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Re: Electric fans/power gain

To: triumphs@Autox.Team.Net
Subject: Re: Electric fans/power gain
From: Gerhard.Wiederholl@t-online.de (Gerhard Wiederholl)
Date: Fri, 27 Mar 1998 21:53:33 +0100
References: <19980327063223004.AAA77@Fatman>
Brad Kahler schrieb:
>
> > >Engineering books state a belt driven fan takes away 6-9% enginepower.
>
> What does the engineering book say about the extra horsepower needed 
> to keep the battery fully charged ?!!?
>
> Brad
>
> 1964 Spitfire4            BFC25720L (After 10 years will be on the road in
>  March!!!)
> 1966 TR4A                CT72398L (Car is finally home and ready for
>  Restoration)
> 1951 Dodge Truck    82217766  B-3-B-108 (Boxes & Boxes of parts right now)

ÐÏࡱ


Sure the electric fan needs  electric power, but

a well designed alternator gives only as much power as presently needed and 
because of this the brake effect on the engine and the discharge to the battery 
while loading is only there when the fan is running. This might be some 5% of 
the time the engine in running as someone on the list stated. I have never 
heared from someone who has converted to an electric fan that this killed the 
Lucas equipment or the battery. 

The other thing to consider is:

The blast of the fan is mainly needed when the car stops or is driven slowly.

This are the circumstances that the factory fan has to be designed for plus 
some 
surplus for long hillups.

If this is true and the fan is designed to cool down the water while idle even 
in Death Valley, then a beltdriven fan at speed becomes the more an airbarke 
the 
less you need it.

I thougt about the size an electric fan should have:

The efficiency of a gas/petrol engine is not better than 30 %, which means that 
70% of the energy in the gas/petrol becomes heat and has to go to the air 
around 
the car via radiator(33%), exhaust(33%) and heat-radiationfrom engine 
-survaces(7%).

If we can imagine that a Triumph drives very slow and the overall engine load 
is 
10hp this would mean that (33%) 3.3 hp or 3.3x736W = 2428 W coolingpower is 
needed. This does not mean that the fan has to have 2428 power/fan-efficiency. 

As I calculated there are 0,94 m3/sec air required. (airheatincrease 20 degrees 
centigrades). Has anyone technical data for fans handy and can say what 
electric 
power is needed to move 0,94 m3/sec through a radiator?...

I guess at 10 hp  overall engineload and 3hp x 80% = 2,4 hp on the wheels my TR 
will go some 10 mph / 16 km/h.  A 0,3 m2 radiator at this speed will catch 1,3 
m3/sec only from the drivingwind.

This makes me think that an electrical fan with only very low power is needed.

What are the other opinions ?


Gerhard

1969 TR 6 

BTW 6-9% loss in enginepower is only for the fan and excludes the belt as the 
belt is needed with or without belt driven fan



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