Entropy of Otto Cycle: Deriving s1=s2

In summary, the conversation discusses how to determine the entropy at the phase of isentropic compression in an Otto cycle. The individual asking the question has already derived the temperatures and pressures at two phases and has a compression ratio of 10. They are wondering if using the table of air properties would be the right way to determine entropy and internal energy. The expert confirms that this is a valid method but notes that the air table typically gives relative entropy rather than absolute entropy. The individual is then seeking confirmation that their equation for determining entropy at phase 3 is correct. The expert confirms that it is and offers additional advice on understanding the reference state for zero entropy. Overall, the conversation highlights the use of equations and tables to determine entropy in an introductory
  • #1
stephen0507
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Homework Statement
Determine s1 and s2 given P1= 90.0, P2= 2260.7, T1=283.1, and T2= 711.2
Relevant Equations
P1= 90kpa
T1= 283.1K
v1/v2=10

P2=P1(v1/v2)^k= 90(10)^1.4= 2260.7kpa
T2=T1(v1/v2)^(k-1)= 283.1(10)^0.4= 711.2K
Hello, just wanted to ask regarding the otto cycle; if we were to find the entropy at the phase of isentropic compression and I was already able to derive the temperature 1 and 2 and the pressures at 1 and 2 and I also have the compression ratio of 10. How do I derive the entropy (s1 and s2)? given the fact that entropy at this phase is s1=s2.
 

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  • #2
Are you sure you are being asked to determine the absolute entropies of the two states? Of what value is that?
 
  • #3
Chestermiller said:
Are you sure you are being asked to determine the absolute entropies of the two states? Of what value is that?
They're asking for the datum of each phase it's more of an introductory class in thermodynamics so I guess they just want us to be familiar with the derivation of each point. So far, I was thinking of using the table of air properties to determine the entropies for each state since I have the temperatures and pressure of both phases 1 and 2. Would this be the right way to determine entropy (s) and internal energy (u)?
 
  • #4
stephen0507 said:
They're asking for the datum of each phase it's more of an introductory class in thermodynamics so I guess they just want us to be familiar with the derivation of each point. So far, I was thinking of using the table of air properties to determine the entropies for each state since I have the temperatures and pressure of both phases 1 and 2. Would this be the right way to determine entropy (s) and internal energy (u)?
Is this course statistical thermodynamics?
 
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  • #5
Are you familiar with the equation for an ideal gas: $$dS=C_p\frac{dT}{T}-R\frac{dP}{P}=C_v\frac{dT}{T}+R\frac{dV}{V}$$
 
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  • #6
yes! I actually wanted to ask about this. So I was able to determine entropy from points 1 and 2 using the ideal gas table for air since the otto cycle goes through an isentropic process is it safe to assume that s1=s2? so the entropy of air I got in phase 1 is the same entropy for the second phase? If so I need to get s3 and I was wondering if I could do this:

s3-s2= Cpln(T3/T2)-Rln(P3/P2)

and solve for s3
 
  • #7
stephen0507 said:
yes! I actually wanted to ask about this. So I was able to determine entropy from points 1 and 2 using the ideal gas table for air since the otto cycle goes through an isentropic process is it safe to assume that s1=s2? so the entropy of air I got in phase 1 is the same entropy for the second phase?
This is OK, but please understand that the air table typically does not give the absolute entropy. It typically gives the entropy relative to some reference state like 25 C and 1 atm. Check the table to see what the use as the reference state of zero entropy.
stephen0507 said:
If so I need to get s3 and I was wondering if I could do this:

s3-s2= Cpln(T3/T2)-Rln(P3/P2)

and solve for s3
Sure.
 
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  • #8
Chestermiller said:
This is OK, but please understand that the air table typically does not give the absolute entropy. It typically gives the entropy relative to some reference state like 25 C and 1 atm. Check the table to see what the use as the reference state of zero entropy.
ok yes understood I have noticed this in different references too, Ill take note of this thank you so much,
 
  • #9
Chestermiller said:
Sure.
Just to confirm I can use the equation that I set up to solve for entropy at 3 correct?
 
  • #10
stephen0507 said:
Just to confirm I can use the equation that I set up to solve for entropy at 3 correct?
Yes.
 
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  • #11
Thank you so much, you helped me with this I will keep note of everything you mentioned. Thanks!
 

FAQ: Entropy of Otto Cycle: Deriving s1=s2

What is entropy and how is it related to the Otto Cycle?

Entropy is a thermodynamic property that measures the amount of disorder or randomness in a system. It is closely related to the Otto Cycle, which is a theoretical thermodynamic cycle used to model the behavior of an idealized spark-ignition reciprocating engine.

How is the entropy of the Otto Cycle derived?

The entropy of the Otto Cycle can be derived using the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. By applying this law to the various processes involved in the Otto Cycle, we can derive the expression for the entropy change.

What is the significance of s1=s2 in the entropy derivation of the Otto Cycle?

The equation s1=s2 represents the isentropic process, where there is no change in entropy. This is important because it allows us to simplify the entropy derivation by assuming that the compression and expansion processes in the Otto Cycle are isentropic.

Can the entropy of the Otto Cycle be negative?

No, the entropy of a closed system like the Otto Cycle cannot be negative. This is because entropy is a measure of disorder and randomness, and it is impossible for a system to have less disorder than a perfectly ordered state.

How does the entropy of the Otto Cycle affect engine efficiency?

The entropy of the Otto Cycle affects engine efficiency by determining the amount of energy that is lost during the various processes. The higher the entropy, the less efficient the engine will be, as more energy is lost as heat. Therefore, minimizing entropy is important for improving the efficiency of the Otto Cycle and other thermodynamic processes.

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