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  #85  
Old 05-29-2010, 12:30 AM
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Re: Homemade CAI for 3.7?

Quote:
Originally Posted by GCL06 View Post
Working on the math as to whether or not the Venturi Effect makes up for the air loss due to the bottleneck effect. Just posting as I go and I'll edit to finish it up (assuming I can work through all the variables). I will be basing calculations on the Spectre P4 Cone Filter (Item#8132) at sea-level at 20C.


Bernoulli's Principle states that the faster a fluid is moving, the lower the pressure. Therefore, fluid in areas of higher pressure will move towards the lower pressure of the faster moving fluid, causing an increase in fluid velocity.

The standard density of air (p) at 20C is 1.2kg/m

Bernoulli's Principle Formula - (p₁-p₂) = [(p/2)(v₂-v₁)] where p=density of the fluid, v₁ is the slower velocity (before bottleneck), and v₂ is the faster velocity (into throttle body).

Replacing p with d to differentiate between pressure and density. p₀ becomes the pressure of 1 atmosphere.
(p₁-p₂) = [(d/2)(v₂-v₁)]

At the surface of the earth air pressure is equal to 1.01x10⁵ N/m

p₀ = 101000
d = 1.2
p₁ = [(101000-p₁) = (.6)(v₁-0)]
.......v₁ = velocity of incoming air at WOT
p₂ =
v₀ = 0
v₁ =
v₂ =


Difference In Pressure
(p₁-p₂) = (1.2/2)(v₂-v₁)
(p₁-p₂) = (.6)(v₂-v₁)


The formula for the Venturi Effect is - Q = (A₁)√{[(2)(p₁-p₂)]/[(p)(A₁/A₂)-1]} = (A₂)√{[(2)(p₁-p₂)]/[(p)(1-(A₂/A₁)]}
where A₁ is the area of the pipe before the bottleneck and A₂ is the area entering the throttle body, and Q is the volumetric flow rate.

A 3" PVC pipe (3.5" including sidewalls) is already a squeeze under the hood so I'll use that over a 4" pipe for the equations.


Non-Bottlenecked
Q₁ = (A₀)√{[(2)(p₀-p₁)]/[(d)(A₀/A₁)-1]} = (A₁)√{[(2)(p₀-p₁)]/[(d)(1-(A₁/A₀)]}

Q₁ = (A₀)√{[(2)(101000-p₁)]/[(1.2)(A₀/28.26)-1]} = (28.26)√{[(2)(101000-p₁)]/[(1.2)(1-(28.26/A₀)]}

A₀ = Area of the filter x % of air flowing through filter


Bottlenecked
Q₂ will represent the amount of air entering the throttle body using the bottleneck.

A₁ = 28.26
A₂ = 12.56
d = 1.2
p₁ =
p₂ =

Q₂ = (A₁)√{[(2)(p₁-p₂)]/[(d)(A₁/A₂)-1]} = (A₂)√{[(2)(p₁-p₂)]/[(d)(1-(A₂/A₁)]}

Q₂ = (28.26)√{[(2)(p₁-p₂)]/[(d)(28.26/12.56)-1]} = (12.56)√{[(2)(p₁-p₂)]/[(d)(1-(12.56/28.26)]}

Q₂ = (28.26)√{[(2)(p₁-p₂)]/[(d)(4.0625)]} = (12.56)√{[(2)(p₁-p₂)]/[(d)(.8025)]}

Q₂ = (28.26)√{[(2)(p₁-p₂)]/[(1.2)(4.0625)]} = (12.56)√{[(2)(p₁-p₂)]/[(1.2)(.8025)]}

Q₂ = (28.26)√{[(2)(p₁-p₂)]/(4.875)} = (12.56)√{[(2)(p₁-p₂)]/(.963)}
you lost me at working on the math.....
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  #86  
Old 05-29-2010, 04:13 AM
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Re: Homemade CAI for 3.7?

Eventhought mine looks nice I like Pretacos idea more.

The aluminum is a big heat exchanger.

I got the tubing from a Dodge Ram 3.7 kit..i plan on a write up soon.
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  #87  
Old 05-29-2010, 04:14 AM
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Re: Homemade CAI for 3.7?

1+1=2
dog = mans best freind
women = trouble
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Old 05-29-2010, 04:48 AM
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Re: Homemade CAI for 3.7?

Quote:
Originally Posted by GCL06 View Post
Sorry to hear =/ At least PVC's dirt cheap!

until it implodes and blows into your intake manifold........
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  #89  
Old 05-29-2010, 01:59 PM
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Re: Homemade CAI for 3.7?

^and that would suck
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Old 05-29-2010, 09:56 PM
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Re: Homemade CAI for 3.7?

You'd be hard pressed to do that...
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Old 05-29-2010, 11:32 PM
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Re: Homemade CAI for 3.7?

^^^^I was speaking of the PVC intakes shown here. PVC isnt made for any kind of heat and will become brittle over time....
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Old 05-29-2010, 11:44 PM
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Re: Homemade CAI for 3.7?

anyone else have pics of their setups?

misfit yours looks clean
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