Exergy Analysis of a Closed System

[Soumalya]

I was going through "Engineering Thermodynamics" by Cengel & Boles studying exergy analysis of a closed(non flow) system.Referring to the attachment as you can see the equation,

δW_HE=δQ(1-T₀/T)=δQ-T₀/T.δQ

should give δQ=δW_HE+T₀dS (using dS=δQ/T)

but in the textbook it's given as,

δQ=δW_HE-T₀dS

Any thoughts on it?

 

[Soumalya]

I have split the attachment and resized it for better view.Hope it's legible!

 

[Useful nucleus]

For some reason this text defines the sign convention such that (-TdS= δQ). You might need to consult earlier chapters to understand why. It is more common to adopt (TdS= δQ)

 

[Soumalya]

For some reason this text defines the sign convention such that (-TdS= δQ). You might need to consult earlier chapters to understand why. It is more common to adopt (TdS= δQ)

I went through the textbook again and it says,

dS=(δQ/T)_{int reversible}

and the direction of entropy transfer depends on the direction of heat transfer i.e, heat transferred to a system increases it entropy while heat transfer from a system decreases it.

Now if I wish to replace δQ/T with 'dS' in any equation is it necessary to take into account the sign convention?

NOTE:The system is rejecting heat and thus direction of heat transfer is such that would result in a decrease in entropy of the system.So for replacing δQ/T with 'dS' should I write δQ/T=-dS?