Heat exchanger efficiency

[Mil75HK]

I've been working on a project to evaluate the performance of an air-handling unit with a counter-flow heat exchanger. When I simulate outdoor air that is hot and humid, in order to investigate the efficiency my results are generally very low.

A typical simulation might be:

 

Tout = 33 deg C, Tin = 25 deg C & RHout = 85%, RHin = 55%

 

I measure all the temperatures and humidities in the ducts and calculated the efficiency and get a low result, usually less then 50%.

I calculated the dew point temperature (Td) to be about 30 deg C in the conditions described.

 

Td = Tout - ((100 - RHout)/5) approximately

 

Is the reason for the low efficiency due to condensation? Do I need to calculate the efficiency differently when dealing with condensation?

 

Trying to determine the reason for the low efficiency, is it due to the small temperature difference between cold and hot air streams, compared to winter conditions, or/and that condensation occurs?

 

Cheers

kep

[CaptainPanic]

Firstly, you mention that the relative humidity is 85% outside (where it's hot and humid), and that the air inside is 55% moist. That does not make any sense to me.

 

You obviously cool the air coming in, which should cause condensation. Then the air coming in (at Tin = 25 deg C) should logically be 100% moist (saturated). the only way to get it below its saturation concentration is to cool it even lower, condensate even more water, and then heat it up again. But that would not make any sense (unless removing the moisture is your goal, rather than reducing the temperature).

 

In addition, the goal of the whole thing is (I think) to cool the air coming in. In the heat exchanger, all your cool air is heated by condensation. So the hot air loses water, but not much heat. Condensation is a problem, but I don't think there is any solution that is also energy efficient.

 

Finally, a gas-gas heat exchanger is always inefficient. And in addition to that, your temperature difference is only 8 degrees. There is sort of an engineering rule of thumb that you need at least 10 degrees difference for sort-of reasonable heat exchanging. It is possible to heat exchange anything, but the size of the required heat exchanger will go up.

 

Do you have any fans to move the air? You want a turbulent flow of air in the heat exchanger. If you have a laminar flow, it's horribly inefficient. You can check this easily by calculating the Reynolds number.

 

Finally, I have never seen a relation to calculate the dew point of water in air. I usually just use the vapor pressure (which can be found easily on internet or handbooks).

Edited November 19, 2008 by CaptainPanic

[Mil75HK]

Thank you for your input CaptainPanic.

The idea is to simulate sub-tropical conditions, with hot and humid outdoor air. The indoor air is controlled to be about 25 °C and 50-60% RH. Of course a cooling device is needed indoors. I'm just evaluating the heat exchangers performance and I'm expecting the efficiency to be quite low. But my question is if there is any way to calculate the effectiveness with consideration to the condensation that occurs? Since the formulas or equations I'm using now are only valid for dry air..

Yes, that is one of the points I'm going to make as well.. That in order to cope with the high latent heat a larger heat exchanger area is needed. This equipment is designed for northern Europe, north America etc..

 

The air-handling unit has two EC fans driving the airflow, but the flow is very low, about 150 m³/h at nominal speed. Going to check if the flow is turbulent or not.. thank you =)

 

I just found the equation here http://en.wikipedia.org/wiki/Dew_point for calculating the dew point temperature. It's only an approximation though..

[CaptainPanic]

I'd express the efficiency in energy / energy (Joule/Joule). that is: the amount of joules that you exchanged related to the maximum. Please note that with cooling and condensation on one side, and only heating on the other side, you can exchange more Joules on one side of your heat exchanger than on the other (unless you plan on evaporating the condensed water again on the other side, which might not even be such a bad idea). If you give the efficiency, you'll have to explain what it means anyway. You can't get away with giving just a number.

 

Have you considered looking up some formulas?

 

http://en.wikipedia.org/wiki/Heat_transfer will give you a start. Disclaimer: formulas can get quite nasty. Eventually you'll also need the links I provided before.

 

Actually, you can just check turbulence with some smoke. Although calculating is cool, measurements and observations are better.

 

The equation for the dew point is still weird. Have a look at http://en.wikipedia.org/wiki/Vapor_pressure - which is, imho, the more common way to look at condensation and evaporation.

Edited November 20, 2008 by CaptainPanic

[Mil75HK]

I love calculations =)

I've used the e-NTU method for my calculations for dry air, where you express the effectiveness exactly as you say: e = Q/Qmax

Where Q is for the air stream that undergoes the largest change.

