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#15
by
Vincent Waldon
on 18 Jun, 2007 22:23
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Yeah, I don't think pressure is the issue here. If you have a pump that makes 10 psi it will make 10 psi into a pop can or a 55 gallon drum... just a matter of time.
In my mind it's more about flow and resistance.... anything you can do to keep resistance down and flow up is good... bigger tubes, smoother corners.
When I worked for a pipeline company I was amazed to see them replacing *some* of a 36" line with 42"... I asked the engineer what was going on... isn't it like a chain (or a wire) where the weakest link sets the pace ?
Not so, says the engineer, it's all about flow, and any time you can increase diameter you will increase flow... even temporarily helps evidently ! That's why a bigger downpipe is still worthwhile after a restrictive manifold.
As I said.. I still have a lot of research to do... !!!
Vince
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#16
by
BlackTieTD
on 19 Jun, 2007 11:03
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I was wondering with this line of questions if this would mean that the amount of boost someone says they are running is really accurate? 20psi at what volume, or what velocity?
these are excellent points. a ported 1.6 engine will flow more air than a stock 1.6 for instance... and so the peak psi readings (all else equal) should be higher on the stock engine (although the potentially increased exhaust gas volume on the ported engine may spin the turbo faster, and create more boost). a cam with some overlap can cause some boost to go in one valve and right out the other.
Yeah, I don't think pressure is the issue here. If you have a pump that makes 10 psi it will make 10 psi into a pop can or a 55 gallon drum... just a matter of time.
ie: lag
there is definitely a 'sweet spot' where you want your tubing diameter to be. too small is restrictive as mentioned, too large creates pointless lag.
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#17
by
OM617
on 19 Jun, 2007 20:16
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According to Corky Bell, Maximum Boost pg 61, 304 MPH or 0.4 mach is the point at which airflow meets increased resistance (drag) and flow losses are experienced.
I found these flow numbers and thought I would pass them on.
0.4 mach = 304 MPH
2" piping
1.57 x 2 = 3.14 sq in
300 cfm = 156 mph = 0.20 mach
400 cfm = 208 mph = 0.27 mach
500 cfm = 261 mph = 0.34 mach
585 cfm max = 304 mph = 0.40 mach
2.25" piping
3.9740625 sq in = 1.98703125 x 2
300 cfm = 123 mph = 0.16 mach
400 cfm = 164 mph = 0.21 mach
500 cfm = 205 mph = 0.26 mach
600 cfm = 247 mph = 0.32 mach
700 cfm = 288 mph = 0.37 mach
740 cfm max = 304 mph = 0.40 mach
2.5" piping
4.90625 sq in = 2.453125 x 2
300 cfm = 100 mph = 0.13 mach
400 cfm = 133 mph = 0.17 mach
500 cfm = 166 mph = 0.21 mach
600 cfm = 200 mph = 0.26 mach
700 cfm = 233 mph = 0.30 mach
800 cfm = 266 mph = 0.34 mach
900 cfm = 300 mph = 0.39 mach
913 cfm max = 304 mph = 0.40 mach
2.75" piping
5.9365625 sq in = 2.96828125 x 2
300 cfm = 82 mph = 0.10 mach
400 cfm = 110 mph = 0.14 mach
500 cfm = 137 mph = 0.17 mach
600 cfm = 165 mph = 0.21 mach
700 cfm = 192 mph = 0.25 mach
800 cfm = 220 mph = 0.28 mach
900 cfm = 248 mph = 0.32 mach
1000 cfm = 275 mph = 0.36 mach
1100 cfm max = 303 mph = 0.40 mach
3.0" piping
7.065 sq in = 3.5325 x 2
300 cfm = 69 mph = 0.09 mach
400 cfm = 92 mph = 0.12 mach
500 cfm = 115 mph = 0.15 mach
600 cfm = 138 mph = 0.18 mach
700 cfm = 162 mph = 0.21 mach
800 cfm = 185 mph = 0.24 mach
900 cfm = 208 mph = 0.27 mach
1000 cfm = 231 mph = 0.30 mach
1100 cfm = 254 cfm = 0.33 mach
1200 cfm = 277 mph = 0.36 mach
1300 cfm max= 301 mph = 0.39 mach
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#18
by
Vincent Waldon
on 19 Jun, 2007 20:55
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Now that's research... thanks for the numbers.
Next question will be... how many CFM does a Garrett T3 pump out at 25 psi on a standard temp and pressure day ??!!
Vince
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#19
by
subsonic
on 19 Jun, 2007 21:04
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Wow, nice bit of info! You should copy that over to the FAQ section for sure.
With those numbers you can see some pretty dramatic velocity drops as the Dia. increases if you keep the same cfm flow rate.
So going to a 3" IC pipe could really increase lag time on the turbo.
When you are doing the math to figure out turbo maps/ surge/ etc.. do people look at the IC loop and factor that in as well? It seams like it could make a big difference.
I am guessing that you will want the flow of air in the turbo-IC-manifold loop to be moving as fast as possible ( up to just under 304mph) to reduce lag. How the hell do you figure out what your cfm requirements will be?
Could you increase velocity right before it enters the manifold by having a slight neck down in the pipe right as it enters the manifold?
Could you figure out a real, corrected boost measurment with this info? Kind of like recalibrating the speedo for larger tires?
