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Post by finiteparts on Nov 27, 2015 13:10:16 GMT -5
Here is the 713C yield plot just in case you didn't want to scrounge through that report...
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Post by madpatty on Nov 27, 2015 15:14:39 GMT -5
Hi Chris,
Grad school is going good but not as cool as i am far from my jet engines hobby.
Need something to start working upon.
And these safety policies here in US are not letting me do anything.
Cheers, Patty
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Post by madpatty on Nov 27, 2015 15:24:40 GMT -5
Hi finiteparts, Another thing that i wanted to ask. The Nozzle areas and turbine inlet/outlet areas are dependent on local conditions(static) there.?? Cheers, Patty
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Post by racket on Nov 27, 2015 17:32:00 GMT -5
Hi Patty A couple of extra points to add to Chris', you'll also need to add on a bit more pressure drop across the stage to account for the energy going out of the turbine wheel , this effectively increases the theoretical PR across the stage to power the comp . The simple reason we need the NGV pressure drop is so the turbine wheel encounters gases going roughly the same speed as the wheel , if the gases were going a lot slower, the wheel would have to speed them up which would require more energy from the PR across the wheel portion of the stage . Horsepower across the stage is a product of gas deflection , the standard velocity triangles will show you quickly enough why we need the NGV . Yep , we static conditions to workout densities for flow areas. I'd strongly suggest you buy this early edition of Cohen and Rogers www.amazon.com/Turbine-Theory-Cohen-G-F-C-Rogers/dp/B0077B8VEC/ref=sr_1_4?s=books&ie=UTF8&qid=1448663282&sr=1-4&keywords=gas+turbine+theory++cohen+and+rogers Its my main referral text, much easier to understand than the later versions. Cheers John
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Post by finiteparts on Nov 28, 2015 13:49:01 GMT -5
Hi Patty, Yep, just like John said, use the static properties to calculate the local gas conditions and flow areas. But to add a touch of complication, the nozzle areas and turbine inlet gas conditions are based on the meridonal flow velocities, while the turbine exit is based on the relative flow velocities. Since the turbine is rotating, to calculate flows within the turbine blades, such as the exducer choke, you have to be in the relative frame of reference and thus the relative flow properties. Obviously this complicates the "choking" calculations for the turbine itself since the choked mass flow changes with turbine inlet conditions and rotational speed. This is similar to what was shown by Dixon for the centrifugal compressor...(discussed here: jetandturbineowners.proboards.com/post/11097). Also, I noticed that you used 0.99 times the turbine work as a means to "approximate" the mechanical efficiency...just to let you know, if you go by available published data, more realistic numbers for full floating journal bearings in larger diesel turbos is around 85-90%...there is even mention of one case measured at 70%. Holset talks about 5% just going into the journal bearings, the rest being absorbed by the thrust bearing and other oil windage. That means that 10-15% of the turbines power was used to churn the oil! Semi-floating bearings are slightly better since they only have one high speed shearing film (even though it is at a higher speed than the full-floating which rotate at approximately 1/3 shaft speed), but using small numbers for mechanical efficiency like 1% or 3% is just not realistic. John, Your statement, "...to account for the energy going out of the turbine wheel...", what energy are you referring to? I think you are meaning that because we are in fact, not adiabatic or isentropic, we will have losses...or maybe you are referring to the mechanical losses through the bearings...but I am not sure and I figured if I wasn't sure then others may also not be clear what is meant. And I absolutely agree with you on Cohen's book...Everyone should buy a copy. I actually like the 4th edition better, but all the editions are excellent. The 4th edition is uploaded online here... earju.com/225768904560/Data/Engineering/Turbo/Gas%20Turbine%20Theory%20-%20Cohen%20&%20Roger%20-%20GearTeam.pdf While I am not a fan of copyright infringement, I do realize that not everybody has access to the same libraries and resources...so I thought I would share that link. Good luck! Chris
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Post by racket on Nov 28, 2015 17:02:56 GMT -5
Hi Chris
With regards bearing power losses beware of the published data if in percentages unless they provide the rpm they're measured at and the sized turbo tested , theres not a huge difference between ball and brass at full power rpm , but at low rpm theres a very big difference .
