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How to calculate thermal insulation savings?

  1. Jul 27, 2010 #1
    Hi

    I am working at a manufacturing plant and we are applying a second layer of glass fibre/alumnium foil insulation to an industrial cylindrical boiler, on top of a previous layer. I need to calculate the approximate savings to be able to give a ROI figure.

    The surface temp of the current boiler insulation is 90c under average load (i am going to assume average load as a constant as it is fairly stable) and the thermal conductivity of the new layer of insulation is 0.033 W/mk at 10c

    The surface area of the boiler is 19m2 around the length of the boiler. (the ends are being insulated by other means).

    I take it once i have calculated a new surface temp i can work that out as equivelant energy from the calorific value of the gas against the internal temp of the boiler?

    This is all the information i have at the moment but i am a bit stuck, obviusly the figures are fairly loose but i have to justify expenditure where possible. this is a cheap saving but it would be useful to know the approximate annual savings.

    if any other figures are needed please let me know.

    thanks in advance.
     
  2. jcsd
  3. Aug 2, 2010 #2

    marcusl

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    We'll ignore the effect of the foil (it reduces radiant heat, but this effect should be small compared to that of the insulation). Obviously you'll need the thermal conductivity of the first layer.

    Heat transfer through insulation is
    Q = (Tinside - Toutside) * A / (t1 / K1 + t2 / K2 +...)
    where Ki is the conductivity and ti the thickness of the ith layer, and A the area.

    The calculation goes like this:
    1. Calc. heat through first layer.
    2. Cacl. heat through first and second layers.
    3. Subtract to find reduction in heat due to second layer.

    Example: assume Tin = 90C, Tout = 20C, and the two layers have identical K but the first is 2.5 cm thick and the outer is 5 cm thick. Then
    Q_onelayer = 70 * 19 / (.025/.033) = 1756 W
    Q_twolayers = 70 * 19 / (.075/.033) = 585 W
    Savings = 1171 W

    If you want to compute monetary savings, then you need to know the heat content of gas and your cost. The latter varies with location, from year to year and, often, season to season.

    I can show an example calculation, but beware--I'm in the US where we still use crazy British units. 1 BTU = 1.06 kJoule, and one watt = 1 Joule / second so you are saving 1.1 BTU/second. In one month (2.6e6 seconds) that's 2.8 million BTU. In the US, one "therm" of natural gas has 100,000 BTU heat content and costs about $0.50 (residential), so you'd be saving about $14/month at my house.
     
    Last edited: Aug 2, 2010
  4. Aug 3, 2010 #3
    Thank you Markusl this is ideal for what I need, and not hard to calculate either (my University engineering math days are way behind me and are but a distant memory!)

    I take it by calculate the temp you mean measure, so I will do that once our summer shutdown is complete and our boiler is up and running again.

    The BTU's are really confusing at first glance but it is nice to have the example, the rest of the data i can obtain easily.

    Thanks again
     
  5. Aug 3, 2010 #4

    marcusl

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    Glad to help. Sorry about the BTU thing, I don't like those units either, but that's how the heat content and cost of natural gas are stated in my monthly utility bills.

    BTW, I forgot to include the boiler efficiency e. Multiply the amount of gas used, and the cost, by 1/e. If the boiler were 80% efficient (e = 0.80) in my example, then the cost savings per month would be 14/0.8 = $17.50 per month.
     
  6. Aug 4, 2010 #5
    great stuff cheers,

    i have another question on the forum now relating to compressors if you fancy having a crack at that! :)
     
  7. Aug 4, 2010 #6

    marcusl

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    No thanks, I know my limits!
     
  8. Mar 16, 2011 #7
    I found you thread and wanted to ask a simple question.
    If you increase the amount of insulation in an attic, what formula would you use to solve the increase in the savings in btu's and $'s?
    In the past I have used Q=U*A*HDD*24*.75, but wanted another opinion.

    Thanks,
    Drew
     
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