Combustion Design

[madpatty]

Hi Chris.

When you say flow speed in the annulus we have to take the liner area into account or its just the U_ref= m/(rho x Aref)?

Secondly, I couldn’t seem to find this pressure loss parameter (K) in Lefebvre’s book but he only talks about pressure loss factor which is (dPt_3 - dPt_4)/Q_ref

Thanks

[finiteparts]

Apr 14, 2020 20:01:53 GMT -5 madpatty said:

Hi Chris.

When you say flow speed in the annulus we have to take the liner area into account or its just the U_ref= m/(rho x Aref)?

Secondly, I couldn’t seem to find this pressure loss parameter (K) in Lefebvre’s book but he only talks about pressure loss factor which is (dPt_3 - dPt_4)/Q_ref

Thanks

In this instance you are looking at the actual annulus  area, so yes, you would subtract out the liner area.  The reference conditions that Lefebvre uses are not to be used with the design procedure where you specify the annulus or liner passage flow speeds.

The K loss parameter is not to be confused with the pressure loss parameter that Lefebvre uses to size the blockage requirement of the combustor liner.  This is one of the complaints that I have about some of Lefebvre's papers and his book.  There are often confusing parameter notations and units that can easily cause errors for newcomer.  I think this is a result of his excellent talent of collecting much of the published knowledge into a single source, but he often leaves the original nomenclature which can get muddled up and confusing.

In my first edition, it is on page 115, under figure 4.3.   K = jet dynamic pressure / annulus dynamic pressure = 1 + delta_pL / q_an

I checked my third edition and you are right, it doesn't seem to be in there.  What a bunch of shit.  Why would they "update" a book by taking stuff out???!!!

I hope that helps,

Chris


  

[finiteparts]

I tried to clean up my equations on here, but they were tough to read, so I just posted images of the pages I had already typed out.  Because I have had so many pages accidentally get deleted or lost in some way or fashion, when I type out larger posts I usually type them out in Word before transferring them to the post.  Hopefully, these will be big enough to be clearly read.

Here is the link to the Peter Stuttaford paper that mention in the write up

asmedigitalcollection.asme.org/GT/proceedings/GT1996/78743/V003T06A026/237662


Here are the pages for working through the Zonal Determination of K-loss Factors:

[finiteparts]

A quick follow up.

To get a K-loss factor down near K = 3, which would give you a Cd ~ 0.46, theta  = 61.1 degrees, you would need to have an annulus velocity of about 216 ft/s for the above conditions.

Taking it further, to get a K = 2, or a Cd ~ 0.36, theta = 50 degrees, the annular velocity would need to be around 265 ft/s.
 
So you can see if you use a annulus velocity based approach to sizing the liner and casing (i.e. the annulus area), then you are probably safe to go to 150 - 160 ft/s without substantial losses in using plain drilled holes. 

If you are concerned about the initial jet angle, you could limit it to being larger than 70 degrees.  This puts you at a a K ~ 4.5 (Cd ~ 0.52) and the annulus speed being around 177 ft/s.

I hope this was helpful.

Chris

[finiteparts]

Apr 19, 2020 14:25:56 GMT -5 turboron said:

Chris, the 2nd edition also uses the name theta. There is no reference to "loading parameter" that I saw. In any case, when I went through the numbers for my combustor using a 2.457/1 pressure ratio, mass flow in lbs/minute and the dimensions in inches I calculated a Theta of .7447 times 10^-4. Figure 5-3 indicates a combustion efficiency of 90%. If I increase the pressure ratio to 4/1 while holding mass flow constant I get a Theta of 2.8772 times 10^-4 which is still a 90% combustion efficiency per Figure 5-3. Do these number seem correct?

Thanks, Ron

Ron,

Let's move this discussion over here so that we don't hijack John's thread.

