Expansions in thermodynamics using differentials

[CAF123]

Homework Statement

A welded railway train, of length 15km, is laid without expansion joints in a desert where the night and day temperatures differ by 50K. The cross sectional area of the rail is 3.6 x 10^-3m².
A)What is the difference in the night and day tension in the rail if it kept at constant length?
B)If the rail is free to expand, by how much does its length change between night and day? (linear expansion coefficient, $\alpha = 8 \times 10^{-6} K^{-1}$, Young's modulus $Y = 2 \times 10^{11} N m^{-2}$)

Homework Equations

Total differential, cyclical relation

The Attempt at a Solution

I have the correct answer to A) and know how the answer to B) is attained, but I don't quite get it.

Expressing the tension F = F(L,T) as a total differential; $$dF = \left(\frac{\partial F}{\partial L}\right)_T dL + \left(\frac{\partial F}{\partial T}\right)_L dT$$

In A), I can cancel the first term on RHS. For B), I cannot.

First attempt

I rearranged the above to obtain: $$\int \left(\frac{\partial F}{\partial L}\right)_T dL = \int dF - YA\alpha(T_2 - T_1),$$using the expression for $Y$.

Simplifying further gives: $$L_2 - L_1 = \exp\left(\frac{\Delta F - YA\alpha(T_2 - T_1)}{YA}\right)$$
I don't know $\Delta F$ so I was unable to proceed here.

Second attempt

Write the differential for L = L(F,T} to give $dL = \frac{A}{YL}dF + \alpha L dT$ This gives the correct answer for B) provided the answer to part A) is used as dF in the above eqn.
Why is it okay to use this dF here? (In part A), that dF is derived assuming constancy in the length of the rail - part B) is concerned with the contrary - i.e dropping that assumption so I don't see how that dF is applicable)
Secondly, why does method 1) fail?

Many thanks.

 

[Chestermiller]

Thermal expansion causes the zero-stress length of an object to increase. When the object is heated, the new zero stress length is equal to $l_o(1+αΔT)$, where l₀ is the length with no temperature change and no loading stress. When stress is also added to the object, the strain must be calculated relative to this thermally expanded length:
$$strain=\frac{l-l_0}{l_0}-αΔT$$
This relation assumes that the strain is small, and neglects second order interactions between thermal expansion and loading stress. With this strain, the stress is given by
$$σ=\frac{F}{A}=Y\left(\frac{l-l_0}{l_0}-αΔT\right)$$
If the rail is kept at constant length, then l = l₀, and
$$σ=\frac{F}{A}=Y(-αΔT)$$
If the rail is free to expand, then σ = 0, and
$$\left(\frac{l-l_0}{l_0}-αΔT\right)=0$$
A good way of thinking about this is to envision it as a two step process...first, heating at zero stress to a new length, and then applying a stress to the thermally expanded object (with strain relative to its thermally expanded length).

 

[SteamKing]

Shouldn't (L - Lo)/Lo = αΔT ?

 

[Chestermiller]

Shouldn't (L - Lo)/Lo = αΔT ?

Oops. Yes. My mistake. Thanks for pointing that out. If I can, I'm going to go back and correct it in my earlier response. Thanks again.

Chet

 

[CAF123]

Thank you Chestermiller, but this problem comes from the second chapter of the book I am referring to, before stress/strain details are considered.
Could you tell me what is wrong with my method 1) and why using dF in B) as found in A) is appropriate?

 

[SteamKing]

In your method 1, you have assumed that F is a function of T and L. However, if T is held constant, then F must equal 0, since there will be no thermal stresses created in the material. I don't follow the rest of your derivation where you calculate L2 - L1 = exp (mess).

 

[voko]

One thing to note is that the notation $\left( \frac {\partial X} {\partial Y} \right)_Z$ is highly misleading and is best avoided altogether. $\frac {\partial X} {\partial Y}$ already means that all variables except $Y$ are held constant while differentiating, so wrapping that in $\left( \ \right)_Z$ is meaningless mathematically, yet leaves the impression that it is somehow different, but it is not.

 

[CAF123]

Hi SteamKing,

In your method 1, you have assumed that F is a function of T and L. However, if T is held constant, then F must equal 0, since there will be no thermal stresses created in the material.

Why is T being held constant here? The difference in T being night and day is a given 50K.

I don't follow the rest of your derivation where you calculate L2 - L1 = exp (mess).

