QUESTION ON A TEST

[solis]

I recently took a Master's test and and had two questions that keep bothering me, I'm doing something wrong, help. this question was on the test,
First question, hope I remember all the parts,
"on a 2 HP 230V 3P motor with a B code letter, what is the locked rotor current?
The answers were:
51A
55A
57.5A
73A (I think)
I hope I am remembering the elements of the question right

#2
3 parallel runs 600 KCMILs conductors, what is the size of the bonding jumper, I said 2/0

thanks in advance

 

[augie47]

On Question #1 none of those answers make any sense for a 2HP
On Question #2, more info is needed. It would depend on what the bond jumper was accomplishing and if it was service or feeder. If all you had was wire size one would assume service bonding in which case it would be a 250 Cu for a main bonding jumper or for one of parallel conduits, a 1/0 Cu

 

[solis]

It was for a main bonding jumper for a service. What is the code reference? I was scrambling all over 250 to find it during the test. Should have got it right, but looks like I didn't. Thanks

 

[Smart $]

It was for a main bonding jumper for a service. What is the code reference? I was scrambling all over 250 to find it during the test. Should have got it right, but looks like I didn't. Thanks

250.28(D) is where you start.

At that size, you'll end up going with 12.5% of ungrounded combined (600?3?12.5%=225).

 

[solis]

Thanks, missed the FPN's on that one, 250.102(C)(1). Had studied that and actually highlighted it, was a miss on my part, thanks again for clearing it up for me

 

[Bill Annett]

Good Morning. In the 2011 Edition of the Ugly?s Electrical Reference Book ( pg44), They have a chart for Maximum Motor Locked-Rotor current in amps, Two and Three Phase, Design B,C and D. The chart shows the 2HP at 230 Volt would have amperage of 50 amps. I hope that this is what you were looking for

 

[solis]

Bill
Thanks, it is, just couldn't get the formula I looked up after to match any answer. Maybe I don't know how to apply the values to the formula I was using HP x 1000 x 3.54(highest value of B code)/ 230x1.73

 

[iceworm]

... 2011 Edition of the Ugly?s Electrical Reference Book ( pg44), They have a chart for Maximum Motor Locked-Rotor current in amps, Two and Three Phase, Design B,C and D. The chart shows the 2HP at 230 Volt would have amperage of 50 amps. ...

Could be. T430.7.B indicates the locked rotor kva per horsepower for a Code B is 3.15 - 3.54

For 230V, 3hp, LR (kva) = 6.30 to 7.08

Locked rotor current (lower) = (6.3 x 1000) \ 230 \ 1.732 = 15.8A
Locked rotor current (upper) = (7.08 x 1000) \ 230 \ 1.732 = 17.8A

However, if UERB is quoting Table T430.251.B, Design B, C, D, then 50A is correct.

The question makes no mention of "Design Letter", only "Code Letter". That seems to indicated T430.7.B applies.

I like augie's answer best.

Personal Opinion: T430.251.B and T430.7.B are evidence the code panel does not have a clear picture of inrush/locked rotor current.

slow poster, edit to add:

just couldn't get the formula I looked up after to match any answer. Maybe I don't know how to apply the values to the formula I was using HP x 1000 x 3.54(highest value of B code)/ 230x1.73

Yes, the way the question was framed, I think you used a good method

ice