# Thermomystery -entropy generation in a closed system

1. Aug 22, 2014

### PatrickAndrews

My question relates to entropy generation in a closed system
ΔS=dQrev/T for a reversible process
ΔS=dQ/T + Sgen for an irreversible process

This seems to suggest that Sgen arises because of the irreversibility of the heat transfer process (eg across a finite temperature difference).

If, however, we have reversible heat transfer to a system, but also some form of internal friction occurring (eg turbulence), can we say something like
ΔS=dQrev/T + Sgen(turbulence)

Also, if we have irreversible heat transfer and also frictional irreversibilities, does something like this apply?
ΔS=dQ/T + Sgen(finite temperature heat transfer) + Sgen(turbulence)

ΔS=Sgen(turbulence) ??

Thanks in advance. I'm finding its surprisingly hard to get a clear view of this apparently simple issue...

Last edited: Aug 22, 2014
2. Aug 23, 2014

### MrMatt2532

Looks like you on thinking about it correctly.

Sgen can be due to internal heat transfer or internal friction (among other things, but these are the most basic I would say). Whenever you have entropy generation you have an irreversible process.

So yes, it is correct to say deltaS=deltaQ/T+Sgen(internal friction)+Sgen(internal heat transfer). And for a isolated or an adiabatic system, you have just deltaS=Sgen(internal friction)+Sgen(internal heat transfer).

Another thing to note is you don't need turbulence to generate entropy due to friction. It happens for laminar flow as well. Basically, whenever you have gradients of any sort you have entropy generation (velocity gradients, temperature gradients, etc.).

3. Aug 23, 2014

### PatrickAndrews

I just reread your answer more closely and noticed the 'internal heat transfer' reference. Just to be clear...am I correct in thinking that an (irreversible) process of heat transfer due to temperature difference at the boundary has no effect on Sgen (Sgen is sometimes referred to as including all irreversibilities). Maybe my confusion is due to the fact that 'at the boundary' is technically outside the system?

4. Aug 23, 2014

### MrMatt2532

Yes, so the deltaQ in the deltaQ/T term refers to heat transfer at the boundary (i.e. between your control volume and some adjacent control volume), which is the reversible part. Note that in practice it is hard to have heat transfer at the boundary without internal heat heat transfer as well. Though it can be done in theory with an infinitely slow process.

5. Aug 23, 2014

### PatrickAndrews

"deltaQ at the boundary...which is the reversible part..."

That is exactly my problem, because my belief has been that heat transfer at the boundary is **irreversible if Texternal<>Tboundary** ie heat transfer via a finite temperature difference occurs.
ie there is an irreversibility at the boundary (to do with entropy transfer) as well as irreversibilities internally(to do with internal entropy generation, dSgen).

I may be making more of a problem here than really exists, but I can't quite understand the fact that dSgen seems not to include the effect of irreversibility at the boundary??

6. Aug 23, 2014

### MrMatt2532

Well part of your issue is that you seem to be thinking about what is happening outside your control volume. You shouldn't even need to consider those things if all you care about is change in entropy inside your control volume.

The basic idea is, inside your control volume, you can increase or decrease entropy reversibly by adding or subtracting heat. Or, you you can increase (one way only, since it's irreversible) entropy from the entropy generation terms: internal heat transfer, or internal friction.

Assume the universe is made up of two blocks that are in thermal contact and are at different temperatures. From either blocks perspective, the heat transfer is a reversible process. But if you are analyzing both blocks together then what was previously external heat transfer is now internal heat transfer, and we have increase in entropy overall.

7. Aug 23, 2014

### PatrickAndrews

Wow...thanks for that insight -especially since it was so quick.
"From either blocks perspective, the heat transfer is a reversible process" I did not realise that reversibility was relative, believing that the definition of irreversibility for a process was that dS universe >0
It seems to me that...
From the cooler block's perspective, dS universe= Q/Tcoolerblock + (-Q/Thotterblock) so dS uni is positive, thus indicating an irreversible process from the perspective of the cooler block. ??

