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From: "Eric Ruttan" <ericruttan@chartermi.net>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
References: <list-764552@logan.com>
Subject: EWP, head or not to head
Date: Tue, 1 Mar 2005 11:24:36 -0500
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Pumping water in a closed loop has to be 10 times less work than lifting it
(Head pressure).

If you think about a pump lifting water X feet, it also has to pump it
through a pipe, but no one rates the length of pipe, because it is
irrelevant compared to the power needed to lift water.

Our closed loop coolant systems will have friction equavalant to some head
pressure, but if it is over 2 inches I'll eat my hat.

A simple test would be to run an ewp in a cooling system, and check the
pumps rpm.  See at what head the rpm's match.  Equal energy means equal
head.

Eric

----- Original Message ----- 
From: "Jim Sower"
> <But the pump has to do the work to overcome the losses around the loop>
>
> I understand that, but I thought that had a different label.  Like head
> pressure to me always represented the pressure produced by a column of
> water - potential energy.  Pressure produced by a pump to pump water
> uphill or round and round real fast I would have labeled "pump pressure"
> or "kinetic pressure" or something like that to connote kinetic energy.
>
> But then I'm not sure how much clarity is gained by the distinct labels
> ... Jim S.
>
> Al Gietzen wrote:
> >Rusty,
> >I only heard today that Al G. had flow data on his dyno.  He might have
> >some ideas around this.  As to head pressure, I believe I was taught
> >that head pressure only exists in an open system.  In a closed system it
> >all cancels out going around the circle.
> >But that was a long time ago ... Jim S.
> >
> >Sorry; but there ain't no free ride; that would be tantamount to
perpetual
> >motion.  Yes, if you add all the losses and offset it with the boost
across
> >the pump, it equals zero.  But the pump has to do the work to overcome
the
> >losses around the loop.  No pressure across the pump - no flow.
> >
> >Al