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Max temp range allowed that's still in range of tolerance?

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data

    A 24" diameter titanium jet engine disk is machined to tolerance or +/- 0.002". Customer will accept +/- 0.003". The part is machined and measured in you plant at 80F. What is the max temp range for the customer to measure the part which you should specify in the contract?


    2. Relevant equations

    Delta L = L initial (alpha)(Tf - Ti) [alpha = coefficient of thermal expansion]

    3. The attempt at a solution

    0.003 = 24" (4.8 x 10^-6)(Tf - 80)--->Tf = 106 F??
    I dont think its this simple. Im pretty sure im wrong. help please!
     
  2. jcsd
  3. May 3, 2014 #2

    AlephZero

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    Your calculation is correct if the disk was exactly 24" diameter. But it was manufactured with a tolerance, so its diameter might not be exactly 24".
     
  4. May 3, 2014 #3
    so do i make two calculations? One with initial length as 24 + 0.002 and the other 24 - 0.002?
     
  5. May 3, 2014 #4

    AlephZero

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    You could say there are four calculations. The disk size might be 24 + 0.002 or 24 - 0.002, and the customer might measure 24 + .003 or 24 - .003.

    But if you think about it a bit, you don't need to do all four. (If you can't see why not, just do all four and see what you get).
     
  6. May 3, 2014 #5
    0.003 = 23.997" (4.8 x 10^-6)(Tf - 80)
    Tf = 106.04

    0.003 = 24.003" (4.8 x 10^-6)(Tf - 80)
    Tf = 106.038

    0.002 = 24.002" (4.8 x 10^-6)(Tf - 80)
    Tf = 97.3597

    0.002 = 23.998" (4.8 x 10^-6)(Tf - 80)
    Tf = 97.3625

    so the range would be from 97 F - 106 F?
     
  7. May 3, 2014 #6

    AlephZero

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    The answer can't be right, because if the customer measured at 80 F they would measure the same size as the supplier.

    If the disk size is 24. + .002 and the maximum the customer will accept is 24 + .003, what is the thermal expansion between those measurements? (It's not .002, or .003).

    Also, for some of the four calculations, the expansion is negative and the temperature will be below 80 F.
     
  8. May 3, 2014 #7
    thermal expansion = .001?

    0.001 = 24.002" (4.8 x 10^-6)(Tf - 80)
    Tf = 88.679?

    0.001 = 23.998" (4.8 x 10^-6)(Tf - 80)
    Tf = 88.33?
     
  9. May 4, 2014 #8

    haruspex

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    That looks reasonable.
    If it were machined to a diameter of 23.998" at 80F but measured at 88.33F, what would it be measured to be?
     
  10. May 4, 2014 #9

    AlephZero

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    Those look right.

    If the size changes from 23.998 to 23.997, the "expansion" is negative and the temperature will be less than 80.
     
  11. May 4, 2014 #10

    -0.001 = 23.998" (4.8 x 10^-6)(Tf - 80)
    Tf = 71.32 F

    thanks!
     
    Last edited: May 4, 2014
  12. May 4, 2014 #11
    -0.001 = 24.002" (4.8 x 10^-6)(Tf - 80)
    Tf = 71.32 F
     
  13. May 6, 2014 #12

    haruspex

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    That isn't what I asked.
     
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