Those numbers look low, especially if you are using full power design conditions.  You should be at or very near 100% efficiency.
In my first edition, the plot that has numbers is done in metric units, so your input values would be incorrect...but that is one of my big complaints about some of Lefebvre's work.  Units are unclear, sometimes variables are unclear or share nomenclature with a different variable altogether.  My guess is that for that plot, the pressure is in atmospheres, the temperature in K, the mass flow is maybe kg/s and the area is likely meters^2.

Now all that being said, be very reluctant to trust the values from those efficiency curves.  They are very generic and different combustor geometries behave very differently.  The loading parameter is used all the time in the combustion industry, but they are not based on generic curves.  They are based on previously tested, very similar or even scaled combustors.   They use them to predict such things altitude light-off capability and emissions at off-design conditions, but again, these are all based on well tested data sets for each combustor design. I would almost venture to state that trying to predict your combustor efficiency from one of those curves is an exercise in futility.  You might be better off drawing some efficiency numbers on a dart board and having a go with that. Ha!!!!

Now, what you can use the theta parameter for is to see trends and how the variables impact the overall design.  The loading parameter is also good for correlating and plotting data the you might collect on your combustor.

I hope that helps,

Chris

[turboron]

Chris, thanks. Zucrow has the same problems as Lefebvre. The Purdue guys all seem to have the same disease. There must be something in the water in Lafayette, Indiana.

I appreciate your comments. All our modern analysis tools help get us in the ballpark. I am still a believer in build, see the results and correct.

Thanks, Ron

[madpatty]

Hi Chris finiteparts,

Thanks for sharing all the theory in this thread for combustion design.

What do you think is the appropriate starting point for secondary zone design.
I see most people start by assuming stoichiometric or near stoichiometric conditions in Primary zone.

But what about secondary zone?
What is the design point for that.

I always assumed ‘for my designs’ that primary and secondary zone combined should form a stoichiometric mixture.

Any literature if you can share will be really helpful.

Cheers

[finiteparts]

Hi Patty,

I would say that you would want to target a secondary zone fuel-air ratio slightly lean to make sure that you have enough O2 to burn out any unprocessed fuel that escapes the PZ. Then also have some additional air to help bring down the bulk gas temperature to a point where the dissociated products would recombine. If you don't reduce the bulk gas temperature enough before you hit it with the large dilution flow, then you will "freeze" any reactions that are still going on and loose efficiency. A typical slow reaction is the burn-out of CO to CO2, thus, when you have poor combustion efficiency, you will have high CO (carbon monoxide) emissions.

This is a huge challenge to ultra-low emissions, lean pre-mixed combustors. Since they run their PZ's at very low FARs in order to keep the peak flame temperatures below levels where thermal NOx forms, they also struggle to burn out the CO. It makes things challenging!

I will search for some literature that gives ideas of FARs for each zone.

- Chris

[finiteparts]

I just dug through my copy of Mellor's, "Design of Modern Turbine Combustors" and in Chapter 4, Will Dodds and David Bahr suggest, based on their experience at GEAE, a secondary zone phi of 0.6 to 0.8 for "modern short length" combustors that you might see in 1990's turbofans. Now, can we use the same approach for a combustor that is only running at a pressure of 3 atm or so instead of 30-40 atms and higher inlet temperatures; I am not sure.

I think that a phi = 0.8 or 0.85 might be a good start, but I am just guessing that right now. I might try to run a chemical kinetics program to see if it can provide an insight on this and follow up later.

Remember phi = FAR / stoichiometric FAR, so if we are using kerosene like fuel with a stoichiometric FAR ~ 0.0658:
--- the SZ FAR = 0.8*0.0658 = 0.05264 lbm fuel / lbm air
So the lbm fuel would be what you injected in the PZ and the air mass flow would be the air added in the PZ plus the air injected through the SZ holes (or if you only run a single row of PZ hole and assume some of that flow is going rearward to the SZ).

I hope that helps,

Chris

[finiteparts]

I thought that I would share a resource for finding the required fuel to air ratio in your combustor.

