Feeder Protection -WA Admin

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snapco

Member
I have worked out the following and asking if it right or wrong on how I arrived at the answer-

Calc feeder protective device- using Inverse Time Breaker

240V 3ph delta system

1 15hp 3ph
2 10hp 3hp
1 50hp 3hp 5 min intermitt duty rated

50hp- 130 amps 130x2.5=325 use 350
15hp- 42 amps 42
10hp- 28 amps 28x2=56 56
total 448

a) 300 b) 350 c) 400 d) 450

I came up with answer 'c'-

I figured that the 5min intermitt only was for sizing the conductors and was not used in branch circuit or feeder device calc's

Could anyone advise me if I am right or wrong as I think it is one I missed on a test- thanks

[ March 24, 2005, 10:35 AM: Message edited by: snapco ]
 
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traderx

Member
Re: Feeder Protection -WA Admin

Hi snapco... this is what I got

50hp at 240v 3P = 130A
130A X .85 (T430.22e) X 250% (430.52)
130 X .85 = 110.5 X 2.5 = 276.25

276.25 add to rest
15hp at 240v 3P = 42
10hp at 240v 3P = 28 x 2 = 56
Total equals 374.25, must step down (240) to 350
which I answer B....

Did I miss any thing any one?
 
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snapco

Member
Re: Feeder Protection -WA Admin

Unfortunatly I think we are both wrong, I would hope that someone could help us out to calc the correct answer as I am also taking the test soon-

Thanks anyone for the info
 
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dillon3c

Senior Member
Re: Feeder Protection -WA Admin

Snapco,
see table 430.22(e) and article 430.62(a)-->In the application of this motor question.

[ March 23, 2005, 11:29 AM: Message edited by: dillon3c ]
 
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snapco

Member
Re: Feeder Protection -WA Admin

Dillon3C,

Am I completely on the wrong path when I think that 430.22(e) is for sizing conductors and is not used for sizing branch-circuit short-circuit and ground-fault protective devices? Is not the protective device based on section IV per 430.52 as stated in 430.62(a)? When determining the feeder it does not reference 430.22(e) at any time. Maybe this is the stumbleing block I am having by looking at it wrong.

Thanks to anyone who can point me in the correct path.
Snapco
 
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luke warmwater

Senior Member
Re: Feeder Protection -WA Admin

Doh.
I see tradrx did.

But why did you step down??

After being jumped around, and sent back to 240.4(B), doesn't that permit you to go to the next higher standard?

My answer: C) 400 amps
 
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traderx

Member
Re: Feeder Protection -WA Admin

Originally posted by luke warmwater:


But why did you step down??
Click to expand...
Look at 430.62... this is dealing with the feeder of the load and not the branch circuit. Correct me if I'm wrong any one

[ March 24, 2005, 01:41 AM: Message edited by: traderx ]
 
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dillon3c

Senior Member
Re: Feeder Protection -WA Admin

Originally posted by luke warmwater:
That points you to 430.52

Does 430.52(C)(1)Exception#2(c)apply to the question?
Click to expand...
No Luke,
It doesn't.Feeder suppling multible motors..Branch-circuit short-circuit ground-fault
protection of table 250.52,you not permitted to round up to next standard size.Not permitted in useing 250.52 exception #2.

In useing 250.62,feeder for multi-motor application,if size calculated in useing 450.52, not standard size,you must(round down)to the next standard size.Not to exceed..250.62-(not greater than)-
 
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snapco

Member
Re: Feeder Protection -WA Admin

It would be nice to get some input from the instructors that monitor this Forum-

Thanks to any of you for helping with this issue that seems to have some questions to still be answered which calc is the correct direction-

Snapco
 
wirenut1980

wirenut1980

Senior Member
Location
Plainfield, IN
Re: Feeder Protection -WA Admin

There is an example in the 2002 NEC Appendix D Example D8. It looks like the number you are looking for is the sum of the largest branch circuit protective device plus the sum of the full load currents of the other motors, and you have to round down to the nearest standard OCPD. I'm going to try and work it out and see what I get. :)
 
wirenut1980

wirenut1980

Senior Member
Location
Plainfield, IN
Re: Feeder Protection -WA Admin

Ok, I get:

50 hp Amps at 130 A from Table 430.150.

