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Convective heat transfer and steady state conduction problem

  1. Nov 12, 2014 #1
    • Homework problems require the use of the template and require some effort on the part of the student
    Hi all, I have a unique problem that I need help with

    A company's curing oven's exposed surface temperature is measured at 80°c when the surrounding air is 23°c. You think it should be insulated and wager to pay for the cost of this yourself if you can keep the savings incurred. Is this a smart move? the oven is 3.7m long and 2.4m diameter. The plant operates 16 hours/day 365 days/year. The insulation to be used is fiberglass (Kins = 0.08 W/m/°c, assumed constant) which costs $2.5/m2 per cm thickness plus $21.5/m2 for labour regardless of tthickness. Convective heat transfer coefficient on the outer surfaced is estimated to be h0= 20W/m2/K. The oven uses natural gas, whose unit cost is £7.1/GJ input and the oven's efficiency is 80%.
    Ignoring radiation heat loss from the outer surface, determine
    A) how much money you will make out of this venture, by installing one inch (1 inch = 2.54 cm) of insulation, if any, and
    B) the thickness of insulation (to nearest inches) that will maximise your earnings


    My thinking for part A is to:

    calculate the area of the oven exposed to the air,
    use q = hA (T2-T1) to obtain the heat transfer rate of the system with no insulation
    Use Rcond = ln(r2-r1)/(2*pi*L*K) to find the heat transfer rate *with* 1" insulation
    find Rconv ( 1/(h*A)
    use Rconv and Rcond to find Q
    calculate loss per year versus installation cost.

    This comes out to be a saving of $3432.67 for the year for me, but given the long answer nature of this question I've undoubtedly gone wrong somewhere and would appreciate other peoples' working if possible :)

    Any input would be greatly appreciated as I'm well and truly stuck
     
    Last edited: Nov 12, 2014
  2. jcsd
  3. Nov 12, 2014 #2
    Show us your detailed calculations of the heat load both with and without the insulation. From your description, it looks like you have the right idea.

    Chet
     
  4. Nov 12, 2014 #3
    Area of the oven exposed to the air;
    (pi x d^2)/2 + (pi x d x L) = 32.42 m^2

    Heat transfer rate of the system with no insulation:
    q=hA(Ts-Tsurr)
    q=20 x 32.42 x ( 80 - 23 )
    q=36.96 KW

    Heat transfer rate of the system with 1" insulation:
    Thermal circuits-
    Rcond= ln(r2-r1)/(2 x pi x L x K)
    Rcond= 5.661x10^-3 K/W

    Rconv=1/(h x A)
    Rconv=1.542x10^-3 K/W

    q=dT/(Rcond + Rconv)
    q= (80 - 23)/((5.661+1.542)x10^-3)
    q=7913 W

    Loss per year, without insulation;
    ( 36.96 x 10^-3/10^9 ) x 7.1 = $5516.7

    Loss per year, if fitted with 1" thick insulation
    ( 7913/10^9 ) x 7.1 = $1181.13

    Total cost of installation;
    assuming $2.5 x 2.54 = $6.35 /m^2 per inch thickness (not calculated by volume of a hollow cylinder)
    Cost by area= A x cost per inch = 32.42 x 6.35 = $205.867
    Cost by labour= 21.5 x 32.42 = $697.03
    Total cost of installation= $205.867 + $697.03 = $902.897

    Total loss with insulation (including installation) in a year= $902.897 + $1181.13 = $2084.03
    Total loss without insulation in a year= $5516.7
    Net savings= 5516.7 - 2084.03 = $3432.67

    Have yet to manage part B unfortunately
     
  5. Nov 12, 2014 #4
    Well, I didn't go through every last detail, but it looks pretty good. However, I have the following comments:
    1. Is that 7.1 British pounds per GJ, of 7.1 $ per GJ?
    2. What did you do about the 80%?
    3. Neglecting the 80% issue, I get the $ amounts for the natural gas that you got, even though the equations in your post are not complete, leaving out the sec/yr part.
    4. In part b, let T be the thickness of the insulation, and calculate the heat transfer and the economics as a function of T. Then determine the value of T that maximizes the return.

    Chet
     
  6. Nov 13, 2014 #5
    Hey,

    Yes that's $/GJ and I'm unsure as to how to factor in the 80% efficiency of the oven. Thanks for pointing that out - I should definitely leave in the sec/yr calculation. This has been a big help!
     
  7. Nov 13, 2014 #6
    I believe I've ran into a snag with the very first calculation - surface area of the cylinder. Somehow I got 32.42m^2 instead of 36.95m^2 so here we go again!
     
  8. Nov 13, 2014 #7
    I used your 32.42, assuming that you got that right. So, just redo the calculation.

    Chet
     
  9. Nov 13, 2014 #8
    yeah that's all good now, I'm still just a tad stumped as to where I throw the 80% efficiency into the mix, though :(
     
  10. Nov 13, 2014 #9
    Efficiency -- if a car is less efficient what does that mean? Same here 80% means what?

    Granted this is homework and that is the stated problem - but beyond the overall cost of running the of the oven, is the cost of removing the ~40KW heat in the factory. For example if the surrounding air is "maintained" at 23 C - then there is a cooling system that has to work harder to remove the heat- it also has an efficiency - and 40KW waste heat is significant.

    Also - had to laugh a little -- Question A = How much you will be paid to do this project? That is what you will "make".... Energy savings do not equal "money made" - unless you have a nice contract with your employer / customer... if you get my point.
     
  11. Nov 13, 2014 #10
    It isn't clear what they mean by this.
     
  12. Nov 13, 2014 #11
    I am quite sure ... 80% of the energy in is converted to "useful" heat. -- this inefficency comes mostly from how good the combustion is and how much heat goes out with the exhaust. So using "£7.1/GJ input and the oven's efficiency is 80%" when you use £7.1 of gas you get 1.0 GJ * 0.80 heat in the oven.

    A "perfect" oven would have 1 GJ of heat for every £7.1 spent.

    --- AND my apologies I overlooked the payment scheme - but the savings over how long...assuming 1 year? etc..
     
  13. Nov 13, 2014 #12
    Yeah assuming one year, sorry for slow replies this part B is taking it's toll on my sanity haha
     
  14. Nov 13, 2014 #13
    It shouldn't. You just do exactly the same thing you did in part a, but retain the insulation thickness as an algebraic variable. Or, just do part a over and over again for a bunch of different thicknesses, and plot a graph of the $$ return as a function of the thickness to find the minimum point.

    Chet
     
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