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

[obscure]

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

 

[Chestermiller]

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?