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technical minded flow map question

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Old Jul 8, 2008 | 11:10 AM
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technical minded flow map question

im taking a s4 n/a motor with a full bridge port and putting twin turbos on it the turbos are garrett T3 50 trims i plan on trying to push them to 15 lbs( i know they are kinda small) im having trouble with the flow map i was wondering if anyone could help. im useing the following formulas. im not a math major or anything like that so plz feel free to correct me if i am wrong

Engine Air Flow Requirements
(CID × RPM x VE%) ÷3456 = CFM

(80 × 7500 x 0.85) ÷ 3456 = 147.56 CFM (assuming 85 % VE)

Pressure Ratio
(15 + 14.7) ÷ 14.7 = 2.020 pressure ratio

Temperature Rise
T2 = T1 (P2 ÷ P1)to the .283 power

Where:
T2 = Outlet Temperature °R
T1 = Inlet Temperature °R
°R = °F + 460
P1 = Inlet Pressure Absolute
P2 = Outlet Pressure Absolute

assuming 75 degree inlet temp T1 = 75 + 460 = 535°R

P2 = 14.7 + 15 = 29.7 psi

pressure will be the outlet pressure of the turbo
T2 = 535 (29.7 ÷ 14.7)0.283 = 652.81°R

convert from degree R to degree F
652.81-460=192.81

which would give me a ideal temp rise of 117.81 degrees F

Adiabatic Efficiency
IOTR ÷ AE = AOTR

Where:
IOTR = Ideal Outlet Temperature Rise
AE = Adiabatic Efficiency
AOTR = Actual Outlet Temperature Rise

assumeing 70% adiabatic efficiency
117.81 ÷ 0.7 = 168.3 °F Actual Outlet Temperature Rise

now i add the temp rise to the inlet temp
75 + 168.3 = 243.3 °F Actual Outlet Temperature

Density Ratio
(Inlet °R ÷ Outlet °R) × (Outlet Pressure ÷ Inlet
Pressure) = Density Ratio

(535 ÷ 652.81) × (29.7 ÷ 14.7) = 1.655 Density Ratio

Compressor Inlet Airflow
Outlet CFM × Density Ratio = Actual Inlet CFM

147.56 CFM × 1.655= 244.211 CFM Inlet Air Flow

convert to lbs/min by

244.211 CFM × 0.069 = 16.85 lbs/min

no using all the information above considering that im useing 2 turbos i cut the flow in half to 8.425lbs/min
when i map it on the map below it seems like one turbo this size would be more efficent then two, or that two turbos will put me in the surge limit



any help with this would be great
thanks
-joe
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Old Jul 8, 2008 | 11:45 PM
  #2  
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Originally Posted by fast87t2
Engine Air Flow Requirements
(CID × RPM x VE%) ÷3456 = CFM

(80 × 7500 x 0.85) ÷ 3456 = 147.56 CFM (assuming 85 % VE)
The 80CID displacement is just Mazda's marketing - the 13B breathes like a 160CID boinger. You need to DOUBLE your result to work with the 3456 (i.e. 2 cubic feet) factor, since the rotary pumps its displacement every revolution.

Sorry, but I stopped reading at this point...
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Old Jul 9, 2008 | 07:12 AM
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so considering that i am useing two turbos i would have to cut the 160cid in half which is 80cid what the above calc are set up for, correct? which would set me almost dead nuts in the 75% island.
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Old Jul 9, 2008 | 08:16 AM
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wouldn't you use 160cid and then half your final lbs/min, like you already have ?

(you halved your cid at the start and again at the end, so ending up with a lesser figure.)
I dont know, just the way I read it

Also you are bridge ported so you will flow more then calced anyway.

Why not do a search and see what other people have done quite a few twin setup about.
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Old Jul 9, 2008 | 06:38 PM
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Feel free to get detailed in your calcs, but there's an easy way to check that you're in the ballpark: expected RWHP divided by 7.5 = lbs/min of air. So your 17 lbs/min turbos will give about 250rwhp on a 13b, which sounds like a bad idea for your bridgeport IMO.
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Old Jul 9, 2008 | 07:18 PM
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according to the map each turbo will make 27 lbs/min max.

27*2=54/1.3=41.5*10= 415 max rear wheel rotary hp.

or 54*14.471= 781 cfm/1.92= 406 max rear wheel rotary hp.

take your pick.

as to the turbos, being T3s... you will need either the .63 or .82 hotside housings w a stage 3 turbine wheel or you will split the housings due to backpressure and heat. do not run the .48 housings.

i currently have over 10,000 miles on my twin setup.
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Old Jul 10, 2008 | 06:01 AM
  #7  
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yeah i was trying to find a .83 hotside, but thanks for all the infomation its has deffinitally helped

--joe
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