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Post by racket on Jun 30, 2016 20:26:52 GMT -5
Hi Ryan
A "freepower" compliant gas producer wheel will easily supply the same thrust as a "pure thrust" type wheel , the slight difference isn't worth worrying about , I'll try and have a compromise design.
Cheers John
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greazy
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Post by greazy on Jun 30, 2016 20:34:46 GMT -5
well a freepower compliant wheel would be just fine if it still whacks out some ok thrust 👍
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Post by racket on Jun 30, 2016 21:38:46 GMT -5
Hi Ryan
Latest rough calcs for a 6 inch wheel ( 152.4mm)
You'll need the NGV angle set at 30 degrees , turb wheel outlet angle 37.5 degrees , blade and NGV length 1.07" - 27.2 mm , NGV throat area 7.7 sq inches , turbine outlet throat area 10.7 sq inches.
Exit axial gas velocity should be ~1,000 ft/sec with ~3.7 psi static pressure in the jetpipe ,requiring a jet nozzle to expand that 3.7 psi to ambient , the 1,000 ft/sec would be an acceptable velocity for entering a freepower wheel .
Hope this helps
Cheers John
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greazy
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Post by greazy on Jul 1, 2016 8:39:57 GMT -5
yea of course it helps thanks. im interested to know how you came to this conclusion? do you only expand enough enpalthy for the compressor wheel to operate or do you add a bit more for insurance and do you aim for a specific "outflow speed and leftover pr" or is that juat a result of expanding less enpalthy.
cheers mate
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Post by racket on Jul 1, 2016 19:59:52 GMT -5
Hi Ryan LOL.............I'll try and explain the process First off I adjust your P2 pressure ratio from 3.36 to 3.2 for the 5% losses across the flametube , 5% is a reasonable number , could be less , shouldn't be more otherwise the FT design needs changing . I used a 75% comp effic for your 3.36 PR and got a temp rise of 159 C degrees, this then requires a drop of 137 degrees through the turbine ( 159 X 0.24 div by 0.28) We then need to find the temp equivalent of the 1,000 ft/sec exhaust velocity from the turb wheel , 1000 X 1000 div by 64.4 div by 1400 div by 0.28 = 40 C degrees This 40 degrees is added to the 137 degrees required by the comp for a total temp drop across the turbine wheel of 177 C degrees , this then needs to be divided by the turb effic of say 80% for a temp of 221 C degrees . Now after some more calcs, that requires a PR of 2.54 across the stage , your ingoing 3.2 PR = 47 psia , divide by 2.54 = 18.5 psia out of the wheel , take off our 14.7 leaves us with 3.8 psi static gauge pressure Density of the gases is then worked out and the cubic feet per second is found for the mass flow of 3.6 lbs/sec , divided by our 1000 ft/sec and an annulus flow area of ~16.5 sq inches is found , its then an easy calc to find the turb disc area simply by taking the 16.5 sq ins from the 28.3 sq in area of the 6" dia wheel Now to get the actual gas angles I need to find the gas deflection in feet per second corresponding with the 137 degrees of energy extracted by the turbine wheel to power the comp , we need to add on say another 10% for "mechanical" losses for the wheel , so lets use a total of 150 C degrees . Mean blade speed is also part of the equation , lets use 5" dia as the mean if blades 1" long , 5" by pi div 12 X 1000 rps =1300 ft/sec . The equation to find the gas deflection is 150 degrees X 32.2 X 1400 X 0.28 divided by our 1300 ft/sec , this comes out at 1456 ft/sec . We then do the velocity diagrams , draw the 1456 line horizontally to scale , marking off the 1300 ft/sec point on that line . If we need an overall 2.54 PR across the stage , we sq root that for a figure of 1.6 PR in the NGV and 1.6 in the wheel , this 1.6 PR gives a velocity of ~1,600 ft/sec . As we want axial gas flow out of the wheel we draw a 1,000 ft/sec line vertically down from the left hand end of the 1456 ft/sec line , our 1,600 ft/sec line is taken from the right hand end of the 1456 ft/sec line at the 1,300 ft/sec point and should come close to meeting the end of the 1000 ft/sec line , its a tad out but don't worry about that , its close enough at this stage . You'll find the angle at the 1,300 ft/sec point is ~37.5 degrees , this is our ouflow angle on the turb blades. The NGV outflow angle is determined by drawing our 1,600 ft/sec line starting from the left hand end of the 1456 line up at an angle so that its other end finishes vertically above the other end of the 1456 ft.sec line , the vertical line should be 800 ft/sec and the angle 30 degrees . Heh heh , clear as mud Could you please PM as I'd like to contribute something else . Cheers John
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greazy
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Post by greazy on Aug 27, 2016 7:54:51 GMT -5
hey john
I have had a good read of the gas turbine theory book and to be honest im getting overwhelmed by it. I need to read and re read it for it to sink in and that may take a while. so im asking you for help im ready to machine the turbine blisk i have sussed out how im going to do it at home just need to confirm a design. these are the specs i plan on using
rpm 65000 PR out of compressor 3.5 comp efficency 78% throughput 1.7kg/s (3.74lb) turbine design inlet temp 1050K the max O.D of blisk is 151mm after squaring up the stock im thinking 23 blades for the turbine 12/13mm bore hole
using your last post with a few alterations most results seem to match by chance ...
