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What size was the SEC cable? I would not rate the service at 150 unless the SEC is sized to support it.

I rated the service at 100Amps due to the SEC size. The clarification I am looking for is, Do we add these tied breakers for total amperage capacity, if the cables were sized properly?

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What size was the SEC cable? I would not rate the service at 150 unless the SEC is sized to support it.

I rated the service at 100Amps due to the SEC size. The clarification I am looking for is, Do we add these tied breakers for total amperage capacity, if the cables were sized properly?

Yes to that last. It looks like the panel is being fed by a pair of 75-amp double pole breakers hooked up in parallel. Effectively a "quad" breaker with 75-amps at each of the 4 poles. That would give you 150-amps available on each leg.

What are you calling the SEC's that you "rated the service at"? I can only see the bottom of one of the main lugs, and not the SEC's, but the split feeders from there to the breakers' lugs look big enough for the 75-amp load each would have to handle.

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150 amps would be my pick. Don't you hate that when they won't just stick a label on it?

If the load on one 75amp breaker exceeds 75 amps and it trips, it also trips the other breaker. How can you call that 150? The panel as shown in the picture will only take 75amp max on either side so that's what it should be rated at.

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150 amps would be my pick. Don't you hate that when they won't just stick a label on it?

If the load on one 75amp breaker exceeds 75 amps and it trips, it also trips the other breaker. How can you call that 150? The panel as shown in the picture will only take 75amp max on either side so that's what it should be rated at.

NO! Because it is NOT a pair of 75-amp SINGLE pole breakers. Look at the wiring to them and it is obvious that they are TWO, DOUBLE-POLE 75-amp breakers being used IN PARALLEL. All things being roughly equal, the load will split fairly evenly between the two breakers. A 240-volt 75-amp load would therefore only put 37.5-amps on each of the four poles. It would take 150-amps (or very close) to first trip any one of the breaker poles which would then internally trip the other pole within the same breaker and then the other two pole breaker by means of the handle tie.

If it helps, try to picture the whole thing as four 75-amp single pole breakers with a long handle tie connecting them all.

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NO! Because it is NOT a pair of 75-amp SINGLE pole breakers. Look at the wiring to them and it is obvious that they are TWO, DOUBLE-POLE 75-amp breakers being used IN PARALLEL. All things being roughly equal, the load will split fairly evenly between the two breakers. A 240-volt 75-amp load would only put 37.5-amps on each of the four poles. It would take 150-amps (or very close) to first trip any one of the breaker poles which would then internally trip the other pole within the same breaker and then the other two pole breaker by means of the handle tie.

If it helps, try to picture the whole thing as four 75-amp single pole breakers with a long handle tie connecting them all.

It may very well be but look at the length of the conductors feeding the main. Poorly done and I don't care if the factory did it, it's wrong. The conductors serving a common leg are not equal in length. There's a way to do it and this is not it. Currents will be imbalanced.

I'd call it a 150 but I doubt you'll get 150 in practice because one will trip before the other.

Marc

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Well, I did say "very close". Marc, I used an online voltage drop calculator (see photo). Given that the voltage drop for a 2 foot 2-AWG Aluminum feeder at 75 amps is only .1% (one tenth of one percent), how much difference do you think a 4"(?) difference will make to the current carrying capacity? I agree there will always be some imbalance. Hell, even if the conductors were the same length, I would expect some other imperfections. However, I suspect in this case it would be a very small amount, and given the topic of the conversation, one that we could largely ignore.

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310.4 only allows 1/0 in alum or copper to be ran in parallel so you would have to have it feeding 240 amp if alum or 300 amps copper

The cable in the photo is a component of a rated assembly, so §310.4 isn't applicable.

§ 310.4 concerns running conductors in parallel to serve large loads because single conductors large enough are either impractical or unavailable.

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Well, I did say "very close". Marc, I used an online voltage drop calculator (see photo). Given that the voltage drop for a 2 foot 2-AWG Aluminum feeder at 75 amps is only .1% (one tenth of one percent), how much difference do you think a 4"(?) difference will make to the current carrying capacity? I agree there will always be some imbalance. Hell, even if the conductors were the same length, I would expect some other imperfections. However, I suspect in this case it would be a very small amount, and given the topic of the conversation, one that we could largely ignore.

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VD is the wrong way to analyze it. Let say that one conductor is twice as long as the other one, just as an example. The conductors are joined on each end so they both have the same voltage across them. The current will be twice as high on the short one than the longer one because it has only half the resistance. That means 2/3 of the total current is in the short one and only 1/3 in the long one. It also means that the total ampacity of that parallel set-up is only 50% higher than the ampacity of either conductor. 150% of 75 is 112 amps which is much less than 150.

Looking at the top of the first photo again...the two horizontal conductors at the top are in parallel. One is not twice the length of the other but it's considerable.

