The ideal gas law for an adiabatic process

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SUMMARY

The discussion centers on calculating the final temperature of air in a diesel engine undergoing an adiabatic process, where the initial temperature is 293K and the volume is compressed by a factor of 15. The correct application of the adiabatic relation T1 * V1^(Ƴ-1) = T2 * V2^(Ƴ-1) yields a final temperature of 866K. Additionally, the pressure increase factor is determined using the equation P1 * V^Ƴ = constant, resulting in a pressure increase factor of 2.95. The importance of using the correct exponent in calculations is emphasized, as using γ instead of γ - 1 can lead to significant errors.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of adiabatic processes in thermodynamics
  • Familiarity with the specific heat ratio (Ƴ) for air
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation and applications of the adiabatic process equations
  • Learn about the specific heat capacities of gases and their implications in thermodynamics
  • Explore the relationship between pressure, volume, and temperature in ideal gases
  • Investigate real-world applications of the ideal gas law in automotive engineering
USEFUL FOR

Students studying thermodynamics, automotive engineers, and anyone interested in the principles of gas behavior in adiabatic processes.

garyd
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Homework Statement



Hi can someone please have a look at this question and let me know if I am on the right track, thanks.

A diesel engine requires no spark plug; instead the air in the cylinder is compresses so highly the fuel ignites spontaneously on injection to the cylinder.

Q. If the air is initially at 293K and then is compressed adiabatically by a factor of 15, what final temperature is attained ?( Just prior to injection, also consider air as an ideal gas)
Variables:
T1=293K
V1= 1
V2=1/15
T2=?

Homework Equations



T1 *V1^(Ƴ-1) = T2* V2^(Ƴ-1)

The Attempt at a Solution


Above relationship applies as process is adiabatic?

293× (1^(1.4-1)/ (1/15^(1.4-1) ))=866K
 
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Looks good to me. Why the doubtful question mark ?
 
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BvU said:
Why the doubtful question mark ?

Confidence, I am learning to trust the equations.

Another part of the question is; by what factor has the pressure increased? Can I use P V^Ƴ=constant, with p1=1, multiplied by ratio of V1/V2, both to the power of Ƴ? ans=2.95
 
Re post 1: T1 *V1^(Ƴ-1) = T1* V1^(Ƴ-1) I read T1 *V1^(Ƴ-1) = T2* V2^(Ƴ-1) which looks good.

Re post 3: numerically I get something different, but then I use ##\gamma##, not ##\gamma -1## as the exponent... Big difference !
 
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Re post 3: numerically I get something different said:
Yes I was using wrong exponent, 44 is the correct factor, thanks.
 
garyd said:
Confidence, I am learning to trust the equations.

Another part of the question is; by what factor has the pressure increased? Can I use P V^Ƴ=constant, with p1=1, multiplied by ratio of V1/V2, both to the power of Ƴ? ans=2.95

Once you have the temperature worked out, just use the ideal gas law: PV=nRT

P1V1/T1 = P2V2/T2

P2/P1 = (V1/V2)(T2/T1)

AM
 
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