View single post by Keinokuorma  
Posted: Fri Jan 18th, 2008 11:30 pm 


Keinokuorma

himeros wrote: I seem to get about 2.0 to 1.8 amps at the compressor during run cycle, on most of the 15 to 20 cu' boxes I have check lately, hardly ever below 1.6 amps. Excuse my possible partial ignorance on the subject, and possible misreadings too... I'm sober after a good while. 115V*2.29A=263,35W  that seems to be arithmetically sound. However, on an inductive (or capacitive, for that matter) load with AC, the product of the measured voltage (U) and current (I) isn't in straight correlation to the Wattage drawn to, or energy consumed in the load, but there is a geometrical factor that needs to be taken into account. On a rather purely resistive load like a light bulb or heating element, U and I are in rather correct sync (excuse the pun), so when U is on the positive side, I is too, and so both are on the negative side the same time too. The product of positive U and I is positive, as well as the product of negative U and I. Therefore you can consider all of the power in the circuit to be positive. P=U*I A load that isn't purely resistive will shift the phase between the U and I curves so, that when you take synchronous momentual readings of voltage and current, the product is sometimes on the negative side  that's parasite power being transferred back to the source. This negative side of parasite power will subtract out its positive counterpart, and the remainder is the true power consumed in the load. The parasite power therefore keeps going back and forth between the load and source. This will appear as a higher working current that you might assume from the Wattage, and although it doesn't add up tp your power bill, it causes the resistive parts of the system to produce more heat. It stresses the wiring and the fuses/breakers as well as the feed transformer, hence requiring the power company to use thicker wire, or facilities producing lots of parasite power, to compensate it. So on an AC circuit, the formula for power is P=U*I*COS(FI) where FI is the phase shift angle between U and I. For pure resistance, FI=0 degrees and COS(FI)=1. For pure inductance or capacitance, FI would be +90 or 90, and COS(FI) would be 0. Therefore a truely pure inductive or capacitive load would be drawing current but not consuming power. A compressor motor would be of inductive nature, and from the POV of the power source, its inductance vs. resistance will vary greatly by how it is loaded. Last edited on Fri Jan 18th, 2008 11:31 pm by Keinokuorma ____________________ "There is no reason anyone would want a computer in their home."  Ken Olson, Digital Equipment Corporation (1977) 