I can write it out more explicitly (and following voko's advice above): $$\int \frac{\partial F}{\partial L} dL = \Delta F - YA\alpha(T_2 - T_1),$$ Using $\frac{\partial F}{\partial L} = YA/L$ this becomes $$YA\int \frac{1}{L} dL = \Delta F - YA\alpha(T_2 - T_1)$$ Solving gives the result in OP.

 

[CAF123]

Hi voko,

One thing to note is that the notation $\left( \frac {\partial X} {\partial Y} \right)_Z$ is highly misleading and is best avoided altogether. $\frac {\partial X} {\partial Y}$ already means that all variables except $Y$ are held constant while differentiating, so wrapping that in $\left( \ \right)_Z$ is meaningless mathematically, yet leaves the impression that it is somehow different, but it is not.

Yes, I understand this, I was just following the notation in my book.

P.S Was my last post in my other thread fine?

 

[voko]

dF = X dL + Y dT. In case A, dL = 0, dT is given, find dF. In case B, dF = 0, dT is given, find dL.

 

[CAF123]

In case B, dF = 0, dT is given, find dL.

dF is zero because the rail is allowed to freely change length as a result of the change in T and since there is no restriction on the final length of the rail, the tension is zero?

 

[voko]

dF is zero because the rail is allowed to freely change length as a result of the change in T and since there is no restriction on the final length of the rail, the tension is zero?

Correct.

 

[CAF123]

Using dF = 0 in the second attempt in the OP gives the correct answer, while in the first attempt, I should set $(\partial F/\partial L) = 0$ but, by doing this, I consequently cancel out the dL and left with 0 = YAα(T₂-T₁), which is nonsense.

 

[voko]

Using dF = 0 in the second attempt in the OP gives the correct answer, while in the first attempt, I should set $(\partial F/\partial L) = 0$ but, by doing this, I consequently cancel out the dL and left with 0 = YAα(T₂-T₁), which is nonsense.

I am not even sure how you set up "integration" in the first attempt. It does not look right at all. dF is a full differential, so its integral does not depend on the path. The other terms are not full differentials, so they are path dependent, and you do not specify the path.

 

[CAF123]

...and you do not specify the path.

Do you mean specify the initial and final points of dL and dT in my integration limits?

 

[voko]

Do you mean specify the initial and final points of dL and dT in my integration limits?

I mean specify the path. The functions X and Y (as in dF = X dL + Y dT) depend on (L, T), and it takes more than specifying a couple of points to integrate them.

 

[CAF123]

I mean specify the path. The functions X and Y (as in dF = X dL + Y dT) depend on (L, T), and it takes more than specifying a couple of points to integrate them.

I don't think I understand exactly what you are saying, but I think what you said stems from the fact I am expressing dL ( not a full differential given that F = F(L,T)) in terms of a full differential (dF) and another non-full differential (dT). Is this so? Expressing L =L(F,T) was easier as dL in this instance was a full differential.

Is my first method perhaps more complicated then? What would I need to do to specify explicitly the path?

 

[voko]

Is my first method perhaps more complicated then? What would I need to do to specify explicitly the path?

As I said, I do not quite follow the method. And the reason is that F and its derivatives are functions of two variables. Integration of such functions generally requires a particular path - one and the same path on both sides of the equation, too. It is not clear at all how you set up your integration, but it seems that $\frac {\partial F} {\partial L}$ is integrated along the L-direction, while the thing on the right along the T-direction, which is not good.

 

[Chestermiller]

Thank you Chestermiller, but this problem comes from the second chapter of the book I am referring to, before stress/strain details are considered.
Could you tell me what is wrong with my method 1) and why using dF in B) as found in A) is appropriate?

I don't understand what you did mathematically in your first attempt, but your starting equation is correct, so let's go from there.

$$dF = \left(\frac{\partial F}{\partial L}\right)_T dL + \left(\frac{\partial F}{\partial T}\right)_L dT$$

Hooke's law assumes that the temperature is held constant while the object is stretched, so
$$\left(\frac{YA}{L}\right)dL = \left(\frac{\partial F}{\partial L}\right)_T dL $$ and so
$$\left(\frac{\partial F}{\partial L}\right)_T =\left(\frac{YA}{L}\right)$$

