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This is the final example of applying property relations to a real situation.
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Thermoelastic effect is the change in temperature of solids upon elastic or reversible deformation
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under adiabatic condition. What we want to know in thermoelastic effect is the temperature change
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under deformation by uniaxial force f. It is dT over df. Here we have blocks of solids.
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Uniaxial force f is acting on a solid. The deformation is assumed within elastic limits.
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When solid is under tension, the solid will be elongated to the direction of the uniaxial force.
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The force, by convention, is positive when it is tension.
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When solid is under compression, the solid will shrink to the direction of the uniaxial force.
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The force, by convention, is negative when it is compression.
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When we consider energy functions, we have to include "work" other than P - V work in this case.
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We are looking for dT over df. Mathematically, (dT over df) at constant S
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can be converted into - ((dS over df) at constant T) over ((dS over dT) at constant f).
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Here the denominator is obtained from heat capacity. Since heat capacity at constant f is
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T times (dS over dT) at constant f, (dS over dT) at constant f can be written as (Cf over T).
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The equation becomes like this. - T times (dS over df) at constant T over (nCf).
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Here n is the number of moles of material and Cf is the heat capacity at constant force.
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To get (dS over df) at constant T, let's start from the thermodyanmic potential dU.
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dU is TdS - PdV, but in this case under the uniaxial force, we need additional work f dl.
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Here the uniaxial work has different sign from the P - V work. By sign convention,
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work is positive when the work is done on the system. For P - V work, if the volume increases
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with a positive dV, the work is done by the system, so P - V work is negative.
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Therefore P - V work on the system is - PdV.
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However, f is positive when it is tensile, so when it elongates with a positive dl,
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the work fdl is done on the system. Thus the uniaxial work is + fdL. Let's examine G now.
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G is H - TS and U + PV - TS. The differential form dG is thus dU + PdV + VdP - TdS - SdT.
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Insert the value for dU then dG becomes - SdT + VdP + fdl.
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Starting from this equation, dG is - SdT + fdl at constant pressure.
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Exactness theorem applies to get a property relation here.
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The property relation is (dS over dL) at constant T is - (df over dT) at constant l.
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Let's look at the left side. (dS over dL) at constant T is mathematically
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the same with (dS over df) times (df over dl), both at constant T.
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The right side (df over dT) at constant l is also mathematically the same with this.
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- (dl over dT) at constant f over (dl over df) at constant T. Then equate those two on the left and right.
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The right side can be written as ((dl over dT) at constant f) times ((df over dl) at constant T).
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Now we see the same things on both sides. Cancel out the same things.
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Then the equation becomes like this. (dS over df) at constant T is (dl over dT) at constant f.
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Now we have a equation for (dS over df) at constant T. Let's turn back to dT over df.
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Insert the equation, just derived here. The equation is now - T times (dl over dT) at constant f over nCf.
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(dl over dT) can be obtained from the linear thermal expansion coefficient, alpha l.
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The equation looks now simple. It is - (T l alpha l) over (nCf).
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Usually, the driving force for elastic deformation is given as a stress not a force.
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The stress sigma is force per area. Force f is sigma times area. So (df over d sigma) is A.
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Since the driving force is given as a stress, we need (dT over d sigma) instead of (dT over df).
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Mathematical manipulation will give us the result.
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(dT over d sigma) equals to (dT over df) times (df over d sigma).
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Here, the relations obtained previously, are applied. (dT over df) is - (T l alpha l) over (nCf).
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(df over d sigma) is A. It is now - (T l alpha l) over (nCf) times A.
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L times A is the volume V. It is now - (T V alpha l) over (nCf).
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V over n is the molar volume V underbar. So finally it is - (T V underbar alpha l) over Cf.
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(dT over d sigma) is now - (T V underbar alpha l) over Cf.
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Heat capacities at constant force, stress, and pressure are essentially the same, so we can change Cf with Cp.
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So the equation becomes - (T V underbar alpha l) over Cp.
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Integrate by assuming delta T is small, then everything on the right side can be regarded as constant.
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The integration is approximated to delta T is - (T V under alpha l) over Cp times delta sigma.