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    • CommentAuthorqeipl
    • CommentTimeDec 18th 2008
     
    Hello,

    I've spent a sizable chunk of today trying to find out if a coil heat exchanger in a cylinder is going to provide a DHW supply at a suitable temperature.
    In the absence of any hard evidence I cobbled together a calculation which has come out with a positive result but I'm not convinced that my high school physics is as sharp as it should be.
    I'm hoping that someone on this forum knows how to answer the question.

    The coil is made from 22mm (ID) smooth copper pipe and is 10m long. I imagine it is a helical coil.
    The maximum available flow rate through a 15mm pipe is around 20 litres/minute (measuring jug under the kitchen tap at full bore), which an online calculator converted to 43l/m for the 22mm pipe.
    The desired delta T is 45 degrees.

    All offers of help gratefully received.
    If you need any more parameters please ask.

    Thanks,

    Malcolm
    • CommentAuthortony
    • CommentTimeDec 18th 2008
     
    i think that the answer is no -- you need a high efficiency coil one with fins on. i m not exactly sure what you are trying to do though.
  1.  
    Hi,
    The smooth coil in most cylinders is about as you describe 10m of 22mm copper. It does not raise the temp in a single pass. If it did then boiler water in at say 70deg would come out at 25deg. The boiler water flows many times over and in each pass will give up about 10 degrees worth of heat. If you look at the spec sheet / brochures for various cylinders from down loads you will find that some quote a power kw for the coil. You can use this as a yard stick for flow temp estimations. But once the physical heat transfer limit is reached it wont transfer any more no matter how you alter the flow rate. After that it would be the finned tube which increases the heat tranfer capacity. Thermal stores (DHW stores such as Albion) whcih use the copper tube quote something like 20+lit/min but that would only be to hot tap temps which is about 38deg, they also assume the incoming is 10deg, so thats only a 20deg rise. Thats what they can claim it. The bigger stores used with larger solar or wood boilers (1000+ lit) use corrogated stainleass at about 1 1/2 inch, which gives a turbulent flow to increase heat transfer. They are longer 20+ meter and also contain 40-60 lit of water. That gives you a big head start when running a bath as its already hot. They will oftern quote the average temp over a period of run off. as it will be cooler water at the end of the run than at the start.
    To do the sort of temp rise you want you would need a plate exchanger. the size / shape would depend on the temp ranges in and out.

    Cheers, Mike up North
    • CommentAuthorCWatters
    • CommentTimeDec 19th 2008 edited
     
    Posted By: qeiplThe maximum available flow rate through a 15mm pipe is around 20 litres/minute (measuring jug under the kitchen tap at full bore), which an online calculator converted to 43l/m for the 22mm pipe.


    I don't follow that bit. The flow rate you need is independant of the pipe diameter. If you need 20 L/min then you need 20 L/min regardless of pipe diameter.

    The way to do the sums is to calculate the amount of power required:

    Power in watts = 4186 per Liter per second per degree centigrade rise

    So to heat 20L/Min (=0.33L/S) by 45 degrees you need

    4186 x 0.33 x 45 = 62KW

    or for a 20C rise..

    4186 x 0.33 x 20 = 27KW

    So thats the coil power you need. Check the power rating allows for scale.

    Aside: These calculation neatly explain why combi boilers aren't the best choice for a house with multiple showers.

    If you want to know where the 4186 comes from ...

    http://en.wikipedia.org/wiki/Specific_heat_capacity
    • CommentAuthorqeipl
    • CommentTimeDec 19th 2008
     
    Tony:
    I suspected that the answer was 'no' (i.e. that my sums had gone awry) but was hoping to be able to confirm it with a calculation.

    Mike:
    Your explanation makes sense. I'll investigate plate heat exchangers.

    CWatters:
    I had managed to work out the power required to heat the water but I was sceptical of my calculation of the heat transfer between the water in the tank and the water going through the coil. I'll follow advice and search for advertised ratings but as a matter of interest do you know how to calculate the heat transfer from tank to DHW via a coil?

    I mentioned the maximum available flow rate, not the required flow rate. I was assuming that the calculation would allow us to vary the flow rate to see the effect on the heat transfer.

    Many thanks to all.
    • CommentAuthorCWatters
    • CommentTimeDec 19th 2008
     
    Posted By: qeipl I had managed to work out the power required to heat the water but I was sceptical of my calculation of the heat transfer between the water in the tank and the water going through the coil. I'll follow advice and search for advertised ratings but as a matter of interest do you know how to calculate the heat transfer from tank to DHW via a coil?


    I'm not 100% certain but I think this is how I would do it...

    The problem is finding figures for the thermal resistance of the pipe wall. I guess you could do the sums assuming pure copper but probably need to include an allowance for scale build up.

