Warning  more SP, VP, temperature and velocity stuff.
Quote:
Originally Posted by ConnClark
If you read up on bernoulli's equation air moving across an opening has nothing to do with it. It is directly related to the velocity of the air and nothing to do with an opening.

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).
Quote:
Actually that is a commercial science kit sold to and used by universities. Yes it is used to demonstrate Bernoulli's principle. It also demonstrate that just like in plumbing of an intake system with an intercooler static pressure rises and drops according to the velocity of the fluid at different points in it, hence the difference in the fluid levels.

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.
Quote:
If what you say is true would you care to explain why the pressure increases again after flowing through the constriction.

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.

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 intercooler in Kelvin; (250F = 394K)
T2 = temp downstream of the intercooler in K; (150F = 338K)
P2 = pressure downstream of the intercooler (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.