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Interstage Air Cooler - Expert Advice Needed


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#1 s udupa

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Posted 02 September 2014 - 05:56 AM

We are planning to replace compressed air cooler installed between 1st and 2nd stages of air compressor as it got corroded. To increase cooling, we are planning to increase area by increasing tube length. Detailed process conditions and calculations are in attached excel sheet. I request expert advice on the new tube length (new H T Area) so that we can float enquiry. Vendors will give mechanical guarantee only.

Attached Files



#2 srfish

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Posted 02 September 2014 - 10:47 AM

If you are replacing the cooler, why wouldn't the vendor only give you a mechanical guarantee when they could give you a quote with a size and price?



#3 Bobby Strain

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Posted 02 September 2014 - 12:16 PM

Looks like you designed it. Why would you expect the vendor to give thermal guarantee to your design?

 

Bobby



#4 PingPong

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Posted 02 September 2014 - 01:16 PM

As far as I understand the calculation is for the existing situation with 4.759 m tube length (area 189 m2 for 666 tubes) and air outlet temperature of 45 oC..

 

In the red notes it is stated that the tube length has to increase to 6.56 m (area 261 m2 for 666 tubes) to lower the air outlet temperature to 40 oC, but no calculation is given for that.

 

However: longer tubes mean that the interstage pressure drop will increase. If the exchanger pressure drop is now 0.19 kg/cm2 then the pressure drop for the new tube length will be approximately 0.26 kg/cm2. As a result the second stage inlet pressure of the compressor will drop by 0.07 kg/cm2. That does not seem much, but it is enough to make that the second stage power consumption will increase despite the lower inlet temperature!

 

To put is simple:

 

- a 5 oC lower inlet temperature on its own would reduce the second stage power by 2.2 % (because the absolute inlet temperature drops by 1.5 % and the molar flowrate drops by 0.7 % due to more water condensation)

 

- a 0.07 kg/cm2 lower second stage inlet pressure on its own would increase the second stage power by 2.8 %

 

- net result: second stage compression power will increase by 0.6 % as a result of the new longer exchanger.

 

I suspect that is not what you had in mind.

 

If you want to reduce second stage compressor power by reducing its inlet temperature, then you have to increase the number of tubes in the new exchanger to limit its pressure drop. However that results in a lower U value, so the new exchanger area then has to be bigger than 261 m2.

 

EDIT 5 september: power change percentages made more accurate based on computer calculation instead of simple manual estimate. Second stage discharge pressure was assumed to be 8.1 kg/cm2a (100 psig).


Edited by PingPong, 05 September 2014 - 11:49 AM.


#5 srfish

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Posted 02 September 2014 - 01:48 PM

A straight line LMTD of 58.6 C. can not be used in an evaluation. The actual LMTD will be less due to the steam condensing at a lower temperature than the inlet temperature. 



#6 s udupa

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Posted 03 September 2014 - 12:57 PM

First of all I thank each of you for your valuable replies.

 

I would like to write following to clarify some of your doubts.

 

1. Different vendors gave different sizes, but not ready to share calculations at proposal stage. So, idea is to freeze suitable size for all. In this case, thermal guarantee will not be vendors'.

2. Straight LMTD 58.56 is used as per D Q Kern's recommendation in "Process Heat transfer ", page from Indian edition is attached. Even I have not understood it properly. Will be thankful if somebody can clarify/explain it.

3. Originally this HE is designed to cool 202 to 40° with water from 32 to 42. But now even after cleaning, it reaches 44-45°C. 

4. Calculation for 261m2 area is attached.

5. I surely consider increased pressure drop in case of increased length as it adversely affects out operating cost.

 

I welcome further comments..

 

 

Attached Thumbnails

  • Kern.png
  • Kern.png

Attached Files



#7 srfish

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Posted 04 September 2014 - 09:11 AM

The two rules that Kern states have to offset each other in order to be valid. They do not in this case. The low dewpoint of 52.4 C. invalidates the premise.



#8 PingPong

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Posted 04 September 2014 - 11:16 AM

Kern's book was an good tool when it came out in 1950, when there was not much else, and calculations were done by hand, in which the designer would add a lot of overdesign.

Nowadays it is outdated and nobody uses it to design actual exchangers. It can of course still be used to learn about the things that matter when designing exchangers, and get a rough idea about impact of varying flow velocities, physical properties, et cetera..

 

Note also that Kern states: "The performance.......... can be predicted rapidly .....".

He does not say: can be calculated accurately.

 

The LMTD can only be calculated accurately from the four inlet and outlet temperatures, when

(1) the enthalpy lines of both streams are straight,

(2) and also the U value is constant over the whole length of the exchanger.

 

In this case neither (1) nor (2) is applicable.

The enthalpy curve of the air is not straight, especially due to the presence of a dew point, as srfish already indicated.

But also the U value is not constant, as it will be different for the part where no water condenses, and the part where water does condense.



#9 s udupa

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Posted 05 September 2014 - 02:25 AM

Thank you.. If possible please suggest reference book/literature to carry out calculation.

 



#10 PingPong

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Posted 05 September 2014 - 07:01 AM

I don't which book would give suitable calculation method for calculating h (and U) for condensing water out of air.

Nowadays designers use software like HTRI for such calculations.

 

To obtain correct LMTD you should split up the exchanger in say 20 parts, each with about 5 % of total duty.

Then calculate for each part the exact duty, the LMTD and the U.A

Add up all 20 calculated Duties and U.A's, divide them and you have the real LMTD.

If you are also able to calculate the correct U for each part, you can calculate the A of each part. Add up all 20 A's and you have the total A required.



#11 s udupa

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Posted 19 September 2014 - 01:05 PM

PingPong,

 

Can you suggest me how you have calculated pressure drop of 0.19 kg/cm2 and 0.26 kg/cm2??

As per Kern's method, I am getting only 0.07 and 0.09 kg/cm2 against your figures..

 

 



#12 PingPong

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Posted 19 September 2014 - 01:28 PM

I did not do pressure drop calculation myself but took data from your spreadsheet.

 

That pressure drop of 0.19 kg/cm2 is mentioned in cell E32 of your own spreadsheet.

 

Increasing tubelength by 38% (from 4.759 to 6.56 meters), while keeping number of tubes the same, will then increase the pressure drop also by about 38 %, hence that 0.26 kg/cm2 number.






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