 

m(hot)*Cp(hot) = Chot or m(cold)*Cp(cold) = Ccold and then determine which one is Cmin

 

e = (Ch*(Th,in - Th,out))/(Cmin*(Th,in - Tc,in))

= (Cc*(Tc,out - Tc,in))/(Cmin*(Th,in - Tc,in))

 

based on hot or cold stream, whichever has the largest change

 

The thing is that I'm doing my MSc thesis in energy engineering and all the measurements are finished. There is unfortunately no way that I can conduct any more tests with the equipment either.

The equipment I've evaluated is a residential AHU with a counter-flow heat exchanger, that is used for energy recovery.

 

So I'm calculating temperature efficiencies and the effectiveness based on the data I acquired. But, I just cant figure out what formulas or equations to use to calculate the effectiveness when condensation occur, since e-NTU is only valid for dry air. I've been trying to find formulas for this online and in books...but no luck..

I know what happens...and I can explain why the efficiency and effectiveness is poor. But I cannot show it by calculations... I think I'll just have to write the why instead of showing with calculations.. =)

Thank you for the help CaptainPanic =)

[CaptainPanic]

Well... the problem challenge that you have, as I said before, is that the energy contents (the energy that can be exchanged) is not equal on both sides. So there would already be 2 ways to define the efficiency (both sides of the heat exchanger).

 

But regardless of that... Normally, a heat exchanger is designed to exactly match certain targets. You're kind of reverse-designing it. You have built a heat exchanger, tested it to see how much energy is being exchanged between 2 flows, and now you try to give it an efficiency.

 

If I design a heat exchanger (for chemical industry) I specify the in and outgoing temperatures of a certain flow, and that exactly determines the heat exchanging surface that I need. Because the heat exchanger then does exactly what I want it to do, it's rather pointless to conclude that it's 100% efficient. So I leave that out

 

In short, I never talk about heat exchanging efficiency. The parameters that are of interest to me are the heat transfer coefficient, and the heat exchanging surface (and of course the temperature difference).

 

The flow of heat, Q is determined by those parameters:

[math]Q = U\cdot{A\cdot{\Delta{T}}}[/math]

[math]Q[/math] = heat flow (W = J/s)

[math]U[/math] = overall heat transfer coefficient (W/m2K)

[math]A[/math] = heat exchanging surface area (m2)

[math]\Delta{T}[/math]=temperature difference (which is actually not constant)

 

I failed to find the overall heat transfer coefficient for air-air atmospheric heat exchangers in my handbook (Perry's chemical engineers handbook). I guess that's because they don't exist (the heat exchangers themselves don't exist).

 

p.s. I think this is engineering. At least, I studied engineering to learn this, and so does Mil75HK.

Edited November 20, 2008 by CaptainPanic
adding formula

[Mil75HK]

I agree...and I feel so stupid..hehe.. I feel there should be a straightforward way to solve my dilemma, but I just cant see it.

I've tried to contact some professors at the uni, but so far no replies, they seem awfully busy. =)

 

Another problem that complicates some of the calculations is that the manufacturer of this AHU wont give me the heat exchanger area or the U-value. The AHU has 3 settings and works as a CAV-unit. It's equipped with two single inlet centrifugal fans with forward-curved impellers and maintenance-free EC motors.

I've tried asking a few times for the U-value at these setting or at least the area, but alas...it's classified.

However, I calculated the UA-value to see what the relationship between the air flow, UA-value, RH and efficiency was. Oh...and the temperature difference between the hot and cold air streams.

So larger temperature difference, higher efficiency...etc..

[CaptainPanic]

Hmm... yeah, if you don't know either the A (area) or the U (overall heat transfer coefficient), then you'll have to regard them as 1 parameter.

 

What's a AHU? And what's a CAV? I love abbreviations, but I never know what they mean.

[Mil75HK]

Oooopps...sorry.. One tend to get a bit blind when working on something in such detail for an extended period of time. =)

 

AHU - Air-Handling Unit

CAV - Constant Air Volume

 

I'm writing the Discussion & Conclusion of my thesis so I just need some of these facts straightened out before I hand it in for the last time =)

 

Thanks for all the pointers, it's really great...makes me look at the problem from different angles.

[CaptainPanic]

Glad to be able to help.

 

As some other forum member says in his signature, if you like my comments, click on the little scales symbol (add to reputation)

[Mil75HK]

Ahhhh...I found it =) Newbie here you know..hehe..