Some Garrett data I have gives a bearing loss for a large T18A sized turbo as ~2.5 HP at 80,000 rpm which represent only a small percentage of the turbines power output, whilst very small turbos often have quite large percentages especially if tested at the low boost levels they might see on an SI engine .
"Energy out of the wheel" ................depending on the configuration of the turbine stage and its wheels exducer the "axial" exhaust gases from the exducer can have appreciable velocity, representing considerable energy which must be produced by a pressure drop somewhere , this pressure drop will change the static pressure/density which in turn needs to be taken into account with designing flow areas .
My early Cohen and Rogers text ( c 1958) has a beautiful worked example starting on page 230 where they have an axial velocity of 1300 ft/sec from the axial turb wheel flowing 40 lbs/sec , that 1,300 ft/sec represents ~68 Centigrade degrees of energy, our usual turbo GT exhaust with its "tighter" exducer angles might only have ~800 ft/sec , but its still worth ~25 degrees of energy , our static temps and pressures will be different if that isn't taken into account , also the overall stage PR needs to have that exhaust velocity "energy" represented within it .
Cheers John
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Post by madpatty on Nov 30, 2015 20:07:22 GMT -5
Hi Racket,
I did a read of the Cohen nd Rogers text and also went through the solved examples:-
Most of it is clear except certain points as there was very little discussed about radial inflow turbines.
Now in our case where we need to calculated the suiatble turbine wheel for our engines:-
1. What blade tip speed we assume at the turbine exducer. Because the tip speed increases with the radius so we do we make the outlet velocity triangle only at the exducer outermost diameter.
2. Now that being said do we assume that all gases exiting the wheel have axial velocity only and no whirl component(that is the gases are travelling straight backwards through the jet pipe) (That means the velocity triangle has three components- U(blade tip speed), C3(actual velocity) and V3(velocity relative to blades).
3. The reference value for the area between the blades at the exducer(which you once calculated in my thread using your garrett turbine should be calculated using V3(relative velocity) and not C3 (as anything in the frame of reference of blades will see that area and will see gases going in direction of V3. (Am i right)??
Cheers, Patty
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Post by racket on Nov 30, 2015 20:53:55 GMT -5
Hi Patty
The Papers I've read on radial inflow turbines using a scroll imply a "mess" coming out of the wheel , there certainly isn't any uniformity, so we can only make assumptions .
I'd use exducer tip speed for the triangle , or perhaps 75% blade length where most of the flow is .
As for power production using gas deflection , I use a "home made" formula where I make some assumptions about the "mean" blade speed .
Yes, for calculating the flow area vs velocity in the exducer we need the gas velocity and density relative to the blades , but because the gas velocity will probably vary at different radii and even between different segments of the wheel fed by different segments of the scroll, its hard to make calculations , its very much "ball park" to start with , then trial and error after that .
Using the slower "exhaust" velocity after the exducer, along with the annulus area, is also a rough guide as is done in the C & R text.
Cheers John
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Post by madpatty on Nov 30, 2015 22:13:16 GMT -5
Hi Racket,
I tried to extract useful points from your post but didnt understand the last paragraph.
What is the meaning :-
"Using the slower "exhaust" velocity after the exducer, along with the annulus area, is also a rough guide as is done in the C & R text."
What slower exhaust velocity are you talking of and which annulus area?
Now that being said, how do you guys calculate how much clipping is required at exducer(obviously the ball park point) given the static density and static temperature at the exducer. Which of the two velocities in the outlet velocity triangle(C3 or V3) is taken into account for the suitable exducer area(area between the blades).
Cheers, Patty
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Post by racket on Nov 30, 2015 22:41:56 GMT -5
Hi Patty
The relative velocity between blade and gas stream within the exducer is high , once the gas leaves the exducer its relative speed to the jetpipe walls is slower, the difference in those speeds represents part of the energy imparted to the turbine wheel .
