DeTomaso Mailing List: June 2001, Message #277

Dave,

All these numbers and figgerin' and fancy talk is gonna get us both thrown
off this list!

But thanks for the quasi - confirmation of the numbers I came up with, looks
like
I wasn't too far off with the equations.

I only picked 2.5 gpm and 5 psi because I knew that would work out to 1/10
of a HP
and that number was part of an earlier post.  I expected even the Pantera
crowd would
recognize intuitively you cannot replace the belt driven water pump with an
electric
pump the size of an electric fuel pump.  Never heard from the guy who
brought up the
subject to begin with.

Since we agree at six grand the water pump could be soaking up 8 hp or more,
any idea what NASCAR guys use for a belt?  I'm still not convinced you could
do
that with a single v-belt around 90 degrees of a roughly six inch diameter
pulley.
Maybe I'll have to break down and get a Gates catalog and do some more
figgerin'.

Changing the subject to computer science, does it help if I hit enter when
the
lines get about this long?  I notice my original msg below is scrambled...

Guy D.


> I think your equation is good but the base assumption of 2.5 gpm for an
> average automotive cooling system is way too low.  And the pressure of 6
psi
> probably only represents idle speed conditions - not driving.  The top of
> the line of water pump marketed by Stewart Components, which is used
> exclusively by almost every single NASCAR team, flows up to 160 gpm in
> racing conditions (8,000 rpm).  Those are firehose type water velocities.
>
> According to coolant flow data compiled by Stewart Components, water flow
> measured in a small block Chevy engine at 1000 rpm is 17 gpm (Edelbrock
wp)
> or 18 gpm (Stewart Stage 1 wp).  The same two water pumps in the same
engine
> generate 102 gpm and 132 gpm at 6000 rpm.  Pump output pressure will be
> equivalent to all the restrictions in the flow line (hoses, radiator,
> cylinder head, block) and in the vicinity of 20psi at 100 gpm.  So, at
idle
> the pressure and flow criteria for a medium size V-8 with a stock water
pump
> would be something like 5 psi and 15 gpm (round numbers) and at 6000 rpm
> around 15 psi and 80 gpm.  Using the same HP equation, the power consumed
by
> the water pump at idle would be about 1/2 HP and at 6,000 rpm about 8 HP.
> These numbers are very similar to waterpump horsepower requirements that
> Stewart Components quote.
>
> So we're not talking about trivial HP power requirements to keep a big V-8
> engine cool.
>
> Dave Bell
>
>
> ----- Original Message -----
> From: "Guy Dellavecchia" <guido_detomaso@prodigy.net>
> To: "Multiple recipients of list" <detomaso@realbig.com>
> Sent: Monday, June 04, 2001 1:45 AM
> Subject: Re: SV: What I learned about cooling systems this week
>
>
> > Andy,
> >
> > You wrote:
> >
> > We're not doing any of the traditional "work" like lifting the water
> > > or accelerating anything of significant mass.  That doesn't sound like
> 10
> > hp
> > > of work to me.  A fraction of a horsepower sounds sufficient.
> >
> > I suspect you are an innocent victim of the public school system, the
> folks
> > at Dyno Jet, Pantera Tech articles or some combination of all three.
> >
> > True, the water is not being lifted nor does it need to be accelerated.
> > But, as with most things, there's no free lunch.  It's going to take
some
> > power to move that water.  How much?
> >
> > Well, having invested all of 5 minutes with a calculator, here's the way
> it
> > looks to me:
> >
> > Power is force X distance / time.
> >
> > Using familiar units, that would be lb-in / sec
> >
> > In the case of the pump, throwing in some arbitrary numbers, say 5 psi
> > across the pump and 2.5 gallons per minute:
> >
> > 5 lbs / in squared X 2.5 gal / minute X 231 cubic inches of water per
> gallon
> > X one min / 60 sec = 48 lb in / sec
> >
> > Since man long ago decided to call 550 lb - in / sec  ONE Horsepower,
48
> lb
> > in / sec is a bit less than 1/10 a horsepower.
> >
> > Now again, I only spent 5 minutes on this and never cracked a textbook (
> nor
> > a 40! ) and I'll die of embarassment if I'm off by a factor of 10 or a
> 100.
> >
> > But my gut thinks this is about right, because these numbers aren't far
> from
> > what an electric fuel pump can do and I would expect that same electric
> fuel
> > pump to soak up about 1/10th of a HP.
> >
> > In conclusion, NO, I don't think 1/10th of a HP is going to pump enough
> > water to cool that 351.
> >
> > Simultaneously, I'm not convinced 10 HP can be transmitted to the stock
> > water pump by a single V-belt wrapped around 90 degrees of pulley.  Not
> > because I did any calculation, but because I have never seen a 10 HP
motor
> > connected to anything with only a single V-belt wrapped around 90
degrees
> of
> > pulley.  So the truth I suspect lies somewhere between these extremes.
> >
> > Hope this helps, but mostly I hope I didn't make a mistake with those
> > numbers!
> >
> > Guy D.
> >
> > ----- Original Message -----
> > From: Andy Poling <andy@realbig.com>
> > To: Multiple recipients of list <detomaso@realbig.com>
> > Sent: Tuesday, May 29, 2001 8:42 PM
> > Subject: Re: SV: What I learned about cooling systems this week
> >
> >
> > > On Sun, 27 May 2001, Tomas Gunnarsson wrote:
> > > > Richard,
> > > >
> > > > 1 hp would require around 60 amps, 12*60 is closer to the 736 W
equal
> to
> > a
> > > > horsepower (here in Europe). But, strangely enough those pumps seem
to
> > work
> > > > anyway!
> > >
> > > I think maybe this is a case of apples and oranges.
> > >
> > > The electric pump impeller can be designed specifically to be
efficient
> at
> > the
> > > speed of the electric motor under load (easily predicted).  The
> mechanical
> > > water pump's impeller, on the other hand, has to work over a huge
speed
> > range.
> > > It's design is therefore necessarily a compromise.
> > >
> > > Also, the way I reckon it, the only "work" done is the creation of the
> > head
> > > pressure needed to force water through the radiator and the rest of
the
> > system
> > > (i.e. countering the friction of the water flowing through the
> restricted
> > > system).  We're not doing any of the traditional "work" like lifting
the
> > water
> > > or accelerating anything of significant mass.  That doesn't sound like
> 10
> > hp
> > > of work to me.  A fraction of a horsepower sounds sufficient.
> > >
> > > Can someone show me where my reasoning is faulty here?
> > >
> > > -Andy
> > > #3822 (making 0 hp at the moment)
> > >
> > > 72 Pantera - Rocky              91 Miata - Steve               84
> RZ350 -
> > Sting
> > >
> > >
> > >
> >
> >
> >
>
>
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