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Post by madrocketscientist on Jan 24, 2018 21:51:26 GMT -5
John, I have found a map for the GT4294 wheel, Would this be suitable? The trim is on the high side of things. What is the effect of machining the compressor to a lower trim, does it reduce the flow and raise the pressure ratio? Shannon.
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Post by racket on Jan 24, 2018 22:05:08 GMT -5
Hi Shannon
Yep , roughly speaking , a lower Trim gets better PRs , but some wheels are designed for lower PRs but with extra map width , we can get away with very slim maps .
The GT4294 is probably a tad bigger than necessary for the 1 lb/sec flow you require , but its certainly better than what you have at present , as long as you design your comp diffuser flow accordingly you'll probably be OK even though you'll be running up closer to the surge line .
You'll need to do the calcs for your NGV and turb wheel flows to check just what can be swallowed by them if the comp is producing a 3.8:1 PR, use a 900 C TIT for calcs, and a 5% pressure drop across the flametube and see how things turn out .
Get back to me with your "numbers" and I'll look them over :-)
Cheers John
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Post by madrocketscientist on Jan 25, 2018 0:44:33 GMT -5
So the lowering the trim generally narrows the map while making it taller at the same time, if I understand that correctly?.. Here are my calcs for the GT4294 compressor RPM 108000 Pressure ratio 3.8 Mass flow 0.45kg/sec Compressor inducer diameter 0.0703m Compressor exducer diameter 0.094m Compressor efficiency (assumed) 74% u = (108000rpm x 0.094m x 3.142) / 60 = 532m/s PR at compressor exit = √3.8 = 1.95 ∆T = 288K / 0.74 x (1.95^0.286 - 1) = 81.9K p = (1.95 x 101300Pa) / (355K x 289J/kg/K) = 1.94kg/m³ C m = 0.45kg/s /(1.94kg/m³ x 3.142 x 0.094 x 0.006) = 131m/s C 2u = 532m/s -(131 / tan(61)) = 459m/s And the Vector diagram Hopefully so far so good? Shannon.
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Post by madrocketscientist on Jan 25, 2018 3:12:03 GMT -5
Here are the gas states I get, hopefully I am right in assuming that the listed pressure ratio in the compressor is what comes out, or does this get reduced due to the 94% efficiency?
Inlet air Compressed air Combustion gas Exhaust gas 288 K 454 K 1173 K 973 K 15 °C 181 °C 900 °C 700 °C 1.013 Bar 3.8 Bar 3.61 Bar 1.013 Bar P = 1.225 kg/m³ 2.95 kg/m³ 1.07 kg/m³ 0.36 kg/m³ ∆H = 0 J/kg 106.0 kJ/kg 741.8 kJ/kg 0 kJ/kg 443.9 m/s
Shannon
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Post by racket on Jan 25, 2018 3:42:25 GMT -5
Hi Shannon
LOL...........its not quite that simple :-(
There are a number of Papers on the Net that try to explain things , its complicated , but a low Trim general gives the inducer an easier time at higher PRs which in turn helps overall efficiency
Once we start getting into higher PRs the Kamps methods start to have shortcomings .
Firstly , theres a matter of "slip factor" which means the airflow doesn't exit the wheel at the same angle as the blading , so if the blading is a 60 degree backswept , then you'll need to roughly take off another 10 degrees .
Also your 131 m/s outflow number will depend on the wheels tip height at the exducer , a "low PR" wheel will have a larger tip height than a "high PR" wheel of the same inducer diameter, your outflow angle is kinda low , a result of running the comp wheel at a flow rate lower than ideal , its designed for 20% more flow
Then theres the matter of PR division between wheel and diffuser , this again can change depending on the wheel design .
With a high PR you'll be experiencing static pressure downstream of the turb wheel , the pressure drop across the stage/wheel is restricted so as not to "overpower" the comp ............the "adjusting" of the stage pressure drop is done by changing the jet nozzle sizing , a larger jet nozzle increases the PR across the stage , a smaller one increases backpressure and reduces the PR across the stage and is generally accompanied by a temp increase as the engine tries to substitute temperature for pressure.
You need to measure the throat area of both NGV and wheel so as to calculate their flow potential at the required pressure ratio across the stage to spin the comp .
Cheers John
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Post by racket on Jan 25, 2018 3:50:15 GMT -5
Hi Shannon
Your jet velocity is low compared to the olympus , its >50 m/s faster.
LOL................I can't work in SI units , I'm a cubic feet/second type of guy .
Theres no point doing the comp calcs until you measure the turb stage throats, they'll determine what happens "up front" :-)
Cheers John
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Post by madrocketscientist on Jan 25, 2018 4:04:35 GMT -5
Thanks John,
I can see how the higher trim would be easier to generate higher pressure ratios, there is more air pushing trying to get further out, with a higher trim, that pressure builds further towards the inducer....
The wheel I was looking at has a 6mm tip height, not sure if that is very small but I have seem some high trim wheels with 8mm tip height!
The slip factor makes sense, air never quite goes where one wants it to go! I will add that extra 10° onto my angles.
So with a compressor map should I be aiming to hit that dotted grey line shown in the map a few posts back for an ideal match?
