Wire sizing peak vs normal current

Status
Not open for further replies.
T

ThatShouldHaveWorked

Member
Location
South West USA
Occupation
Engineer
I have an industrial machine coming. The spec sheet has a listed current consumption of 25A with a peak of 110A for 15 seconds.

If we ignore all the corrections and adjustments for simplicity. Logically I should be able to size the wire to the breaker and not the peak.

For the life of me I can't find anything in the NEC that address peak vs normal current as it relates to conductor sizing.

Thoughts?

Thank you
 
Welcome to the forum.

I'm also not sure how to address this. 15 seconds is much longer than a typical motor starting current.

Does the manufacturer recommend a particular breaker size?
 
It looks like the nameplate breaker size is 30A.

However the suppling electrical panel takes Square D - EDG breakers and the trip curve shows I can be a little under half the peak for 15s to avoid nuisance tripping. So I'm planning to use a 60A breaker.
 
So, the question is whether the long starting current will heat up wire sized for 60a. I'm sure it would wire sized for 25a.
 
I second Larry's question about whether the manufacturer specified a certain breaker.
I agree that you should be able to size the wire's ampacity based on the breaker rating in your application.
 
210407-1537 EDT

ThatShouldHaveWorked:

You are list as an engineer, and I assume electrical. Thus, I would assume you studied thermal and electrical problems and considerations.

First, uninsulated copper wire in free air will have some nominal current rating based on material, somewhat by diameter, ambient temperature, and desired maximum operating current. This is a steady state test.This same wire will have a much higher fusing current value, and is dependent upon ambient temperature.

When you add insulation to the wire you do at least two major things. Change allowed maximum steady-state temperature, and the insulating effect of the insulation.

To some extent the trip-time characteristics of a breaker or fuse are related to wire tolerance to peak current. Generally there are tables that relate wire size for a particular breaker. Thus. it is probably safe to assume wire sized to a particular breaker will be adequately protected.

If you look at what happens to wire vs current there are some items of importance.

Temperature rise above ambient is about proportional to power dissipated in the wire per unit length. If 1 W produces a rise of 1 C, then 10 W produces about 10 C rise.

For most wire as temperature rises so does resistance, and therefore at constant current the resistance and power dissipation both go up.

Since wire has mass it also has a thermal time constant. So some fuses are rated based on I^2*t, and so you can apply the same concept to wire.

QO breakers also have an instantaneous trip characteristic. I believe this is about 6 times rating, This is not really instantaneous, but probably means about 1/2 to 1 cycle at 60 Hz.

Another factor to consider is the repetition rate of the load. A 50% duty cycle would likely be a big problem.

If your steady-state current is 25 A, and starting current is 110 A, then then the energy ratio for the 15 S is (110/25)^2 = 19.4 . Depending upon the thermal time constant of the wire this can produce lot of excess temperature rise.

Another factor to consider is the type of load and the actual voltage at the load. A motor at the end of a long line can be a big problem, just a resistor and it may not matter.

.
 
How do you know the peak current will exist at 100% for the entire 15 sec starting period?

Common motor starting current, therefore running overload protection, is 6 times for 20 sec in the US, and 10 sec in the IEC world.
 
jim dungar said:
How do you know the peak current will exist at 100% for the entire 15 sec starting period?

Common motor starting current, therefore running overload protection, is 6 times for 20 sec in the US, and 10 sec in the IEC world.
Click to expand...

It's an industrial machine. The peak is 110A for 15 seconds because that is what the OEM says it is. It will draw 25A the remainder of the time when operating because that is what the OEM says it will. Additionally given that the blower is under enough load to require Wye-Delta starting 110A for 15 seconds does not seem unrealistic to me.
 
gar said:
210407-1537 EDT

ThatShouldHaveWorked:

You are list as an engineer, and I assume electrical. Thus, I would assume you studied thermal and electrical problems and considerations.

First, uninsulated copper wire in free air will have some nominal current rating based on material, somewhat by diameter, ambient temperature, and desired maximum operating current. This is a steady state test.This same wire will have a much higher fusing current value, and is dependent upon ambient temperature.

When you add insulation to the wire you do at least two major things. Change allowed maximum steady-state temperature, and the insulating effect of the insulation.

To some extent the trip-time characteristics of a breaker or fuse are related to wire tolerance to peak current. Generally there are tables that relate wire size for a particular breaker. Thus. it is probably safe to assume wire sized to a particular breaker will be adequately protected.

If you look at what happens to wire vs current there are some items of importance.

Temperature rise above ambient is about proportional to power dissipated in the wire per unit length. If 1 W produces a rise of 1 C, then 10 W produces about 10 C rise.

For most wire as temperature rises so does resistance, and therefore at constant current the resistance and power dissipation both go up.

Since wire has mass it also has a thermal time constant. So some fuses are rated based on I^2*t, and so you can apply the same concept to wire.

QO breakers also have an instantaneous trip characteristic. I believe this is about 6 times rating, This is not really instantaneous, but probably means about 1/2 to 1 cycle at 60 Hz.

Another factor to consider is the repetition rate of the load. A 50% duty cycle would likely be a big problem.

If your steady-state current is 25 A, and starting current is 110 A, then then the energy ratio for the 15 S is (110/25)^2 = 19.4 . Depending upon the thermal time constant of the wire this can produce lot of excess temperature rise.

Another factor to consider is the type of load and the actual voltage at the load. A motor at the end of a long line can be a big problem, just a resistor and it may not matter.

.
Click to expand...

As for duty cycle It will hit that peak 3 to 10 times per hour.

As I said before logically there shouldn't be a problem sizing the wire to the breaker (with appropriate corrections to limit thermal buildup and voltage drop). I just couldn't find anywhere in the code that says anything about it. I'm ok with it being silent about it but wanted to see if I was missing something.

Yes electrical engineering. I am also a journeyman electrician if that makes a difference to you.
 
ThatShouldHaveWorked said:
It's an industrial machine. The peak is 110A for 15 seconds because that is what the OEM says it is. It will draw 25A the remainder of the time when operating because that is what the OEM says it will. Additionally given that the blower is under enough load to require Wye-Delta starting 110A for 15 seconds does not seem unrealistic to me.
Click to expand...
It does not seem unreasonable to me either, in fact it is slightly less than what is allowed by a standard Class 20 thermal overload relay. Typical motor starting curves are not square shaped they usually taper off as the load accelerates. However the transition from wye to delta may cause a second current spike.

Molded case breakers listed to UL 489 will properly protect conductors chosen and applied per the NEC. If your protection needs to be upside due to the starting duty then so should your conductors.
 
Hey,
Say something, short-time current ratings define different circuit breaker performance characteristics. ... because they do not need to withstand a high current for an extended time delay...ha, i need to organize the language..
 
This is one of the drawbacks to ordering something from overseas and not requiring it to be UL listed. UL would REQUIRE that the mfr make a CLEAR indication of FLC, or if there are combination loads, then values of MCA (Minimum Circuit Ampacity) and MOP (Maximum Overcurrent Protection). In this case that is going to have to be up to you, and the calculations for those values are;

MCA = 1.25 x (Load 1 + Load 2 + Load 3 + Load 4 etc.)
MOP = (2.25 x load 1) + load 2 + load 3 + load 4 etc.

Unfortunately, you don't have all of the necessary data you will need from that chart you posted. I'm not ever really sure it is a piece of equipment with multiple loads.
 
Status
Not open for further replies.
Top