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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