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Thermal Expansion of a circular steel plate

  1. Apr 17, 2014 #1
    1. The problem statement, all variables and given/known data
    A circular steel plate of radius 15 cm is cooled from 350 C to 20 C. By what percentage does the plates area decorate ?


    2. Relevant equations

    A=∏r^2
    Af = Ai (1+2∂ΔT)
    specific heat of steel = 12 x 10^-6

    3. The attempt at a solution
    r = 15 cm = .15 m
    Ai = .070685 m^2

    Af = Ai (1+2∂ΔT)
    = (.070685m^2)(1 + 2 (12 x 10^-6)(330 C))
    = 23.3 m^2?

    What am I doing wrong? The final area does not look right at all. Thank you.
     
  2. jcsd
  3. Apr 17, 2014 #2

    Simon Bridge

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    This would be an approximate form. Is the approximation valid for this problem?

    Did you use the linear or volume coefficient for ∂?
    Where did you get the value from?

    If the temperature decreased, is ΔT positive or negative?
     
  4. Apr 17, 2014 #3

    rl.bhat

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    = (.070685m^2)(1 + 2 (12 x 10^-6)(330 C))
    It should be
    = (.070685m^2)[(1 + 2 (12 x 10^-6)(330 C)]
    Now try.
     
  5. Apr 17, 2014 #4

    Simon Bridge

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    ... you've got one extra parenthesis.

    "(.070685m^2) ( 1 + 2 (12 x 10^-6)(330 C) )" is fine as it is written.

    But (.070685m^2)(1 + 2 (12 x 10^-6)(330 C)) ≠ 23.3m^2 ... well spotted.

    ... so I should add "check your arithmetic" to my list :)
     
  6. Apr 17, 2014 #5

    rl.bhat

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    Yes. You are right.
     
  7. Apr 17, 2014 #6
    Thank you for your help!

    So, Af = .0712 m^2
    Then percent decrease should be (.0712 - .070685)/.070685 or 72%. Does this make sense?
     
  8. Apr 17, 2014 #7

    Simon Bridge

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    This is not correct.
    1. (.0712 - .070685)/.070685 ≠ 0.72

    2.
    The percentage change in area A would be: $$p=100\frac{A_f-A_i}{A_i}$$ ... you swapped final and initial over.
    If ##A_i > A_f## then the negative percentage tells you the change was a decrease.

    Go back to your original calculation - is your final area bigger or smaller than the initial area?
    Now compare with the question: should the plate be bigger or smaller after cooling?

    Hint: ##\Delta T = T_f-T_i##
     
    Last edited: Apr 17, 2014
  9. Apr 18, 2014 #8
    That helped. Thank you!
     
  10. Apr 18, 2014 #9

    Simon Bridge

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    No worries.

    For the future: it is best practice to do all the algebra before you plug in the numbers.

    Considering you have the same trouble with arithmetic that I do, I figure you'd do better to do what I do and avoid arithmetic as long as possible.

    For example - the final answer you want is a percentage, so derive the equation for the percentage first: $$\begin{align}A_\% &= 100\frac{A_f-A_i}{A_i}\\ &=100\frac{A_i\big(1+2\alpha\Delta T\big)-A_i}{A_i}\\
    &=200\alpha\Delta T\end{align}$$ .. so you could have avoided a bit of work.

    I know the algebra looks more intimidating than the numbers, but it is much easier to troubleshoot.
     
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