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From: "Eric Ruttan" <ericruttan@chartermi.net>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
References: <list-769911@logan.com>
Subject: Re: EWP Test Results/DRAG and hat eating
Date: Sat, 5 Mar 2005 08:48:17 -0500


In a prior thread titled "EWP, head or not to head", see partial copy at
bottom of this message, I suggested that the head of a closed system cannot
be that high.

If I understand Bill Schertz post correctly, he measured the psi of an
engine in our operating range at 9 psi.  Or perhaps 5 psi?

And if head (in feet) = pressure(psi) *2.31/s.g. then 9*2.31/1=20.79 source
http://tinyurl.com/53elw
if it is 5 psi then it is 11.55 feet head.

Can someone show me...
If I did the above correct?
How much power it takes to move 20GPM 20 feet, assuming 0 pumping loss?
If bills data was at temp?  Will the reduced viscosity have a large effect?

I have selected the tastiest hat I own, but I don't want to get started
prematurely.

I thank you for your assistance.
Eric


----- Original Message ----- 
From: "Ernest Christley"
> Tracy Crook wrote:
>
> > Ernest,
> > I had no idea that water viscosity changed that much with
> > temperature.  Is the same true of 50 - 50 glycol & water?
> >
> > Tracy
>
>
> It appears that just about all liquid follow an inverse log function.
>
> Andrade Viscosity Correlation:  ln(u,cps) = 5338/(T(K))-12.63 (Valid
> between -10C to 50C)
>
> http://www.lyondell.com/html/products/techlit/2570.pdf
>  http://hypertextbook.com/physics/matter/viscosity/
>
> Ernest (Google is my friend)
>

Addendum from "EWP, head or not to head"

What flavor hat?

I *measured* the flow generated by the Mazda pump *attached* to the engine.
I did this in open loop format, but measured the pressure into the pump and
the pressure out of the pump.  I then did a curve fit of the data, and
produced the following graph.

Note that at 2448 rpm, the pressure available to drive the fluid through the
cores is zero when the flow is 20 gpm -- track that curve back to zero flow
and you can see that at that rpm the pump is generating 5 psi. Think about
it. It means that at 20 gpm, it takes 5 psi to drive the water through the
*core* of the engine and the water pump.

At 3730 rpm, the pressure drop *through the engine* is ~9 psi.
Bill Schertz
KIS Cruiser # 4045
----- Original Message ----- 
From: "Eric Ruttan" <ericruttan@chartermi.net>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Sent: Tuesday, March 01, 2005 10:24 AM
Subject: [FlyRotary] EWP, head or not to head


> 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