I searched the CD and found this data on the ability of
air to absorb heat.
I wrote a little program to calculate the area required
at 100 MPH and 120 MPH as a function in the increase in
air temp passing through the rad. Not sure about the
specific heat constant so by all means check my calculations.
As you can see in the ideal case where the air is increased
by 60 degrees exiting the rad the area required is about 200 square
inches or about 10 by 20 roughly 63% of the frontal area of a 320
square inch rad. Sounds real high so there could be something
wrong. Sounds like it is off by a factor of two.
Tracy promised to measure the air temp and air velocity
exiting his rad.
Paul Lamar
I found the mistake. It had to do with actual airflow through
the rad. You must measure the actual air velocity at a point just
inside the duct opening. You also need to measure the static
pressure at the same point and correct the air velocity measurement
for the static pressure. In other words use a pito tube system.
If the rad is too thick this number will be below aircraft
forward speed due to external diffusion. This also assumes no leaks
around the radiator.
For a 15" by 20" rad 103 square inches (80 MPH) is of course 34% of core
area which is close to the rule of thumb of 30%. This assumes of
course the rad achieves a 60 degree increase in the air temp.
A high number by radiator standards IMHO. The jury is still
out on that and I also must see some measurements
on those numbers as well to have any confidence in
these calculations.
Two round holes 4.5 inches in diameter have an area of
of only 32 square inches so they appear to be inadequate
up to real high speeds.
Two round holes 6 inches in diameter have an area of
of only 57 square inches so they too appear to be inadequate.
Even for an air velocity into the rad duct of 120 MPH.
For an inlet velocity of 80 MPH the opening should be
6.6 by 15 inches or 103 square inches. Assuming an
optimistic 60 degree F air temp rise.
Now we need some real world numbers from those flying
airplanes to get a precise handle on this.
Paul Lamar
**********************************************************
Assumptions:
6000 BTU per minute waste heat
Specific heat of air = 0.01988 BTU per cubic foot per deg
20% humidity 100 degree day
Flow through rad is 7039 FPM or 80 MPH
____________________________________________
|air temp | airflow req | area | area |
| rise F | CFM | SF | SI |
--------------------------------------------
10 30303 4.31 620
15 20202 2.87 413
20 15152 2.15 310
25 12121 1.72 248
30 10101 1.44 207
35 8658 1.23 177
40 7576 1.08 155
45 6734 0.96 138
50 6061 0.86 124
55 5510 0.78 113
60 5051 0.72 103
Assumptions:
6000 BTU per minute waste heat
Specific heat of air = 0.01988 BTU per cubic foot per deg
20% humidity 100 degree day
Flow through rad is 8820 FPM or 100 MPH
____________________________________________
|air temp | airflow req | area | area |
| rise F | CFM | SF | SI |
--------------------------------------------
10 30303 3.44 495
15 20202 2.29 330
20 15152 1.72 247
25 12121 1.37 198
30 10101 1.15 165
35 8658 0.98 141
40 7576 0.86 124
45 6734 0.76 110
50 6061 0.69 99
55 5510 0.62 90
60 5051 0.57 82
Assumptions:
6000 BTU per minute waste heat
Specific heat of air = 0.01988 BTU per cubic foot per deg
20% humidity 100 degree day
Flow through rad is 10560 FPM or 120 MPH
____________________________________________
|air temp | airflow req | area | area |
| rise F | CFM | SF | SI |
--------------------------------------------
10 30303 2.87 413
15 20202 1.91 275
20 15152 1.43 207
25 12121 1.15 165
30 10101 0.96 138
35 8658 0.82 118
40 7576 0.72 103
45 6734 0.64 92
50 6061 0.57 83
55 5510 0.52 75
60 5051 0.48 69
***************************************************
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