12 replies to this topic

[jimbo28]

Has anyone tried using the Equation in Chapter 5 of "Pressure Vessel Design Manual" by Moss to calculate the pressure drop in a helical tank coil?

 

I believe there is something wrong with the equation as the results yield an extremely high value.

 

There is a worked example in text but no reference to where the equation comes from or its valid range of parameters.

 

This is just a reality check. I have found equations in the literature that yield reasonable results but it bugs me that I can not figure out what is wrong in Moss.

 

He gives DPcoil=DPstraightline*(RE(di/dc)^2)^.5

 

di= ID of pipe (in)

dc= centerline diameter of coil (in)

DP= psi

[latexman]

I don't have Moss' book.  Crane TP410 has a helical coil method too.  I know for a fact that one is pretty close to my equipment.

[breizh]

http://www.cheresour...op-calculation/

 

Consider the resource above

 

Breizh

[Napo]

Jimbo 28,

 

Your report is in Pressure Vessel Design Manual by Moss D., in 3rd ed., Procedure 6-8: Design of pipe coils for heat transfer, p. 335 and followings or in 4th ed., Procedure 5-6, p. 326 and following. The problem is the same, the estimation of correction factor of “f” for the coil.

 

I think there is an error in the mathematical expression:

 

f coil/ f straight = [Re (d/D)^2]^0,5

 

f coil/ f straight = [Re (d/D)^2]^0,5 = Re^0,5 (d/D), this is a simplification and it is a possible indication that the main root is different to 2 (= ½).

 

d = pipe diameter

D = coil diameter

 

If we have the following information (Moss, Sample Problem 2):

 

Re = 54961

d = 3.826 inch

D = 98 inch

f straight = 0.0218

 

f coil/ f straight = [Re (d/D)^2]^0,5 = Re^0,5 (d/D) = (54961)^0.5 (3.826/98) = 9.1526

 

For other estimation methods we have:

 

f coil/ f straight = 1 + 3.54 (d/D) = 1 + 3.54(3.826/98) = 1.1382

f coil/ f straight = e^(d/D) = e^(3.826/98) = 1.0398

f coil/ f straight = log (Re^0.25) = log (54961^0.25) = 1.1850

 

f coil White = 0.32 Re^-0.25 + 0.048 (d/D)^0.5 = 0.32 (54961)^-0.25 + 0.048 (3.826/98)^0.5 = 0.0304

f coil White/f straight = 0.0304/0.0218 = 1.3945

 

f coil Ito = 0.304 Re^-0.25 + 0.029 (d/D)^0.5 = 0.304 (54961)^-0.25 + 0.029 (3.826/98)^0.5 = 0.0256

f coil Ito/f straight = 0.0256/0.0218 = 1.1743

 

If you take for example the correct value 1.1382, the possible solution to the original equation will be:

 

f coil/ f straight = [Re (d/D)^2]^x = 1.1382 x = ln (1.1382)/ ln (83.7708) = 0.0293

1/x = 34.13, approximately 34

f coil/ f straight = [Re (d/D)^2]^(1/34) = [54961 (3.826/98)]^(1/34) = 83.7708^(1/34) = 1.1391

 

Napo.

 

 

[ankur2061]

Jimbo28,

 

Refer the link below:

 

http://www.thermoped...om/content/639/

 

Regards,

Ankur.

[breizh]

Hi Jimbo 28,

Reading this paper ( part for Newtonian fluid) you may find pointer .

Your reference (Moss)  is giving similar numbers using equation 2,3,4 references 8,9,10 of the paper attached.

 

Note : All those correlations are subject to the calculation of the Dean number .

 

Another great resource is "pipe flow a pratical and comprehensive guide " page 166 -168

 

For a constant pitch helix : K=N*( f*[(2*pi*r)^2+p^2]^.5 / d +0.20+4.8*f) -Head loss coefficient

with N :number of coils ,

[(2*pi*r)^2+p^2]^.5 being the centerline length of a 360 helical coil ,

f : friction factor straight pipe ,

d : diameter of the pipe

r : radius of the coil

 

Hope this helps

 

Breizh

Edited by breizh, 24 November 2015 - 07:13 PM.

[Napo]

Jimbo28,

Breizh´s information is important, but equations (2), (3) and (4) are for laminar (viscous) regime.

In this information the only equation for turbulent regime is:

fc = fs + 0.01 (d/D)^0.5

In the Moss, Sample Problem 2, I got:

Re crit = 2100 x [1 + 12 (3.826/98)^0.5] = 7079. 205 --> turbulent
fc = 0.0218 x 0.01 (3,826/98)^0.5 = 0.0238
then,

fc/fs = 0.0238/0.0218 = 1.0906

this valor is similar to the other estimations.

Napo.

[Napo]

Jimbo28,
 
The correct expression is:
 
f coil/ f straight = [Re (d/D)^2]^0,05
 
I found this in the book "Boundary-Layer Theory", by Schlichting H., 7th ed., McGraw-Hill, 1979, USA, p. 627.  
 
Napo.

[christopherchoa]

Jimbo28,

 

The equation given by Napo is correct. If you have AFT fathom, the program has function for coils. Looking at the help file, it says the same thing pointing to the paper of Ito, 1959.

[jimbo28]

Thanks to everyone for the thorough and detailed replies. I have used some of the referenced equations for my work and the results seem reasonable. I will advise as to how accurate the numbers are once we get or vessel up and running.

 

Napo I am confused by your last post, the equation you posted is the equation used is Moss which you previously seemed to agree was in error.

Can you elaborate?

 

Thanks Again

[Art Montemayor]

Jimbo:

 

If you could come back to this thread and report the resulting, actual pressure drop across the coil after getting the vessel up and running, I would certainly appreciate it.

 

Information like this is valuable as field experience and for design engineers.  The pressure drop resulting in a coil should, according to expectations from reported research work, be lower than the value resulting from an equivalent straight run of the same internal diameter of pipe.  What would be interesting would be to confirm the expectations.

 

[Napo]

Dear Jimbo,

In the original equation of Moss book the exponent is 0.5, this value is wrong.

The correct value is 0.05, this is the value that I put in my last post (0,05).

The pressure lost in the coil pipe will be great than straigh pipe.

Napo.

[jimbo28]

My apologies, I did not catch the shift in decimal point. Mystery solved!