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30131Re: SS wall heat conduction math

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  • dearknarl
    Jun 1, 2005
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      Hate to bring bad news, but I'm guessing the conduction of heat
      through the steel will not be your limiting factor, and your 120kW
      sounds about right.

      It will more likely be the convection of the water bath and the wash
      inside the boiler, more so the water bath once it is boiling, that
      will limit heat transfer.

      Agitation of both will provide better heat transfer. Trying to
      calculate heat transfer for such a system would be a nightmare.

      I'd say your best bet is an experiment, unless you're a hardcore nerd
      that wants to create a computer model, or a hardcore physicist/maths
      guru who has way too much spare time =). Let us know how you get on.

      knarl.

      --- In Distillers@yahoogroups.com, Andrew Forsberg <andrew@u...>
      wrote:
      > Hi all,
      >
      > The internet has provided me with far too many conflicting rates and
      > formulae for calculating whether an external water bath for a
      stainless
      > steel boiler will or will not provide sufficient heat.
      >
      > I'd very much appreciate it if an engineer on the list would give
      this a
      > once over:
      >
      > T1 = 105 degC (water + table salt jacket)
      > T2 = 78 degC (mash)
      > Heat transfer coefficient = no idea... Ignoring it for the moment.
      > K = approx 14 W(m2 . K) (I have seen more figures for SS than I
      care to
      > poke sticks at. This is the most conservative, and is for 270K, so
      > anything better would be super).
      > Q = Watts
      > A = Area = .64m2
      > D = thickness of the SS wall = approx 2 mms
      >
      > Q/A = (T1 - T2) / (D . 1/K)
      >
      > Q/A = (27) / (.002 / 14)
      > = 27 . 7000
      > = 189,000
      >
      > Q = 189,000 . 0.64
      > = 120.96 kW
      >
      > So, what I'd like to know is -- will a 2mm stainless steel wall for
      an
      > area of .64m2 allow heat to pass through at a rate of up 120kW for
      that
      > temperature differential? Even approximately? As long as it's higher
      > than 3 or 4kW I'm happy. Unfortunately some formulae I've found
      > suggested the figure would be closer to 800W...
      >
      > Cheers and thanks,
      > Andrew
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