I was playing around with some numbers for FI applications and it seems incredible.
An engine spinning at 6000rpm will have 100 revolutions per second, 50 comprssion strokes and 50 exhaust/refill strokes. The engine is a 348 cu in motor which is roughly 21.75 cu in per cylinder. On a NA engine that breathes at Atmosphere it will ingest 7.6 cu ft of air per second. However, with boost that increases slightly. CHANGED: So the enigine is ingesting the equivalent of 10.184 ft3/s
Question:
Can a 2.5" charge pipe efficiently flow 7.6 cu ft of "compressed" air (or 10.184 cu ft of uncompressed air) per second (@ 5 psi), to keep up with what the engine is ingesting @ 6000 rpm?

Who amongst us can compute flow dynamic of air in a pipe? Bong pipes excluded

I can't wait for business to pick up again.

 

I do not believe the 5 psi of boost is a five times multiplication of atmospheric pressure. Isn't the 5 psi of boost equal to the 14.696 psi of atmosphere PLUS the additional 5 psi, which result in a total of 19.696 psi?

 

Gee, that would have been good information years ago when I use to smoke out of my bong.

 

try (14.7 + 5)/14.7=1.34 atm

 

Okay so the equivalent air volume with 5 psi of boost is 10.184 ft3/sec?

 

Yes roughly you are looking at N/A comsumption * pressure ratio = boosted consumption.

There are of course other factors which some will point.. temperature, etc.

The pressure ratio is what RJW showed you above.

I have put together a spreadsheet for all of this so calculate compression ratio and consumption when I was going to source my own turbos.

 

Known Variables

CID = Engine Displacement
RPM = RPM
Volumetric Efficiency (naturally aspirating) = VE - in %
Engine Air Flow Requirements (naturally aspirating) = ((CID*RPM)/3456)*(VE/100) - in CFM
Air mass conversion (cubic foot to mass) = CF1
CF1 = 0.075-0.069 (temperature based)
Mass flow calculation Min. Qm1 = CFM*CF1 – in lbs/min

Pressure Ratio

Target Boost (PSI) TB = 5,10 what ever desired in PSI
Inlet Pressure / Atmospheric Pressure P1 = 14.6960
Absolute Boost Pressure P2 = P1+TB
Pressure Ratio PR1 = (P1+TB)/P1 – in x.xx:1

Temperature Rise Ideal

Inlet Temperature T1 = your inlet temp in F
Temperature Conversion to ºRankine CF2 = 459.69
Outlet Temperature T2 = ((T1+CF2)*(PR^0.283))-CF2
Temperature Rise T3 = T2 - T1 – in degrees F

Temperature Rise Estimate Based on Adiabatic Efficiency

Adiabatic Efficiency AE = blower efficiency in % 58-72?????
Estimated Outlet Temperature Rise (adiabatic) T4 = T3/(AE/100) – in degrees F
Estimated Total Outlet Temperature T5 = T1+T4 – in degrees F

Heat Exchanger

System Heat Exchanger Efficiency HXE = - in %
Heat Exchanger Discharge Temperature HXT = ((T5-T1)*((100-HXE)/100)))+T1 – in degrees F
Heat Exchanger Pressure Drop PD = 0.5, 1 guess also include piping loses in PSI
Heat Exchanger Discharge Pressure Ratio PR2 = (P2-PD)/P1

Density Ratios

Compressor Outlet Air Density Ratio DR1 = ((T1+CF2)/(T5+CF2))*PR1
Charge Cooled Outlet Air Density Ratio DR2 = ((T1+CF2)/(HXT+CF2))*PR2

Compressor Inlet Airflow

Compressor Inlet Airflow Estimate I_CFM1 = CFM*DR1 – in CFM
Compressor Inlet Mass Flow Min. Qm2 = I_CFM*CF1 – in lbs/min
Charge Cooled Compressor Inlet Airflow Estimate I_CFM2 = CFM*DR2 – in CFM
Charge Cooled Compressor Inlet Mass Flow Min. Qm3 = I_CFM*CF1 – in lbs/min

These are generic and simplified calculations but should get you into ballpark.