For the Compressor PR division I am 'assuming' a 50% split but like you say this is depending on wheel design, and also I guess, on the diffuser setup! I will keep hunting for a suitable wheel, I am making up a spreadsheet as I go so I can change these numbers on the fly and get faster results.
The areas I have for the NGV and turbine wheel are; NGV outlet area 0.00242m³ Turbine outlet area 0.00279m³
I will do the calcs with the 3.61:! (95% or 3.8:1 pressure ratio) drop across the ngv/turbine.
Shannon.
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Post by madrocketscientist on Jan 25, 2018 6:09:51 GMT -5
Here is what I get for the turbine stage, PR x.95 = 3.61:1 Assuming 50% drop across NGV and Turbine √3.61 = 1.9:1 ∆h ngv & ∆h turbine = 1100J/kg/K x 1301K x(1-1.9^-0.286) = 244kJ/kg c = 0.75 x √2 x 244000J/kg = 658m/s (assumes 75% efficiency, I have no idea what this should be?) ∆T ngv & ∆T turbine = 0.75 x 244000J/kg / 1100 = 164K (again assuming 75% efficiency) Taking the EGT from the Olympus of 700°C (973K) NGV inlet temp = 1301K TIT = 1137K p ngv = (1.9 x 101325Pa) / (1137K x 287J/kg/K) = 0.590kg/m³ Free section area A ngv= 0.45kg/s / (658m/s x 0.59kg/m³) = 0.00116m² A ngv= 0.00116m²/(sin30° x 0.95) = 0.00244m² (95% for blade thickness factor) Actual NGV area = 0.00242m² p turbine = 101325Pa / 973K / 287J/kg/K = 0.363kg/m³ Free section area A turbine = 0.45kg/s / (658m/s x 0.363kg/m³) = 0.00188m² A turbine= 0.00188m²/(sin37° x 0.95) = 0.00348m² Actual turbine area = 0.00279m² The Kamps book has this thing where you find the free section area of the NGV and turbine and then use the blade angles to find the frontal areas. Using that method it seems that the NGV is the correct size but the turbine blades are slightly undersized for the 230N thrust. I may not have these calculations correct. Comments John? Shannon.
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Post by racket on Jan 25, 2018 17:54:41 GMT -5
Hi Shannon
Theres a problem :-(
a 1.9:1 PR will choke a nozzle , thats OK for the NGV , but with the turb wheel theres a "carry over" of "pressure" in the form of the axial velocity component , this then has the effect of reducing the pressure drop required to choke the wheel , the result is a static pressure higher than ambient at the wheel outlet, this static pressure increases the density which allows a greater mass flow , the remaining static pressure is then expanded in the jet nozzle to further increase the gas speeds exiting the wheel .
Thanks for the throat areas , I'll run some numbers .
Cheers John
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Post by madrocketscientist on Jan 25, 2018 18:42:53 GMT -5
I am still hunting through compressor maps... The Holset HX55 seems to be a closer match, but it would need to be machined down in diameter as the exducer is 99mm. The trim is very low at 45. Shannon
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Post by racket on Jan 26, 2018 16:41:34 GMT -5
Hi Shannon
Yep , the HX55 is a closer match .
With regards the larger diameter , this will reduce the rpm for a given PR , .................have you looked into the bearing situation , availability , cost , supply etc ?
When I ran your numbers using my "methods??", things came out OK , the ~1,000 ft/sec of axial velocity entering the wheel is worth a few psi of dynamic pressure , so theres only a need for an extra ~1.7 PR of drop to choke the wheel , so a few psi of static pressure at the wheel exit/jetpipe on top of the axial velocity exiting the wheel .
My thrust level was only ~10% below the Olympus , but then I always calculate conservatively .
Cheers John
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Post by madrocketscientist on Jan 26, 2018 17:52:13 GMT -5
Thanks John, The HX55 is the closest match I can find so far, the wheels are available with 4 inducer sizes (71mm,67mm,65mm,63mm) there are also billet wheels available with much smaller inducer sizes but I think these have a much too small inlet area for a turbine. I made a spreadsheet with the trims for various inducer sizes, the tip height changes with each inducer size too. I have no idea what trim the compressor map was from, I did spend quite a while looking to try to find out. There might be a bit of guesswork to get just the right wheel? For Bearings I have some GRW D6000's, 10x26x8mm full compliment ceramics. I purchased these through Xicoy, not cheap but seem to be the most reliable option for microturbines. From my measurements, these are the same ones that the Olympus uses. I also have some cheap ceramic bearings off ebay that I disassembled and ground the inner race, then reassembled into full compliment bearings. I will use these for setup etc and keep the GRW's for the final fitting. I must be doing something wrong in my calcs if your numbers come out fine.. Shannon.
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ripp
Veteran Member
I'm sorry, I don't speak english, so I torment you (and myself) with a translation program,Sorry
Joined: January 2013
Posts: 231
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Post by ripp on Jan 27, 2018 4:37:14 GMT -5
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Post by madrocketscientist on Jan 27, 2018 4:41:28 GMT -5
Thanks Ralph, That thread is a very helpful resource! I missed the Compressor dimensions on the page, there is a lot to take in! Shannon.
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Post by racket on Jan 27, 2018 19:04:03 GMT -5
Hi Shannon
I've been meaning to ask ................why do you need 2 "small" engines rather than one larger one ??
Cheers John
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