Sender: <marv@lancaironline.net> (Marvin Kaye)
To: flyrotary
Date: Sat, 19 Oct 2002 23:36:49 -0400
Message-ID: <redirect-1840532@logan.com>
From: Wschertz2@aol.com
Subject: Electric Water Pumps
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Some Calculations -- if that is allowed.

Todd's installation looks very neat and tidy, but I am concerned with whether 
it will perform as well as desired. Before being dismissed, please look over 
the string of calculations that I went through.

According to the information that I have, the energy in the fuel burned by a 
Mazda engine is distributed approximately like this:

8% to the oil
17% to the coolant
28% to the shaft (drives the prop)
47% to the Exhaust

What this means is that for 1 horsepower to the prop, we burn 1/.28 = 3.57 HP 
worth of fuel.

Full throttle climb out, lets assume we have done the intake and exhaust very 
well, and get 180 HP climb out.  This will require burning enough fuel to 
generate

180/0.28 = 642 HP worth of fuel.

That 642 HP worth of fuel generates heat, 28% is converted to shaft work 
(desirable), and 17% is rejected through the coolant, so the coolant has to 
take away 

642*.17 = 109 HP  We can convert the HP to a heat rate,

109 HP = 4626 Btu/min that has to be rejected by the coolant system.

The heat is carried away by the coolant.  Using a 50-50 mix of Ethylene 
Glycol water, the heat capacity of the mix is 

Cp = 0.56 Btu/(degF-# )

Q = m*Cp*DeltaT

m is mass flow rate, if the pump puts out 80 liter/min,

80 * 1000 grams/liter = 80000 grams, = 176#/minute

So, Q = (176#/min)*(0.56 Btu/degF-#)*DeltaT = 4626 Btu/min

>From this we can solve for Delta T

Delta T = 4626/(176*0.56) = 47 deg F

So to remove the waste heat from the engine, operating at 180 horsepower, 
there needs to be a 47 Degree temperature rise across the engine and a 47 
Degree temperature drop across the radiator.

Lets pick a upper temperature for the engine of 190F, this means that the 
water going into the engine (coming from the rad) needs to be 190 - 47 = 143 
Deg F.  Can we do this? I doubt it, because on a 100 degree day, this would 
only give a 43 degree difference to get the heat from the rad to the air.

One other point -- I believe that Powersport reported that they had to get 
the flow rate up to ~ 50 gpm to avoid overheating.  That is consistent with 
the above calculations, 80 liter/min is 21 gallons/min, doubling the flow 
would mean you could remove the heat with only a 23.5 degree rise, giving us 
more margin to work with.

At cruise, (say 60% power), the standard pump (80 liter/min) would require a 
temperature rise of

47 * 0.60 = 28 deg F, which seems more reasonable and do-able.

All of the above calculations are based on the data that 80 liters per minute 
is delivered by the pump.  However we know that as you insert hoses, 
radiators, and engine blocks into the circuit, the head pressure will rise, 
reducing the flow rate.  How fast it drops off is only determined by testing, 
both the pump, and the rest of the circuit.

Bottom Line,
I would hope that Todd's installation is satisfactory, however I am willing 
to predict that on a hot day he will be limited to 50-60% power for any 
sustained period of time, with the current pump characteristics.  I am deeply 
interested in the results from the tests.

Bill Schertz
KIS Cruiser #4045