group of motors

Status
Not open for further replies.
E

electricrw

Member
240v 3phase system. Question is: What is the size of the feeder inverse time circuite breaker on a group of motors.
moters are 25hp 10hp 10hp 50hp. The 50hp is a 5min intermittent duty cycle.

25hp= 430.250 = 68amps
10hp= 430.250 = 28amps
10hp= 430.250 = 28amps
50hp= 430.250 = 130amps 5min intermittent= 85%

when answering this question I keep getting mixed up with 430.24 exception 1 and 430.52 exception 1 and 430.53c4. please help with the answer and maybe explain my code refrence. But would just be happy with answer
thank looking for answer based on 05 code
 
Last edited:
50HP starts every 5 minutes ?

50HP starts every 5 minutes ?

NEMA MG10 States a 50HP is good for 3.5 starts per hour. If you are starting the 50HP over and over I hope this motor is on a drive. If not as the motor restarts over and over O/L' and the CB's trip unit heats up often tripping OL's and causing CB to trip. Have you looked at the NEC Annex D, Example D8 Also see 430.62... I'm using the 08 NEC, but think this hasn't changed much. There was a change to 430.52(C), But as I believe I understand it you should be in 430.53 (motors in groups on one branch circuit)
 
I'll take a stab at it:

10 HP-28 amps
10 HP-28 amps
25 HP-68 amps
50 HP-110.5 amps 130 x 85%=110.5

430.52
Largest Motor FLC--110.5 x 150%=166 amps--->175 amp brkr.

430.62(A)
175+68+28+28=299-->250 amp brkr.
 
Cow said:
I'll take a stab at it:

10 HP-28 amps
10 HP-28 amps
25 HP-68 amps
50 HP-110.5 amps 130 x 85%=110.5

430.52
Largest Motor FLC--110.5 x 150%=166 amps--->175 amp brkr.

430.62(A)
175+68+28+28=299-->250 amp brkr.
Click to expand...

Thanks for the answer. How did you know the question was asking for a wound rotor motor. I think I got this problem wrong thinking to use 250%
 
In class I was taught to use wound rotor. Doesn't necessarily mean it's right, it's just the way I was taught.;)
 
5min intermittent duty motor

5min intermittent duty motor

Is a 5 min intermittent duty motor considered a wound roter motor thus using 150%

240v 3phase system. Question is: What is the size of the feeder inverse time circuite breaker on a group of motors.
moters are 25hp 10hp 10hp 50hp. The 50hp is a 5min intermittent duty cycle.

25hp= 430.250 = 68amps
10hp= 430.250 = 28amps
10hp= 430.250 = 28amps
50hp= 430.250 = 130amps 5min intermittent= 85%

when answering this question I keep getting mixed up with 430.24 exception 1 and 430.52 exception 1 and 430.53c4. please help with the answer and maybe explain my code refrence. But would just be happy with answer
thank you, looking for answer based on 05 code
 
Let me try this i am not sure if this will match up right anyway here it goes;...


68+ 28+ 28= 124a. (*)


130a X .85 =110.5a but i will add 125% factor to this 138.125a

now this what i did i will take the largest motor {68 amp } X 1.25% = 85a


now i will sum it up

85+28+28+138.125 = 279.125a

i know it sound little high but keep in your mind i did took one largest motor in the cont group and figure that one out first then took the 50 HP figures [ i ran by like induction motor due OP did not mention which type of motor ] so i readjust the figures like that then add all the motor loads [ the 125% is allready includeing both large motors] and came up 279 amp and you can use 300 amp breaker for this sum of motor.

hope that should ring the bell right if i am wrong let me know

Merci, Marc
 
I have been given 2different answers 3different ways. I want to thank you all, but need right answer. I talked to another master electrician friend of mine and thaught I would run his thaughts with you all. Please evaluate this answer;

largest motor
50hp= 130amps x .85 for 5min intermittent =110.5a
430-52
largest motor 110.5 x 250%=275a
can go up 430.52C1exc1
so over current device for this motor is 300amps
then 430.24 exc 1. Must compare the largest of the other motors x 125% with the largest intermittent duty and pick the largest. that would be 68 x 125% =85a is smaller

So must pick 300 then add the rest =
300+68+28+28= 424a must go down 400amp answer
Does this sound good? All other answers from this forum are below.

----------------------------------------------------------------
240v 3phase system. Question is: What is the size of the feeder inverse time circuite breaker on a group of motors.
moters are 25hp 10hp 10hp 50hp. The 50hp is a 5min intermittent duty cycle.

25hp= 430.250 = 68amps
10hp= 430.250 = 28amps
10hp= 430.250 = 28amps
50hp= 430.250 = 130amps 5min intermittent= 85%

when answering this question I keep getting mixed up with 430.24 exception 1 and 430.52 exception 1 and 430.53c4. please help with the answer and maybe explain my code refrence. But would just be happy with answer
thank looking for answer based on 05 code

----------------------------------------------------------------

10 HP-28 amps
10 HP-28 amps
25 HP-68 amps
50 HP-110.5 amps 130 x 85%=110.5

430.52
Largest Motor FLC--110.5 x 150%=166 amps--->175 amp brkr.

430.62(A)
175+68+28+28=299-->250 amp brkr.
---------------------------------------------------------------

10 HP-28 amps
10 HP-28 amps
25 HP-68 amps
50 HP-110.5 amps 130 x 85%=110.5

430.52
Largest Motor FLC--110.5 x 150%=166 amps--->175 amp brkr.

430.62(A)
175+68+28+28=299-->250 amp brkr.

------------------------------------

68+ 28+ 28= 124a. (*)


130a X .85 =110.5a but i will add 125% factor to this 138.125a

now this what i did i will take the largest motor {68 amp } X 1.25% = 85a


now i will sum it up

85+28+28+138.125 = 279.125a

i know it sound little high but keep in your mind i did took one largest motor in the cont group and figure that one out first then took the 50 HP figures [ i ran by like induction motor due OP did not mention which type of motor ] so i readjust the figures like that then add all the motor loads [ the 125% is allready includeing both large motors] and came up 279 amp and you can use 300 amp breaker for this sum of motor.


----------------------------------------------------------------
 
Status
Not open for further replies.
Top