PDA

View Full Version : Howto: Calculate coaxial heat exchangers







JKProSoft
25-04-2006, 03:02 PM
Can someone point out some technincal articles on how to calculate heat transfer in coaxial heat exchangers using R404A refrigerant and seawater.

Lc_shi
26-04-2006, 01:10 AM
you mean the tube in tube heat exchanger?

nh3simman
21-03-2007, 09:43 PM
The same basic principles apply to all heat exchangers.

Best is the e-Ntu method. Reason is that there is no iteration so is ideal for hand calculation.

There are 3 simultaneous equations to solve.

1. Heat transfer from hot fluid to cold fluid.
Q = e Qmax
e = effectiveness
Qmax = (mc)min x ITD
ITD = inlet temperature difference
(mc)min = min capacity rate
read e from e-Ntu curve
where Ntu = UoAo/(mc)min

2. Hot fluid energy loss
Q = (m Cp dT)hot

3. Cold Fluid energy loss
Q = (m Cp dT)cold

Note that Q is the same in each equation. Implies that heat lost from hot fluid = heat gain by cold fluid.

This is standard textbook stuff, you will find it in Holman or any other heat transfer text book. There is a very good book that is available for free download by Prof Lienhard at MIT.
http://web.mit.edu/lienhard/www/ahtt.html (http://web.mit.edu/lienhard/www/ahtt.html)


The South African Institute of Refrigeration and Air Conditioning (SAIRAC), has a technical data CD that will cost you about $100.
www.sairac.co.za

nh3simman
22-03-2007, 05:37 AM
Best is the e-Ntu method. Reason is that there is no iteration so is ideal for hand calculation.

There are 2 heat exchanger solve methods:

1. Effectiveness method
2. LMTD method

They are theoretically equivalent, so you can use either.

I'm not fan of LMTD because it requires iteration. It goes like this:

Q = Uo Ao LMTD

Seems simple on the face of it but the LMTD demands the knowledge of both fluid leaving temperatures.

See the paradox here?

If you know the leaving temperature of any of the fluids then you have the answer.

The LMTD method is good for selectors where the temperatures are specified and you need to calculate the coil size.

Where the LMTD method fails, is on cooling coils where you have condensation. This means 2-phase heat transfer and so the temperature difference is not the driving force (the driving force in this case is the enthalpy difference). On the other hand, the effectiveness method does not suffer from this problem since Qmax can easily by defined in terms of the enthalpy difference.

eg. for DX coil
Qmax = (mc)min x (hai - hswi)
hswi = enthalpy of saturated air at suction temp.


I have published an article on this subject that can be read on my web site. It is freely available to read on-line. If you would like to read it, I will send you a link privately (apparently you get suspended by ANDY if you post links)

Josip
22-03-2007, 02:39 PM
Hi, nh3simman :)



I have published an article on this subject that can be read on my web site. It is freely available to read on-line. If you would like to read it, I will send you a link privately (apparently you get suspended by ANDY if you post links)

I think you can make a link in your profile (go to User CP - Home page URL) to your web page and we all can read your article;)

Best regards, Josip :)

Brian_UK
22-03-2007, 07:48 PM
..........I have published an article on this subject that can be read on my web site. It is freely available to read on-line. If you would like to read it, I will send you a link privately (apparently you get suspended by ANDY if you post links)You will suspended by ANY moderator for posting adverts, links we don't mind but you were pushing your site quite a lot.;)

ernestlin
23-03-2007, 07:30 AM
eg. for DX coil
Q=eQmax
Qmax = (mc)min x (hai - hswi)
hswi = enthalpy of saturated air at suction temp.

Good information, simman.
I regard this method just be suitable to estimate, not the exact calculation. how to get the value of "e"? it is based on many factors such as tube material, flow configuration etc. This kinds of calculation always depend on some empirical formulas.

nh3simman
23-03-2007, 01:46 PM
Good information, simman.
I regard this method just be suitable to estimate, not the exact calculation. how to get the value of "e"? it is based on many factors such as tube material, flow configuration etc. This kinds of calculation always depend on some empirical formulas.

Hi ernestlin,

The effectiveness is no estimate, in fact it can be derived directly from the LMTD method.

e comes from the e-Ntu curve.
where Ntu is the number of transfer units.
Ntu = UoAo/Cmin
Ao and Cmin are known values.

The unknown is Uo (the overall heat transfer coefficient), and here I must agree with you.

Uo can be calcuated from the many well known correlations or it can be based on manufacturers data.

But the uncertainty in Uo applies equally to the e-Ntu method or the LMTD method.

My point is that in calculating the duty of an existing heat exchanger, the e-Ntu method has huge benefits in terms of calculation speed. In addition, the e value is bounded by 0-100% so your calculations cannot diverge.

In any event, you can easily calculate with one method and compare with the other.

chillman
07-01-2011, 02:12 PM
Hi nh3simman,

I know this is an old thread, but I've recently been looking into coaxial heat exchangers, could you please go through an example to give a better understanding,
please don't take offence but I'm a hands on kind of guy and nomenclature's make my brain basically freeze when I dont have a grasp of where the figures are coming from.

Say if we have a 3/4" tube for the refrigerant side and a 1/2" tube as the inner water side, using any flow of choice on both the refrigerant side and water side, and all other necessary values, how can we work out the capacity of a 1 meter length?
in lamens terms would be helpful


I would really appreciate your knowlege input on this as I've been struggling to find answers for quite some time

NoNickName
07-01-2011, 04:45 PM
Have you tried coolpack? You can easily calculate heat exchange on liquid or gas pipes.

chillman
10-01-2011, 08:39 AM
Hi NoNickName,

Just downloaded coolpack, using the liquid and gas pipe modules, but I don't see how it can help me because I want to run refrigerant on the outside tube and water on the inside, so the outside tube volume would be the difference beween the two and the surface area would be the water side tube.

I was hoping nh3simman would go through a working example since he obviously knows what he's doing

Most school testbooks didnt just give you a formula, they gave you an example to show how to use it.

Would be great to set up a little excel spreadsheet where I could plug in my figures and play around with the values, rather than re start each time.

NoNickName
10-01-2011, 09:33 AM
It works, if you know how to bend it to your needs. Calculate the two things separately.

chillman
10-01-2011, 11:17 AM
so whats step 1 ? :)

I've never used coolpack before,
where would you start?

NoNickName
10-01-2011, 12:08 PM
Step 1 is data input. Please give input.

chillman
10-01-2011, 01:10 PM
say we're working with a condenser,
the condenser outer (refrigerant) tube is 1 1/8" with 1mm Wall thickness.
and the inner (water) tube is 5/8" also with 1mm Wall thickness.

the water in temperature is at 30degC
with a flow of about 1L a second.

on the gas side, 404A the inlet temperature is 60 deg at 400psi (+-2700kPa)

the overall tube assembly length is 5 meters.

what else do you need?