# Exergy Analysis of a Closed System

1. Aug 21, 2014

### 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,

δWHE=δQ(1-T0/T)=δQ-T0/T.δQ

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

but in the textbook it's given as,

δQ=δWHE-T0dS

Any thoughts on it?

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• ###### Exergy Analysis of a Closed System.jpg
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2. Aug 21, 2014

### Soumalya

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

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• ###### Exergy Analysis of a Closed System.jpg
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3. Aug 21, 2014

### 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)

4. Aug 21, 2014

### Soumalya

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 transfered 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?

Last edited: Aug 21, 2014