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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
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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
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