pressure ratio starts at 3.5 after 5% loss = 3.325 i was working on a 78% comp efficency ( you chose 75% should i be using a lower efficency?) gives temp rise of 160°c drop requiring a drop through the turbine of = 137°c
temp for 1000 f/s out of the turbine (an acceptable inlet speed for a freepower) is 40°c
40 + 137 = 221 @ 80% efficency thats the total temp drop required through the turbine
this needs 2.6pr drop leaving 3.8psi static guage pressure!
here is where im stuck. is density calculated on the 1050K before entry to the turbine or 829K the leftover after expanding the required 221K
density for 1050K @ 3.74lb = 178 cu/f density for 829K @ 3.74lb = 140.6 cu/f
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Post by racket on Aug 27, 2016 17:25:15 GMT -5
Hi Ryan
With an overall PR of 2.6 , we square root it for a PR of ~1.61 in the NGV and another 1.61 PR in the turb wheel .
At NGV throat the PR is ~2.065 or 30.35 psia and at the turb outlet its ~1.28 or 18.8 psia
Our temperature drops are ~110 C degrees in each ( 221 total drop) , so the NGV throat temp is 1050 - 110 = 940 K
Density at NGV throat is 144 X 30.35 psia div by 96 X 940 K = 0.0484 lb/cu ft or 20.64 cu ft / lb , multiply by 3.74 lbs/sec = 77.2 CFS , with our gas velocity of ~1660 ft/sec we need an area of ~0.0465 sq ft or ~6.7 sq inches , add on 10% for boundary layer and we need ~7.4 sq inches of throat area .
At the wheel exhaust throat , temp is 1050 minus 221 = ~830K , pressure is 18.8 psia , so density is 144 X 18.8 psia divided by 96 X 830 K = 0.0339 lb/cu ft or 29.4 cu ft /lb , X 3.74 = ~110 CFS , divide by our ~1660 ft/sec = 0.066 sq ft or ~9.54 sq ins , add on our 10% gives us a throat area of ~10.5 sq ins .
Keep those questions coming :-)
Cheers John
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greazy
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Post by greazy on Aug 27, 2016 17:40:13 GMT -5
just to clarify the throat is the 2 dimensional area at the exit of its respective stage. ?
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Post by racket on Aug 27, 2016 17:55:00 GMT -5
Yep , the sum of all the individual throats , so allowances then need to be made for vane or blade thicknesses at that exit point to design the actual annulus area and vane/blade angle.
The vane/blade angles also need to produce the required gas deflection to satisfy the energy/horsepower requirement of the compressor wheel and other losses.
Cheers John
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greazy
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Post by greazy on Aug 27, 2016 18:16:12 GMT -5
would it be safe to assume that i could up the combustion temp to 1100K since the heat drops by 110 degrees through the ngv it would give about 717°c turbine inlet temp. the material optimum range is 700 to 750°c
or is it just me getting cocky and trying to extract too much from the turbine
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Post by racket on Aug 27, 2016 18:39:41 GMT -5
What are you using for material
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greazy
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Post by greazy on Aug 27, 2016 19:11:08 GMT -5
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greazy
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Post by greazy on Aug 27, 2016 19:12:44 GMT -5
type udimet in Google and the first link is the pdf with info of the material the forum wont let me post the link
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Post by racket on Aug 27, 2016 19:38:01 GMT -5
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greazy
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Post by greazy on Aug 27, 2016 20:01:43 GMT -5
ok ill re hash my calculations tonight with 1150k
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