Calculate OCPD size for largest motor.
130 A*0.85=110.5 A from table 430.22(E)
110 A*250%=276.25 A from Table 430.52.
Size down to next standard OCPD from 430.62(A) Exception 1.
Size down to 250 Amps OCPD

Add in FLC of other 2 motors
250 A + 42 A + (28*2) A = 348 A

But I am confused as to whether I go up or down to the next standard size breaker. Because 430.62(A) Exception seems to be talking about the branch circuit OCPD rating and not the feeder OCPD rating. If I size down I get a). If I size up I get b). :confused:
 
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snapco

Member
Re: Feeder Protection -WA Admin

Thanks first for the feedback and help- I am still a little confused and do not want to sound ignorant- in the example in Annex D example 8 in calculating the Feeder scgfp they refer to using the figure from the bcscgfp and not conductor ampacity from 430.22- you do not use 430.22 to determine overload protection or branch circuit protection, why would you use it to determine feeder protection? Maybe I just do not "get it" and that is why I struggle with what appears to be a simple question-
 
charlie b

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Semi-Retired Electrical Engineer
Re: Feeder Protection -WA Admin

We have one feeder that supplies four motors. We are told to use the ?inverse time breaker? part of the table. We are not told whether that applies to the feeder protection only (i.e., the focus of the question) or to all four individual motors. This impacts whether Exception 1 to 430.62(A) applies.

For the moment, I will assume that the four motors have instantaneous trip circuit breakers. More to the point, I will assume that the 50 hp motor has an instantaneous trip circuit breaker. Given this assumption, I would do the calculation as follows:
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">50 hp @ 230 volts is 130 amps (same as others have shown)</font>
<font size="2" face="Verdana, Helvetica, sans-serif">NOTE: I would ignore the ?5 minute intermittent? information. I agree with snapco?s original statement that this applies only to sizing the branch circuit conductors. It is what the engineering world calls a ?confusion factor.? My evidence is that Table 430.52 gives percentages of the ?full load current,? not the ?full load current as adjusted for continuous or intermittent duty or for any other reasons.?
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">130 times 250% is 325 amps.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Now I apply Exception 1 to 430.62(A). For the purposes of the calculation, I have to use a number not higher than 325. I can?t round up to 350 (the next higher size). But I don?t need to round down to 300 (the next lower size) yet. You don?t do any rounding until you have finished the calculation. That is a basic engineering principal that I believe applies here.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Add the 42, the 28, and the other 28, and you get 423 amps.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">430.62(A) says I can?t be higher than that 423 amps. Therefore, I select the next available lower value: 400 amps.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">My answer is ?C.?
 
wirenut1980

wirenut1980

Senior Member
Location
Plainfield, IN
Re: Feeder Protection -WA Admin

snapco, I believe 430.22 does apply even though it is refering to conductor sizing. The reason I think it applies is because when you size an OCPD, you are sizing it to protect the conductors. And sizing the conductors comes from the expected/estimated load to be put on them.

Here is something I noticed in the example. It seems to apply 430.62(A) exception 1 to the feeder OCPD, but they way I read it I was thinking it applied to the branch circuit OCPD.
 
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dillon3c

Senior Member
Re: Feeder Protection -WA Admin

wirenut,
In your calculation,you missed adding a 10hp motor.There are two 10hp motors and one 15 hp,along w/the 50hp duty rated per 430.22(e)

table 430.250
50=130amperes
15=42 amperes
10=28 amperes
10=28 amperes

table 430.22(e)
130amperes x 0.85=110.5 x 2.50=276.25
42amperes
28amperes
28amperes
------------
374.25
article 430.62= 350 ocp device

edited: in all due respect,not being an engineer,have no knowlege of the (confusion factor) just test answer factor...

[ March 24, 2005, 12:01 PM: Message edited by: dillon3c ]
 
charlie b

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Semi-Retired Electrical Engineer
Re: Feeder Protection -WA Admin

Originally posted by wirenut1980: . . . when you size an OCPD, you are sizing it to protect the conductors. And sizing the conductors comes from the expected/estimated load to be put on them.
Click to expand...
That is not true for the OCPD of a motor circuit. If I can select a circuit breaker for more than double the full load amps of a motor, then that circuit breaker is not going to protect the conductors from an overload. Protect from a fault? Yes. Overload? No. Rather, the overload protection (i.e., for the conductors) comes from the motor?s control circuit (e.g., from thermal overloads).
I believe 430.22 does apply even though it is referring to conductor sizing.
Click to expand...
I disagree. You get the value of a motor?s ?full load current? from Tables 430.147 through 430.150. Table 430.22(E) does not alter that value. What Article 430.22(B) does is to say that if you are sizing a conductor, you select it with an ampacity no smaller than the percentage (of ?full load current?) as shown in Table 430.22(E). I say again, this does not change the value of the ?full load current.?

When you work with Table 430.52, you start with the ?full load current.? Nothing in Table 430.52 refers you back to Table 430.22(E), nor to any other place in the code. I reiterate my view that the 85% factor (for the 5 minute intermittent motor) does not come into the present calculation (of the feeder OCPD).
 
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