# Exergy of combustion products, steam and the loss of exergy

1. Oct 31, 2015

### Absentee

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. The attempt at a solution

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?

2. Oct 31, 2015

### Simon Bridge

Do you need the derivation of the equation you used?

3. Nov 1, 2015

### Absentee

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 alot.

4. Nov 4, 2015

### Simon Bridge

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 text book... you can also look them up online.
You should not be guessing.