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Re: Electric Cooling Fans

To: "INTERNET:DANMAS@aol.com" <DANMAS@aol.com>, list <triumphs@Autox.Team.Net>
Subject: Re: Electric Cooling Fans
From: David Massey <105671.471@compuserve.com>
Date: Tue, 12 Aug 1997 13:15:26 -0400
>In a message dated 97-08-10 16:54:47 EDT, Jody007@aol.com writes:

>> It's said that you'll gain about 18
>>  horse power by removing the fan.

>That's very interesting. That implies that the alternator is more than
100%
>efficient. In other words, it produces more energy that it takes to run
it.
>Assuming, of course, that you are getting the same cooling with the
electric
>fan that you got with the mechanical unit.

>Dan Masters,
>Alcoa, TN

Dan,  

You are forgetting that the engine driven fan is waaaay oversized at any
speed except at idle when it is needed the most.  An electric fan runs at
the same speed all the time and consumes the same amount of power hether
the car is idling or at redline.  And if you use a thermostatic switch the
fan is off most of the time anyway.

Not so for the fan on the crankshaft.  It turns at engine speed.  It is
sized to give adequate air flow at idle and the rest of the time it pulls
much more air than is needed.

Consider this:  I have a fan (from a VW Rabbit) that draws 10 Amps at 12
volts.  That is 120 watts.  at 645 watts per horsepower and 50 % efficency
thats about 0.1 HP.  If an engine driven fan is sized for the same air flow
at idle then the load on the engine at idle is 0.1 Hp.  The power required
to drive a fan is porportional to the Cubed of the speed.  which means the
fan that requires 0.1 Hp at at 800 RPM will require: 
        0.1 x 6^^3 HP at 4800 RPM or 0.1 x 216 or 21.6HP!     
At Red line.  At normal driving speed (eg: 2400 RPM) it is more like 2.7
HP!

The electric fan draws 10 amps from the alternator.  Assuming a 50%
efficiency the engine load is 2 x 120 watts / 645 = .37 HP!  Even if the
fan runs all the time there is a significant net savings!


Dave Massey
St. Louis, MO

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