubing for bio-pharm use) over clean commercial steel pipe. The frictional losses will be significantly less for pharma pipe/tubing than for the standard chemical/petrochemical industry steel pipe.
BTW, I am doing a project for vegatable oil tank farm expansion and we are using a smaller diameter 304 SS pipe than I would normally choose for our flow rates (asked for by the client) because in their experience, there is essentially very little effective friction at all, the stuff just glides right down the pipe!
An please don't get me wrong, I am a big advocate of using the 2-K or 3-K method over CRANE for the same reasons you have been pointing out.
@katmar: My example is to show how one can manipulate, or try to manipulate absolute roughness to suite the type of piping material they have rather than just blindly following the table on A-26 that most people do. And yes, for this instance, I don't see anything wrong with using a lower K value for polished stainless steel pipe (actually tubing for bio-pharm use) over clean commercial steel pipe. The frictional losses will be significantly less for pharma pipe/tubing than for the standard chemical/petrochemical industry steel pipe.
BTW, I am doing a project for vegatable oil tank farm expansion and we are using a smaller diameter 304 SS pipe than I would normally choose for our flow rates (asked for by the client) because in their experience, there is essentially very little effective friction at all, the stuff just glides right down the pipe!
An please don't get me wrong, I am a big advocate of using the 2-K or 3-K method over CRANE for the same reasons you have been pointing out.
There seems to be a good deal of confusion about the relationship and use of the friction factor and the K-factor. Inspection of equations 2-1, 2-2, 2-3 and 2-4 on page 2-8 of Crane TP 410 (see below) should clarify the issue.
Equation 2-1, hL = v2/2g, this is the velocity head of a flowing fluid.
Equation 2-2, hL = K v2/2g, this defines the K-factor as the number of velocity heads lost due to a valve or fitting.
Equation 2-3, hL = (f L/D) v2/2g, this is the Darcy equation.
Equation 2-4, K = (f L/D), this is the familiar K-factor equation.
It should be completely evident that the link between the friction factor and the K-factor is established by virtue of the Darcy equation and that f is the Darcy friction factor. The purpose for inventing a K-factor and doing this algebraic manipulation is to develop a dimensionless group to be used for model scaling and to generalize experimental findings based on a limited number of experimental results. It should be further noted that the friction factor, f, applies only to straight pipe; there is no friction factor associated with any valve or fitting.
The Darcy friction factor can be obtained using your favorite charts or equations. For the special case of fully developed turbulent flow, fT, which represents the maximum possible friction factor, the von Karman Rough Pipe Law should be used, namely 1/[√]fT = 2Log (3.7D/k) or fT = 1/[2Log (3.7D/k)] 2, where k is the pipe roughness. There is no special significance given to the fT values provided on page A-26. Indeed, the first paragraph on page 2-10 states that these fT values are provided for “convenience.”
K-factor values for valves and fittings are determined experimentally by measuring the head lost due to the valve or fitting during a flow test. Once the K-factor value is determined for a valve or fitting it can be equated to a hypothetical pipeline (having particular values for the parameters f, L and, D) by using equation 2-4. The values for the parameters of the hypothetical pipe are not set in stone.
For example:
A flow test for a 2” valve produced a pressure loss equal to 3 velocity heads under fully developed turbulent flow. What is the length of 2.067” ID hypothetical pipe that would cause a 3 velocity head loss if the hypothetical pipe roughness were 0.0015”?
Using the rough pipe law, fT = 1/[2Log (3.7*2.067/0.0015)] 2 = 0.01819
Using Equation 2-4, LT = (3* 2.067)/0.01819 = 340.9” or 28.4 feet
LT is the equivalent length of the valve under fully developed turbulent flow.
Equation 2-1, hL = v2/2g, this is the velocity head of a flowing fluid.
Equation 2-2, hL = K v2/2g, this defines the K-factor as the number of velocity heads lost due to a valve or fitting.
Equation 2-3, hL = (f L/D) v2/2g, this is the Darcy equation.
Equation 2-4, K = (f L/D), this is the familiar K-factor equation.
It should be completely evident that the link between the friction factor and the K-factor is established by virtue of the Darcy equation and that f is the Darcy friction factor. The purpose for inventing a K-factor and doing this algebraic manipulation is to develop a dimensionless group to be used for model scaling and to generalize experimental findings based on a limited number of experimental results. It should be further noted that the friction factor, f, applies only to straight pipe; there is no friction factor associated with any valve or fitting.
