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Date: Tue, 11 Jul 2006 22:21:18 -0600
From: Bob White <rlwhite@comcast.net>
To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
Subject: Re: [FlyRotary] Re: -  EWP Power
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On Tue, 11 Jul 2006 22:52:32 -0500
"Bill Schertz" <wschertz@comcast.net> wrote:

> Charlie England wrote:
> > On to the pump:
> > Ignoring pump & motor efficiency, consider the formula for 1 horsepower 
> > (~746 watts, or 62 amps @12 volts): 33,000 lbs lifted 1 foot in 1 minute. 
> > 40 gallons of water is ~320 lbs. At 1 foot of 'head', that's 0.0097 
> > horsepower. Someone please check my math & be sure I haven't displaced a 
> > decimal point; it's late & I'm sleepy. Anyone know the approximate 'head' 
> > pressure in feet across the pump in a car engine? Surely we can do a 
> > sanity check on approximate hp required to pump the water if we know the 
> > 'head' pressure in feet.
> >
> > Charlie
> 
> 
> -----------------------
> Charlie, et.al.
> Attached is a plot of measured flow vs pressure data for the Mazda water 
> pump, and for A/C cores used as radiators. I have published this before, but 
> some salient points.
> 
> 1. The curve labeled 'one core' was measured by flowing water through a core 
> and measuring the pressure drop across the core. The data points (shown) 
> were then curve fit to get the load curve.
> 
> 2. The curve labeled 'two parallel cores' is a calculated curve based on the 
> flow through a single core. (If you have two cores, for a given pressure 
> across the cores, you will get two times the flow). To check this, I made 
> one run with the cores connected in parallel and that is shown as the point 
> labeled 'Real Rad Test'. Since it comes close to the projected parallel core 
> lines, I am very comfortable with the calculated curve.
> 
> 3. The pump performance curves are made at three different rpm's of the 
> pump. I was using a 1 horsepower table saw motor, and used different size 
> pulley's to get the different rpm's.  The pump is *pumping through the 
> engine*, and the pressure was measured at the pump inlet and outlet, to get 
> the value shown on the graph. I made several runs,  and plotted the data to 
> get a curve-fit for each curve.
> 
> At zero flow, the pump generates the highest pressure at any given rpm. As 
> flow is increased, the pressure that is available to push the water through 
> the cores is lowered, this is because of the friction of pushing the water 
> through the engine block, so at 2448 rpm (on the pump) it generates ~5psi, 
> and all of that pressure capability is expended on pushing 20 gpm through 
> the block when the valve is wide open.
> 
> The proper way to use a chart like this, is to look for the place where the 
> load curve (lines curving up to the right) intersect the pump curve (line 
> curving down to the right). For a given RPM and load curve, that 
> intersection is where the system will operate. For example, at 5594 RPM, 
> with parallel cores, one should expect about 32 gpm flow rate through the 
> cores, and about 9psi pressure drop across the cores.
> 
> The pump (at 100% efficiency) is absorbing horsepower sufficient to raise 32 
> gpm through 19psi (~10 lost in the engine core and ~9 lost in the 
> radiators).
> 
> Increasing the speed of the pump increases the pressure, and the horsepower 
> requirements, when I tried to go to a larger pulley to go to 8,000 pump rpm, 
> my circuit breaker would trip, and I could not take data. Since I don't plan 
> on operating at those rpm's, I terminated the tests. If you drive the pump 
> faster, more of the energy is diverted into heating the water, the increase 
> in flow may be minimal after a certain point.
> 
> 
> Bill Schertz
> KIS Cruiser # 4045 
> 

Bill's charts and an analysis he did of Todd Bartram's cooling system
are available on the wiki. http://www.rotarywiki.org  I always refer
back to these when I have cooling questions.

The hp requirement increases with the cube of the rpm on centrifugal
pumps.

Doesn't the pump turn faster than the engine?

Bob W.

-- 
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