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### Topic: Shell and tube heat exchanger design  (Read 26956 times)

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#### gunasankar

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• Mole Snacks: +0/-0 ##### Shell and tube heat exchanger design
« on: March 02, 2009, 10:12:11 PM »
:-\I would like to prepare the excel spread sheet to size a shell and tube exchanger for quick estimation. The calculation is based on "Process Heat Transfer" by Donald Q.Kern. To find out the logarithmic mean temp difference I need the correction factor, which should be find out from the respective graph. Similarly the correction factor for caloric temperature also should be find out from the respective graph.

Is there any numerical equation available to determine the LMTD correction factor as well as Caloric temperature correction factor rather than graph?

#### saqi78

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• Mole Snacks: +1/-0 ##### Re: Shell and tube heat exchanger design
« Reply #1 on: March 28, 2009, 10:24:11 AM »
Dear
I am not a chemical engineer, but as I know may be true or false, you can check
For correction factor Ft, the formula used is Ft=R/S
where R= (T1-T2)/t2-t1 and
S=(t2-t1)/(T1-t1)

Where
T=Hot
t=cold
1=Inlet temperature
2=outlet temperature

You can check it from Heat Transfer book of Prof. Binay K Dutta

#### gunasankar

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• Mole Snacks: +0/-0 ##### Re: Shell and tube heat exchanger design
« Reply #2 on: March 30, 2009, 12:55:07 AM »
Dear Saqi78,

The formula what you have mentioned is correct. Using R value & S value we have to find the FT value from the graph of S Vs Ft. Instead of using the graph I would like to get the LMTD correction factor FT directly by using any numerical correlation. Also for caloric temperature correction factor I need the numerical correlation.

Regards
Guna

#### qwisp

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• Mole Snacks: +0/-0 ##### Re: Shell and tube heat exchanger design
« Reply #3 on: April 02, 2009, 08:56:52 AM »
As far as I know, it's different for different exchanger types, e.g. 1 shell, 1 pass, 1 shell 2 pass, etc.
But from Coulson & Richardson Volume 6, Thrid edition (p656):

Quote
For a 1 shell : 2 tube pass exchanger, the correction factor is given by:
Ft=(((R^2+1)^0.5)ln((1-S)/(1-RS)))/((R-1)ln((2-S(R+1-(R^2+1)^0.5))/(2-S(R+1+(R^2+1)^0.5))))
The derivation of this equation is given by Kern (1950).

It then shows the graph, and says that you can find other graphs in TEMA standards and books by Kern (1950) and Ludwig (1973).

Your best bet might be trying to find the book by Kern. From the reference list:
Kern, D.Q. (1950) Process Heat Transfer (mcGraw-Hill)

Hope this helps Chris

#### gunasankar

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• Mole Snacks: +0/-0 ##### Re: Shell and tube heat exchanger design
« Reply #4 on: April 03, 2009, 01:35:26 AM »
Dear Chris,

Thank you very much.

#### sourabh.mahawar

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« Reply #5 on: July 28, 2010, 05:48:04 AM »
dear using graph for correction factor is the simplest methord for finding it..for finding u had to solve some stiff integration which is done by numerical integration...and these graphs are result of solving these integrations only...so its better to solve your problem graphically only...