Voltage drop calculation

[Ivan Young]

Thank you George, point well taken
Volta, thanks for your explanation. You are saying that in a continuous because it is going to run for more than three hours more heat is going to be generated, so in order th keep the wire from overheating and damaging the insulation. Also if you install a 40 amp breaker you cannot size your wire to 30 amps but to the increased amperage of 40 amps. How am I doing so far?
Ivan

 

[Volta]

Thank you George, point well taken
Volta, thanks for your explanation. You are saying that in a continuous because it is going to run for more than three hours more heat is going to be generated, so in order th keep the wire from overheating and damaging the insulation. Also if you install a 40 amp breaker you cannot size your wire to 30 amps but to the increased amperage of 40 amps. How am I doing so far?
Ivan

Right on with the 40 amp OCPD, meaning the wire has to be rated at least the larger of: (30x1.25 = 37.5) or (35.5 amps per 240.4(B) and 220.5(B) ).

But the heat is being generated in the OCPD more so than the wire, as far as I know. Or at least, the wire should draw some of the heat away from the device.

 

[Ivan Young]

Thank you volta:bye:

 

[Volta]

Thank you volta:bye:

Sure thing!

 

[ActionDave]

Thank you George, point well taken
Volta, thanks for your explanation. You are saying that in a continuous because it is going to run for more than three hours more heat is going to be generated, so in order th keep the wire from overheating and damaging the insulation. Also if you install a 40 amp breaker you cannot size your wire to 30 amps but to the increased amperage of 40 amps. How am I doing so far?
Ivan

Your doing good, just remember that there are to exceptions to the rule that the wire ampacity needs to match the breaker.

 

[kwired]

Right on with the 40 amp OCPD, meaning the wire has to be rated at least the larger of: (30x1.25 = 37.5) or (35.5 amps per 240.4(B) and 220.5(B) ).

But the heat is being generated in the OCPD more so than the wire, as far as I know. Or at least, the wire should draw some of the heat away from the device.

That is exactly why 125% ampacity is used for sizing conductors with continuous load. The conductor is a heat sink for the overcurrent device. NEC states that for 100% rated devices you do not have to increase conductor size. This is because the overcurrent device has some other way to deal with the heat, or in the case of electronic trip devices there is not the same heating effects as there is in a thermal trip device.

 

[PaulWDent]

The wire is not just sized to the load. It is sized to the breaker too

The wire is not just sized to the load. It is sized to the breaker too

First off if the circuit protection is 30 amps then the load must be less than 30. If it is a continuous load then the largest connected load on the circuit is 24 amps. 24*1.25= 30 amps. The wire must be sized to the load. So if it is a 24 amp load then you need #10 wire.

First the breaker has to be sized to greater or equal to 125% of the expected load current to avoid running too close to its tripping point.

The wire ampacity has to be greater than the greater of:

(i) the expected load current, multiplied by 125% if that load lasts more than three hours; and

(ii) the next standard breaker size down on the one selected

The reason the wire has to be sized to the breaker and not just the load is that the normal load current is irrelevant if you have a fault condition

 

[kwired]

First the breaker has to be sized to greater or equal to 125% of the expected load current to avoid running too close to its tripping point.

The wire ampacity has to be greater than the greater of:

(i) the expected load current, multiplied by 125% if that load lasts more than three hours; and

(ii) the next standard breaker size down on the one selected

The reason the wire has to be sized to the breaker and not just the load is that the normal load current is irrelevant if you have a fault condition

Not entirely true. For some loads, especially inductive loads, the overcurrent device can be allowed to be a much higher rating then it would be for same amount of resistive load. The breaker will protect against short circuits and ground faults just fine at the higher level, overload protection is still required and is normally provided by a separate device that does not need placed at the point of origin of the circuit.