 #1
binbagsss
 1,277
 11
Okay, I am considering a cycle, where the working fluid is an ideal gas, with heat capacities Cv and Cp, the cycle consists of: isochoric increase in volume, adiabatic expansion back to initial pressure and a isobaric compression back to initial conditions.
Questions:

q1) I am asked to draw an TS diagram. The diagram indicates the entropy at the end returning to its initial entropy. I am not sure why this is:
Thoughts
 for there to be no change in entropy, a process must be reversible. I can see that we have returned to the initial conditions and I know this is the definition of reversible.
 But am I then correct to say that, whilst the individual processes are not reversible, the cycle, considered as a whole, is reversible?*
q2) For the adiabatic reaction, it is drawn at constant entropy. I know this equality only holds for a reversible process, and increases if not. I'm struggling to see how you justify that this is a reversible adiabatic reaction?
q3) The expression for the efficiency includes ΔQ. (whilst I am able to attain the correct value by plugging in values attained for Qh and Ql  heat input and output respectively (more information is specified than I have said at the start of the post), I am not able to justify why the following is flawed:
 Any change in heat comes from the isocharic and isobaric processes.
 We must have that ΔSp + ΔSv = 0 (for the cycle to be constant entropy consistent with the diagram), where Sp is the entropy of the isobaric process and Sv the isocharic
 I then know that [itex]\int[/itex]dS=[itex]\int[/itex]dQ/T** for a reversible process only
 Whilst I know that these individual processes are not reversible , st ** can not be applied, surely they can be considered reversible together? ( consistent with the diagram)
 So could you not apply **to the processes together? (which yeilds efficiency=0 as ΔS=0 => ΔQ=0...complete nonsense)
Many thanks to anyone who can help shed some light on this, greatly appreciated ! :)
Questions:

q1) I am asked to draw an TS diagram. The diagram indicates the entropy at the end returning to its initial entropy. I am not sure why this is:
Thoughts
 for there to be no change in entropy, a process must be reversible. I can see that we have returned to the initial conditions and I know this is the definition of reversible.
 But am I then correct to say that, whilst the individual processes are not reversible, the cycle, considered as a whole, is reversible?*
q2) For the adiabatic reaction, it is drawn at constant entropy. I know this equality only holds for a reversible process, and increases if not. I'm struggling to see how you justify that this is a reversible adiabatic reaction?
q3) The expression for the efficiency includes ΔQ. (whilst I am able to attain the correct value by plugging in values attained for Qh and Ql  heat input and output respectively (more information is specified than I have said at the start of the post), I am not able to justify why the following is flawed:
 Any change in heat comes from the isocharic and isobaric processes.
 We must have that ΔSp + ΔSv = 0 (for the cycle to be constant entropy consistent with the diagram), where Sp is the entropy of the isobaric process and Sv the isocharic
 I then know that [itex]\int[/itex]dS=[itex]\int[/itex]dQ/T** for a reversible process only
 Whilst I know that these individual processes are not reversible , st ** can not be applied, surely they can be considered reversible together? ( consistent with the diagram)
 So could you not apply **to the processes together? (which yeilds efficiency=0 as ΔS=0 => ΔQ=0...complete nonsense)
Many thanks to anyone who can help shed some light on this, greatly appreciated ! :)