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Friction Factor for Extremely Low Renyolds Num. 2
- Thread starter RJB32482
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Assumptions
Chemical
From a practical standpoint I would say yes, because it will not result in a large pressure drop. However, the plots of f vs Re that I have seen normally end around Re = 100.
So you Darcy-Weisbach f would be correct.
Just a further note, I obtain a Re of about 0.1 for your conditions, assuming that the fluid is Newtonian.
So you Darcy-Weisbach f would be correct.
Just a further note, I obtain a Re of about 0.1 for your conditions, assuming that the fluid is Newtonian.
- Thread starter
- #4
Thanks for the Reply.
For Newtonian Flow I Used the Equation in Crane
Re=(50.6*flow*density)/(d*viscosity)
flow=GPM density=lb/ft^3 d=inches viscosity=centistokes
So Re=(50.6*0.35*60.43)/(1.049*338.8)= 3.01
Then f=64/Re=64/3.01=21.3
So should I use the 2 K Method in the laminar flow part? I have access to that. Does everyone agree with my logic? Thanks
For Newtonian Flow I Used the Equation in Crane
Re=(50.6*flow*density)/(d*viscosity)
flow=GPM density=lb/ft^3 d=inches viscosity=centistokes
So Re=(50.6*0.35*60.43)/(1.049*338.8)= 3.01
Then f=64/Re=64/3.01=21.3
So should I use the 2 K Method in the laminar flow part? I have access to that. Does everyone agree with my logic? Thanks
Assumptions
Chemical
I believe that your flow rate given in the equation is different than that in your initial listing.
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- #7
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- #8
The tips given by Assumptions are what you should be concentrating on. He/she has told you where you are wrong, and you need to go back and check all your data and your units. They are seriously mixed up somewhere.
0.8 GPH is not 0.35 GPM. Given the data in your first post your calculation of Re is wrong and Assumptions' Re calc is correct.
The reason you have laminar flow is not because of high viscosity, but because of very low flowrate. A viscosity of 350 cP is not unusually high. Given the data in your first post, your velocity is only 0.3 ft/minute. Your pressure drop will be around 0.1 PSI per 100 ft, even though your friction factor (Darcy-Weisbach) is about 550.
Under these circumstances pressure drop seems irrelevant, but we do not know the full picture.
0.8 GPH is not 0.35 GPM. Given the data in your first post your calculation of Re is wrong and Assumptions' Re calc is correct.
The reason you have laminar flow is not because of high viscosity, but because of very low flowrate. A viscosity of 350 cP is not unusually high. Given the data in your first post, your velocity is only 0.3 ft/minute. Your pressure drop will be around 0.1 PSI per 100 ft, even though your friction factor (Darcy-Weisbach) is about 550.
Under these circumstances pressure drop seems irrelevant, but we do not know the full picture.
- Thread starter
- #9
Using 64/Re is correct for laminar Newtonian fluids as stated in other replies.
The 2k method is also mentioned. I understand that you refer to the 2k method for fittings. The usual fixed k method will give too low numbers for the fittings.
You can use the following website: http//:Then choose pressure drop > Pipe with fittings
This calculation uses the 2k method.
Regards - Wim
The 2k method is also mentioned. I understand that you refer to the 2k method for fittings. The usual fixed k method will give too low numbers for the fittings.
You can use the following website: http//:Then choose pressure drop > Pipe with fittings
This calculation uses the 2k method.
Regards - Wim
- Thread starter
- #11
OK Here's my calculation for just the straight pipe:
Re=(50.6*60.43 lb/ft^3*0.35 GPH)/(0.622in*338.8cp)=3.01
f=64/Re=64/3.01=21.3
K=f*L/D=(21.3*51 in)/(0.622 in)=1746.5 (51 in is length of pipe)
delta P= (18*10^-6*K*density*Q^2)/(d^4)
=(18*10^-6*1746.5*60.43 lb/ft^3*(0.35 GPM)^2)/(0.622 in)^4
=1.55 psi
Looks Good? Thanks. The application is for the inlet side of a PSV.
Re=(50.6*60.43 lb/ft^3*0.35 GPH)/(0.622in*338.8cp)=3.01
f=64/Re=64/3.01=21.3
K=f*L/D=(21.3*51 in)/(0.622 in)=1746.5 (51 in is length of pipe)
delta P= (18*10^-6*K*density*Q^2)/(d^4)
=(18*10^-6*1746.5*60.43 lb/ft^3*(0.35 GPM)^2)/(0.622 in)^4
=1.55 psi
Looks Good? Thanks. The application is for the inlet side of a PSV.