The resource can be found here in NASA Tp-2359:

ntrs.nasa.gov/search.jsp?R=19850001757&hterms=tp-2359&qs=N%3D0%26Ntk%3DAll%26Ntx%3Dmode%2Bmatchallany%26Ntt%3Dtp-2359

Using Figure 3(e) on page 17, we can find the required ideal FAR for a given temperature rise based on the initial air inlet temperature.

As an example, let's say we have a compressor discharge temperature of around 309 deg F ( 426 K) and we want to get the turbine entry temperature up to 1650 deg F (1172 K).  That is a delta T = T4 - T31 = 1172 - 426 = 745 K

So if we  plot this on the Figure, we get:



If we start from the temperature rise of 745 and come over to the 400K initial temperature curve, then we drop down to find that we need an overall FAR ~ 0.020.

So, now that we know the overall FAR  for our design point, let's say that we are anticipating that at the design point the engine will flow 1.2 lbm/s.  It is easy to find that the amount of fuel that we need is:

-----------   mass flow of fuel = FAR * mass flow of air     =     0.020 * 1.2 lbm/s  ------> fuel flow = 0.024 lbm/s

Since kerosene has a volumetric density of around 6.82 lbm /gal, we need a fuel flow rate at this design condition of:

----------  volume flow rate = (0.024 lbm/s)  / (6.82 lbm/gal)  = 0.00352 gal / s *3600 s/hr ----->  fuel flow = 12.67 gal/hr

I hope that helps people work through this easier themselves.

- Chris

[finiteparts]

In addition to providing us a means to calculate the required FAR, the NASA TP-2359 document can helps us get a very good estimate for the ratio of specific heats for us to use in the power calculations in the turbine section.

If we look a Figure 4(f), we see how the ratio of specific heats (Cp/Cv) called gamma changes due to temperature and FAR.



From the above example, we found that the overall FAR we needed at this design point was around 0.020 and we were assuming a turbine inlet temperature of 1650 deg F (1172 K).  So if we start on the x-axis with the turbine inlet temperature (Mixture Temperature) of 1172 K and we go upward to the FAR = 0.020 curve, we see that the gamma at that condition is 1.31.

Likely in the past, it is common to assume a gamma ~ 1.33, which is pretty close.  But with this method, you can get a very correct gamma value with very little effort and help to minimize the errors in your calculations.

- Chris

[madpatty]

May 2, 2020 12:36:41 GMT -5 finiteparts said:

Hi Patty,

I would say that you would want to target a secondary zone fuel-air ratio slightly lean to make sure that you have enough O2 to burn out any unprocessed fuel that escapes the PZ. Then also have some additional air to help bring down the bulk gas temperature to a point where the dissociated products would recombine.

- Chris

Hi Chris.

I agree with the lean (fuel lean) mixture in the secondary zone. But this always confuses as to how we are calculating equivalence ratio in the secondary zone.
According to my combustion class, we should calculate equivalence ratio according to unburnt fuel (fuel leftover from primary zone) in the secondary zone and not according to total fuel.
That being said, I am not sure how efficient a ‘normal’ primary zone is but just for the sake of this discussion let’s say we are able to burn ~90% fuel to respective end products completely.

Now we only need air corresponding to the leftover fuel that managed to escape PZ and some air to fully oxidise the monoxides.

I always assumed we need very less fresh air in SZ.

BUT i just realised the way you calculated phi and FAR in SZ with total air in PZ+SZ in denominator takes care of this reduced fresh air requirement in SZ.

Food for thought.

Not intended to confuse anyone reading this😊

Thanks

[CH3NO2]

May 2, 2020 21:13:23 GMT -5 madpatty said:

May 2, 2020 12:36:41 GMT -5 finiteparts said:

Hi Chris.

I agree with the lean (fuel lean) mixture in the secondary zone. But this always confuses as to how we are calculating equivalence ratio in the secondary zone.
According to my combustion class, we should calculate equivalence ratio according to unburnt fuel (fuel leftover from primary zone) in the secondary zone and not according to total fuel.

Not intended to confuse anyone reading this😊

Thanks

As long as each zone is given the equivalence ratio it needs, what difference will it make how the equivalence ratio is calculated? 