8. Aug 23, 2014

### MrMatt2532

Again, I think you mostly have it but you need to be careful. When you are analyzing the universe (or an isolated system), there is no external heat transfer, and you can only say that change in entropy of the universe (or the isolated system) is due to internal entropy generation. Basically, realize there are three forms of the second law:
isolated system (universe): ΔS=Sgen
closed system: ΔS=∫δQ/T+Sgen
open system: ΔS=∫δQ/T+mass_flow_in*s_in-mass_flow_out*s_out+Sgen

Lets take this two block universe. You can say, for the first block, ΔS1=integral of δQ/T=m1*c1*ln(Tf/T1). And for the second block, deltaS2=integral of δQ/T=m2*c2*ln(Tf/T2), where T1 and T2 is the initial temps of the blocks, c is the specific heat, and m is the mass of the blocks and Tf is the final temperature that the system reaches. Now lets say m1=m2 and c1=c2. Then Tf=(T1+T2)/2 and ΔS universe is equal to ΔS1+ΔS2=m*c*(ln(Tf/T1)+ln(Tf/T2))=m*c*ln((T1+T2)^2/(4*T1*T2)) Now plug in any two temperatures into the expression ln(((T1+T2)^2/(4*T1*T2)) and you can see ΔS universe is always positive.

9. Aug 24, 2014

### Staff: Mentor

MrMatt2532 has given really good answers.:thumbs: For additional perspective, see example 11D.1 in Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 11.

Chet

10. Aug 24, 2014

### PatrickAndrews

Perspective is the key point

Ok, I think I get this now. For a closed (dm/dt=0, but not isolated) system, I've drawn the attached diagram.

From the system's perspective, the external temperature(s) from which heat is transferrred into the system is unknown. This means that the dS-system equation is 'unaware of' any (external) irreversibilities to do with finite differences that may be driving heat transfer.

Please do let me know if I'm still in error, but this now makes sense -for which I'm hugely grateful.

(thanks also for the earlier reminder that all gradients, eg velocity gradients in laminar flow, being important to Sgen).

#### Attached Files:

• ###### Sgeneration.png
File size:
11.3 KB
Views:
177
11. Aug 24, 2014

### MrMatt2532

I think your diagram looks ok, except for a somewhat minor point: Instead of ΔS, it should be dS. In this form it will tell you the instantaneous change in entropy in your system. Alternatively, you need integrals on the other side of the equation. In this form it will tell you your change in entropy for your system over some integration time.

12. Aug 24, 2014

### PatrickAndrews

Thanks for that. Clarity achieved.
Cheers

13. Aug 24, 2014

### Staff: Mentor

The external temperature from which the heat is transferred can be known and measured. This is just the temperature at the interface (boundary) between the system and the surroundings. At this interface, the temperature variation is continuous, and the system temperature matches the surroundings temperature. But, within the system, for an irreversible change, the temperature distribution within the system is not generally uniform, and there can be significant local temperature gradients and heat fluxes. This gives rise to entropy generation.

See my PF blog in my PF personal area. Unfortunately, the PF blogs disappear on 9/2.

You may also find it of interest to read some of the latter posts in the thread https://www.physicsforums.com/showthread.php?t=750946. This thread has the unusual name, Question About Flow Between Parallel Plates. However, it quickly evolves into an extended discussion of thermo. One of the models Red_CCF and I study in this thread involves a spring-damper analog to gas compression within an insulated cylinder. The spring represents the p-V elastic behavior of the gas, and the damper represents the viscous dissipation provided by the gas. This model is an interesting example to study on its own, because it exhibits the important features of gas behavior, and can be used to study how the dissipative damper affects the total work done. We also apply it to studying a sequence of discrete constant force compressions, and how the total work approaches quasistatic behavior as the total pressure change is held constant while the size of each of the compressions is reduced and the total number of compressions increases. The results of the modeling are very revealing.

Chet

Last edited: Aug 24, 2014