Re: elec heater calc
Originally posted by spsnyder: If we use Phase A as an example, it would have the first and third heater on it.
If a single load touches two phases (and the other two loads do the same), then you are describing a DELTA configuration, not a WYE.
The current from the first and third heater would be 14.4 amps each, or 28.8 amps, correct?
Sorry, not correct. If you are adding up apples, oranges, ducks, telemarketers, or single-phase amps, then 14.4 + 14.4 will equal 28.8. Not so for three phase amps. That is because just calling it ?14.4 amps? is not a complete description. What you are adding is the value ?14.4 amps at an angle of zero degrees with respect to the Phase A Voltage? to the value ?14.4 amps at an angle of 120 degrees with respect to the Phase A Voltage.? The result is ?25.0 amps at an angle of 60 degrees with respect to the Phase A Voltage.? When you factor in the 125% for the heater load, this becomes ?31.2 amps at an angle of 60 degrees with respect to the Phase A Voltage.?
Current per leg is not affected by 1.73 in a wye connection. I line = I phase for wye. Only voltage. E line = E phase x 1.73 ie 208V = 120 x 1.73.
If Bill did choose to construct his heater bank in a WYE configuration, then you would be right in saying that line current and phase current would be the same. However, we started with a 3000 watt heater. If you connect it in a WYE configuration, then the voltage it would see is not 208, but rather 120. The difference between the 208 and the 120 is, as you have already pointed out, tat factor of 1.732 (i.e., the square root of 3). The current in each heater would be 25 amps (i.e., 3000/120 = 25). Multiplying this by the 125% factor for the heater load brings you back to 31.2 amps. Bottom line, when you deal with three phase, there will be no escape from that strange ?square root of three? factor.