170225-0927 EST
wyreman:
Referring to your plots in post numbered #60 there is some interesting information.
Now that we know that each 100 A pulse corresponds to travel from one floor to the next it appears the motor is on for 10 seconds for this travel, and then the elevator stops for 10 seconds at each floor.
Disregarding the transient current pulses the current for the elevator rising looks like about 100 A at start, rises by about 20% to a peak over 5 seconds during travel and acceleration, settles back down to about 100 A while finding its floor level (possibly some of this last time period is non-acceleration, just linear up motion). This is a consistent pattern. Similar for all three phases.
The large transient current pulses, about 3 times the needed current to run the elevator are random, but occur at turn on of the motor. These are some sort of turn on transient. They simultaneously occur on all three phases and are about the same magnittude for each phase. Their apparent magnitude differences and non-existrence on some elevator cycles could result from the data logger, but I doubt it. This could be determined with a Fluke 27 or other similar min-max meter. A 20 second time period for the elevator cycle is plenty of time to read and do the min-max reset.
As I roughly calculated before the transformers have about an 800 A full load capability, and one might expect a 3% voltage drop at full load. 3% of 120 V is 3.6 V. 3.6*300/800 = 1.4 V out of 120 V is the voltage change that might be expected at the transformer for a 300 A transient load. This is to address some comments by others.
There is certainly a greater voltage drop on one phase vs the other two.
Some have mentioned the load is a delta, but from what I conclude from this thread the source is a wye and the sourcre being a wye is the important factor in looking at the problem.
Power input to the elevator is about 3*120*100 = 36 kW. Power factor won't be 1, but probably not 0.6 . About 10*36/3600 = 1/10 of a kWh is used to raise the elevator about 10 ft. Use half of that as an estimate, or 1/20 kWh. 40,000 ft-pounds of work equals about 0.015 kWh, So this systrem is likely quite inefficient.
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