As an armchair inspector with just that photo, I'd still call it a 150A service, but if I were inspecting it, I'd take out my measuring tape and android.

Marc

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Well, I did say "very close". Marc, I used an online voltage drop calculator (see photo). Given that the voltage drop for a 2 foot 2-AWG Aluminum feeder at 75 amps is only .1% (one tenth of one percent), how much difference do you think a 4"(?) difference will make to the current carrying capacity? I agree there will always be some imbalance. Hell, even if the conductors were the same length, I would expect some other imperfections. However, I suspect in this case it would be a very small amount, and given the topic of the conversation, one that we could largely ignore.

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VD is the wrong way to analyze it. Let say that one conductor is twice as long as the other one, just as an example. The conductors are joined on each end so they both have the same voltage across them. The current will be twice as high on the short one than the longer one because it has only half the resistance. That means 2/3 of the total current is in the short one and only 1/3 in the long one. It also means that the total ampacity of that parallel set-up is only 50% higher than the ampacity of either conductor. 150% of 75 is 112 amps which is much less than 150.

....

Marc

Marc, I'm not an electrical engineer, but I have to call BS on that.

First of all, voltage drop is due to resistance and I'm sure that it corresponds nicely with the current carrying capacity.

Next, let's say I punch two holes exactly the same size in a bucket of water and then attach two hose of the same diameter, but one say 4' long and the other 2'. There is some resistance in the hoses and the 4' long one will have twice as much of that, but it's not the same as pinching one hose to half the cross section. You are effectively saying that just having double the length of hose will shut down the flow by half. That's not true and I guarantee you that you would have a hard time actually measuring the difference in flow rate from the bucket.

Conductors act somewhat similarly to the hoses. Doubling the length may double the resistance but that initial resistance is a percentage of the capacity of the wire (based on the other variables as well). In my sample, with those loads and lengths, it would seem that the 2' wire would retain 99.9% of its capacity while the 4' one is reduced to 99.8%. That is an awful long way from your two third/one thirds.

I know I'm simplifying things a bit, but I'll bet you a case of beer that I'm a lot closer to the facts than you are. Where's that Doug Hanson?

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Wait a minute. If matto rated the service at 100 amps due to the size of the SEC, and the conclusion that the dual feed 75's equal 150, then there would be a potential to overload the SEC. You could have 70 amps running through each main breaker totaling 140 amps and thereby overloading the SEC which is only intended to support a max of 100amps.

What is the actual size of the SEC? It better be at least 2/0. If it's 1/0, #1 or #2, there's a potential problem there.

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Well, I did say "very close". Marc, I used an online voltage drop calculator (see photo). Given that the voltage drop for a 2 foot 2-AWG Aluminum feeder at 75 amps is only .1% (one tenth of one percent), how much difference do you think a 4"(?) difference will make to the current carrying capacity? I agree there will always be some imbalance. Hell, even if the conductors were the same length, I would expect some other imperfections. However, I suspect in this case it would be a very small amount, and given the topic of the conversation, one that we could largely ignore.

Download Attachment: icon_photo.gif Voltagedrop.jpg

98.98?KB

VD is the wrong way to analyze it. Let say that one conductor is twice as long as the other one, just as an example. The conductors are joined on each end so they both have the same voltage across them. The current will be twice as high on the short one than the longer one because it has only half the resistance. That means 2/3 of the total current is in the short one and only 1/3 in the long one. It also means that the total ampacity of that parallel set-up is only 50% higher than the ampacity of either conductor. 150% of 75 is 112 amps which is much less than 150.

....

Marc

Marc, I'm not an electrical engineer, but I have to call BS on that.

First of all, voltage drop is due to resistance and I'm sure that it corresponds nicely with the current carrying capacity.

Next, let's say I punch two holes exactly the same size in a bucket of water and then attach two hose of the same diameter, but one say 4' long and the other 2'. There is some resistance in the hoses and the 4' long one will have twice as much of that, but it's not the same as pinching one hose to half the cross section. You are effectively saying that just having double the length of hose will shut down the flow by half. That's not true and I guarantee you that you would have a hard time actually measuring the difference in flow rate from the bucket.

Conductors act somewhat similarly to the hoses. Doubling the length may double the resistance but that initial resistance is a percentage of the capacity of the wire (based on the other variables as well). In my sample, with those loads and lengths, it would seem that the 2' wire would retain 99.9% of its capacity while the 4' one is reduced to 99.8%. That is an awful long way from your two third/one thirds.

I know I'm simplifying things a bit, but I'll bet you a case of beer that I'm a lot closer to the facts than you are. Where's that Doug Hanson?

I=E/R where I is the current, E is the voltage and R is the resistance. There's no percentage value in that formula.

Remember, I is a indianeer. I is a smart man.

Marc

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I=E/R where I is the current, E is the voltage and R is the resistance. There's no percentage value in that formula.

