For centrifugal pumps, flow is directly proportional to speed and power
increases as the cube of the speed. Thus, if a pump runs at twice its
normal speed it consumes eight times the power.
At idle, the engine is producing minimum BHP, therefore the rejected heat
will also be less, although probably a greater percentage of the input
energy than at full load.
-----Original Message-----
From: Randall Young <randallyoung@earthlink.net>
To: 'Michael Marr' <mmarr@idcnet.com>
Cc: Triumphs (E-mail) <triumphs@autox.team.net>
Date: Wednesday, March 17, 1999 7:20 PM
Subject: RE: Electric Waterpumps
>Nice analysis !
>
>One quibble : your illustration is for minimum flow to keep the engine cool
>at full output.
>Now, we need the flow required at idle to size the pump (which I suspect is
>worst case, but I'm not certain), then the power vs rpm curve for a
>centrifugal water pump extrapolated to redline (or maybe a little beyond
>for us "boy racer" types) <g>
>
>I suspect the pump on your Audi was required to cool the turbocharger
>housing after shutdown. One of the bugaboos of turbos on production cars
>is that if they are shut off suddenly after high load operation, the
>exhaust heat will cook the oil in the turbo bearings, leading to coke
>formation and failure. AFAIK, this is the main reason to run water through
>the turbo housing in the first place.
>
>Was the electric water pump the only one ? or did it have a separate, belt
>driven pump ?
>
>Randall
>
>On Wednesday, March 17, 1999 4:12 PM, Michael Marr [SMTP:mmarr@idcnet.com]
>wrote:
>>
>> Seems to me that my 1989 Audi 200 Turbo (God, I loved that car...) had an
>> electrically driven water pump for the purpose of maintaining flow
>through
>> the block after shutdown.
>>
>> As for power consumed, the equation is:
>>
>> Horsepower = (TDH X Flow)/(3960 X Pump Efficiency)
>>
>> where TDH is the total discharge head (in feet) and the flow is measured
>in
>> gpm. Required flow rate at full engine load (max heat rejection) can be
>> calculated from:
>>
>> Heat rejected = mass flow X specific heat X temperature rise
>>
>> where specific heat = 1, for water. The temperature rise is the
>temperature
>> difference between the cooled water supplied to the block and the heated
>> water leaving the block. Thus, FOR THE SAKE OF ILLUSTRATION ONLY(!!!),
>> let's assume that an engine discharges heat from the block equivalent to
>33%
>> of it's energy input, and that the engine is 33% efficient. Thus, for a
>100
>> BHP engine operating at full output, the heat equivalent of 100 BHP is
>> rejected to the cooling system. This is equivalent to approximately
>254,600
>> BTU/Hr. If we assume a 20 degree temperature difference across the
>block,
>> then a water mass flow of about 12,700 lbs/hr, or a little over 25 gpm,
>is
>> required. If we assume that the pressure drop through the cooling system
>is
>> around 5 psig, or 11.5 ft head, and we'll assume a pump efficency of 65%,
>> then the power consumed by the pump is around 0.11 BHP. Not much, eh?
>>
>> As a sanity check for this calculation, I looked in my TR2 and 3 factory
>> manual and found an interesting graph that showed that the engine
>consumed a
>> constant 0.5 (imperial) pints/BHP/hour of fuel at full load. This is
>> equivalent to about 1,000,000 BTUH input, assuming fuel at 20,000 BTU/lb,
>> which means the engine is somewhere between 25 and 30% efficient. This
>will
>> change the heat rejected to the cooling jacket, but I can't see the water
>> pump consuming much more than .5 BHP, in a TR3. OK, I'm ready for the
>> responses...
>>
>> -----Original Message-----
>> From: Dave Massey <105671.471@compuserve.com>
>> To: TR List <triumphs@autox.team.net>
>> Date: Wednesday, March 17, 1999 10:19 AM
>> Subject: Electric Waterpumps
>>
>>
>> >
>> >All this talk about electric water pumps raises a few questions:
>> >1) If the pump and drive pulleys are sized for adequate coolant flow
>> >at idle will the pump caviate at red-line?
>> >2) How much power does the pump use?
>> >
>> >The Pump question is much like the fan question with the exception that
>> >there is no induced coolant flow form the forward movement of the
>vehicle
>> >as there is with the air. So if the pump were driven from an electric
>> >motor
>> >what speed would be used?
>> >
>> >The Ideal setup would be to use a variable speed electric drive using
>> >coolant temperature to determine the pump speed.
>> >
>> >On start-up a cold engine requires no pump at all. As the engine begins
>to
>> >
>> >warm up the pump would be required to allow the heater to work. Once
>> >warmed up, the engine would need a modest amount of coolant flow at
>idle.
>> >At speed, underload the coolant flow requirements would increase but
>> >not as a function of speed of the engine but as a function of the load
>and
>> >the heat generated.
>> >
>> >I could design such a system for, oh, about $10,000 complete with
>working
>> >prototype. ;-)
>> >
>> >Talk about gilding the lilly. Or is it Rube Goldgerg?
>> >
>> >Seriously, though, one advantage to an electric driven pump is that it
>can
>> >continue to run after the engine is shut off and cool down the motor
>> >between
>> >runs (eg:at an autocross) a la John Lye.
>> >
>> >Dave Massey
>> >St. Louis MO USA
>>
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