How Do You Calculate Work and Entropy Generation for Reversible Processes?

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SUMMARY

The discussion focuses on calculating work and entropy generation for two reversible processes involving atmospheric air transitioning from 300K and 100kPa to 600K and 200kPa. The first scheme employs reversible adiabatic compression followed by internally reversible isobaric heating, while the second scheme reverses this order. Key equations utilized include the heat transfer equation Q=m.Tb(s2-s1)-Tb.sigma and the energy balance equation W=Q+m(h1-h2). The participants seek clarity on the entropy changes associated with adiabatic reversible compression and the overall process path.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the laws of thermodynamics.
  • Familiarity with ideal gas behavior and properties.
  • Knowledge of entropy and its calculation in thermodynamic processes.
  • Proficiency in applying mass and energy balance equations in steady-state systems.
NEXT STEPS
  • Study the principles of reversible adiabatic processes in thermodynamics.
  • Learn about entropy generation calculations in various thermodynamic cycles.
  • Explore the implications of heat transfer with reservoirs in thermodynamic systems.
  • Investigate the differences between isobaric and adiabatic processes in energy systems.
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Students and professionals in thermodynamics, mechanical engineers, and anyone involved in energy systems analysis and optimization.

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Homework Statement


Atmospheric air at 300K and 100kPa is to be delivered to a line at 600K, 200kPa. Two possible schemes for doing this are suggested. The first scheme involves reversible adiabatic compression and then internally reversible isobaric heating. .The second scheme involves internally reversible isobaric heating first and then reversible adiabatic compression. In both cases heat exchange takes place with a reservoir at 900K. Assume an ambient condition of 100 kPa, 300K.

a) Determine work and entropy generation for both of the suggested ways.

Homework Equations


Q=m.Tb(s2-s1)-Tb.sigma sigma= entropy

The Attempt at a Solution


heat transfer only occurs at Tb, ignore KE Pe, air is ideal gas

mass and energy balances at steady state W=Q+m(h1-h2)
entropy balance gives Q/Tb+m(s2-s1)+sigma

I don't understand the difference betwwen 1 and 2
 
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What is the change in entropy for any adiabatic reversible compression? What is the change in entropy in going from 300 K and 100 kPa, to 600 K and 200 kPa, irrespective of the process path?
 

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