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Motor contribution current from LV motors. 1
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Additional info: I calculate SC at 110 kV bus, many LV motors 0.4 kv connected to transformers of 10/0.4 kV, then transformers 35/10 kV and then 110/35 transformers.
Group of LV motors 0.4 kV => transformers 10/0.4 kV => transformers 35/10 kV => transformer 110/35 kV.
Group of LV motors 0.4 kV => transformers 10/0.4 kV => transformers 35/10 kV => transformer 110/35 kV.
Dear Mr. beyond86 (Electrical)(OP)10 Dec 22 12:07
" #1....When we perform SC calculation, should taking into account motor contribution current from LV motors (0.4 kV or less) or only from MV motors?"
1. All motors LV and MV (except those connected to simple VFD ) do contribute "motor-feed back" during short-circuit. Therefore, motor contribution current shall be taking into account on SC calculation. There are conditions and recommendations, see 3.
"....#2. I calculate SC at 110 kV bus,.... many LV motors 0.4 kv connected to transformers of 10/0.4 kV, then transformers 35/10 kV and then 110/35 transformers...."
2. All 0.4kV motors do contribute to short-circuit current during short circuit. But are you considering its contribution to 110kV bus ? The answer is NO. See 3.
"....#3. In which standard I can find information about this? "
3. In the IEC world, IEC 60909 covers this very well.
Che Kuan Yau (Singapore)
" #1....When we perform SC calculation, should taking into account motor contribution current from LV motors (0.4 kV or less) or only from MV motors?"
1. All motors LV and MV (except those connected to simple VFD ) do contribute "motor-feed back" during short-circuit. Therefore, motor contribution current shall be taking into account on SC calculation. There are conditions and recommendations, see 3.
"....#2. I calculate SC at 110 kV bus,.... many LV motors 0.4 kv connected to transformers of 10/0.4 kV, then transformers 35/10 kV and then 110/35 transformers...."
2. All 0.4kV motors do contribute to short-circuit current during short circuit. But are you considering its contribution to 110kV bus ? The answer is NO. See 3.
"....#3. In which standard I can find information about this? "
3. In the IEC world, IEC 60909 covers this very well.
Che Kuan Yau (Singapore)
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I doubt if such a standard exists. For details of low voltage induction motors short-circuit contribution see IEEE 551/2006.
Based on ANSI IEEE 551/2006 IEEE Recommended Practice for Calculating Short-Circuit Currents in Industrial and Commercial Power Systems
Chapter 6 Calculating ac short -circuit currents for systems with contributions from induction motors
IEEE 141/1993 chpt. Low-voltage duties.
The standards for interrupting equipment allow a modified subtransient reactance for a group of low-voltage induction and synchronous motors fed from a low voltage substation. If the total of motor horsepower ratings at 480 or 600 V is approximately equal to (or less than) the transformer self-cooled rating in kilovolt-amperes, a reactance of 0.25 per unit based on the transformer self-cooled rating may be used as a single impedance to represent the group of motors.
IEC 60909-0 Chapter 3.8.2 [see attachment]
Based on ANSI IEEE 551/2006 IEEE Recommended Practice for Calculating Short-Circuit Currents in Industrial and Commercial Power Systems
Chapter 6 Calculating ac short -circuit currents for systems with contributions from induction motors
IEEE 141/1993 chpt. Low-voltage duties.
The standards for interrupting equipment allow a modified subtransient reactance for a group of low-voltage induction and synchronous motors fed from a low voltage substation. If the total of motor horsepower ratings at 480 or 600 V is approximately equal to (or less than) the transformer self-cooled rating in kilovolt-amperes, a reactance of 0.25 per unit based on the transformer self-cooled rating may be used as a single impedance to represent the group of motors.
IEC 60909-0 Chapter 3.8.2 [see attachment]
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According to IEC 60909-0 Chapter 3.8.2 we may neglect the motors' contribution if
ΣPrM/ΣSrT=<0.8/(c*100*ΣSrT/sqrt(3)/UnQ/I"Q-0.3)
Let’s take an example
one two-winding transformer of 1000 kVA 480 V secondary 5% short-circuit voltage supplying 3 motors of 75 kW , 4motors of 50 kW and 10 motors of 25 kW.
ΣPrM=3*75+4*50+10*25=675 kW
ΣSrT=1000 kVA
Neglecting the System impedance and taken only transformer impedance the short-circuit current will be 24 kA. For low-voltage c=1.05 then
I”Q=24 kA UnQ=480 V 0.8/ (c*100*ΣSrT/sqrt(3)/UnQ/I"Q-0.3)= 0.8/(1.05*100*1000/sqrt(3)/480/24-0.3)= 0.161215 but ΣPrM/ΣSrT=675/1000=0.675
In this case we cannot neglect the motors' contribution. However, if we take only 1*75+1*50+1*25 kW then ΣPrM/ΣSrT=150/1000=0.15 and we may neglect the motors.
ΣPrM/ΣSrT=<0.8/(c*100*ΣSrT/sqrt(3)/UnQ/I"Q-0.3)
Let’s take an example
one two-winding transformer of 1000 kVA 480 V secondary 5% short-circuit voltage supplying 3 motors of 75 kW , 4motors of 50 kW and 10 motors of 25 kW.
ΣPrM=3*75+4*50+10*25=675 kW
ΣSrT=1000 kVA
Neglecting the System impedance and taken only transformer impedance the short-circuit current will be 24 kA. For low-voltage c=1.05 then
I”Q=24 kA UnQ=480 V 0.8/ (c*100*ΣSrT/sqrt(3)/UnQ/I"Q-0.3)= 0.8/(1.05*100*1000/sqrt(3)/480/24-0.3)= 0.161215 but ΣPrM/ΣSrT=675/1000=0.675
In this case we cannot neglect the motors' contribution. However, if we take only 1*75+1*50+1*25 kW then ΣPrM/ΣSrT=150/1000=0.15 and we may neglect the motors.
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