    I would calculate an average for the delta T across the pipe wall. For example if the water in the tank is say 80C then the delta T at the cold end of the coil might be 70 (eg 80-10) and at the hot end it would be 35 eg (80-45). So the average is (70+35)/2 = 55C. Aside: You should be easy adjust this if the top of the coil is in water that's hotter than the bottom of the coil (eg in a stratified thermal store).

    Then if:
    Dt = average delta T across wall of pipe in Centigrade
    C = thermal conductivity of copper = 401 W·m−1·K−1
    Wt = wall thickness in meters
    A = area of pipe wall in meters
    P = Power in watts

    then I think the equation would be..

    Dt = (P x Wt)/(A x C) ......... I've checked that the units balance.

    but we want the area of the copper pipe so rearrange to get....

    A = (P x Wt)/(C x Dt)

    Then it's easy to get the length pipe from the area.

    Problem is I do think allowing for scale could be critical because I suspect the conductivity of scale is a lot less than copper.
    • CommentAuthorCWatters
    • CommentTimeDec 19th 2008
     
    http://www.energymanagertraining.com/Documents/lecture8.doc

    Page 6-7 says the conductivity of scale is between 0.08 and 2.33 compared to 400 for copper!

    Lets assume it's 1.0 W·m−1·K−1, then unless I've made a mistake, a layer of scale only 1/400th of the pipe wall thickness would double the length of pipe required. Means you probably need to estimate how much scaling might occur over the life of the pipe. You could probably do the sums assuming the pipe is made of scale and forget the copper exists by comparison.

    I was sceptical of my calculation of the heat transfer between the water in the tank and the water going through the coil.


    Perhaps this is why?
    • CommentAuthorEd Davies
    • CommentTimeDec 19th 2008
     
    The conductivity of the copper and/or scale does indeed put an upper limit on the rate of heat transfer but I'd have thought that the transfer of heat from the pipe water to the copper and from the copper to the tank water would also need to be considered.
    • CommentAuthorCWatters
    • CommentTimeDec 19th 2008
     
    Yes I've no idea how to account for the interfaces. I guess we could hope the flow is turbulent inside the pipe but that's perhaps unlikely for the water on the outside.
    • CommentAuthorqeipl
    • CommentTimeDec 20th 2008
     
    You've all confirmed that I shouldn't waste any more time trying to do the calculations.
    Thanks for the advice and the sums.
    • CommentAuthorchuckey
    • CommentTimeDec 20th 2008
     
    I am not sure that the waterflow in the pipe will be turbulent, also what percentage of the water will actually be transferring its heat into the copper walls? To ensure tubulence, if a stainless steel spring was wound into the pipe, which would then grip the inside walls of the pipe and so provide "roughness" without having to crawl up it and physically roughen its surface? But it still begs the quastion how much of the water would actually touch the copper during its trip.
    Frank
    • CommentAuthortony
    • CommentTimeDec 20th 2008
     
    these coils work OK for heating water in the cylinder so will work just as well the opposite way round.
    • CommentAuthorCWatters
    • CommentTimeDec 21st 2008
     
    I was thinking inside out might be better. Eg hot water on the outside of the pipe with the water being heated on the inside...not that it changes things much.
    • CommentAuthorcookie
    • CommentTimeJan 22nd 2009
     
    hmmm... I was looking to make my own coil heat exchanger for the DHW for a thermal store I'm planning (to save costs and eliminate the need for an additional pump)

    I'd really like to know what length of tube also / do the calculation. I had thought about using a small manifold to split my 22mm say into four 10mm coils then joining them back up. Is it the surface areas in contact with the waters that increase the energy transfer?

    Cheers Cookie
  2.  
    Hi,
    Couple of points. The DHW preparation tubes in a large europeon thermal store are made from corrugated stainless to promote the turbulence. The pipes are about 35mm dia so hold quite a bit of water – maybe 30-40-60 lit. They could be 20-30m or more long.
    Cheers, Mike up North
  3.  
    This is what you need to do.
    The overall heat transfer coefficient (h) for coils in water is 0.4 - 0.57 kW/m2.k. Source - Perry's Chemical Engineering Handbook

    If you have a 22mm coil, 10metres long, thats a heat transfer area of 0.69m2.

    The heat transferred will be given by

    Q = h * A * DeltaT.

    Delta T is a bit tricky, as you need to a) average it, and b) iterate it.
    As has been pointed out, to heat 20l/min by 20C needs 27kW.

    Doing it the other way round, A delta T of 30C on this coil, will transfer about 10kW, So that would heat your 20 l/min by about 7.5C.

    Which isn't much!
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