- #1
Keldric
- 6
- 0
1.Homework Statement .
Can someone give me more accurate mathematical models on thermal expansion? for example, linear expansion.
2.Relevant equations
L=L0(1+α∆T)
this equation is not accurate. for example.
When an object 100m long at 0oC (example: aluminum, α=2.4x10-5) is heated to 100oC, then:
L0=100 and ∆T=100
L= 100.24
When it is cooled back to 0oC, then:
L0=100.24 and ∆T=-100
L=99.999424
There is a little discrepancy but it still is not accurate.
another is this:
First situation: aluminum (100m long at 0oC) is heated to 200oC, then measure the length.
L=100.48
Second situation: aluminum (100m long at 0oC) is heated to 100oC, the length is measured. then from 100oC it is heated again to 200oC then the length is measured again.
at 0oC, L=100
at 100oC, L=100.24
at 200oC, L=100.480576
Again, there is a dicrepancy at 200oC . one is 100.48 and another is 100.480576.
I made a formula. can you test its validity?
L=L0eα∆T
Can someone give me more accurate mathematical models on thermal expansion? for example, linear expansion.
2.Relevant equations
L=L0(1+α∆T)
this equation is not accurate. for example.
When an object 100m long at 0oC (example: aluminum, α=2.4x10-5) is heated to 100oC, then:
L0=100 and ∆T=100
L= 100.24
When it is cooled back to 0oC, then:
L0=100.24 and ∆T=-100
L=99.999424
There is a little discrepancy but it still is not accurate.
another is this:
First situation: aluminum (100m long at 0oC) is heated to 200oC, then measure the length.
L=100.48
Second situation: aluminum (100m long at 0oC) is heated to 100oC, the length is measured. then from 100oC it is heated again to 200oC then the length is measured again.
at 0oC, L=100
at 100oC, L=100.24
at 200oC, L=100.480576
Again, there is a dicrepancy at 200oC . one is 100.48 and another is 100.480576.
The Attempt at a Solution
I made a formula. can you test its validity?
L=L0eα∆T