Assumptions
Chemical
RJB32482
Your units for flow rate are mixed up.
Your origional post has 0.8 GPH
Your last calculation has 0.35GPH for the Re calculation.
Your delta P calculation has 0.35 GPM.
A good way of checking things is to make sure that the units come out correctly. Otherwise you might not be off by 10% but mabe 1000 times.
Your units for flow rate are mixed up.
Your origional post has 0.8 GPH
Your last calculation has 0.35GPH for the Re calculation.
Your delta P calculation has 0.35 GPM.
A good way of checking things is to make sure that the units come out correctly. Otherwise you might not be off by 10% but mabe 1000 times.
- Thread starter
- #13
Assumptions
Chemical
OK
I get the same Re and the same f
However, I do not get the same delta P. I looked in Crane and could not find the equation that you have used for your delta P calc.(0.1PSI)
Or list the reference for your delta P calculation.
I get the same Re and the same f
However, I do not get the same delta P. I looked in Crane and could not find the equation that you have used for your delta P calc.(0.1PSI)
Or list the reference for your delta P calculation.
- Thread starter
- #15
Crane is the book I used for the delta P Calclation. It is used in one of the example problems (I don't have the book here, its at work). It is used in one of the examples in chapter 4 (I believe its an oil example).
It reads:
deltaP=(18*10^-6*K*density*Q^2)/(d^4)
Thanks.
It reads:
deltaP=(18*10^-6*K*density*Q^2)/(d^4)
Thanks.
Assumptions
Chemical
OK
I checked the Crane reference.
Your value for K contains an error for "d".
You repeated the error for "d" in the calculation for delta P.
Hope that this helps a bit.(My comments pertain only to delta P calcs.) Aparently you have changed to 1/2".
If 1/2" is used Re changes and just by chance f (Fanning) is 3 while f (D-W) is 13.
Anyway. Everything should be on the same basis If 1/2" is used then your Re is incorrect according to my check.
Again. Check dimensions and units and consistency.
I checked the Crane reference.
Your value for K contains an error for "d".
You repeated the error for "d" in the calculation for delta P.
Hope that this helps a bit.(My comments pertain only to delta P calcs.) Aparently you have changed to 1/2".
If 1/2" is used Re changes and just by chance f (Fanning) is 3 while f (D-W) is 13.
Anyway. Everything should be on the same basis If 1/2" is used then your Re is incorrect according to my check.
Again. Check dimensions and units and consistency.
- Thread starter
- #17
Assumptions
Chemical
No. Your method is correct.
I do not get the same Re in your calculation.(Assuming the flow was in GPM)
For your own calcs write down the inputs and stay with those same values throughout.
I am not checking all simple calcs. I have been working from your inputs.
In a few years from now if you have to go back to the calculations above do you think that you could show an observer how the inputs have been definitively defined, other than well it seems to be obvious. When you enter something, keep in the back of your mind, is this really correct. In a couple of years from now, will I be able to follow the calculations, that I have assumed is obvious to everyone.
Hope that this helps.
I do not get the same Re in your calculation.(Assuming the flow was in GPM)
For your own calcs write down the inputs and stay with those same values throughout.
I am not checking all simple calcs. I have been working from your inputs.
In a few years from now if you have to go back to the calculations above do you think that you could show an observer how the inputs have been definitively defined, other than well it seems to be obvious. When you enter something, keep in the back of your mind, is this really correct. In a couple of years from now, will I be able to follow the calculations, that I have assumed is obvious to everyone.
Hope that this helps.
sailoday28
Mechanical
Search for "creeping flow" For example
fluid at very low Reynolds number. In the flow of fluids, a Reynolds number (density • length • velocity/viscosity) describes the relative importance of inertia effects to viscous effects. In creeping flow the Reynolds number is very small (less than 1) such that the inertia effects can be ignored in comparison to the viscous resistance. Creeping flow at zero Reynolds number is called Stokes flow
fluid at very low Reynolds number. In the flow of fluids, a Reynolds number (density • length • velocity/viscosity) describes the relative importance of inertia effects to viscous effects. In creeping flow the Reynolds number is very small (less than 1) such that the inertia effects can be ignored in comparison to the viscous resistance. Creeping flow at zero Reynolds number is called Stokes flow
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