To calculate inlet pipe velocity, take your FT^3 or appropriate CFM and divide by cross-sectional area in FT^2.
TYPO^^^^ corrected sorry then divide by 60 to get ft/sec

As far as restriction, it’s more of a pressure drop do to frictional/heating losses. Those calculations are allot more involved but I may be able to post simplified version latter. A general rule is stay below 250 ft/sec (this is my opinion). On inlet side 200 ft/sec would be better.

I’m Slower

Mike

EDIT--------------------------------------------------------------------------------------------------------------------------
PM me with email and I can send excel spreadsheet.

 

I think you want to redo your math. The amount of air consumed @ 100% fill @6000 RPM's is 10.0 cu' sec @ 1 atm, or X 1.34 for 5Psi boost I had to correct this also I multiplied by 1.34 twice

 

348 cu"
6000 rpm
100 rev/s
50 rev/s filling

50x348=17,400 in3 / 144 in = 120.833 ft3/s

120.833 x 1.34 (5psi)= 161.917 ft3/s

Dinner date has come. I'll finish this thought later.

 

Look at www.forcedinductions.com/consumption346.htm and you will see my calcs will be very close that is why I included lbs/min. Also with supercharger VE is typically 100%, with turbo it depends on backpressure.


Mike

 

Alright, so the engine is consuming 170 ft3/second of air at 6000 rpm. Can a 2.5" charge pipe supply that much air at 5-6 psi or would it need to be bigger?

 

Methinks that your math is still off base...

170 cu.ft per second= 60 x 170 or 10200 cubic feet per minute.

Think about that?

A sewer pipe can flow 10200 CFM...

 

I think I've got that part correct. An engine spins 6000 rpm filling only hlf the time. The engine displacement is ~350 cu in. That 50 complete fills per sec at 350 cu in or 17,500 cu in of air per second times 1.34 because it is pressurized to 5 psi under boost. Extrapolated to cubic ft just divide by 144 gives you 162.84 cubic feet of air per second. Remember that liquid flows slower than air and that same sewer pipe you speak of could flow a heck of a lot more air than it could water and sludge.

 

To convert cubic inches to cubic feet = divide by 1728...hello?

I don't mean to be an a$$ but...

Keep plugging away, you'll get it...

 

I forgot a factor of 12.

dejvue, I think I've got that part correct. An engine spins 6000 rpm filling only hlf the time. The engine displacement is ~350 cu in. That 50 complete fills per sec at 350 cu in or 17,500 cu in of air per second times 1.34 because it is pressurized to 5 psi under boost. Extrapolated to cubic ft just divide by 1728 gives you 10.13 cubic feet of air per second.

Better !

 

I ran some numbers for you. Using your above values, LS1 ~346 CID at 6000 RPM with good breathing engine at 5 PSI will likely consume 725 CFM or 53.4 lbs/min. This assumes good IC and efficient layout. Also it is widely accepted that 10.5-10.7 is a good mass flow to FWHP multiplier.

With above you can expect around 560 FWHP at 5 pounds of boost (53.4*10.5).

The restriction you are asking about if it is on inlet then for supercharger it is 725 CFM, on twin turbo divide by 2 or 362.5 CFM (ft^3).

2.5 inch ID pipe = 4.9087 in^2 or 0.0341 ft^2

So single inlet pipe velocity is;
(725/0.0341)/60 or 354 ft/sec (not good)

TT inlet pipe velocity is;
(362.5/0.0341)/60 or 177 ft/sec (much better)


Losses (pressure drop) per linear foot (ft) of pipe based on velocity

100 ft/sec loss PSI = 0.0036
150 ft/sec loss PSI = 0.0098
200 ft/sec loss PSI = 0.0203
250 ft/sec loss PSI = 0.0358
300 ft/sec loss PSI = 0.0570
350 ft/sec loss PSI = 0.0845
400 ft/sec loss PSI = 0.1189

*** Note this assumes laminar non turbulent flow ****
To estimate total losses add up linear length and for each turn multiply (1-2) ft worth of pressure drop.

As I mentioned in above post on inlet you want to stay below 200 ft/sec and on charge air (compressed) you want to stay below 250 ft/sec.