Your combustion teacher probably had a good reason for it but it seems either way will give the same numbers.

Thanks,
Tony

[finiteparts]

May 2, 2020 21:13:23 GMT -5 madpatty said:

Hi Chris.

I agree with the lean (fuel lean) mixture in the secondary zone. But this always confuses as to how we are calculating equivalence ratio in the secondary zone.
According to my combustion class, we should calculate equivalence ratio according to unburnt fuel (fuel leftover from primary zone) in the secondary zone and not according to total fuel.
That being said, I am not sure how efficient a ‘normal’ primary zone is but just for the sake of this discussion let’s say we are able to burn ~90% fuel to respective end products completely.

Now we only need air corresponding to the leftover fuel that managed to escape PZ and some air to fully oxidise the monoxides.

I always assumed we need very less fresh air in SZ.

BUT i just realised the way you calculated phi and FAR in SZ with total air in PZ+SZ in denominator takes care of this reduced fresh air requirement in SZ.

Food for thought.

Not intended to confuse anyone reading this😊

Thanks

The purpose of the secondary air is not really to burn out residual unburned fuel. If you notice, I stated that the primary reason is actually to locally reduce the temperature to a point where the dissociated products recombine.  At high temperatures, certain molecules have so much internal energy that they can break their chemical bonds.  Typically, we a talking about N2, O2, CO2 and maybe some H20.  It takes energy to break apart chemical bonds, thus this dissociation reduced the energy released from the combustion of the fuel and air.  If these radicals were allowed to freeze, then the energy that they robbed from the process will be uncovered.

So you push in a small amount of air to locally cool the gas, which means that these dissociated products reduce their energy to a point where they are no longer have enough energy by themselves and must recombine.  In the industry, we often call it burning out CO, which is a bit of an incorrect statement.  Now don't get me wrong, CO is a fuel and in the presence of O2 and enough heat, it reacts exothermicly to produce CO2, so there is an oxidation pathway.  I was talking about the high temperature dissociated products, recombining to form CO2.  Slightly different and I apologize for using misleading lingo.

Lefebvre covers this pretty well in his book in the chapter on Aerodynamics.  

Good luck,

Chris

[finiteparts]

I received a request to explain the pressure loss through the combustor a little more thoroughly, so I will try to back up few steps and do my best to explain this a little more completely.

I think the first thing that we should talk about is pressure, or more specifically, the different way we talk about pressure. At the most fundamental level, pressure is a force applied to a unit area, such as pushing with 1 lbf (lbf is pound force, while lbm is pound mass) over an area of let’s say 1 in^2. That is the definition of 1 psi. If we are dealing with a gas, such as air (really a composition of gases), then the reason that a force is applied to a surface is because molecules are flying around and when they hit a solid wall, they bounce away just like a rubber ball. As you get billions or more of these molecules bouncing off the wall, they start to create a substantial force due to the sheer number of these occurrences and this is what we call pressure.

Ok, now you probably have heard the terms total and static pressure on this site quite often. As you can probably surmise, the static pressure is the pressure that we experience when there is no bulk motion. But when there does start to be bulk motion of the gas (flow), there is now a kinetic component to the fluids energy and this kinetic energy makes up part of what we call total pressure.

The total pressure of a flowing gas is the static pressure of the gas plus the kinetic energy of the gas plus the elastic energy of the gas. If the fluid wasn’t a gas, but a liquid, in which the compressibility is almost insignificant, then you could say that the total pressure (Pt) is the static pressure (Ps) plus the kinetic energy…

Pt = Ps + 0.5*density*velocity^2

In this incompressible formulation, the pressure due to the kinetic energy of the fluid is typically termed dynamic pressure. People often think that they can use this formula for compressible flows also and the term dynamic pressure often gets thrown around incorrectly for compressible flows. Now, some engineering standards do a better job of naming this as the kinetic pressure, which I feel is much less confusing.

