Linear expansion equation

  • #1

Main Question or Discussion Point

Lfinal-Linitial=a*Linitial*(Tfinal-Tinitial)
???
lets assume the linear expansion coefficient for sth is (0.0001) kelvin^-1, its (1) meter long and its at (10) kelvin.
now lets calculate its length when its at (20) kelvin in 2 different ways:
first way: L1-1 = 0.0001*1*10
so: Lfinal=1.001
second way:first lets find its length when its at (15)kelvin:
L2-1=0.0001*1*5
so L2=1.0005
and then lets find its length when it goes from (15) kelvin to (20) kelvin:
L3-1.0005=1.0005*0.0001*5
so: Lfinal=1.00010025!!
so if the linear expansion equation is true then:
1.00010025=1.0001
the problem is that the equation doesnt spot the difference between the increase of length that is occurred by increase of mass and the increase of length that is occurred by increase of temperature .
for example :we can have 1 meter and 2 meters long pieces of iron both at 0'c.if first piece(1 meter long one)lengthen (a)meters for every 1'c increase ,second piece will lengthen (2a)meters for every 1'c increase.but a 2 meters long piece of iron that is at 100'c won't lengthen (2a)meters for every 1'c increase.
we can fix that error very easily but why they have defined the equation like this when the fixed form aint very complex ?
 

Answers and Replies

  • #2
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but a 2 meters long piece of iron that is at 100'c won't lengthen (2a)meters for every 1'c increase.
Why not

A 2m piece at 0C heated to 1C

Lfinal-Linitial=a*2*(1-0) = 2a

A 2m piece at 100C heated to 101C

Lfinal-Linitial=a*2*(101- 100) = 2a

They both expand exactly the same amount
 
  • #3
Why not

A 2m piece at 0C heated to 1C

Lfinal-Linitial=a*2*(1-0) = 2a

A 2m piece at 100C heated to 101C

Lfinal-Linitial=a*2*(101- 100) = 2a

They both expand exactly the same amount
because the formula isn't right.
if not,there won't be any differences between a 2 meter long piece of iron that has (n) particles,and a 2 meter piece of iron that has (n/2) particles and is warmer...
 
  • #4
5,439
7
What do particles have to do with it?

I simply substituted into your formula, now you say your formula is wrong?
 
  • #5
AlephZero
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In the real world, the values of an expansion coefficient are quoted relative to a fixed reference temperature.

If your (0.0001) K^-1 is relative to say 10 degrees K, you have to calculate all the expansions as changes from the 10K reference temperature.

So if you have a 1m long object at 20K and you heat it to 30K, the calculation would be

Let the reference length be L at 10K. Length at 20K = 1m
L x (1 + 0.0001 x (20-10)) = 1
L = 1 / 1.001 = 0.999001 m
Length at 30K = 0.999001(1 + 0.0001 x (30-10)) = 1.000999m
Increase in length from 20K to 30K = 1.000999 - 1 = 0.000999m.

Or alternatively, for the exact same material, the expansion coefficient might be given as 0.0000999 K^-1 relative to 20K, which would give you the same answer with a simpler calculation.

In practice, this is not very important except for large changes of temperature (hundreds or thousands of degrees) because the "errors" are tiny. But kudos for realizing this is not quite as simple as it first appears to be :smile:
 
  • #6
AlephZero
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Why not

A 2m piece at 0C heated to 1C

Lfinal-Linitial=a*2*(1-0) = 2a

A 2m piece at 100C heated to 101C

Lfinal-Linitial=a*2*(101- 100) = 2a

They both expand exactly the same amount
That is wrong, because it is not self-consistent.

If you heat a 2m piece from 0 to 2 degrees, by your calculation you get an increase in length of 4a.

But, if you heat it from 0 to 1 you get an increase of 2a,

And if you now heat it from 1 to 2, its length is now (2+2a) not 2, so you get an increase of
2a + 2a^2
And the total increase in length from 0 to 2 is 4a + 2a^2.

They can't both be right (which is what the OP had realized).
 
  • #7
In the real world, the values of an expansion coefficient are quoted relative to a fixed reference temperature.

If your (0.0001) K^-1 is relative to say 10 degrees K, you have to calculate all the expansions as changes from the 10K reference temperature.

So if you have a 1m long object at 20K and you heat it to 30K, the calculation would be

Let the reference length be L at 10K. Length at 20K = 1m
L x (1 + 0.0001 x (20-10)) = 1
L = 1 / 1.001 = 0.999001 m
Length at 30K = 0.999001(1 + 0.0001 x (30-10)) = 1.000999m
Increase in length from 20K to 30K = 1.000999 - 1 = 0.000999m.

Or alternatively, for the exact same material, the expansion coefficient might be given as 0.0000999 K^-1 relative to 20K, which would give you the same answer with a simpler calculation.

In practice, this is not very important except for large changes of temperature (hundreds or thousands of degrees) because the "errors" are tiny. But kudos for realizing this is not quite as simple as it first appears to be :smile:
tnx alphazero,,
then:
the fixed form will be :Lfinal-Linitial=a*Linitial*(Tfinal-Tinitial)/(((Tinitial-Treference)*a)+1)
and its not very complex (is it?).so,why they have defined the equation as: Lfinal-Linitial=a*Linitial*(Tfinal-Tinitial)?
 
  • #8
Mapes
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There's really no "they," there's only "us" - people trying to work efficiently with equations that are neither too hard nor too simple.

A better equation is

[tex]d\epsilon=dL/L=\alpha\,dT[/tex]

where [itex]\epsilon[/itex] is the strain. Now let's say you start every day by integrating this to give

[tex]\Delta L=\int_{T_1}^{T_2}\alpha L\,dT[/tex]

which is a relatively hard integral to solve because both [itex]\alpha[/itex] and [itex]L[/itex] are temperature dependent. But you might assume that they don't change much with temperature, so you'll take them out of the integral and integrate as follows:

[tex]\Delta L\approx\alpha L_1 \int_{T_1}^{T_2}dT=\alpha L_1\Delta T[/tex]

This equation is much easier to use, but you need to remember that it's only an approximation that required a couple of assumptions. As a result, the answers won't match perfectly if you compare two successive temperature increases of [itex]\Delta T/2[/itex] with a single increase of [itex]\Delta T[/itex]. Does this make sense?
 
Last edited:
  • #9
Mapes
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I should add, you could also solve the differential equation

[tex]d\epsilon=dL/L=\alpha\,dT[/tex]

by integrating [itex]dL/L[/itex] to give

[tex]\ln\left(\frac{L_2}{L_1}\right)=\int_{T_1}^{T_2}\alpha\,dT\approx\alpha\Delta T[/tex]

This approximation will work for successive temperature increases - try it. You might notice that when [itex]L_2\approx L_1[/itex], the logarithm is just about equal to [itex](L_2-L_1)/L_1[/itex]. It's easier to do the second calculation by hand or in one's head, so that's the form that most people use, perhaps forgetting or not knowing that it's just an approximation.
 

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