Heat Exchanger Tube/Shell side Pressure Drop

[AndreChE]

I have to calculate the velocity of CW in tubes and the pressure drop in the tube/shell side of several heat exchangers. I used

http://www.processassociates.com

but I can't get the value of pressure drop. An example: An heat exchanger with 4 tube passes, 1232 tubes, 6m each tube length and ID of 23.3mm with a flowrate of 238 m3/h of CW (density 993.184 kg/m3 and viscosity of 0.0010565 Pa.s. The calculated velocity is 0.503392 m/s (checked by myself), Reynolds Re= 11026.512(too) and pressure drop of 60.370 mbar (??). I need to know the equation for calculate this pressure drop. I think my error is on tube length. What is the correct formula to calculate tube side pressure drop and the tube legth to consider? Another...formula to shell side pressure drop?

 

[22082002]

Hi,
I am not trying to answer your query.
I just followed the same steps,
your velocity and reynolds no. are correct.
But I get pressure drop of 246mbar for same confiuration.
Please check.
I feel the exchanger has very low tubeside velocity, shouldnot be less than 1.1 m/sec to have self cleaning.

Regards,
Saa

 

[AndreChE]

Yes, I know that the velocity is too low in the tubeside. That's the reason of the study. What was the equation used for your 246mbar DP?

 

[AndreChE]

Ok. The problem is...no longer a problem. I calculated the pressure drop in the tubes and the pressure drop for return losses. The total deltaP is about 270mbar (232+38 mbar). The result from

http://www.processassociates.com

is, in my opinion, the deltaP per pass of the heat exchanger. The problem of low velocity is because of low CW pressure in the plant (4.3 bar for 6 bar design)....next step...

Thanks a lot and regards
AndreChE

 

[AndreChE]

Sorry but there was an error in my last calculation. The total deltaP is 70,6 mbar, using the theory listed in Process Heat Transfer, Kern, taking into account tube and return losses. It seems to me that 200-300 mbar is a too high value, reason for why I re-checked my calculations.

The value from

http://www.processassociates.com

probably doesn't take into account return losses.

AndreChE

 

[22082002]

Hi ANdreChE,
Thanks for letting me know return losses are also so significant

Regards,
SAA

 

[AndreChE]

Hello SAA

Did you recalculate deltaP? Still with 200-300 mbar? Have you followed Kern, Process Heat Transfer method? I'm with 70,6mbar and I think this is the correct value but it could be some mistake in my calculation.

Regards,
AndreChE

 

[garyloz]

Hi AndreChE
For correct calculation of pressure drop and velocity of gas flow with taking into account of the tube length try to use the materials of papers:
"On static head in the pipe flow element"

http://arXiv.org/abs/physics/0301070

and
"Saint-Venant –Wantzel’s formula modern form"

http://arXiv.org/abs/physics/0302038

for prediction of flow regime (laminar turbulent)you may be use
the result of work: "The laminar flow instability criterion and turbulence in pipe"

http://arXiv.org/abs/physics/0303071

Sincerely
garyloz

 

[mharris]

The calculation is fairly straight forward. You simply take the number of tubes and use this deterimine the average flow of CW through a single tube. Then, use the Bernouli equation to determine your pressure drop. For example if you have 100 gpm and 25 tubes, your average flow would be 4 gpm per tube. Use the 4 gpm to calculate your dp. This is your pressure drop through the tube section. (it does not matter how many tubes, except to get your average flowrate per tube) A good resource would be the Fluid and Particle Dynamics section of Perry's, Crane's Tehnical paper No. 410, or Rules of Thumb for Chemical Engineers (chapter on heat exchangers). Per the Rules of Thumb for Chemical Engineers the number of velocity heads for other losses include:
1) entering plus exiting the exchanger = 1.6 velocity heads (use the velocity in the pipe leading to and from the exchanger)this is usually negligible.
2) entering plus exiting the tubes = 1.5 velocity heads * number of tube passes (use velocity in the tubes)
3) end losses in tubeside bonnets and channels = 1.0 velocity head * number of tube passes (use velocity in the tubes)

**A velocity head = deltaH = (velocity)^2 / (2g) where deltaH is head loss in feet of fluid. See Perry's for further info OR Crane's tech paper 410.

***The three losses described above are additive to the loss you determine from the flow through the single tube using average velocity through one tube.