If we had a piston in a cylinder that was filled with air at atmospheric pressure, and then we put our rock on top of the piston so that it would compress the gas in the cylinder a bit, it should now be clear how the gas has absorbed some of the rocks potential energy to convert it into an increase in the gas pressure. The compression of the gas can be thought of as elastic energy in the gas and it is this compressibility that adds complication to higher speed flows.

For the flow of a gas such as air, as the fluid moves at higher and higher bulk velocities, the compressibility becomes more significant. It is often used as a rule of thumb that as long as the flow is less than a Mach number of 0.3, it can be considered incompressible…but what is often left out of that statement is that it can be considered incompressible because the error in the calculated fluid properties is relatively small. So, by definition, using the equation,

Pt = Ps + dynamic pressure

is always wrong for compressible flows. So what is missing? As we saw above, there is an elastic energy, which is captured as a change in the internal energy of the fluid that also plays into the total pressure,

Pt = Ps + dynamic pressure + pressure due to change in internal energy

Sometimes, the dynamic pressure + the pressure due to the change in internal energy is called the recovery pressure, which I am also in favor of because it makes things much clearer when speaking about compressible flows.

Ok. Hopefully that helps to understand the two basic flavors of pressure, static and total. Now I should also tell you that the static pressure is what objects feel as the fluids flow by. Total pressure is an idealized notion of what the pressure could be if you were able to bring the fluid to a complete stop without inducing any losses into it. In this world, nothing is loss-free…there is always some amount of loss…like the statement, “ there is no free lunch”, you always have to give something up to get something. When we try to bring fluids to rest, there is always some small amount of turbulence that dissipates a small amount of the fluids energy off as heat. The main thing to remember is that total pressure represents and ideal potential to do work and static pressures are what the stuff in contact with the fluids, “feels”.

Now we should be able to get down to the meat of the pressure drop. When you stick your hand out a car window, you feel a force pushing your hand backwards. This is due to the pressure recovery in the flow on the forward side of your hand and the back side of your hand is at a lower pressure due to the dissipation of the fluids recovery pressure through turbulence and ultimately, heat. This means the backside of your hand is closer to the local static pressure while the front of your hand is closer to the total pressure of the flowing air.

A higher pressure on the front of you hand means that it has a higher force and pushes your hand backwards (remember pressure is force over an area, and we assume that areas of the front and back of your hand are the same)…which is called a drag force. From this, you should be able to see that the drag force against you hand is cause by the loss of total pressure through turbulent dissipation as the flow comes by your hand.

Now, if we jump to a combustion chamber, you can probably guess that having more blockage to the flow causes higher drag. More blockage means smaller or fewer holes through the combustion liner. In a combustor, we are not really concerned about the drag it produces in the flow, but we are concerned about how well the flow mixes around inside the combustion chamber. As you might have guessed also, if we want more turbulent mixing within the combustor, we need more turbulence or as we saw above, more drag.

But, it should also be remembered that we spend a lot of energy in the compressor get the total pressure of the gas increased up to the level that it is at the entrance to the combustion chamber, so we don’t want to waste it. Well, it has been determined that around 3-5% total pressure drop through the combustion liner gives a very high level of mixing in the combustor, and any more is just taking more energy from the overall cycle and hurting the engine efficiency, so that is why we usually target 3 to 5% dP/p. What is meant by dP/p? That is the ratio of the total pressure of the air entering around the outside of the combustion liner (Pt3) minus the total pressure of the air inside the combustor (Pt4), divided by the upstream total pressure ,

dP/p = (Pt3 –Pt4)/Pt3

since the flow velocity in the combustor is usually really low, the total pressure inside the combustor is nearly equal to the static pressure (Ps4) in there, so it is often written as,

dP/p = (Pt3 – Ps4)/Pt3

So if you wanted a dP/p = 5%, then

dP/p = 5% = 5/100 = 0.05 = (Pt3-Ps4)/Pt3

or

Ps4 = Pt3(1-0.05) = Pt3*0.95


I hope that helps and please feel free to ask any more questions if something is unclear.

- Chris