Manipulation of a Thermodynamic Relation

[Chetty]

Homework Statement

In the text "Elements of Gasdynamics" by Liepmann & Roshko, the first chapter exercise asks for confirmation of the following relation.

Relevant Equations

Equation is below

$(\frac {∂p} {∂ρ})_s=ϒ(\frac {∂p} {∂ρ})_T$

The variables are p for pressure, ρ for specific mass density and γ is ratio of specific heats. I am able to show that the relation is valid for a perfect gas but cannot show its validity in general.

The closest I get is $dp=(\frac {∂p} {∂ρ})_s dρ+(\frac {∂p} {∂s})_ρ ds=(\frac{∂p} {∂ρ})_T dρ+(\frac {∂p} {∂T})_ρ dT$

For the perfect gas, the first part with constant 's' implies an isentropic relation for a partial derivative of $\frac p {ρ^γ} =const$ and the second part is the partial derivative of the perfect gas expression $p=ρRT$.

 

[Chetty]

In the first equation, the last term is the partial of p relative to ρ holding temperature T constant.

 

[Chetty]

This is my latest result, not what I was hoping for but maybe all that is achievable. Set ds=0 and used full expressions for the specific heats. Anyone have any suggestions for improvements? The text does not ask for a derivation but only to show that the expression is valid. The expression below still satisfies the perfect gas example but hope to justify the inital expression to be valid in general.

$(\frac {∂P} {∂T})_ρ =-(\frac {∂ρ} {∂T})_p (\frac {∂P} {∂ρ})_T$

 

[Chestermiller]

Start with $$dS=\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP$$and $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$

 

[Chetty]

I think I may have it but feel a little uneasy with the resulting expression:$(\frac {∂P} {∂T})_ρ=-(\frac {∂P} {∂ρ})_T (\frac {∂ρ} {∂T})_P$

Working all the expressions comes down to the above and if true then I have the proof. The letters and symbols after the parenthesis are the variables to be held constant during the partial derivative. Incidently the above expression is still satisfied for a perfect gas.

 

[Chestermiller]

I think I may have it but feel a little uneasy with the resulting expression:$(\frac {∂P} {∂T})_ρ=-(\frac {∂P} {∂ρ})_T (\frac {∂ρ} {∂T})_P$

Working all the expressions comes down to the above and if true then I have the proof. The letters and symbols after the parenthesis are the variables to be held constant during the partial derivative. Incidently the above expression is still satisfied for a perfect gas.

This equation is a mathematical identity.

 

[Chetty]

Found a better answer.
Starting with:
$$dP=(\frac {∂P} {∂ρ})_s dρ+(\frac {∂P} {∂s})_ρ ds=(\frac {∂P} {∂ρ})_T dρ+(\frac {∂P} {∂T})_ρ dT$$
Write the sum of partials of enthalpy in variables P and T
Use this in the the expression of dh-vdp=0 (dq=ds=0 for isentropic)
From this extract $(\frac {∂h} {∂P})_T -V =-C_P \frac {dT} {dP}$
Write the expression for the difference in heat capacities and substitute from the above
$C_v -C_p=[(\frac {∂h} {∂P})_T -V ] (\frac {∂P} {∂T})_v=-C_p \frac {dT} {dP} (\frac {∂P} {∂T})_v$
Now use the expression ϒ=Cp/Cv to remove Cp and Cv.
Expand dP with the partials of dP in variables ρ and T
Combine everything for the final desired result
$$(\frac {∂P} {∂ρ})_s =ϒ (\frac {∂P} {∂ρ})_T$$

 

[Chetty]

Now I have a different question:
In the text by Zemansky, in chapter 10, exercise problem 10.14 the following question is asked;
"Two identical bodies of constant heat capacity are at same initial temperature Ti. A refrigerator operates between these two bodies until one body is cooled to the temperature T2. If the bodies remain at constant pressure and undergo no change of phase, show that the minimum amount of work needed to do this is
$$W_(min)=C_p (\frac {Ti^2} {T2} +T2-2*Ti)$$
I can see that Ti-T2 is the amount of heat removed from the body being cooled but the entropy would involve a log function and no log function is shown. Any one have any ideas?

 

[Chetty]

I can rewrite their answer so it makes a little more sense but not sure why the log function is not used for the entropy?

$$W_(min)=C_p\left[\frac {T_i(T_i-T_2)} {T_2}-(T_i-T_2)\right]$$

 

[Chestermiller]

Now I have a different question:
In the text by Zemansky, in chapter 10, exercise problem 10.14 the following question is asked;
"Two identical bodies of constant heat capacity are at same initial temperature Ti. A refrigerator operates between these two bodies until one body is cooled to the temperature T2. If the bodies remain at constant pressure and undergo no change of phase, show that the minimum amount of work needed to do this is
$$W_(min)=C_p (\frac {Ti^2} {T2} +T2-2*Ti)$$
I can see that Ti-T2 is the amount of heat removed from the body being cooled but the entropy would involve a log function and no log function is shown. Any one have any ideas?

