return line to gas tank

Greg Hermann bearbvd at sni.net
Thu Oct 22 19:04:44 GMT 1998


>>If the pump draws 8 amps (@ 12 volts), and all of the heat from that winds
>>up in the fuel (which it basically does , sooner or later), that is 96
>>watts. Which converts to 327 BTU/hour. which is enough heat to raise the
>>temperature of 10 gallons of gasoline about 7 degrees over the course of
>>one hour if none of that heat is dissipated anywhere.
>
>The power which goes into heat is significantly less than the power the
>pump uses
>to push fuel. The way to determine how much electrical input power is
>converted to
>heat is to convert the output flow of the pump to work. Then, calculate the
>efficiency of the pump based on input power (96 Watts). The amount of
>energy not
>directly used for pumping should be a fair approximation of  how much heat is
>added to the fuel by the pump.
>
>It is not simple to do, but I would guess that the pump must be 75-80%
>efficient.
>Any thoughts on this?

Hey--give an A+ to the guy who caught the trick question!! But note that I
said "sooner or later", and was running a quick, worst case calc to
disprove Tom's theory. What I meant by sooner or later was that some of the
work put into flow energy (and static pressure potential energy)  in the
fuel by the pump comes back as fuel heating where the throttled flow
through the pressure regulator occurs. More heating happens (but does not
stay in the system) with the throttled flow through the injectors.  More
heating of the fuel occurs from typical frictional flow losses in the
supply and return fuel lines. The IR losses in the pump motor windings go
straight into the fuel, and I would guess to be about 25% of the input.
mechanical (friction) losses in the pump (analogous to the heating of lube
oil) I would guess to be about another 15-25 % of input. But no matter--it
ain't the pump which is heating up a tank full of gas!!

Regards, Greg





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