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#20
by
burn_your_money
on 19 Jun, 2007 21:08
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How the hell do you figure out what your cfm requirements will be?
RPM X displacement X PSI?
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#21
by
subsonic
on 19 Jun, 2007 21:24
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Ummmm...I don't know :?
Could you run a turbo into various dia. pipes and into a flow meter to get info as to what the tubo is actually putting out for cfm at various levels of psi input? OR..perhaps we could have one of our physics friendly contributers whip up a formula to help solve this?.
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#22
by
bevboyy
on 20 Jun, 2007 06:26
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I never found it of any real benefit to use a larger diameter than what is the inlet of your intercooler and well as the outlet. The largest outlet I have seen is 2.25" off my TC intercooler. I have always used OEM intercoolers of various makes, bu then I am also keen to keep my projects cost down to a reasonable level..
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#23
by
lord_verminaard
on 20 Jun, 2007 09:34
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I think you're pretty close there. From what I see it, and the way I read it in "maximum boost" you can figure out CFM this way:
First, figure out the pressure ratio you are running according to how much boost you are using. For example, say you are running 8 psi:
(14.7+8)/14.7 = PR of 1.544
14.7 being 1 atmosphere, or 1 bar.
Now, you can figure out CFM with this:
(cidxrpmx0.5xVE)/1728
Which: cid= cubic inches of displacement, rpm= maximum rpm, 0.5 is the 4-stroke engine cycle only firing once per 2 revolutions, and VE is volumetric efficiency of an engine in percentage. The 1728 converts cubic inches to cubic feet.
So, (using my Jeep's 4.0 since I just did this on paper the other day for fun) we have:
(244x5000x0.5x0.8)/1728 = 282.4 CFM.
So, say I wanted to turbo my jeep's I-6 with 8 psi, I simply multiply 282.4 CFM by the PR that I figured earlier:
282.4x1.544 = 436.03 total CFM. You can divide this by 14.47 to convert to lbs/min which is what a lot of companies use on their compressor maps.
Hope this helps!
Brendan
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#24
by
larry104
on 20 Jun, 2007 09:45
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I think you're pretty close there. From what I see it, and the way I read it in "maximum boost" you can figure out CFM this way:
First, figure out the pressure ratio you are running according to how much boost you are using. For example, say you are running 8 psi:
(14.7+8)/14.7 = PR of 1.544
14.7 being 1 atmosphere, or 1 bar.
Now, you can figure out CFM with this:
(cidxrpmx0.5xVE)/1728
Which: cid= cubic inches of displacement, rpm= maximum rpm, 0.5 is the 4-stroke engine cycle only firing once per 2 revolutions, and VE is volumetric efficiency of an engine in percentage. The 1728 converts cubic inches to cubic feet.
So, (using my Jeep's 4.0 since I just did this on paper the other day for fun) we have:
(244x5000x0.5x0.
/1728 = 282.4 CFM.
So, say I wanted to turbo my jeep's I-6 with 8 psi, I simply multiply 282.4 CFM by the PR that I figured earlier:
282.4x1.544 = 436.03 total CFM. You can divide this by 14.47 to convert to lbs/min which is what a lot of companies use on their compressor maps.
Hope this helps!
Brendan
Yep, that is correct. And if you do the numbers for a 1.6 L (98 cu. in. diesel) at 4,500 rpm, Flow rate = 115 CFM. Say you run 15 psi boost. Then, PR = 2.0. Total CFM = 230.
I ran 2 in. piping to my GTD intercooler, which is the factory size. The intake inlet necks down to 1.5 in. so I used a transition adapter. I have no noticable lag.
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#25
by
veeman
on 20 Jun, 2007 13:12
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One reason people use silicone is that it handles heat very well. If you were running a lot of boost, the air might get too hot for rubber tube or connectors.
Don't forget that intercooler piping often needs some degree of flex to allow for movement with the engine to a certain extent.
Lots of people use "hump" hoses to allow the piping to move a bit when the engine is shifting back and forth.
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#26
by
Mark(The Miser)UK
on 20 Jun, 2007 18:38
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A summary of the engine revs required to shift 300CFM cubic feet of air (atmospheric equivalent) per minute at the following turbo pressures:
0psi requires 10700rpm
8psi requires 6900rpm
12psi requires 5900rpm
16psi requires 5120rpm
20psi requires 4530rpm[/i]
Worked example for 0; & 8psi
2" pipe cross sectional area is 3.14" sq this is 156mph
300cfm
300 x12^3 =518400" [ cubic inches !]/min
518400 x 2.54^3 =8,495,054cc/min
8495054/(1588/2) =10,700rpm [4 stroke, STP]
@8psi... (14.7 +8.)/14.7 = 1.54 compression ratio
10700/1.54 = 6929rpm
So 2" pipe 156mph flow STP [300cfm STP] is 6929rpm engine
E&oE :wink:
Does this seem about right-ISH Someone thumb through this stuff :shock:
Seems most of the flow rates are unattainable.
Don't understand why drag only increases at the highest speeds...Surely Reynolds #'s come in to play way before these speeds and create turbulence. Intercoolers require turbulence to work efficiently though.
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#27
by
Tintin
on 22 Jun, 2007 19:56
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