Adiabatic process - Diesel engine

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SUMMARY

The discussion focuses on the adiabatic process in a diesel engine, specifically calculating the final temperature after compressing air by a factor of 15. Using the ideal gas law and the equation T2/T1 = (V1/V2)^(k-1), where k is the specific heat ratio (1.4 for air), the final temperature is determined to be approximately 866 K. Additionally, the pressure increase factor is calculated to be around 44 times. These calculations are confirmed as correct by participants in the discussion.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with the ideal gas law and its applications
  • Knowledge of specific heat ratios, particularly for air (k = 1.4)
  • Ability to perform calculations involving temperature and volume ratios
NEXT STEPS
  • Study the derivation and applications of the adiabatic process in thermodynamics
  • Learn about the ideal gas law and its implications in real-world applications
  • Explore the concept of isentropic processes and their significance in engine efficiency
  • Investigate the relationship between pressure, volume, and temperature in gas systems
USEFUL FOR

Students studying thermodynamics, mechanical engineers, and anyone interested in the principles of diesel engine operation and efficiency calculations.

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Homework Statement


A diesel engine requires no spark plug. The air in a cylinder is compressed so
highly that the fuel spontaneously ignites when sprayed into cylinder. If the air is
initially at 20°C and is then compressed by a factor of 15 in volume, what final
temperature is attained (before fuel injection)? Assume the compression is adiabatic,
and treat the air as an ideal gas. Take delta = 1.4 for air. By what factor does the
pressure increase?


Homework Equations


PV^ delta = constant
TV ^delta-1 = constant.
Delta = Cp/Cv.

Internal energy = Work done.

The Attempt at a Solution



I am totally lost with this problem, I know what an adiabatic process is, how to derive the delta, etc but I can't seem to match that with the above question.

Any tips ?
 
Last edited:
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I think I have got an idea...
We know TV^delta -1 = constant...
Could I use the following relation :
T1 V1 ^delta -1 = T2 V2 ^delta-1 To get my t2 value?No that doesn't seem right...

How about this :
TV^{\gamma-1} = k.

So if I plug in my initial temperature value as : 293.15 k and volume assumed to be at 1 , that will give me a value = k.

For the final temperature:
T_{2}V^{\gamma-1} = k
so t2 * (1/15)^ 1.4-1 = 5863/20
which gives me t2 = 866k.

=] ?
 
Last edited:
Your answer is correct. Assume isentropic so exponent is specific heat ration. Your equation is:

T2/T1 = (V1/v2)^(k-1)

You know the volume ratio, T1, and you know k.
 
lawrencec said:
your answer is correct. Assume isentropic so exponent is specific heat ration. Your equation is:

T2/t1 = (v1/v2)^(k-1)

you know the volume ratio, t1, and you know k.

thanks you very much! : ))))))))))))))))))))))))))))))
 
Also at the end of the question it asks by how much factor has hte pressure risen.. which according to my calculations is ~ 44 times.

Is this correct too ?
 
Yes, that is what I get for two significant figures.
 
Thanks for your help! :))
 

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