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Originally Posted by instarx
Warning - more SP, VP, temperature and velocity stuff.
That's twice now you have condescendingly told me to either "read up on" the issue or to use a computer program so I would understand what is going on. I didn't go to grad school for nothing and I understand the underlying principles just fine thank you. (Yes, it's been 15 years since I used any of it, so I made a few thoughtless errors earlier, but I do know absolutely that the velocity of air in a duct does not change the static pressure in that duct - but it is not intuitively obvious that it does not so I understand your confusion).
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Lets see what nasa has to say on the subject (See attached). Oops, static pressure does change.
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NO, NO, NO! It does not demonstrate that at all - that contraption only demonstrates the Bernoulli effect. Also, you are thinking "I know the velocity is fast in certain areas, and the pressure is lower there, so velocity lowers pressure" - but that is false logic.
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Nope, my logic is totally valid
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Easily. You have to abandon the idea that the device shown is measuring static pressure - it isn't. It is measuring the relative strengths of the Bernoulli effect at each pressure tap. The last manometer has a lower liquid level (higher pressure) because the velocity of the air moving across its pressure tap is lower than the velocity of the air moving across the tap in the restricted section. I have no idea what the relative static pressure values are in that tube from looking at the manometers, and neither does anyone else.
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Nasa and I do
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But enough of this VP versus SP versus Bernoulli artifacts stuff - here is the answer to the original issue:
The gas law germane to this problem is
(P1*V1)/T1 = (P2*V2)/T2
simplifying and solving for P2 we get an estimation of the pressure drop caused by any drop in temperature:
P2 = (P1*T2)/T1
where
P1 = pressure upstream fo the intercooler (assuming a 10 psi boost, 25 psi)
T1 = temp upstream of the inter-cooler in Kelvin; (250F = 394K)
T2 = temp downstream of the inter-cooler in K; (150F = 338K)
P2 = pressure downstream of the inter-cooler (the unknown in our problem)
V1 and v2, the volume of the contained air, can be ignored because they remain essentially constant
P2 = (25 psi * 338K)/394K = 21.4 psi
Now since 25 psi - 21.4 psi = 3.6 psi, an intercooler operating at 25 psi that lowers the temperature of its inlet air 100F will also drop the downstream pressure 3.6 psi! That's very close to what the original poster said he was measuring. Therefore, the pressure drop he is seeing across his IC is almost certainly caused by the drop in temperature and NOT from a physical restriction in the IC itself.
If you don't trust my math, here is a little Gas Law calculator I found that let's you solve for various unknowns, including P2:
COMBINED GAS LAW CALCULATOR
Pressure can be in any units, but T must be in degrees Kelvin.
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These equations are only valid for a closed system. This is your source of error. You cannot just discard the volumes in this equation with out making the assumption the volume on one side of the equation is identical to the volume of the other side of the equation.
With this assumption the density of the air would remain constant. This runs totally contrary to the purpose of an intercooler as an intercooler is used to increase intake charge density.
Do you own an intercooler ?