Ek's Home   |   Forum   |   Chat   |   Electrical Links   |
	Subject - 3 phase heater loses one element jamacian If you had a 16kw 480 volt 3 phase heater delta wired. Lose one element. How does the current react on all three legs? Amps up? Down?All legs? 2 up one down? Need to know. Have several guesses already. Thanks kbsparky My guess would be that two phases would show ½ load, 3rd phase would show normal load. Assuming that there are 3 equal sized elements in the first place. lctrc789 Delta loads are somewhat tricky and not normal, An example would be : let's say that Line 1 has a current flow of 100 amps line 2 has 75 amps and line 3 has 50 amps the amount of current flow would be 125 amps total on the neutral. The formula for Delta is L 1 + (L2 - L3) or 100 = 75-50 100 + 25 Neutral current is 125. I hope this helps. kbsparky Why would there be any load on the neutral When dealing with a delta 480 Volt system, there is no neutral present in the branch circuit supplying a unit heater as stipulated in this example lctrc789 KB, you could be right I was thinking the same why would there even be a neutral but perhaps it is a trick question to stump us,. jamacian No neutral. No Tricky question. This is on a wet sump on a large refrigerated ammonia system. Want to put a current detector on one line. Current drops cut out heater. If all three lines drop, (my theroy) I can do it with one current detector. If I am wrong I need three that will cut out on rise or fall. I wanting someone with an answer thats fairly sure. Really didnt want to give my theroy as it might tip the scales. Thanks James PEI You lead us to believe you may lose an element....or did you mean a phase (as in a conductor). I'd like to suggest a zero sequence relay. Basically, this works as a CT wrapped around all 3-phase conductors back at the source (could be at the load but the CT is only about 3-inch ID and all conductors have to fit thru). A special relay senses the current in the CT, as the current in the CT increases, the relay trips at a predetermined level. Okay, let's back up and explain what goes on. If all 3-phases are balanced, no EMF field is created in the CT and thus no current is generated. When 1 phase drops out, the other 2 phases cause a current in the CT which then trips the relay. Some additional circuitry allows adjustment because all 3 phases will not naturally be balanced perfectly, so an adjusment is needed to set an acceptable trip point and you won't have nuisance tripping. Oh yeah, why is it called zero sequence - because in a perfectly balanced 3 phase system, the sum of all currents in the 3 is equal to zero. This current is denoted as script "i" subscripted zero, to differentiate it from i sub a, i sub b and i sub c (the phase currents). bigbillnky Goodmorning I will try to see if I can help with this. If the heaters are connected closed Delta, the current on 2 phases will decrease. The third phase current will remain unchanged. This is due to the fact in a closed Delta configuration, each phase voltage is dropped thru 2 resistances in parallel. When one element burns open, 2 of the phases are then dropped thru only one resistance. We know from theory that total resistances in parallel are always less than the value of 1 resistance. When one of the resistances (element in this case) is removed from the circuit, then total resistance will increase causing a reduction in phase current(current and resistance are inversely proportional). If the elements are Wye connected, the current on one phase would be zero, and the current on the other 2 phases would decrease by about 1/3. The reason is that when a 3 phase heater is Wye connected, each phase is connected to 1 element in series AND 2 elements in parallel. If you are unsure why this is, simply draw out a Delta and a Wye diagram. Now give each leg of each configuration a resistance value. Now calculate the resistance each PHASE voltage is dropped thru in each configuration. Now open any one leg of each configuration and recalculate to see the answer. One other note, this information applies ONLY TO RESISTIVE LOADS. Inductive loads have reactances that change connected impedances, so the calculations vary a great deal from purely resistive loads. lucky1122 Im going to take a crack at this one . In a delta connected circuit feeding a 16 KW heater connected in delta config if the heater was working properly with all resistances equal the current in the resistance alone would be 16KVA x1000=16,000/480v x 1.73 =19.26A however the current in the line (and in each leg they would be equal when the circuit is working properly) the current would be 33.3 A.Now lets assume you are talking about an open resistor in this case it is my guess that the one leg of the delta that is connected to the good resistors would read 33.3A while the other two which are common to the bad resistor would read only the current in the respective resistors which would be apprx. 19.26 A . The reason being that in a delta configuration the line current is always 1.73xthe current in the coilseeing that one leg is always common to both.