Hi paul,
I conducted more test and found the following that is also in Kays of
London. I have increased the intake duct to 230mm long from the start of
the core with vast improvement.
I now have 50 knots air speed at the the opening and 11 knots at the
back of the core to 6 knots at the front. With a new shape duct that I
will build tomorrow I think I can get equal airspeed out of the radiator
core. One thing that seems to be a common factor is 50 knots in and a
core to intake size has a scale factor. For example core length 460mm

[18"],

intake is 80mm [3.14"] and this equals a 5.76 ratio. The average outlet

air

speed is 8.6 knots which is close to the to the inlet speed x 5.76.
The relationship has been the same at all airspeeds that I have tested.
In the morning I will build the new shape and retest.
Regards Don

-----------------------------------------------------------
I don't exactly understand what you are doing here. I need more
details. How wide is the rad?

Cold volume of air going in should equal cold volume of air
coming out all other things like pressure being the same.


If the blower opening is a round 5 inches in dia. (127 mm)  the area
is 19 square inches or 19/144 or .132 square foot.
If the air velocity coming out of the blower really is 40 MPH
(50 kts roughly) in a sealed system  that would be
about 58 feet per second. The volume of the air coming out
of the blower is therefore 7.66 cubic feet per second.

If the scoop opening is 23 square inches or 23/144 or .160 square feet
the velocity going in there is 7.68/.160 or 48 foot per second
or 32 MPH.

If the wedge opening going into the rad is 54 square inches (3" X 18")
or .375 square feet the air velocity going in there is 7.68/.375 or
20.5 foot per second or 14.5 MPH.

Lets assume for the moment the rad is 18 inches (460 mm) wide.

If the rad core is 18 inches by 18 inches the area is 324 square inches
or 2.25 square foot the velocity of the air coming out of the
back side of the rad should be 7.68/2.25 or 3.4 foot per second
or 2.33 MPH at best on average.

Help. Did I make a mistake here somewhere? Where am I going wrong.

This assumes no losses or leaks anywhere (a completely
sealed system). Assume the free stream air velocity coming out of
the blower is 50 Kts. If the system is sealed the air velocity
coming out of the blower is not 50 Kts but something lower
due to resistance in the ducts and in the rad core. Since it is near
impossible to seal the anemometer inside the duct.... measurements
are better made with simple pito tubes (sealed inside the sealed
duct system), static pressure taps and a water manometer.

The pito tubes must be referenced to the local static pressures
to accurately measure the velocities in the duct.

Therefore you need at least one pito static tube
at the outlet of the blower and one U tube water manometer.

No problem using the anemometer on the back side of the rad
open to the atmosphere to measure the air flows coming out of there.

What is the thickness of the rad as that will give us an indication
of its resistance to air flow?

Paul Lamar

Don,
I follow your maths but not the rational behind it (except that I think you
mean inlet speed/5.76).
If your saying "all thing being equal" and the inlet is widened!?

George ( down under)


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