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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
----- 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
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