How do you account for initial conditions when coordinating an across the line motor start ?

[bdn2004]

We've got an MCC that's rated for 1200Amps at 480V. There are a number of 200 Hp (FLA = 240A) motors connected to it.

Say the system runs normally at 800 Amps. When looking at the coordination curve (example attached), when the motor starts it shows it's 6X FLA for the start and the long starting time.... - but it doesn't show that 800A that's already on the line anywhere. Does that need to be accounted for when coordinating with the Main breaker ?

 

[xnuke]

You can't determine what will happen with the system if you only model individual components. Model the MCC with incoming feeder and all motors attached, add all overcurrent protection devices with appropriate settings, then look at how the main overcurrent protection device coordinates with downstream overcurrent protection devices and make sure the main overcurrent protection device won't trip on the load current that it sees normally. At that point, settings or breakers may need to be changed based on the analysis.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[che12345]

Dear Mr bdn2004

1. Say the system runs normally at 800 Amps. Then start [one] 200hp motor which would draw
6x240A=1440A for 5s . This is on top of the normal 800A i.e. (800A+1440A)=2240A for 5s.
After that the system runs at (800A+240A)=1040A; which is lower then the main breaker rated 1200A; i.e. no issue.

2. I reference to a ["typical" thermomagnetic MCCB In=1200A "hot trip condition"], the
tripping times are approx 1.5In 100s, 2In 40s, 3In 13s ...
Attention: use the actual breaker ["hot trip condition"] characteristic curve given
by the manufacture.

3. As for your case, during starting of a 200hp motor, the current would be (2240A/1200A)
= 1.26 time of In; for a duration of 5s.

4. Reference above 2, there should be no issue.

Che Kuan Yau (Singapore)

 

[7anoter4]

In my opinion, the 2240 A current from 0 to 0.08 sec it is the induction motor instantaneous peak
inrush current for 5 cycles. This current decreases fast. No other load is involved here only the induction motor.
After then for another 4 sec the starting [locked rotor] current [constant-more or less] and finally the rated current.

 

[che12345]

Dear Mr 7anoter4

1. Mr bdn2004 states that " Say when the system runs normally at 800 Amps...
....when the motor starts ...." It is explicit that the question is when the
system runs normally at 800 Amp as the (base load); when the motor starts on top
of the (based load).
2.1 "In my opinion, the 2240 A current from 0 to 0.08 sec it is the induction
motor instantaneous peak inrush current for 5 cycles " is irrelevant.
2.2 "No other load is involved here only the induction motor" This is NOT what
Mr bdn2004 is concerned.
3. With due respect Mr 7anoter4
a) can you kindly clarify whether [is there any issue] with the MCCB In=1200A with
base load of 800A and then start DOL (one) 200hp motor.
b) if [there are problems], please advise your reasoning.

Che Kuan Yau (Singapore)

 

[7anoter4]

Thank you, Che Kuan Yau.
From the attached curve: motor CNPC-501-4 200 HP 480 V[constant]
Since the supply voltage is constant it is not the case of the actual motor on site but it is a curve produced in laboratory conditions.
By the way, it's only a coincidence if Istart and Iload gives the result 2240 A[taking absolute values]. Actually Istart =1440 A cosfi 0.3[?] and Iload=800 A cosfi=0.9[?] then total current it is only 2027.9 A. Even if we take 1200 A as full current steady state on MCC [cosfi=0.9] and we subtract one motor rated current and we add the motor starting current the total current will be 2161.9 [still not 2240 A]