Diesel Engine Problem - Find total work

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SUMMARY

The discussion focuses on solving a diesel engine problem involving the calculation of total work during the thermodynamic cycle. Key equations used include the adiabatic compression formula T2=(V2/V1)^(1-(Cp/Cv)) yielding T2=987.7K, and work calculations for different stages of the cycle. The work done during the adiabatic compression (w1) is calculated as 8601.3 J, while the work during the combustion phase (w2) is expressed as w2=-Pmax(V3-V2) with Pmax=66.29 bar. The discussion highlights the importance of recognizing that the combustion phase is not isothermal, which is crucial for accurate analysis.

PREREQUISITES
  • Understanding of the diesel thermodynamic cycle
  • Familiarity with adiabatic processes and equations
  • Knowledge of specific heat capacities (Cp and Cv)
  • Ability to interpret temperature-entropy plots
NEXT STEPS
  • Study the principles of the diesel cycle in thermodynamics
  • Learn about adiabatic and isothermal processes in detail
  • Explore the derivation and application of the first law of thermodynamics
  • Investigate the use of temperature-entropy diagrams for thermodynamic analysis
USEFUL FOR

Students and professionals in mechanical engineering, thermodynamics enthusiasts, and anyone involved in the analysis and optimization of diesel engine performance.

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Homework Statement



The problem is in the attached document

Homework Equations



attached in pdf document

The Attempt at a Solution



1->2 adiabatic compression

T2=(V2/V1)^1-(Cp/Cv) = 987.7K
w1=nCvdT = (1mol)((3/2)*8.314J/molK)(987.7K-298K) = 8601.3 J

2->3 w2=-PdV=-Pmax(V3-V2) = 66.29bar(V3-0.1L)

3->4 w=nCvdT = (1 mol)(3*8.314J/molK)(T4-T3)

4->1 w=0
dH=nCpdT?
 

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"2->3 w2=-PdV=-Pmax(V3-V2) = 66.29bar(V3-0.1L)"

This relationship is for a reversible isothermal process. 2-3 is the combustion portion of the cycle and is not isothermal. This can be noted on a temperature-entropy plot for the diesel cycle.
 

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