The coefficient of thermal expansion is defined by the change in length of the object as a function of temperature at zero constant tensile force (or any other constant tensile force). Thus, with pure thermal expansion, you have dF = 0, and thus
$$0 = \left(\frac{\partial F}{\partial L}\right)_T dL + \left(\frac{\partial F}{\partial T}\right)_L dT$$
or, equivalently,
$$\left(\frac{\partial L}{\partial T}\right)_F=-\frac{\left(\frac{\partial F}{\partial T}\right)_L }{\left(\frac{\partial F}{\partial L}\right)_T}$$
The left hand side of this equation is αL. So,
$$\left(\frac{\partial F}{\partial T}\right)_L =-αL{\left(\frac{\partial F}{\partial L}\right)_T}=-YAα$$
Substituting back into your original equation gives:
$$dF = YA(d\ln L -αdT) $$
I hope this helps.
Chet

 

[CAF123]

$$\left(\frac{\partial L}{\partial T}\right)_F=-\frac{\left(\frac{\partial F}{\partial T}\right)_L }{\left(\frac{\partial F}{\partial L}\right)_T}$$

This was derived using the cyclical relation, right?

Substituting back into your original equation gives:
$$dF = YA(d\ln L -αdT) $$

This means $F = YA(\ln(L_2 - L_1) - \alpha(T_2 - T_1))$. For part A), the first term is zero on the RHS and so $F = -YA\alpha(T_2 - T_1)$. For part B), the LHS is zero but when I solve for $L_2 - L_1$, I obtain the incorrect answer. Why is this so?

 

[CAF123]

As I said, I do not quite follow the method. And the reason is that F and its derivatives are functions of two variables. Integration of such functions generally requires a particular path - one and the same path on both sides of the equation, too. It is not clear at all how you set up your integration, but it seems that $\frac {\partial F} {\partial L}$ is integrated along the L-direction, while the thing on the right along the T-direction, which is not good.

I see, so what I was trying to do simply made no sense?

 

[voko]

I see, so what I was trying to do simply made no sense?

That depends on what you were really trying to do. It may be that you had some clever idea, but did not express it mathematically correctly. Then you need to explain exactly what you had in mind. Some vague idea about integration won't do it because of the complications we have already discussed.

 

[CAF123]

That depends on what you were really trying to do. It may be that you had some clever idea, but did not express it mathematically correctly. Then you need to explain exactly what you had in mind. Some vague idea about integration won't do it because of the complications we have already discussed.

Essentially what I was trying to do was given $dF = X dL + Y dT,$ and rearranging so as to obtain $dL: dF - YdT = X dL.\,\, dF = 0$ so $XdL = - YdT$ where $$X = \frac{\partial F}{\partial L}\,\,\,\,\,\,\text{and}\,\,\,\,\,\,Y = \frac{\partial F}{\partial T}$$

I was then attempting to integrate them like so: $$\int_{L_1}^{L_2} X dL = - \int_{T_1}^{T_2} Y dT,$$ but because dL is not a total differential in this instance I may not do so.

 

[voko]

Essentially what I was trying to do was given $dF = X dL + Y dT,$ and rearranging so as to obtain $dL: dF - YdT = X dL.\,\, dF = 0$ so $XdL = - YdT$ where $$X = \frac{\partial F}{\partial L}\,\,\,\,\,\,\text{and}\,\,\,\,\,\,Y = \frac{\partial F}{\partial T}$$

Using the given approximations for $X$ and $Y$, this gives you $dL$, why go further?

I was then attempting to integrate them like so: $$\int_{L_1}^{L_2} X dL = - \int_{T_1}^{T_2} Y dT,$$ but because dL is not a total differential in this instance I may not do so.

You can get around that by saying that $L = L(T), \ X = X(L), \ Y = Y(T)$ (X and Y are the usual single-variable approximations), then you have $X dL = X(L(T)) L'(T)dT = - Y(T) dT$, which you can integrate, and you will end up with that same integral. But note there is no mention of $dF$ as you had originally. But I think integration is wholly unnecessary here anyway, using the differential approximations should be enough.

 

[CAF123]

Using the given approximations for $X$ and $Y$, this gives you $dL$, why go further?

Okay, that makes sense. What does the negative sign mean though? I would have $dL = -\alpha L dT$ The answer is that the rail expands by 6m, however, the above eqn implies dL/dT < 0, i.e the length decreases with temperature.

You can get around that by saying that $L = L(T), \ X = X(L), \ Y = Y(T)$ (X and Y are the usual single-variable approximations), then you have $X dL = X(L(T)) L'(T)dT = - Y(T) dT$, which you can integrate, and you will end up with that same integral.

