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Protection riddle No.73 - Motor protection
I know locked rotor current, of an induction motor, how can be calculated/chosen the suitable setting current of 50 (instantaneous over current ANSI code) of specified motor protective relay?
Author : H.S - From: Qatar
 
#1
Thu, June 24th, 2010 - 17:04
The theorem of constant flux linkages applies to each induction motor stator phase separately and so one phase will usually exhibit a higher dc offset current than the other two phases. To see why this is so, assume that the closing times for all three phases of the circuit breaker are identical (pole closure times differing by 1 ms or less) and that a pole closure results only rarely in an initial voltage of zero on any terminal of the motor. If no initial terminal voltage is zero, then the initial voltages on the three terminals will all differ, and the resulting dc offset current in one phase will be higher than in the other two. This dc offset current has a time constant of a few cycles or less. Even so, this dc offset current can add substantially to the maximum rms current.



There are other factors to be considered that can cause higher levels of transient inrush current. In above Eq. , Vt is normally assumed to be 100% voltage. It can be substantially less when starting a large induction motor across-the-line on a heavily loaded auxiliary electrical supply network. On the other hand, it can be as high as 110% voltage when starting a small induction motor on a lightly loaded electrical auxiliary network.
Therefore:



The “Guide for Induction Motor Protection,” ANSI Std. C37.92 n (1972) uses a similar analysis for determining locked rotor protection settings.



where:
Kerr=1.10, including the relay trip device tolerance of 10%
Ksf=1.25, including a safety factor of 25%.
The safety factor appears to compensate for the low level of 1.5 used as the dc offset current multiplier as well as to provide for other contingencies not specifically identified in the equation.
One might ask what limits the setting that can be applied to the instantaneous fault protection relays. Clearly, this setting should not be any higher than necessary because of the possibility of resistive faults that will not be cleared if the fault current magnitude is less than the instantaneous fault protection relay pick-up setting. It is this consideration that accounts for the following portion of Section 430–52(a) of the National Electric Code (1993):
Where the setting specified in Table 430–152 is not sufficient for the starting current of the motor, the setting of an instantaneous trip circuit breaker shall be permitted to be increased but shall in no case exceed 1300% of the motor full load current.

Section 7.2.10.4.2 of ANSI/IEEE Std. C37.96–2000, IEEE
Guide for AC Motor Protection, also makes reference to the 1300 percent of the motor full load current restriction. If the locked-rotor current is six times the full load current, the maximum short circuit current pick-up level to ILRA ratio is 2.165, but if the locked rotor current is seven times the full load current, the maximum ratio is only 1.857. This ratio is less than the value of 2.065 computed in above Eq. , which means that the relay setter will have to settle for a lower safety factor.
The induction motor application engineer should also be aware of the fact that higher transient inrush currents have been measured during testing than are predicted using the analysis presented above. Buckley describes this phenomenon. In essence, this means that higher transient inrush currents are being measured than were calculated by the motor designers. These higher transient inrush currents can be attributed, in some cases, to eddy currents that tend to decrease the leakage inductance at high rotor slip frequencies and saturation that tends to reduce stator and rotor leakage reactances (and thus increase stator and rotor currents). It is also possible that there are subtransient current characteristics in rotors with double squirrel-cages. It is difficult to quantify these additional phenomena; however, the multiplier of 2.065 in above Eq. when applied to the calculated ILRA may be too low when these additional phenomena are present.
 
Author : Hamid - From: Iran
 
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