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Spiral heat exchanger - pressure drop?

  1. May 4, 2015 #1
    Hi, I hope this is the right section.
    I'm trying to figure out a rough estimate of the pressure drop in a spiral heat exchanger. The fluids both being water.
    My guess it it would be in terms of the surface area of the inside of the heat exchanger, and the flow rate, but I can't seem to nail anything down.
    I should be able to figure out the actual heat conduction from one fluid to another on my own. I just need the pressure drop. I've found relationships between pressure drop and flow, but those are for round pipes, and don't seem to be applicable here. I could be wrong.

    Any advice you can give me would help a lot (this isn't homework). It doesn't need to be particularly precise, but if it was, that would be great. I'd be glad to have a ballpark figure.
  2. jcsd
  3. May 4, 2015 #2
    You could put a lower value for the heat exchanger assuming it simply consists of two long tubes. First calculate the Reynolds number and Fanning friction factors
    From Geezer ChE. Assume a general approach: two helical coiled pipes (or tubes)of L1 and l2 just calculate the delta_p using Reynolds Number, friction coefficient and pipe(tube) length and diameter. Then add the delta_p for the fittings. You will also need to consider the effect, if any, of the working temperatures on the fluid properties. I would suggest using the mean temperature for each of the working fluids of the fluid properties.

    See this link http://www.engineersedge.com/fluid_flow/pressure_drop/pressure_drop.htm or general delta_p calculations
  4. May 4, 2015 #3
    Geezer ChE had a Brain Frt or senior moment.
  5. May 5, 2015 #4
    Thanks for the reply, Geezer ChE,

    That's an idea, it would be an extremely long tube though; I'm not sure if it will get me within the right order of magnitude (maybe? At least it probably couldn't be worse in real life so I'd have a conservative estimate). Unless Laminar flow monkeys with things (wouldn't that effectively stop heat exchange?).
    But then I'm also not sure how to calculate heat conduction, since it goes from a simple gradient of materials, to a tube where heat needs to be conducted from the center out, radially.

    Warm water (average temperature) = 5.6 x 10^-7
    8 liters/second
    1 cm diameter
    12.73 meters/second

    Re = 227,364.203571 seconds
    No risk of laminar flow there, I suppose

    If absolute roughness is 0, lambda is 0.0152513496

    But I guess it would be something like copper...

    1.5 * 10^-6 m

    Which would make lambda 0.0164518291


    Pressure drop = Lambda * length/diameter * density/2 * flow velocity^2

    Knowing what it is per meter might be useful, since it's linear with respect to length anyway, that can be converted into whatever length is really needed.

    0.0164518291 * 1m/1cm(100) * 1,000kg/m^3 * 162.0529 m2/s2 = 266,606 Pa
    So, a little less than the entire pressure of a garden hose being lost each meter.

    Since the diameter is small enough, and the flow is not laminar, I hope I can ignore the insulation of the water itself and assume those molecules that pick up heat from the sides get spirited away toward the center.
    And of course I need to pay that cost twice (for each direction).

    With a contact area of pi*10^-2 m^2 per meter
    An average difference in temperature here of let's say 20 degrees

    Copper thermal conductivity is around 400 W/(mK)
    At 2mm thickness of copper, and contact with the copper for 0.07855459544 seconds

    400 W/(mK) * 0.07855459544 s * 20 degrees * (pi*10^-2 m^2) / 2mm = 9,871.46161774 J (transferred from one stream to the other)

    At 8 liters per second, and 0.078... s = 0.628436764 liters (or kg)

    If a joule raises the temperature of 1g of water by 0.24K

    9,871.46161774 / 628 * 0.24 = 3.77253310232 degrees

    That's not very efficient, given that pressure drop. And I'm left with my hot water still pretty hot, and my cold water still pretty cold.

    Did I do all of that right?
    My assumption is that I improve this with more pipes, and lower velocity in each pipe (which should reduce the drag non-linearly). Is that a correct assumption?
    Or I could just keep making the pipe longer

    I think I can safely ignore the fittings as a drop in the bucket, with regard to this estimate, since their contribution should be pretty minimal (they're usually very large openings), it's the actual heat exchange surface that does most of the restricting. Or am I underestimating fittings?
  6. May 5, 2015 #5
    I tried a longer pipe:

    12.73 m pipe
    3,393,894.38 Pa drop
    0.3999247448 m^2 area
    1,599,698.976 Joules transferred
    to 8 liters

    47.99096928 degrees
    That's enough heat transfer to reduce nearly boiling water to nearly freezing, and increase nearly freezing water to nearly boiling (which is just slightly more than I intend, but a conservative estimate).

    Energy required to pump:
    3,393,894.38 Pa * pi * 0.5cm^2 * 12.73 m / 1 s = 6,786.51172 watts

    *2 (both ways) is 13.573 kW (at least, assuming a 100% efficient pump) to run the pump to compensate for the pressure drop.
    Unless I'm calculating that incorrectly.

    Can anybody tell me if this is roughly correct?
    Also, sorry my significant figures are all over the place, I'm just looking for a ballpark estimate, so I did a lot of copy&paste of these numbers to avoid transcribing them incorrectly.
  7. May 5, 2015 #6
    See if your method yields similar results to this,
    http://www.nt.ntnu.no/users/skoge/prost/proceedings/icheap8-pres07/pres07webpapers/84%20Picon-Nunez.pdf [Broken]
    Last edited by a moderator: May 7, 2017
  8. May 5, 2015 #7
    Thanks for the link. Although I have to admit, I think it's going over my head.

    If I go by this (listed in the table, no idea if it's comparable)
    2.824E+3 kg/hr
    6,894757 bar

    47 liters/second
    689,475,700,000 pascals (or are they using commas and periods inconsistently?)

    117,357,565,957 Pa for each 8 liters per second

    That pressure drop is about 200,000 times larger than mine.
    And doesn't really sound plausible, so I'm probably reading this incorrectly. A pressure drop like that would make the heat exchanger less efficient than just heating your liquid up again with the energy you would have used to obtain such insane pressures (which I don't think are even achievable - almost 700 GPa, that's twice the pressure at the core of the Earth).

    If they're using a comma instead of a period, then the pressure drop I estimated is almost 30 times too large. More believable than the alternative, but I have no idea if that's right because I think I'm misreading this stuff.

    One other thing that really confuses me in that chart is that for inlet and outlet temperature, both are getting colder.
    Last edited by a moderator: May 7, 2017
  9. May 5, 2015 #8


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    This P = 6.894757 bar (i.e. about 7 bar)

    On page 4 of the paper, second paragraph under Results and Discussion, the pressure 6.894 bar is mentioned.

    I think in that one instance with the pressure, there was a typo or some sort of editing problem. Otherwise, why would anyone use a spiral heat exchanger if it required overcoming pressures greater than those at the center of the earth?
  10. May 5, 2015 #9
    Thanks SteamKing, I missed that. I thought it was probably an editing problem.
    So, then my result for a simple pipe was 30 times too high. If I read the rest correctly.
    Probably because of the very high flow rate through such a small pipe. More pipes with a slower flow rate each might give me a better result, closer to that of a spiral heat exchanger, since the contribution of velocity is exponential.
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