"Thanks for reading this far." YOU MEAN THERE'S MORE RON???
Ron Soave wrote:
> 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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