Quote:
Originally posted by myv65
Doubling the delta-P doubles the energy to ~1.6 watts.
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Is this right? If so, I'm missing something.
Doubling the pressure increases the flowrate. (Generally as the square root of the pressure change.)
So if the power dissipation (X) in the first case is:
X1 = dP1 * Q1
then the power dissipation in the double pressure case is:
X2 = ( 2 * dP1) * ( Q1 * 2^.5 )
X2 = ( 2 * 2^.5) * ( dP1 * Q1 )
X2 = ( 2.828 ) * ( X1 )
???