So $$X(L(T)) L'(T)dT = - Y(T) dT \Rightarrow \frac{\partial F}{\partial L(T)} \frac{dL}{dT} dT = -\frac{\partial F}{\partial T} dT \Rightarrow \frac{\partial F}{\partial L(T)} dL = -\frac{\partial F}{\partial T} dT$$ How should I proceed with the integration now?

Edit: I can rewrite that as $$\frac{YA}{L(T)} dL = YA \alpha dT \Rightarrow dL = \alpha L(T) dT$$, but this did not require any integration.

But note there is no mention of $dF$ as you had originally.

Is that because you didn't give it an explicit functional dependence as you did for L, X and Y?

 

[Chestermiller]

This was derived using the cyclical relation, right?



This means $F = YA(\ln(L_2 - L_1) - \alpha(T_2 - T_1))$. For part A), the first term is zero on the RHS and so $F = -YA\alpha(T_2 - T_1)$. For part B), the LHS is zero but when I solve for $L_2 - L_1$, I obtain the incorrect answer. Why is this so?

The equation you are referring to was derived directly from the previous equation, with the understanding that F is constant.

To answer your second question, you integrated incorrectly:

$(F_2-F_1) = YA(\ln(L_2 / L_1) - \alpha(T_2 - T_1))$

So, in part B,

$\ln(L_2 / L_1) = \alpha(T_2 - T_1)$

 

[voko]

Okay, that makes sense. What does the negative sign mean though? I would have $dL = -\alpha L dT$ The answer is that the rail expands by 6m, however, the above eqn implies dL/dT < 0, i.e the length decreases with temperature.

Should $X$ be positive or negative? Think Hooke's law.

So $$X(L(T)) L'(T)dT = - Y(T) dT \Rightarrow \frac{\partial F}{\partial L(T)} \frac{dL}{dT} dT = -\frac{\partial F}{\partial T} dT \Rightarrow \frac{\partial F}{\partial L(T)} dL = -\frac{\partial F}{\partial T} dT$$ How should I proceed with the integration now?

Edit: I can rewrite that as $$\frac{YA}{L(T)} dL = YA \alpha dT \Rightarrow dL = \alpha L(T) dT$$, but this did not require any integration.

In fact, this is equivalent to $\int dL/L = \alpha \int dT$

Is that because you didn't give it an explicit functional dependence as you did for L, X and Y?

This is because we know $dF = 0$ identically in this case.

 

[CAF123]

I think all my confusion with B) not working has all stemmed from integrating dL/L incorrectly. My second attempt in the OP was fine, it was just I did an integral wrong.

However, there is a couple of things that are still not clear:

..$\frac {\partial F} {\partial L}$ is integrated along the L-direction, while the thing on the right along the T-direction, which is not good.

Is this not what I have in the equation $\frac{\partial F}{\partial L} dL = - \frac{\partial F}{\partial T}dT$ which eventually led onto the correct answer?

You can get around that by saying that $L = L(T), \ X = X(L), \ Y = Y(T)$ (X and Y are the usual single-variable approximations), then you have $X dL = X(L(T)) L'(T)dT = - Y(T) dT$

Why is this necessary? Is there any particular reason why doing $$\int_{L_1}^{L_2} \frac{\partial F}{\partial L} dL = \int_{T_1}^{T_2} \frac{\partial F}{\partial T} dT,$$ replacing the integrands with their equivalent approximations and then integrating?

 

[CAF123]

The equation you are referring to was derived directly from the previous equation, with the understanding that F is constant.

I get to here by rearranging the previous equation: $$\frac{dL}{dT} = -\frac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial L}}$$

Did you convert the LHS to a partial because given F = F(L,T) we can express L = L(F,T), and so L also depends on F?

 

[voko]

Is this not what I have in the equation $\frac{\partial F}{\partial L} dL = - \frac{\partial F}{\partial T}dT$ which eventually led onto the correct answer?

In the original post, you had an integral of dF, and it was not clear at all what that integral was.

Why is this necessary? Is there any particular reason why doing $$\int_{L_1}^{L_2} \frac{\partial F}{\partial L} dL = \int_{T_1}^{T_2} \frac{\partial F}{\partial T} dT,$$ replacing the integrands with their equivalent approximations and then integrating?

Because the non-approximated X and Y are functions of two variables, and to integrate them you need a path. Once a path a selected, the 2D-functions are reduced to 1D, which is what happens above, too, albeit via a different approach.

