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Water to air heat exchange - imperial units doing my head in

  1. Jul 28, 2014 #1
    Hey guys, newbie here and first post. I've got a (hopefully) simple question, but for some reason my brain just isn't working. I'm too used to working with metric, and even though it should be fairly simple, working with imperial units just hurts.

    Anyway, here's the problem: I have a specific volume of air that I need to move, and at the same time I need to heat the air to a specific temperature. I'm using a water to air heat exchanger to heat the air, and found what I think is a suitable product.

    The heat exchanger product page lists:
    size (in inches)
    BTU (is this BTU or BTU/h?)
    volumetric flow rate of air (CFM)
    pressure drop of air (this really confuses me - is it 81% drop of original pressure? ie: the pressure is now 19% of original??)
    pressure drop of water (ditto as above)
    volumetric flow rate of water (GPM)

    I have T1 and T2 of the air that I wish to heat, and I know the velocity that I want the air to be after the heat exchange (I picked the heat exchanger based on the volumetric flow rate - assuming it remains constant through the heat exchange?). I also have the energy required to heat the air from 20 deg C to 32 deg C (the final temp).

    How do I figure out the energy in to the water? Sorry if this is a stupid question, but it seems like US methods are completely different to what I learned in University
  2. jcsd
  3. Jul 28, 2014 #2


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    Can you post this HE product page or give a reference?
  4. Jul 28, 2014 #3
    Yes, here's the product page: http://www.northlanddistrib.com/Water-To-Air-Heat-Exchangers_c_423.html

    Basically I want to produce a flow of hot air at 32deg C, assuming T1 is 20 deg C and a volumetric flow rate of approx 1 m^3/s by blowing it through the HE. Trying to figure out how to work out the energy needed to be put into the water.
  5. Jul 28, 2014 #4
    The decimals are not % loss. The website is terribly laid out for just having the values (and not being consistent with them...), but if this is meant to be installed with HVAC ducting, the .81 for air pressure loss would generally be read as inH20 (inch water column), which is what they are using here, according to the spec sheet. They use Ft. H20 (feet head) for the water pressure drop.

    Have you read the spec sheet? http://www.northlanddistrib.com/WA16X18-Heat-Exchanger--40000-106000-Btu_p_67.html

    T2air, and T1water required, are determined by the airflow across the unit. Make sure your system can achieve these requirements (800-1200 SCFM of air at roughly 7-10 ft/s for the 16x18 model)

    Also, check the performance characteristics of your selected model, 2100 CFM (1m^3/s) is very high. In this case, it would be best to contact the vendor and discuss your particular application.
    Last edited: Jul 28, 2014
  6. Jul 28, 2014 #5
    I didn't even see the specs there :blushing:

    It's more for really rough order of magnitude to get an idea of what we're dealing with. 1650 CFM should be just on the lower limit of what we can get away with, so I'm trying to aim for that.

    It doesn't help that I'm completely unfamiliar with any of these units of measurement. Maybe I'll try and use the dimensions provided of the HE and see if I can work out what I need from there instead of relying on the empirical data?

    I should be able to use the surface area and thermal conductivity of aluminium to work out the total thermal resistance, correct? Then from there I'll have the overall heat transfer coefficient and since I know how much energy I want to put into the air, I can then use that to find out how much energy needs to be in the water?
  7. Jul 28, 2014 #6
    I think you'd get further calling the vendor...

    You could take the operating data given in the spec sheet and work backwards to find the expected conditions at your operating points, but you'll likely be making lots of assumptions.
  8. Jul 28, 2014 #7
    Yeah it's more for ballpark figures, but I take your point... contacting them would probably be the easiest way. I just wanted the satisfaction of working it out myself lol. Thanks for the help!
  9. Jul 28, 2014 #8
    Go for it, then, by all means! Even if only to assure yourself that you know what the vendor is talking about. Should be a good exercise to get yourself a pretty darn good understanding of this type of exchanger.

    Getting the surface area of these can sometimes be tricky business though.
  10. Jul 28, 2014 #9
    Yeah I've already run into a wall with that lol. But I figure 12 fin/in by 3 rows of fins by the height of the HE by the OD of the tube should be close enough to give me an approximate surface area. Now figuring out the rest is hurting my brain :S Thermodynamics were never my strongest subject. It's really interesting, but it's just layers upon layers and layers.
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