If T* is denoted as the final temperature of the heated body, in terms of T*, what is the total change in entropy of the two bodies. Assuming that the refrigerator operates in a cycle, what is the change in entropy of the refrigerator working fluid? What is the change in entropy of the refrigerator working fluid plus the two bodies? If minimum work is obtained when the process is somehow operated reversibly, the change in entropy of the working fluid plus the two bodies must equal zero. What value of T* is required for this to happen? How much heat is transferred from the refrigerator to the hot body? How much heat is transferred from the cold body to the refrigerator? What is the amount of work done by the refrigerator?

 

[Chetty]

I think I understand all that you high lite, but I am trying to understand Zemansky's answer. If I understand correctly, a refrigerator is operating between two identical bodies and is extracting heat from one and adding the extracted heat to the other body. The fact that this is under constant pressure conditions would seem to indicate that the two bodies will adjust their volumes to accommodate the constant pressure requirement as heat is added to one and taken away from the other. As these are bodies and not reservoirs the temperatures of both bodies vary and it would seem necessary to use the logarithm of the ratio of starting and ending temperatures multiplied by the constant pressure heat capacity. None of this appears in Zemansky's solution. It appears that the entropy change of the refrigerator working fluid is zero for a reversible cycle and if not how would you compute any entropy change if you do not know what the working fluid is? What am I missing here?

 

[Chestermiller]

I think I understand all that you high lite, but I am trying to understand Zemansky's answer. If I understand correctly, a refrigerator is operating between two identical bodies and is extracting heat from one and adding the extracted heat to the other body. The fact that this is under constant pressure conditions would seem to indicate that the two bodies will adjust their volumes to accommodate the constant pressure requirement as heat is added to one and taken away from the other. As these are bodies and not reservoirs the temperatures of both bodies vary and it would seem necessary to use the logarithm of the ratio of starting and ending temperatures multiplied by the constant pressure heat capacity. None of this appears in Zemansky's solution. It appears that the entropy change of the refrigerator working fluid is zero for a reversible cycle and if not how would you compute any entropy change if you do not know what the working fluid is? What am I missing here?

I don't see what your difficulty is. The solution method involved in the combination of hints I gave in post #10 does involve the constant pressure heat capacity multiplied by the logarithm of temperature ratios.

 

[Chetty]

My difficulty is with my post #9 showing Zemansky's complete solution to this problem. It shows not logarithms of temperature ratios.

 

[Chestermiller]

My difficulty is with my post #9 showing Zemansky's complete solution to this problem. It shows not logarithms of temperature ratios.

That is his solution for the work, not for the entropy. The part of the solution involving entropy does involve logarithms of temperature ratios. If you follow the procedure I outlined, you will be using entropy changes which involve log's of temperature ratios, and will end up with a final answer for the work which matches his.

 

[Chestermiller]

If you ever come back here again, here is a hint for you: The change in entropy of the two bodies is
$$\Delta S=MC_p\ln{(T_2/T_i)}+MC_p\ln{(T^*/T_i)}$$

 

[Chetty]

Thanks for the hint. In your absence I had moved on with the problem and agree with your hint. Turns out the change in entropy you refer to is zero and this determines the T star.

$$T^*=\frac {T_i^2} {T_2}$$

Doing the heat balance $$W=C_p[T^*-T_i]-C_p[T_i-T_2]=C_p[\frac {T_i^2} {T_2}+T_2-2T_i]$$
Which is the answer given by Zemansky
Thanks for your followup.

 

[Chestermiller]

Thanks for the hint. In your absence I had moved on with the problem and agree with your hint. Turns out the change in entropy you refer to is zero and this determines the T star.

$$T^*=\frac {T_i^2} {T_2}$$

Doing the heat balance $$W=C_p[T^*-T_i]-C_p[T_i-T_2]=C_p[\frac {T_i^2} {T_2}+T_2-2T_i]$$
Which is the answer given by Zemansky
Thanks for your followup.

I would write it as $$W=C_p\frac{(T_i-T_2)^2}{T_2}$$

 

[Chetty]

I agree that is more succinct and directly shows the positive nature . I provide the answer as provided by Zemansky. I believe Zemansky likes to present his answers in disguise so the student has to work to understand.

 

[Chetty]

Got another one. In Elements of Gasdynamics the author describes what he calls The Canonical Equation of State where $(\frac {∂E} {∂S} )_v=T$ and $(\frac {∂E} {∂V} )_s=-P$
He does a simple one for a perfect gas and uses the enthalpy(T,V) for the Canonical Equation of
state. Now he asks to find the Canonical Equation of State for E(V,S) corresponding to Dieterici's equation of state.

$$P= \frac {RT} {v-β}*exp( \frac {-α} {RTv})$$

I tried using this in the pdv term but could not integrate it for insertion in the ds=Q/T equation. Any ideas?

 

[Chestermiller]

Got another one. In Elements of Gasdynamics the author describes what he calls The Canonical Equation of State where $(\frac {∂E} {∂S} )_v=T$ and $(\frac {∂E} {∂V} )_s=-P$
He does a simple one for a perfect gas and uses the enthalpy(T,V) for the Canonical Equation of
state. Now he asks to find the Canonical Equation of State for E(V,S) corresponding to Dieterici's equation of state.

$$P= \frac {RT} {v-β}*exp( \frac {-α} {RTv})$$

I tried using this in the pdv term but could not integrate it for insertion in the ds=Q/T equation. Any ideas?

Please submit this as a separate thread.