The annulus is the cross sectional ring of the exducer at right angles to the shaft
We could have a gas velocity of 2,000 ft/sec between blades if the exducer is choked , but if the blade is traveling at high velocity in a direction away from that of the gas stream , then the exit velocity is reduced relative to its surrounding which aren't moving .
Its the final static pressure at the exducer throats thats used for determining the clipping required, the static pressure will depend on the PR across the stage to that point ,the actual velocity of the gases at the exducer throat is uncertain unless its running choked , and then its roughly the local speed of sound at the temperature of the gases at the throat.
Theres lotsa unknows and is the reason why trial and error is often the only way to determine the outcomes .
If there is a lot of clipping then the power producing capability of the wheel is reduced due to the reduction in gas deflection ( use velocity triangle) , so there has to be a compromise between opening the wheel for flow but keeping sufficient power production so as not to require very high temperatures to compensate , and is the reason why I still don't know if my 12/118 engine will work with the radical clipping I've done on its turbine wheel , I only have the results from my TV84 experiments as a guide to a positive outcome .
Cheers John
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Post by madpatty on Dec 30, 2015 22:23:21 GMT -5
Hi Racket nd experts,
Is it possible to use RR Allison C18 3rd stage turbine wheel as the main gas producer's turbine(obviously if throat areas allow) or it can't be used due to the labyrinth seal ring at its OD.
If somebody can tell what is the max. rpm of this wheel whilst in its original engine?
Are these wheels suitable for only power turbine application from the materials perceptive?
Most of the initial stage wheels(gas producer turbines) have open blades(no OD labyrinth seal ring).
Cheers, Patty
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Post by racket on Dec 31, 2015 1:39:40 GMT -5
Hi Patty
The 3rd stage wheel only spins to a ~35,000 rpm in its original C18 application, the C20 ~32,000 , this is too slow for a gas producer unless its diameter was ~200mm , but then the mass flow would be so high and the horsepower requirement beyond what the wheel is designed for .
The 3rd and 4th stage wheels are freepower wheels , they're relatively "light weight" wheels designed to be run with stage inlet temps of <800 deg C , way too cool for a gas producer , the material is good , but because of weight of that labyrinth tip seal , the blades would be over stressed at higher temps , they are what they are .
If you want a gas producer wheel , use the 1st and 2nd stage wheels , but neither are really suitable , the first stage wheel has limited flow capacity at lowish pressure ratios being designed for a 7:1 PR at the inlet , at a 3.5 PR the flow would be <2 lbs/sec , the second stage wheel is a bit better but the flow is still relatively small for such a large diameter wheel, finding a suitable diametered compressor wheel would be difficult .
I've had a number of these wheels for years and not been able to find a use for them , the second stage wheel would make a nice ~80 hp freepower wheel , but its flow can easily be matched by a large turbo wheel .
Cheers John
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Post by madpatty on Mar 20, 2016 22:31:04 GMT -5
Hi Racket,
I need some advice about using direct injection(spray nozzles) in the annular combustor i am currently using.
I had faced enough problems regarding evaporators and fuel syringes so i am thinking of replacing the evap system with fuel nozzles because i know i simply cannot get away with evap system.
Can you tell if that can be done with annular combustors. If so which type and how many fuel nozzles can be used?
I am ready to do any necessary changes to the whole system.
Thanks.
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Post by racket on Mar 21, 2016 0:14:30 GMT -5
Hi Patty
You might just have as many problems with spray nozzles in the small size you'd be needing , also the number of nozzles will need to be fairly high so as provide adequate coverage of the flametube cross section.
Not an easy job , I think I'd stick to evaporators :-(
Cheers John
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Post by madpatty on Mar 21, 2016 7:54:17 GMT -5
Hi Racket,
Actually the problem is i cannot get all that silver solder and right sized needles in my locality.
The setup i can use is of very poor quality. I had to buy copper capillaries, then cut them down to required size and then gas-weld them to manifold.
During the high temperatures of gas welding those capillaries frequently loose their shape and strength.
Sometimes get blocked etc.
So this is the reason i want to go with fuel nozzles.
I can increase the diameter of flametube if required. Is it possible then?
Thanks.
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