How Do You Solve Thermal Expansion Equations Under Time Pressure?

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In summary, the brass plug will fit into the ring at a temperature where the diameter of the brass plug is the same as the inside diameter of the ring.
  • #1
Jayhawk1
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https://www.physicsforums.com/showthread.php?p=543646#post543646

I asked this question earlier and I didn't get any significant help, atleast beyond what I already knew. I do not know how to calculate this, nor do my classmates who I have seen also posed this question to the forum. Please help! I know the concept, but I can not calculate it. I have worked on it for a few days now, and it is due in about an hour.
 
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  • #2
You are given the coefficient of linear expansion of iron so you can calculate how much the linear dimensions of the ring have increased and so the volume at each temperature. From that you can calculate the inside diameter of the iron ring (which is now smaller!). You are given the coefficient of linear expansion of bronze so you can calcuate how much the linear dimension (in particular, the diameter) of the brass plug increases with each degree and so what the diameter is at any temperature. Now find the temperature at which those two diameters are the same.
 
  • #3
Jayhawk1 said:
A brass plug is to be placed in a ring made of iron. At room temperature (20oC), the diameter of the plug is 8.755 cm and that of the inside of the ring is 8.745 cm. They must be brought to what common temperature (in Co) in order to fit? The coefficient of linear expansion for brass is 10x10-6(Co)-1 and that for iron is 12x10-6(Co)-1.

I don't know what to do with this... I know I have to subtract and divide some of these numbers, but I have no clue as to how to do it exactly. I need help please! Thank you.,

Diameter of brass plug as a function of the temperature change is

[tex] D_b = D_{b0}(1+10*10^{-6}/C^o*{\Delta}T)[/tex]

Diameter of iron ring as a function of the temperature change is

[tex] D_i = D_{i0}(1+12*10^{-6}/C^o*{\Delta}T)[/tex]

Set the two diameters equal and solve for the temperature change. Add the change to the starting temperature.
 
  • #4
HallsofIvy said:
You are given the coefficient of linear expansion of iron so you can calculate how much the linear dimensions of the ring have increased and so the volume at each temperature. From that you can calculate the inside diameter of the iron ring (which is now smaller!). You are given the coefficient of linear expansion of bronze so you can calcuate how much the linear dimension (in particular, the diameter) of the brass plug increases with each degree and so what the diameter is at any temperature. Now find the temperature at which those two diameters are the same.


The diameter of the inside of the ring actually gets larger. If it were a disk, all the linear dimensions would increase. The diameter of the "missing" part of the disk that makes it a ring will expand in proportion to the outer diameter.
 
  • #5
Thank you but...

Hahaha... I'm sorry, but now I'm just really confused. I tried making the two equations equal, but I got a very very small number. Can someone please clarify for me? I appreciate all your help so far.
 
  • #6
Jayhawk1 said:
Hahaha... I'm sorry, but now I'm just really confused. I tried making the two equations equal, but I got a very very small number. Can someone please clarify for me? I appreciate all your help so far.

[tex] D_b = 8.755cm(1+10*10^{-6}/C^o*{\Delta}T)[/tex]

[tex] D_i = 8.745cm(1+12*10^{-6}/C^o*{\Delta}T)[/tex]

[tex] 8.755cm(1+10*10^{-6}/C^o*{\Delta}T) = 8.745cm(1+12*10^{-6}/C^o*{\Delta}T)[/tex]

[tex] 8.755cm-8.745cm = 8.745cm(12*10^{-6}/C^o*{\Delta}T) - 8.755cm(10*10^{-6}/C^o*{\Delta}T)[/tex]

[tex] (8.755cm-8.745cm)*10^6C^o = (8.745cm*12 - 8.755cm*10){\Delta}T[/tex]

Can you finish this?
 

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