Recent content by Chetty

It appears that this response is referring to another source. Is that at the listed website cisco?

In Elements of Gasdynamics the author describes what he calls The Canonical Equation of State where (∂E∂S)v=T and (∂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)...

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...

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.

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}...

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

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...

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

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...

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...

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...

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...

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

##(\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...

I just read your write up on entropy and have another question. When I asked earlier about the change in entropy for the piston problem we were working on, you mentioned that entropy was a state variable which seems accurate and I interpret it to mean that for an equilibrium condition there is...