Surface temperature of cooling coil

[MrReds]

Hallo to everybody !

I am posting also to this forum following question:
I am trying to calculate a cooling coil (it's fin and tube heat exchanger).
I know inlet conditions, water inlet and outlet conditions (or, alernatively, water mass flow), but I am looking for a formula to get the fin temperature.
This temperature is what I need to get the working line of the coil.
Please, can anybody give me a formula to calculate this temperature (I suppose it would be function of fin pattern, air inlet temperature and somehow linked to fluid temperature), or suggest me a book or some literature where to find this formula ?

Thanks in advance

 

[IRstuff]

Please don't double post

TTFN

 

[stanlsimon]

James Threlkeld's 'Thermal Environmental Engineering' has a formula for both dry cooling and cooling with dehumidification. In the second edition see page 254.

 

[MrReds]

Dear Mr. Stanlsimon, please could you write the formulas in a reply of this post ? Maybe you can find me the easier solution !
Thanks in advance

 

[imok2]

The capacity of an evaporator is a function of two variables; the change it produces in the air’s temperature and the sensible heat ratio (SHR) of the heat transfer process . Mathematically, the sensible heat ratio is equal to the amount of sensible heat removed by the evaporator divided by the total heat (sensible & latent) removed by the coil. This ratio is calculated using the formula : SHR = Sensible heat/sensible+latent heat
Whenever a heat transfer process has a sensible heat ratio that is less then one, it indicates that some of the heat absorbed by the refrigerant (water) is used to condense some of the moisture out of the air. So when he temperature of the air drops below the water vapor mixed with the air condensation will occur. This requires latent heat removal and is known as the dew point.
I.m not sure I answered your question But if your looking for the rule of thumb for an air conditioning evaporator TD
it's usually about 16-22*F Remember, Evaporator TD is the the mathematical difference betwen the air entering the evaporator and the the saturation temperature of the refrigerant(water) inside the coil. EXample let's say air on =75*F and you have a 20*F TD, then air off should be about 55-56*F.

 

[AbbyNormal]

Search around for "Apparatus Dew Point" as well. I usually think of this as the average surface temperature of the tubes and the fins.

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[SAK9]

Reds,

u need to refresh your heat Transfer funda to solve this.

Use the following eqn to get the LMTD for the coil,
(Coil capacity)Q=U*A*LMTD.U Heat transfer cofft)and A (heat transfer area)and would need to be read from the coil charts.

After solving for LMTD,insert the value into the LMTD equation.The known values in this equation are air entering and leaving temperatures.Treat the coil temperature as constant thru out the section for simplicity.

The same eqn can be appiled to the water side heat transfer as well.

 

[willard3]

If you really want to know the ....fin temperature.....as you posted, this also will require you to get back to basic heat transfer.

Fin temperature will vary along its length.

Try "Elements of Heta Transfer", Jacob and Hawkins, Third Ed, chapt 9.

 

[AbbyNormal]

I always wondered about LMTD with cooling. For a chiller barrel with next to no superheat, I can see it being applied with a refrigerant and the saturation temperature and the change of the water temperature.

For a chilled water cooling coil, I can only see dealing with the senible component on the air side but how is the latent heat dealt with? Change in mass flow rate on the air side.

With a DX coil you have some superheat, negigible compared to the overall heat transfer, but a surface temperature change no the less, and you also have phase change on the air side.

So to me when it comes to cooling with dehumidifacation, the best use of LMTD is to explain why it is important to have the return water header upstream of the supply water header with respect to air flow.

Take the "V" out of HVAC and you are left with a HAC(k) job.

 

[sterl]

Anything that reasonably models the mechanisms of heat transfer from moist air to fin below dewpoint to tube below dewpoint will at best represent what is happening at one fin in the whole bundle. To analytically undertake the bundle, you would have to deal with the models: Both Fin and Tube Below Dewpoint; Fin above but Tube below dew point; Fin and Tube both above Dew Point; then drain filming from the cold tube to the marginally warmer fin; and then undertake it for the refrigerating fluid's conditions as well, that is change in evaporating temperature coinciding with pressure difference; change in tube temperature with (apparent) superheat; and so on.

Because the analytical model is so multifaceted, the commercial manufacturers of such coils have atmospheric chambers and similar lab setups.

The mechanisms are all treated in heat transfer books, and if you know a whole lot about the coil geometry, and have a lot of available computer time, the outcome might be significant.

 

[stanlsimon]

I was reticent to show the formula because I didn't know if you are doing dry cooling or cooling with dehumidification. Also these formulae don't give you the surface temperature directly, they give you an overall U value which gives you heat transfer which in turn could be used to find fin surface temperature.

Mr. Threlkeld's approach seems to be to first of all to calculate the resitances to heat flow. These would include the reciprocal of the inside of pipe convective heat transfer coefficient times the incremental area of pipe wall. There also should be a fouling factor resistance on the inside of the pipe. Then you have the resistance of the pipe wall, a simple calculation involving the conductivity of the pipe, then fin resistance, then the outside film heat transfer coefficient which is in the case of dehumidification related to the dry heat transfer coefficient by the relationship hw/hd=0.626(face velocity)**0.101 .

Once you have all the resistances added up take the reciprocal of that and that gives you a U value. Since we are assuming steady state heat transfer then the same amount of heat transfers across each element of resistance. Therefore the temperature differences across each resistance will be proportional to the amount of each resistance. At that point the solution to the fin temperature questions will be a matter of simple algebra.

Of course this would apply primarily to a small increment of tube and fin but Mr. Threkeld uses the logarithmic means of air enthalpies, some tables and some iteration to encompass real world situations. If you don't have his book in your library you really should get a copy. I have tried to condense over thiry pages into this post.

Most coil manufacturers can give you a heap of empirical data on their products, this would be very beneficial.


Hope this helps.

 

[MrReds]

I'd like to thank everybody who has answered to the post!
I'll keep care of every suggestion.

Please, Mr. Stanlsimon, where can I find the book you are quoting ?

Many thanks

 

[stanlsimon]

Prentice-Hall of Engelwood Cliffs, New Jersey was the publisher. Library of congress card number 77-105450

Also the University of Minnesota Institute of Technology Engineering Library might be able to help you.

 

[MrReds]

Many thanks Mr. Stanlsimon,

please, do you think that ASRHAE handbook can be useful the same ?

Thanks in advance

 

[stanlsimon]

I looked in my ASHRAE book and it had a few formulas, but nothing like Threlkeld.