Motor contribution

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chris kennedy

chris kennedy

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Miami Fla.
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Retired electrician
How exactly do motors contribute to fault current? I understand spinning motors become gennys upon power loss but when a fault occurs how is this based? Duration?

Thanks
 
A fault is broken down into 3 pieces
a rough description
asymmetrical or sub-transient first cycle or so
transient next few
Symmetrical steady state >3-4 cycles
durations depend on system x/r, etc

breakers are usually sized on steady state i
a peaking factor can be applied to the ss to approximate peak/sub-transient

to you question
a motor stores reactive (and mechanical) power and dumps it into the fault
think of it as the opposite of starting current

many times only peak let thru of the transformer is used
motor contributons are ignored
but if motor kva/xfmr kvs is high, say 0.25+ it may need considered
if done by hand I add 4 x motor kva to the fault kva
this may be dependent on fault location, etc

duration is seldom considered since it is assumed the cb will trip during transient or steady state 2-6 cycles

so only magnitude is considered
must withstand peak/sub-transient
interrupt steady state
 
What about different types of motors, water and fire pumps(fire pumps unlikely to be running at time of fault, no?) and AC compressors?
 
chris kennedy said:
What about different types of motors, water and fire pumps(fire pumps unlikely to be running at time of fault, no?) and AC compressors?
Click to expand...

Just think of it as a water system with many sources/tanks. Those tanks connected to the point where a pipe break occurs will surely contribute water that gushes out! Those tanks with no water in them (motors not running or off) will not.:)
 
A lot depends on fault type/location
ground, line and load, load alone, line alone
line-line, same combinations, both ends, line end, load end
many permutations

sometimes fault current will flow thru the motor to the fault
eg line breaks, line is open, load side is grounded
 
When a motor loses power due to a three phase fault, the rotor continues spinning due to its own inertia. During normal operation the rotor winding will have a voltage on it, either from induction in the case of an induction motor or from the exciter in the case of a synchronous motor. When the fault occurs, this voltage cannot be removed instantaneously due to the time constant of the circuit. This combined with the fact that the rotor continues to spin results in generator like behavior for the first few cycles after a fault.
 
A good primer

excerpt
Assuming a motor contribution of fourtimes rated full load current is acceptable. Thestandard arrived at this value by assuming themotor contribution of 3.6 times rated current camefrom 75% induction motors and 4.8 times ratedcurrent from 25% synchronous motors

http://apps.geindustrial.com/publibrary/checkout/SC-Motor?TNR=White%20Papers%7CSC-Motor%7Cgeneric


Hmmmm
never knew how the 4 was arrived at
learn something everyday

 
Ingenieur said:
A good primer

excerpt
Assuming a motor contribution of fourtimes rated full load current is acceptable. Thestandard arrived at this value by assuming themotor contribution of 3.6 times rated current camefrom 75% induction motors and 4.8 times ratedcurrent from 25% synchronous motors

http://apps.geindustrial.com/publibrary/checkout/SC-Motor?TNR=White%20Papers%7CSC-Motor%7Cgeneric


Hmmmm
never knew how the 4 was arrived at
learn something everyday

Click to expand...


So when figuring motor contribution what values do we use?
 
chris kennedy said:
So when figuring motor contribution what values do we use?
Click to expand...

I use 4 for all motors
my systems the motors are always running and fed from a common feeder

A rough example for illustration
let's assume
20 mva xfmr
pu Z 5%
fault kva 400 mva

10000 hp connected/running or ~ 10 mva (1 hp ~ 1 kva)
so motor contribution 40 mva

total fault 440 mva
Only a 10% increase for 50% motor load
that is why it is often ignored

may need considered for large motor feeders
 
IEEE Standard 141 IEEE Recommended Practice for Electric Power Distribution for Industrial Plants
4.2 Sources of fault current 4.2.3 Induction machines:
"A squirrel-cage induction motor will contribute current to a power system short circuit. This is generated by inertia driving the motor in the presence of a field flux produced by induction from the stator rather than from a dc field winding. Since this flux decays on loss of source voltage caused by a fault at the motor terminals, the current contribution of an induction motor to a terminal fault reduces and disappears completely after a few cycles. Because field excitation is not maintained, there is no steady-state value of fault current as for synchronous machines...
induction motors are assigned only a subtransient value of reactance X"d “
“This value varies upward from the locked rotor reactance to account for the decay of the motor current contribution to the short circuit.”
4.5 Detailed procedure:
Xpu=per-unit reactance (base kVA)/kVA rating
IEC 60909-0 CH.3.8 Asynchronous motors
The impedance ZM = R M + jXM of asynchronous motors in the positive- and negative-sequence systems can be determined by:
ZM=1/(ILR/IrM)*UrM/SQRT(3)/IrM=1/(ILR/IrM)*UrM^2/SrM [Formula 26]
where:
UrM is the rated voltage of the motor;
IrM is the rated current of the motor;
SrM is the rated apparent power of the motor [SrM = PrM/((eff* COS(fi))]
ILR/IrM is the ratio of the locked-rotor current to the rated current of the motor.
 
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