# Heat Availability

1. Apr 8, 2005

### Clausius2

When studying about Availability $$A$$ function, defined as:

$$A=U-U_o+P_o(v-v_o)-T_o(S-S_o)$$ being: (1)

$$U$$ Internal energy
$$P$$ Pressure
$$v$$ Specific volume ($$1/\rho$$)
$$T$$ Temperature
$$S$$ Entropy

Subindex "o" refers to the Dead State: the thermodynamic state of the atmosphere.

Thus, the maximum work available from a thermodynamic closed system submerged into an atmosphere $$P_o,T_o$$ is:

$$W_{max}=A$$

In general, the nominal work extracted would be less, due to irreversibilities:

$$W=A-T_o\Delta S_{universe}$$

Let me do the next: take differential in expression (1):

$$dA=dU+P_odV-T_odS_{sys}$$ (2)

in addition to 1st principle and Entropy definition:

$$\delta Q=dU+\delta W=TdS_{sys}-TdS_{universe}$$ (3)

substituting (3) in (2), and defining the useful work as the net work substracting the work done by the atmosphere $$\delta W_{useful}=\delta W-P_odV$$, we have:

$$dA=-dW_{useful}+\delta Q\Big(1-\frac{T_o}{T}\Big)-T_odS_{universe}$$

Let's take a look at this last expression. We see effects of extracting work, global irreversibilities and extracting heat enhances a loose of Availabilty and so a loose of mechanical energy available for doing work.

The question: Imagine the system is absorbing heat due to some process of burning such it happens in an internal combustion engine at combustion stage, or merely there is an inflow of heat due to some thermal contact. Assume there is no global irreversibility. Only some percentage of Heat would be available for doing work, in particular this fraction would be $$\Big(1-T_o/T\Big)$$. On the contrary the part of heat unavailable for doing work is called "Anergy" and would be $$QT_o/T$$. The question is: Where does this Heat unavailable for doing work go? What is its mission?

Thanks for clarifying this concept, I really don't know which is the physical mission of this unavailable heat, assuming there is no irreversibility, as we have seen this term can be dropped without affecting the answer of my question.

2. Apr 9, 2005

### marlon

Clausius,

The energy that is NOT available for work is T_0 multiplyed by the entropy-change of the universe. This arises because of the fact that the processes involved are IRREVERSIBLE (during such a process, some of the energy will be lost, that is the clue). The unavailable heat goes to the UNIVERSE, yielding the rise in its entropy. You cannot assume 'no irreversibility'

In order to turn an irreversible process into a reversible process you will need to convert this aneregy to energy that is available for work. You can only obtain the maximal amount of work in a reversible process.

Given the fact that in nature, there are more irreversible processes as there are reversible ones, more and more energy becomes unavailable for work. This is Kelvin's principle of energy degradation.

regards
marlon

Last edited: Apr 9, 2005
3. Apr 10, 2005

### FredGarvin

It seems a bit counter intuitive to talk about availability and irreversibility in the same topic. Shouldn't they go hand in hand? Availability is the taking into account of unusable energy, second law efficiency and such. Where does it go? I like Marlon's answer. I was going to say that it probably goes to supporting the dead state's conditions, but that doesn't seem right to me now.

4. Apr 11, 2005

### Clausius2

I don't agree. Imagine a Carnot engine. Ideally it is reversible. In spite of it, there is some amount of heat not available for doing work. This unavailable amount is $$1-T_c/T_h$$ times the heat absorbed. The question where does this heat go is simple in this case: it goes to the cool source.

But in this example I have got a thermodynamic system, which can be an isolated mass. There is an inflow of heat (reversibly). Some of this heat could be eventually extracted for being available for doing work, but the rest it cannot.

I've got it :surprised , just after writting my last words I have realised. Imagine there is an isolated mass of air at some temperature and pressure. Due to an infinitesimally slow flux of heat I am able of heat up the mass by means of a heat inflow$$\dot Q$$. The mass has reached a temperature $$T$$ while the external atmosphere is at temperature $$T_o$$. According to the last equation I wrote above, only a part ($$(1-T_o/T)\dot Q$$) of this heat will be eventually available for doing work, i.e. can be transformed in mechanical energy in some way. One of this ways would be by means of a Carnot engine. If we put a Carnot engine between the mass and the atmosphere, it actually could be extracted only $$(1-T_o/T)\dot Q$$ units of work, couldn't be?.

Hey! This last consideration is just the demonstration of the heat availability: when applying heat to some thermodynamic system, this energy only can be recovered as mechanical energy in some part, and this part is given by a Carnot engine working between the thermodynamic system and the atmosphere assuming an rreversible process. If it is irreversible it must be substracted also some amount due to global irreversibilities $$T_o\Delta S_{univ}$$ to the last quantity.

Thanks!. Any comment?

Now the question where does this heat go has non sense. The heat $$\dot Q$$ will increase the thermal energy of the mass, but the unavailable part will be eventually wasted at the atmosphere due to the 2nd principle.

Last edited: Apr 11, 2005