The Darcy friction factor can be obtained using your favorite charts or equations. For the special case of fully developed turbulent flow, fT, which represents the maximum possible friction factor, the von Karman Rough Pipe Law should be used, namely 1/[√]fT = 2Log (3.7D/k) or fT = 1/[2Log (3.7D/k)] 2, where k is the pipe roughness. There is no special significance given to the fT values provided on page A-26. Indeed, the first paragraph on page 2-10 states that these fT values are provided for “convenience.”
K-factor values for valves and fittings are determined experimentally by measuring the head lost due to the valve or fitting during a flow test. Once the K-factor value is determined for a valve or fitting it can be equated to a hypothetical pipeline (having particular values for the parameters f, L and, D) by using equation 2-4. The values for the parameters of the hypothetical pipe are not set in stone.
For example:
A flow test for a 2” valve produced a pressure loss equal to 3 velocity heads under fully developed turbulent flow. What is the length of 2.067” ID hypothetical pipe that would cause a 3 velocity head loss if the hypothetical pipe roughness were 0.0015”?
Using the rough pipe law, fT = 1/[2Log (3.7*2.067/0.0015)] 2 = 0.01819
Using Equation 2-4, LT = (3* 2.067)/0.01819 = 340.9” or 28.4 feet
LT is the equivalent length of the valve under fully developed turbulent flow.
- Thread starter
- #37
vzeos, I am pleased to see that you used bold letters to emphasize your statement "there is no friction factor associated with any valve or fitting". This was the crux of my argument that Crane have confused the issue by associating fT with the K values, leading even competent engineers to believe that they should adjust the K values according to their actual friction factors.
We engineers learn best by example, so let me also give an example to illustrate the point. Equation 2-16 on page 2-12 gives the head loss through a bend as
ht = hp + hc + hl
where
ht = total loss through bend
hp = excess loss
hc = loss due to curvature
hl = loss due to length
For a standard radius bend the average flow length through the bend would be 1.5 times the diameter, and it could be argued that hl is therefore influenced by the actual friction factor in the fitting. Page A-29 shows that the equivalent length of a standard radius bend is 14 diameters. We can therefore say that 1.5/14, or approximately 11% of the head loss through the bend is influenced by the friction factor. In an extreme case the actual friction factor may vary by 50% from the given value of fT. In this extreme case the overall pressure loss would change by 50% of 11%, or 5.5%. This is well within the confidence limits of the values for K, and it is therefore meaningless to adjust the K value according to the friction factor. In most real life cases the actual variation would be much less than this 5.5%
Katmar Software
Engineering & Risk Analysis Software
We engineers learn best by example, so let me also give an example to illustrate the point. Equation 2-16 on page 2-12 gives the head loss through a bend as
ht = hp + hc + hl
where
ht = total loss through bend
hp = excess loss
hc = loss due to curvature
hl = loss due to length
For a standard radius bend the average flow length through the bend would be 1.5 times the diameter, and it could be argued that hl is therefore influenced by the actual friction factor in the fitting. Page A-29 shows that the equivalent length of a standard radius bend is 14 diameters. We can therefore say that 1.5/14, or approximately 11% of the head loss through the bend is influenced by the friction factor. In an extreme case the actual friction factor may vary by 50% from the given value of fT. In this extreme case the overall pressure loss would change by 50% of 11%, or 5.5%. This is well within the confidence limits of the values for K, and it is therefore meaningless to adjust the K value according to the friction factor. In most real life cases the actual variation would be much less than this 5.5%
Katmar Software
Engineering & Risk Analysis Software
I think we all agree that the friction factor ‘f’ only applies to pipe; we’ve been saying this throughout this thread, haven’t we, so no news here. I also think we all agree that we do not adjust the K value for the fitting according to the actual pipe friction factor, again no news here. And katmar@, I think we'll just have to agree to disagree about CRANE making this concept confusing because I don’t see the problem with relating K to fT; it is up to the engineer to be competent enough to understand the difference between fT and just f and how to apply them. Pardon my bluntness but I can't excuse graduate engineers from not knowing basic hydraulics.
I must disagree with vzeos’ statement, “There is no special significance given to the fT values provided on page A-26. Indeed, the first paragraph on page 2-10 states that these fT values are provided for “convenience.”” First, I read the referenced paragraph differently and believe the word “convenience” as used in the above quote is being taken out of context. I believe the authors are saying they present this table as a convenience so you don’t have to go looking things up in the graph they present on page A-23, not that they have no real significance. Second, the significance of fT is that the experiments were carried out very specifically with flow in the zone of full turbulence; or perhaps I’m misinterpreting the meaning of vzeos’ statement? To me the CRANE engineers are trying to distinguish between this friction factor and the friction factor in the pipe, this is the significance of fT.