Yes Marc, we know about Ohm's Law, but you seem to have overlooked the critical fact that the resistance needed to get 75 amps in a 240 volt circuit lies mostly (and I do mean mostly!) elsewhere.

I am going to use stranded 2AWG copper conductor for the calculations and we will assume the breakers trip at exactly 75-amps.

On a 240V circuit at 75-amps, the total resistance would be 3.2 ohms;. R=3.2=240/75. Correct?

From http://www.stealth316.com/2-wire-resistance.htm we find that at 75 amps the resistance of 2 feet of the wire is 0.0003 ohms;. 4 feet doubles the resistance to 0.0006 ohms;. So, to maintain the equation at 75 amps, the resistance of the entire circuit, not including those two lengths of wire, would need to be 3.1997 ohms; and 3.1994 ohms; respectively.

R=3.2=(3.1997+0.0003)=240/75 and R=3.2=(3.1994+0.0006)=240/75

Still following?

Or, to put it another way, the available current before tripping the breakers, again not including those two lengths of wire, is (roughly) 74.99 and 74.98 amps.

The difference caused by the conductor lengths at the panel is a minuscule PERCENTAGE that could have, should have, been largely ignored in providing a simple answer to the initial question on that panel's main breaker set-up. That being 150-amps.

Remember, I is a indianeer. I is a smart man.

Really?

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This has strayed far from the realm of useful information.

Marc, you're flogging minutiae that might matter in a mathematical equation, but meaningless for HI's. The OP doesn't need Peck-ism's.

I'd call it a 150 amp service.

It really doesn't matter within the scope of this thing we do. What's important is grounding, bonding, and everything else that keeps things from burning down or folks getting electrocuted.

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Yes Marc, we know about Ohm's Law

Do you?

...but you seem to have overlooked the critical fact that the resistance needed to get 75 amps in a 240 volt circuit lies mostly (and I do mean mostly!) elsewhere.

That is not relevant at all to the question at hand. The question regards how a current in the SEC's will divide between the two parallel conductors.

I am going to use stranded 2AWG copper conductor for the calculations and we will assume the breakers trip at exactly 75-amps.

On a 240V circuit at 75-amps, the total resistance would be 3.2 ohms;. R=3.2=240/75. Correct?

From http://www.stealth316.com/2-wire-resistance.htm we find that at 75 amps the resistance of 2 feet of the wire is 0.0003 ohms;. 4 feet doubles the resistance to 0.0006 ohms;. So, to maintain the equation at 75 amps, the resistance of the entire circuit, not including those two lengths of wire, would need to be 3.1997 ohms; and 3.1994 ohms; respectively.

R=3.2=(3.1997+0.0003)=240/75 and R=3.2=(3.1994+0.0006)=240/75

Still following?

Or, to put it another way, the available current before tripping the breakers, again not including those two lengths of wire, is (roughly) 74.99 and 74.98 amps.

The difference caused by the conductor lengths at the panel is a minuscule PERCENTAGE that could have, should have, been largely ignored in providing a simple answer to the initial question on that panel's main breaker set-up. That being 150-amps.

I follow everything you've said. It's just not relevant.

Remember, I is a indianeer. I is a smart man.

Really?

Marc

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This has strayed far from the realm of useful information.

Marc, you're flogging minutiae that might matter in a mathematical equation, but meaningless for HI's. The OP doesn't need Peck-ism's.

I'd call it a 150 amp service.

It really doesn't matter within the scope of this thing we do. What's important is grounding, bonding, and everything else that keeps things from burning down or folks getting electrocuted.

I'll knock it off now though I disagree with you Kurt. Wide discrepancies in the length of SECs in parallel can definitely alter the total ampacity of the connection from that of matching lengths. Also, magnetic linkage between individual conductors involving parallel configurations can alter currents in each of them to varying degrees. It's called current transformation.

Now I'm done with this thread.

Marc

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That's a good thing, but I'm not quite.......

"Wide discrepancies" has nothing to do with what the OP was talking about, nor does it matter in this case. Extrapolating to engineer-ese doesn't do anything to answer the question. Understand the difference between useful information and arguing technical fine points to be "right".

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Marc does have a point here about parallel conductors. If a parallel installation looks too neat, there IS something wrong. In order to get the conductors the same length, it will end up pretty ugly.

However, that isn't what we have here. Marc, your numbers would only make sense if the total length of those two pairs of conductors were so different from each other all the way back to the utility transformer. That isn't the case. One of those 2-pole breakers sees X amount of resistance back to the transformer, and the other one sees 99.99% that much resistance. It isn't enough to make a difference. Factory-made quad breaker setups always have these slight differences in conductor length. They wouldn't be listed and labeled if it made a difference.

The amount of VD within those final lengths of service conductors are indeed going to be very different; they just still don't amount to nearly enough in such a short distance to have any consequential effect.

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