Sizing service entrance conductors

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patricknola

Electrical contractor/Generac Dealer
Location
new orleans, La, United States
Occupation
Electrical contractor.
Hey Guys, I 'm calculating the conductor size for a new service drop estimate. The customer wants a 200 amp service, it's a 120/240 volt system. So it's been a second since I've done this calculation. So the problem I'm having is that I thought if it is a 200 amp service I would divide 200 by 2 since there are 2 hot wires coming from utility. Each ungrounded conductor from the service drop would have to be sized to carry 83% of 100 amps times 2 for the two ungrounded conductors from the utility. I've gone online and used a conductor calculator to check my calculations but when I enter the parameters the calculator is telling me I need to fun 4/0 wire when i thought I would need to run #1 wire( 2 times for 2 ungrounded conductors). Thanks in advance.
 
Hey Guys, I 'm calculating the conductor size for a new service drop estimate. The customer wants a 200 amp service, it's a 120/240 volt system. So it's been a second since I've done this calculation. So the problem I'm having is that I thought if it is a 200 amp service I would divide 200 by 2 since there are 2 hot wires coming from utility. Each ungrounded conductor from the service drop would have to be sized to carry 83% of 100 amps times 2 for the two ungrounded conductors from the utility. I've gone online and used a conductor calculator to check my calculations but when I enter the parameters the calculator is telling me I need to fun 4/0 wire when i thought I would need to run #1 wire( 2 times for 2 ungrounded conductors). Thanks in advance.

83% of 200 amps... 2/0 copper or 4/0 aluminum
 
Look at it as 200 amps at 240v (2 hot wires) or 200 x 240 = 48000 Watts

If you use a dp 100 amp breaker you would have 100 x 240 v = 24000 watts. Each pole of the breaker can handle the amperage shown. A dp 60 amp breaker can handle 60 amps on each leg.
 
I am state licensed. It's been a second since I've done this and I thought I was thought you take total amps and divide it by 2 since the total amp load is divided amongst the 2 ungrounded hot wires coming in.
If you were given 24000 watts, all composed of 120V loads, you can take the current of 200A and assume that the 120V loads are equally distributed between L1-N and L2-N. If you were told that the load consists of 200A at 240V (all L1-L2) then you would have 48,000 watts instead and you would have 200A on L1 and 200A on L2.
When working out panel loading, a standard technique is to take the wattage of a 240V load and assign half that wattage to L1 and half the wattage to L2. Then, after adding up all of the wattages on, say, L1, you divide that total wattage by 120V to get the current on L1.
 
If you were given 24000 watts, all composed of 120V loads, you can take the current of 200A and assume that the 120V loads are equally distributed between L1-N and L2-N. If you were told that the load consists of 200A at 240V (all L1-L2) then you would have 48,000 watts instead and you would have 200A on L1 and 200A on L2.
When working out panel loading, a standard technique is to take the wattage of a 240V load and assign half that wattage to L1 and half the wattage to L2. Then, after adding up all of the wattages on, say, L1, you divide that total wattage by 120V to get the current on L1.
So are you telling me i was right to rate each of the 2 service conductors at 100 amps each? Because I'm being told to rate each service conductor at 200 amps each. Big price difference.
 
So are you telling me i was right to rate each of the 2 service conductors at 100 amps each? Because I'm being told to rate each service conductor at 200 amps each. Big price difference.


No that is not what he is saying. The wires must be sized for 200 amps
 
So are you telling me i was right to rate each of the 2 service conductors at 100 amps each?
Absolutely not! This is an all-too common misconception. The 200 amps that leave the source on Phase A return to the source on Phase B. It is the same amps. They do not add in the fashion of "2 apples plus 2 apples equals 4 apples." The two phases are 180 degrees apart from each other. That means when Phase A is at a positive peak, Phase B is at a negative peak. That is another way of saying that the current on A returns on B.

 
Absolutely not! This is an all-too common misconception. The 200 amps that leave the source on Phase A return to the source on Phase B. It is the same amps. They do not add in the fashion of "2 apples plus 2 apples equals 4 apples." The two phases are 180 degrees apart from each other. That means when Phase A is at a positive peak, Phase B is at a negative peak. That is another way of saying that the current on A returns on B.

Thank you for the explanation.
 
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