Exergy of combustion products, steam and the loss of exergy

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Homework Help Overview

The discussion revolves around the calculation of exergy related to the combustion of biomass in a steam generator. The problem involves understanding the heat transfer from biomass to steam and the implications of surrounding temperatures on exergy calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants express confusion about the necessity of using surrounding temperature in exergy calculations for both combustion products and steam. There are inquiries regarding the connection between biomass combustion and the equations provided.

Discussion Status

Some participants are exploring the derivation of the equations used and questioning the rationale behind modeling biomass as a hot reservoir and the surroundings as a cold reservoir. There is an ongoing examination of the percentage loss of exergy and its calculation method.

Contextual Notes

Participants note the need for clarity on the definitions and derivations related to exergy and efficiency, as well as the implications of the heat transfer process in the context of the problem.

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Homework Statement



Biomass with a heating value of 15 MJ/kg burns in steam generator at a temperature of 1200K and all of the heat is given to the steam at a temperature of 600K. The ambient temperature is 300K. Calculate exergy of combustion products, steam and the loss of exergy.

Homework Equations


[/B]
ex = q * [1 - (Tsurr / T)]

The Attempt at a Solution


[/B]
I don't really 'get' what's really going on here. Can somebody explain me why should I be using surrounding temperature to calculate exergies for both combustion products and steam? I can't make connection of the fact that burning biomass is giving heat to the steam with these equations.

I guess it turns out the solution should be:

ex1 = 15 MJ/kg * [1 - (300K / 1200K)] = 11.25 MJ/kg
ex2 = 15 MJ/kg * [1 - (300K / 600K)] = 7.5 MJ/kg


However, I don't know why. It turns out the solution to the loss of exergy is 33% (As stated in solutions)

(11.25 MJ/kg - 7.5 MJ/kg) / 11.25MJ/kg = 0.33 ?

But why? Can somebody explain me what is going on?
 
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Do you need the derivation of the equation you used?
 
Simon Bridge said:
Do you need the derivation of the equation you used?

I believe it is derived from efficiency of a ideal Carnot engine, am I right?

( Q(HR) - Q(CR) ) / Q (HR)

With HR being hot, and CR being cold reservoir.

However, If biomass gives all of the heat to the steam, why am I even calculating 'exery between biomass as hot, and surroundings as cold reservoir'?

And why is percentage of loss of exergy calculated in that way?

Thanks a lot.
 
Heat flows from the hot reservoir to the cold one, doing work on the way. In this case, the work gets extracted between the biomass and the surroundings, therefore it is sensible to model the biomass as the _____ reservoir and the surrounding as the _____ reservoir (fill in the blanks).

The percentage loss of something is ##100(x_i-x_f)/x_i## by definition. Therefore the percentage loss of exergy is... (use your knowledge of exergy to derive the formula you need.)

You should have a derivation, and definitions, in your notes or in your textbook... you can also look them up online.
You should not be guessing.
 

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