Before the next salvo, I want to wish all happy holidays and a happy and healthy (and profitable) New Year.
I must disagree with vzeos’ statement, “There is no special significance given to the fT values provided on page A-26. Indeed, the first paragraph on page 2-10 states that these fT values are provided for “convenience.”” First, I read the referenced paragraph differently and believe the word “convenience” as used in the above quote is being taken out of context. I believe the authors are saying they present this table as a convenience so you don’t have to go looking things up in the graph they present on page A-23, not that they have no real significance. Second, the significance of fT is that the experiments were carried out very specifically with flow in the zone of full turbulence; or perhaps I’m misinterpreting the meaning of vzeos’ statement? To me the CRANE engineers are trying to distinguish between this friction factor and the friction factor in the pipe, this is the significance of fT.
Before the next salvo, I want to wish all happy holidays and a happy and healthy (and profitable) New Year.
Pleckner,
I agree that you are misinterpreting the meaning of my statement. I said “no special significance”; you restated it as “no real significance.” The reason that I said no special significance is that there is a notion out there that if you use the Crane K-factor formulas you must use the fT values from table on page A-26. This is not true. The fT values on page A-26 apply to clean commercial steel pipe and are assembled there for the users convenience. If you are using cast iron pipe or plastic pipe or any other pipe material, you need to use an fT value appropriate to your pipe diameter and your pipe roughness.
On a more interesting subject. I read your article and I’ve been working with your example. I found that if you calculate the equivalent lengths using f instead of fT, the calculated dP using the Equivalent Length method is the same as the dP you get using the K-factor method. I have posted the numbers I calculated below. Please check my math.
[tt]Equivalent Lengths Using f
3" pipe 31.5
(2) 90deg Long Radius Elbow 6.167
(1) Branch Tee 9.250
(1) Swing Check Valve 7.709
(1) Plug Valve 2.775
(1) 3” x 1” Reducer4 496.088
TOTAL 553.489[/tt]
dP=0.00000336{(0.02985)(553.489)(63,143)2/(( 112.47)( 3.068)5)} = 7.2398
Also, if you use f instead of fT to calculate equivalent lengths, Leq= KD/f, then the equivalent length dP formula can be algebraically transformed into the K-factor dP formula. This implies that it would be correct to use the flowing friction factor, f, in the K-factor formulas instead of fT when appropriate. Please check my math below.
[tt]dP= 0.00000336{fLeqW2/([ρ]d5)}
= 0.00000336{f(KD/f)W2/([ρ]d5)}
= 0.00000336{(Kd/12)W2/([ρ]d5)}
= 0.00000028{KW2/([ρ]d4)}[/tt]
Happy holidays!
I agree that you are misinterpreting the meaning of my statement. I said “no special significance”; you restated it as “no real significance.” The reason that I said no special significance is that there is a notion out there that if you use the Crane K-factor formulas you must use the fT values from table on page A-26. This is not true. The fT values on page A-26 apply to clean commercial steel pipe and are assembled there for the users convenience. If you are using cast iron pipe or plastic pipe or any other pipe material, you need to use an fT value appropriate to your pipe diameter and your pipe roughness.
On a more interesting subject. I read your article and I’ve been working with your example. I found that if you calculate the equivalent lengths using f instead of fT, the calculated dP using the Equivalent Length method is the same as the dP you get using the K-factor method. I have posted the numbers I calculated below. Please check my math.
[tt]Equivalent Lengths Using f
3" pipe 31.5
(2) 90deg Long Radius Elbow 6.167
(1) Branch Tee 9.250
(1) Swing Check Valve 7.709
(1) Plug Valve 2.775
(1) 3” x 1” Reducer4 496.088
TOTAL 553.489[/tt]
dP=0.00000336{(0.02985)(553.489)(63,143)2/(( 112.47)( 3.068)5)} = 7.2398
Also, if you use f instead of fT to calculate equivalent lengths, Leq= KD/f, then the equivalent length dP formula can be algebraically transformed into the K-factor dP formula. This implies that it would be correct to use the flowing friction factor, f, in the K-factor formulas instead of fT when appropriate. Please check my math below.
[tt]dP= 0.00000336{fLeqW2/([ρ]d5)}
= 0.00000336{f(KD/f)W2/([ρ]d5)}
= 0.00000336{(Kd/12)W2/([ρ]d5)}
= 0.00000028{KW2/([ρ]d4)}[/tt]
Happy holidays!
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