Another way to look at it is on NA LS1 (stock cubes) with 78mm TB engine can consume 570 CFM (around stock).

So it looks something like this;
TB = 7.4065 in^2 or 0.0514 ft^2
(570/0.0514)/60 or 185 ft/sec

If you are looking at 2.5 inch inlet piping for TT your are fine as you will be flowing slower (less losses which is restriction) then stock C5 across TB.

Your charge air density ratio should be around 1.2707 (this is pressure ratio factoring in temperature rise after IC). Your approximate CFM to engine looks something like this;

725/1.2707 or 570 CFM (inlet CFM divided by density ratio)

Hope this helps.

Mike

 

A cubic ft (ft^3) in inches = 12*12*12 or 1728.

Because we own 4-stroke not 2-stroke 1728*2 = 3456, this has same effect as dividing RPM by 2.

I wouldn’t mislead you.


Mike

 

Good work Mike. You may have answered my question in a round about way. The TT I'm referring to is the TTi stage 1. Look at this photo and tell me that 2.5" intake is sufficient to flow enough air at 5000-6000 rpm. That's the point the dyno starts to look weak. I know, there are isssues with the exhaust flange being too small on the T2 manifold but what if...?
1) the back pressure from the intake stalling/slowing the turbine and preventing the exhaust from exiting?



There are two 2" pipes feeding into a Y pipe with a 2.5" (probably a 2.38 ID) common feed. Based on square area alone I find a 19% reduction and at a minimum a 2.83" ID pipe should have been used but why not 3.5" to equal the MAF and reduce the speed and turbulence.
In your explaination, this scenario is accellerating the air well above 250CFM and making it less efficient. It may good for more HP to enlarge that pipe and remove the cats to free up any un-needed back pressure.

Merry Christmas

 

Well done Mike,

The only thing I would add (which was allready commented on) is thats assuming 100% VE. The LSX engines are typically in the 87-90% range. This will give you a bit of headroom on the max consumption of the motor. Typical NA consumption for 346 would be 36 lb/min and 48 lb/min at 5 psi boost.

Happy Holidays.

Phil

 

I’ll give this another try. First I use slightly different more complicated spreadsheets I built and MathCad routines, so my results are slightly different then simplified calculations I posted above. I’m better with superchargers because backpressure with turbos is harder to predict (at least for me, need more practice) and it’s effect on mass flow.

I assumed you had an intercooler which increases density ratio (colder denser air) which will result in more mass flow being ingested by engine.

As mentioned above inlet side piping is fine. On inlet side air separation inside pipe while turning is concern, but generally acceptable below 200 ft/sec. Don’t get me wrong slower is better.

Onto your discharge side, I’m just going to use 2 inch and 2.5 inch as your ID because I don’t know thickness. First compressor efficiency directly effects pipe velocity, hotter less dense air travels faster then cooler (mass flow stays the same but, CFM goes up with heat).

As a guess assuming 68% compressor wheel efficiency total flow maybe in 655-680 CFM range. Divide by 2, so lets call it 340 CFM a side.

2 inch pipe velocity;
3.1416 in^2 or 0.0218 ft^3
(340/0.0218)/60 or 260 ft/sec
Not too bad, run is short and losses reasonable. Only bad bend is turbo scroll discharge, but fitment dictates it.

2.5 inch pipe velocity;
4.9087 in^2 or 0.0341 ft^2
(680/0.0341)/60 or 332 ft/sec

So yes you do have a velocity spike but not that bad because it’s short. Then you have a “dump diffusion” loss (2-7% as a guess, would have to calculate). This is where air suddenly dumps into larger volume (it separates and generates vortices). You don’t have the room taper or flare pipe to MAF, the taper has to be a shallow angle to prevent separation.

I don’t think you can improve on it enough to justify cost and trouble. You may be able to recover 0.1 - 0.2 PSI, hard to say maybe more. If you turn up boost it will help more.

Ideally if you do want to do this, then dump into 3.5 inch pipe that connects to MAF. Reasoning is your dump losses are going to be lower do to lower inlet velocity and you have more room to establish laminar flow across MAF.


Mike