Clarification on Brayton Cycle 1

[Ferex]

Hi everybody, my question is very simple, I think but I can't have a point on it:
in brayton cycle there's the isobar step, during the combustion process... everywhere I read that this step is isobar, but I am guessing how can it be isobar if temperature of the gas rises because of combustion, it could be isobar if I have an increase in volume, but the expansion is only in the turbine, in the combustion chamber I have only a little increase in volume...
Could someone tell me where do my thoughts fail??
Many thanks

 

[ivymike]

well, the combustion is continuous, so it was going on before the air got there and it continues after it leaves. the compressor brings the air up to combustion pressure and it stays there. (I think you're incorrect about expansion happening only in the turbine.)

 

[FOURe]

Hi there:

My suggestion is just to look at the typical temperature -- entropy (T -- s) diagram for the Brayton Cycle operation.

​

Also, to help you with the understanding of isobaric process when combustion takes place -- combustion takes place at constant pressure, one can assume that the combustor chamber cross sectional area increases as the working fluid passes through it resulting in maintaining the constant pressure during the combustion process.

However, in actual combustor operation, there are at least two air streams brought to the combustor -- primary air stream for intensifed combustion and secondary cold air stream to reduce the working fluid temperature before entering the gas turbine ...

Thanks,

Gordan

PS

Here is a typical Wikipedia combustor article:

http://en.wikipedia.org/wiki/Combustor

http://www.engineering-4e.com

 

[Ferex]

Thanks guys... so do you confirm that during combustion the chamber cross-section area increases?
Link very interesting, I read brayton cycle on wikipedia but not the combustor...
On that link cross-section area seems to reduce after burning.

 

[FOURe]

Ferex:

Yes, you are right that the real combustor exit cross sectional area reduces because the secondary cold air stream gets injected and as a result the working fluid temperature goes down while its density increases -- according to the mass flow rate equation, the cross sectional area should get reduced at the exit of the real combustor having the working fluid velocity constant/unchanged.

I just tried to explain what happens for both cases: ideal and real -- so that it gets easier to understand what is going on.

Thanks,

Gordan

http://www.engineering-4e.com

 

[Ferex]

Ok, I keep reasoning, fuel entrance and consequent burning give the fluid great energy: how this energy is stored? what parameter does this energy change? It's not pressure because it remains constant, is it velocity? I am not sure because the enlargement of the cross-section slows down the flow.... what do you think?

 

[FOURe]

Ferex:

During combustion, fuel chemical energy gets released and as a result the working fluid temperature goes up as well as enthalpy.

OK, here are a few typical plots regarding the basic Brayton Cycle operation -- thermal efficiency, specific power output and power output:

​

Thanks,

Gordan

http://www.engineering-4e.com

 

[Ferex]

Many thanks, you gave me answers for my question.
Sadly I understand that I have a big lack of knowledge, things studied and then forgotten...
I will try to fill these voids.

 

[Tsiolkovsky]

Im afraid that Ferex has been answered wrong in this thread.

The real reason for the isobaric addition of heat (and the removal of heat) during the Brayton cycle (and in all control volume cycles) lies in the fact that there is NO WORK TRANSFER during heat addition or removal. If you look at the equation for work in a control volume cycle, for there to be no work during heat addition/removal, the pressure term in the equation must stay constant. If the pressure term is a function of specific volume, then there would be work transfer, which there is not. A physical interpretation of this is knowing that pressure equally distributes throughout the process irrespective of the cross sectional area distribution, despite whatever temperature rise/fall there is.

 

[ajphoneix]

Hi everyone

I also had the same difficulty as with ferex on understanding the reason for assuming constant pressure in brayton cycle. I agree with all the information Tsiolkovsky gave, but still if you try to imagine the real actual process, where in the chamber air and fuel are combusted, everything goes crazy, shouldn't the pressure increase? I mean thats the ponit of combusting the fuel isn't it?to increase both T and P.right?I know it depends on the volume and that it is confined but this is also in the real case.

 

[btrueblood]

The pressure can't increase, or it will stall the compressor upstream. Which does happen from time to time, but controls types are given the job to make it not happen. In reality, for the flow to keep going the way it is designed to go, the pressure will have to drop, however slightly, thru the burner.

To go further along what Tsiolkovsky wrote, combustion is idealized in 1D cycle analysis to take place at a Mach no. of essentially zero; doing so results in zero losses due to heat addition. In reality, the combustion takes place at a finite, but low, Mach no., and typically the Mach no. increases along the length of the burner cans (i.e. the flow is expanding slightly during the process).

 

[btrueblood]

"if you try to imagine the real actual process, where in the chamber air and fuel are combusted, everything goes crazy, shouldn't the pressure increase?"

Hmmm, you may be stuck thinking about otto-cycle engines too. A real-world example of constant pressure burning is a fire in your fireplace. Heat is added, but there is no physical containment of the working fluid (air), so it expands (density decreases), and the hot air rises, entraining more air, etc. Take the same process, but make it use pressurized air - now you need a burner "can" to hold the air in, but the can will need a special shape, perhaps one like a bell, to let the gas have room to expand without appreciably increasing its velocity.

 

[Tsiolkovsky]

"I mean thats the ponit of combusting the fuel isn't it? to increase both T and P.right?"

AHA! But that's exactly where you are wrong! I dont blame you for thinking so for it seems intuitive at first. Im going to further clarify on my earlier reasoning which only constituted a mathematical approach.

Lets take a control mass example first (such as an Otto cycle). A piston free to slide within a cylinder. When heat is transferred to this system, The pressure stays constant aswell! Why was no one flailing their arms in the air when they heard this? The reason for the isobaric process in the cylinder is that the piston keeps rising until the pressure beneath the piston equals the pressure above the piston. If this were not the case, the piston would continue to accelerate up or down due to unequal forces (pressures). But no, it reaches a stable equilibrium where the pressures equal. If the piston were to be LOCKED by a pin within the cylinder, then the pressure would rise because no work would be given off (boundary movement associated work). And because there is a failure to produce work, this energy must reveal itself as a rise in pressure. The mathematical representation of this control mass cycle is the same I provided previous post except the P and V terms are swapped around (the v is the integral)

Once again in control volume cycles no pressure rises because the transfer of energy to the system reveals itself as a pure temperature rise. ""Yes but just HOW doesnt the pressure rise physically looking at it from a laymens point of view"" you ask? The answer is SIMPLE. The inlet to the boiler and the exit of a boiler, similarly the inlet of a condensor and the exit of a condensor are physically connected and plumbed to each other directly. we know from fluid mechanics that any fluid equally distributes the pressure of its volume equally and entirely across the entire volume which it occupies. So if one end of the plumbing is cooler due to heat transfer, its pressure drop in that portion would immediately distribute itself into the other end of the boiler and condensor!

We see that our intuition, that heat transfer causes pressure rise, is flawed only because we think in terms of a piston/cylinder arrangement in which the piston is locked in place.


Damn I LOVE Thermodynamics.

 

[Tsiolkovsky]

I must also add very importantly:

When we talk about no pressure rise or fall during combustion or condensation within boilers (burners) or condensors we only talk of just that: DURING the process of heat addition/removal. Pressure across that stage and that stage alone. Say stage 1-2 from the T-s diagram posted in the beginning. From the entire cycle point of view it HAS increased pressure because it occurs at a higher/lower isobaric line than 1-4.