 

[CAF123]

Because the non-approximated X and Y are functions of two variables, and to integrate them you need a path. Once a path a selected, the 2D-functions are reduced to 1D, which is what happens above, too, albeit via a different approach.

So why, if I express F = F(L,T) and then write $dF = \frac{\partial F}{\partial T}dT + \frac{\partial F}{\partial L}dL$ I can then say $$F_2 - F_1 = \int_{T_1}^{T_1} \frac{\partial F}{\partial T} dT + \int_{L_1}^{L_2} \frac{\partial F}{\partial L}dL$$ replace each integrand with the approximation and then integrate? (as in part A) for example)

I recall you mentioning that this was because dF was a total differential in this case and so this implies the change in tension was path independent. What is it about dF being a total differential implies path independency?

 

[voko]

So why, if I express F = F(L,T) and then write $dF = \frac{\partial F}{\partial T}dT + \frac{\partial F}{\partial L}dL$ I can then say $$F_2 - F_1 = \int_{T_1}^{T_1} \frac{\partial F}{\partial T} dT + \int_{L_1}^{L_2} \frac{\partial F}{\partial L}dL$$

More verbosely, this is: $F_2 - F_1 = \int_{T_1}^{T_2} \frac{\partial F(L, T)}{\partial T} dT + \int_{L_1}^{L_2} \frac{\partial F(L, T)}{\partial L}dL$. What is the value of L in the first term on the right, and the value of T in the second term? Observe L and T are not variables on the left.

 

[CAF123]

More verbosely, this is: $F_2 - F_1 = \int_{T_1}^{T_2} \frac{\partial F(L, T)}{\partial T} dT + \int_{L_1}^{L_2} \frac{\partial F(L, T)}{\partial L}dL$. What is the value of L in the first term on the right, and the value of T in the second term?

In the first term on the RHS, we consider how F varies with T only, so the value of L in first term: L₁. Similarly for the second term on the RHS, we consider only changes in L so the value of T in the second term is T₁.

Edit: So this means each integral on the RHS is an integral over a single variable but how does that help me? For example, I can express $\frac{\partial F}{\partial T} dT = - \frac{\partial F}{\partial L}dL$ as $\frac{\partial F}{\partial T} dT + \frac{\partial F}{\partial L}dL = 0$ and each integral on the LHS are also integrals over a single variable.

Observe L and T are not variables on the left.

Yes, $F_2$ is the value of F at the state $(T_2, L_2)$ and $F_1$ the value of F at the state $(T_1, L_1)$

 

[Chestermiller]

I get to here by rearranging the previous equation: $$\frac{dL}{dT} = -\frac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial L}}$$

Did you convert the LHS to a partial because given F = F(L,T) we can express L = L(F,T), and so L also depends on F?

This is a standard result in working with partial differential equations. Check your math textbook on pde's. The left hand side is a partial derivative because F is being held constant (recall, dF = 0). So we are getting the rate of change of L with respect to T at constant F (i.e., the partial derivative of L with respect to T at constant F).

 

[voko]

In the first term on the RHS, we consider how F varies with T only, so the value of L in first term: L₁. Similarly for the second term on the RHS, we consider only changes in L so the value of T in the second term is T₁.

Then the RHS is not an integral over a single path such as L = L(T), so the equation cannot hold unless you show it results from a single-path integral and some other assumptions.

Edit: So this means each integral on the RHS is an integral over a single variable but how does that help me?

You cannot just integrate things independently over different paths and claim the equation still holds. In this case, where the rail is allowed to expand freely, you have some unknown path L = L(T), and you have to integrate the whole thing over this path - or some other path, but then you have to show the equivalence.

Because $dF$ is a full differential, you can say that you can integrate over any path and the result will be the same, provided the endpoint are the same. One such path could be made of two straight-line segments: $(L_1, T_1) \rightarrow (L_1, T_2) \rightarrow (L_2, T_2)$. Physically, this corresponds to heating the rail while holding it in place, then to letting it expand at constant temperature - we can expect that this should have the same end result. But this corresponds to $$ F_2 - F_1 = \int_{T_1}^{T_2} F_T(L_1, T) dT + \int_{T_1}^{T_2} F_L(L_1, T) L'(T) dT + \int_{L_1}^{L_2} F_T(L, T_2) T'(L) dL + \int_{L_1}^{L_2} F_L(L, T_2) dL \\ = \int_{T_1}^{T_2} F_T(L_1, T) dT + \int_{L_1}^{L_2} F_L(L, T_2) dL $$ Observe this is different from what you had earlier.