The measure of how effective a heat exchanger is
called, you guessed it, effectiveness. What that is
is a ratio of how well the unit rejects heat divided
by how well it could theoretically reject heat:
eff = (T_hot_in - T_hot_out)
______________________
(T_hot_in - T_cold_in)
Basically, to get a perfect radiator the outlet
temperature of the hot side (in our case, coolant) is
so cooled by the air that it leaves at the same
temperature as the air coming into the radiator (this
would happen in an infinitely large radiator). The
thing is, as intuition would dictate, the more cold
flow, the higher the effectiveness. The effectiveness
also varies with hot flow (there is a separate curve
for each different hot flow). See the graph below for
the trend (as cold flow (the X-axis) increases, the
effectiveness (Y-axis) goes up.
E | *
f | *
f | * Hot Flow (Coolant)
n | *
e | *
s | *
s | *
---------------------------------------------------
Cold (air) Flow (lb/min)
When people talk about "slowing down the coolant
flow", what they really mean is that the hot flow is
less on this curve for your radiator, and with less
hot flow, the "new" effectiveness curve will be
higher. So, therefore, for a given cold flow (air),
the outlet temp of the coolant will be lower. The
problem is in the block, the coolant is now the cold
flow. In there, the heat rejected by the coolant is
now in direct proportion to the massflow of the
coolant. Less coolant flow, the hotter the block
gets, and the hotter the coolant will leave the block.
So your temperature change in the RADIATOR will be
great, but the whole temperature of the cooling system
as a whole is elevated, and you can cook your engine.
Not a bad argument for why you want a temp sensor in a
head or block and not the radiator. So....in the
world of radiator cores, lets say that on a 90 degree
F day your thermostat is wide open and the coolant
temp on the hot side is 220 degrees, and your
effectiveness is .8.
.8 = (220 - T_hot_out)
_______________
(220 - 90)
So your hot outlet temp is 116F. Now, if you drop the
air flow by 20% due to more fins/tubes restricting
airflow, your effectiveness might go down about as
much (it won't, more on that later). So effectiveness
is now .64. So,
.64 = (220 - T_hot_out)
_______________
(220 - 90)
So now your hot outlet temp is 137F. So that's for
your first pass through your calculation. Since the
mass flow of the coolant thru the block is still the
same, your temp change in the block will be the same,
104 degrees (from the initial case: 220F - 116F). So
now, by the time you exit the block your temp is 137 +
104 = 241F. You've boiled over. It is not quite that
simple in real life - as you add fins, you also
increase the effectiveness. So there is a break even
point where you stop adding more fins/tubes because
the increase in effectiveness is offset by the loss of
airflow.
Thanks for reading this far.
=====
Ron Soave
"I swear I found the key to the Universe in the engine of an old parked car." -
B. Springsteen
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