1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermal expansion of lead sphere

  1. Jan 2, 2005 #1
    Hi guys and girls,

    I've been going thorugh some problems in cutnell & johnson on heat and temperature and I have become stuck on this question. Can anybody help with this? Its from Cutnell and Johnson 5th p375 Q18*.

    A lead sphere has a diatmeter that is 0.05% larger than the inner diameter of a steel ring when each has a temperature of 70.0 oC. Thus, the ring will not slip over the sphere. At what common temperature will the ring just slip over the sphere?


    Cheers, any help is much appreciated.

    Craig
     
  2. jcsd
  3. Jan 2, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Consider the coefficient of linear expansion of each material: [itex]\Delta L = \alpha \Delta T L_0[/itex]. What [itex]\Delta T[/itex] is required to make the diameters equal?
     
  4. Jan 2, 2005 #3
    OK, I have tried, but my maths is not very good. Can you give me a hint please? So far I have got the same point but I am not sure how to work the equations.

    Thanks
     
  5. Jan 2, 2005 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Here's a hint: Call the diameter (at 70 degrees) of the steel ring D; then the diameter of the lead sphere is 1.0005 D. Now find the diameter of the steel ring as a function of [itex]\Delta T[/itex]:
    [tex]D_s = D + \Delta L = D + \alpha_s \Delta T D[/tex]

    Now write a similiar equation for the diameter of the lead sphere. Set them equal and solve for [itex]\Delta T[/itex].
     
    Last edited: Jan 2, 2005
  6. Jan 2, 2005 #5
    ok, so

    [tex]L_L=[/tex] diameter of lead at new temp
    [tex]L_S=[/tex] diatmeter of steel at new temp

    at [tex] 70^oC[/tex] [tex] L_o_L=1.0005L_o_S[/tex]

    so at the new temp,

    [tex] L_S=L_o_S+ \Delta L_S =L_o_S+ \alpha_SL_o_S \Delta T[/tex]
    and
    [tex] L_L=L_o_L+ \Delta L_L =L_o_L+ \alpha_LL_o_L \Delta T[/tex]

    and we want [tex] L_S=L_L[/tex]

    so
    [tex] L_S=L_L[/tex]
    [tex]L_o_S+ \alpha_SL_o_S \Delta T=L_o_L+ \alpha_LL_o_L \Delta T[/tex]
    since
    [tex] L_o_L=1.0005L_o_S[/tex]
    then
    [tex]L_o_S+ \alpha_SL_o_S \Delta T=1.0005L_o_S+ \alpha_L1.0005L_o_S \Delta T[/tex]

    ok, bit stuck now, am I on the write track? if so what do i do next?
     
    Last edited: Jan 2, 2005
  7. Jan 2, 2005 #6
    Now you want to cancel out whatever you can, like your L_os, substitute in the co-efficients of Linear thermal expansion, and solve for Delta T, after that you should be able to find what temp they mesh well together...;)
     
  8. Jan 3, 2005 #7
    ok thanks guys i have sovled that one,
    BUT i have another question.

    how do you work out the relative proportions of an alloy if you are given the co-efficient of that alloy?

    cheers
    C.
     
  9. Jan 3, 2005 #8
    Hello,

    I agree with the finding of the new diameter of the steal ring with the following formula for lenear expansion.

    [tex] L_S=L_o_S+ \Delta L_S =L_o_S+ \alpha_SL_o_S \Delta T[/tex]


    However, inorder to find the new diameter of the lead sphere I think one need to consider the volume expansion of the sphere.

    [tex] V_L=V_o_L+ \Delta V_L =V_o_L+ \beta_LV_o_L \Delta T[/tex]

    Then find [tex] L_L[/tex] using

    [tex] V_L=\frac{4}{3} \pi (\frac{L_L}{2})^3[/tex]

    Gamma.
     
  10. Jan 3, 2005 #9

    Doc Al

    User Avatar

    Staff: Mentor

    No. All linear dimensions (such as diameter) will expand per the linear expansion coefficient.
     
  11. Jan 3, 2005 #10
    Dr. Al,

    Let's find the change in radius in both ways.

    Say the initial and final radii of a sphere is ,[tex]r_1[/tex] and ,[tex]r_2[/tex] respec.

    Linear expansion gives,

    [tex] r_2=r_1(1+ \alpha\Delta T)[/tex] ---(1)

    Considering the volume expansion (assuming a uniform sphere),

    [tex] r_2^3=r_1^3(1+ 3\alpha\Delta T)[/tex]

    [tex] r_2=r_1(1+ 3\alpha\Delta T)^\frac{1}{3}[/tex] ----(2)


    What am I missing?

    Regards,

    Gamma.
     
  12. Jan 3, 2005 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Right.
    This is an approximation. Nothing wrong with it; it's just not needed since all we care about in this problem is the linear dimension.
    Do a Taylor series expansion of this. You'll find that:
    [tex] r_2=r_1(1+ 3\alpha\Delta T)^\frac{1}{3} \approx r_1(1 + \alpha\Delta T)[/tex]
     
  13. Jan 3, 2005 #12
    Sorry, I don't get it.

    Why do approximate while there is an exact way? Also, how can we neglect the higher order terms for not small enough [tex]\Delta T[/tex]

    :confused:

    Gamma
     
  14. Jan 3, 2005 #13

    Doc Al

    User Avatar

    Staff: Mentor

    I hope you realize that:
    [tex] r_2^3=r_1^3(1+ 3\alpha\Delta T)[/tex]
    Is itself an approximation that ignores higher order terms!

    Also, recognize the circularity of your approach: You start with linear expansion, use that to find volume expansion, then take the cube root to get back to linear expansion: right where you started! :smile: All you need is linear expansion.
     
  15. Jan 4, 2005 #14

    Thanks for the clarification...


    :approve:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Thermal expansion of lead sphere
  1. Thermal Expansion (Replies: 6)

Loading...