increased wire size now trips cct brkr?

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switchleg45

Member
Location
Ft Myers,FL
Now I have another conundrum. My neighbor friend is a pool repair guy and he installed a couple of pool heaters at two different residences in which the pool panel originally only had the 6A, 240V pump motor and a 12V, 300W transformer powering the 12V,100W pool light and was fed with a #10-3 NM cable. Now a 38A, 240V pool heater is installed. Both..-same builder, same house. He asked my opinion and of course I said the feeders had to be up-sized to at least #8. So..that is what we agreed to and did..replaced the #10-3 with #8-3. Now his 2P40A QO circuit breaker in the pool panel trips each time the motor is running and the heater kicks on. ?:?
He was actually running the same set up on the #10's for about the previous 48 hours. Now that I upsized the conductors the OCPD decides to trip. ?? All I can think of is that possibly the #10 was absorbing some of the in-rush power?..acting as some sort of 'buffer'? and dissipating it as heat?
 

iwire

Moderator
Staff member
Location
Massachusetts
You are on the right track, the impedance in the smaller conductors was causing voltage drop reducing the current the heater draws and reducing motor start up current.

That said, 8 AWG with a 40 for a panel supplying a 38 amp heater and a 6 amp motor?

Why would it not trip?
 
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augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
Forgive me for answering your question, with questions but my curiosity has me......:D

(a) FL allows a pool sub-panel to be fed by NM cable ?
(b) Do I read correctly.. a 38 amp pool heater, a 6 amp pool pump and the light are to be fed by the 40 amp feeder ?

(others posted as I was pondering :) )
 

G._S._Ohm

Senior Member
Location
DC area
He was actually running the same set up on the #10's for about the previous 48 hours. Now that I upsized the conductors the OCPD decides to trip. ??
Knowing the distance you can calc. the difference in impedance for each wire size but I think it's doubtful that just this wire change caused this.

"A Post Hoc is a fallacy with the following form:
  1. A occurs before B.
  2. Therefore A is the cause of B."
For instance, the breaker may now be tripping prematurely or failed to trip earlier when it should have.

Make some measurements. This change may be revealing a problem that was always there but never came up.
 
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jmellc

Senior Member
Location
Durham, NC
Occupation
Facility Maintenance Tech. Licensed Electrician
You are on the right track, the impedance in the smaller conductors was causing voltage drop reducing the current the heater draws and reducing motor start up current.


Doesn't Ohm's Law tell us that with decreased voltage, we have increased current? Seems with a voltage drop, the loads would pull more current. I'd like to see the #10 conductor & see if it shows any burning.
 

jumper

Senior Member
Doesn't Ohm's Law tell us that with decreased voltage, we have increased current? Seems with a voltage drop, the loads would pull more current. I'd like to see the #10 conductor & see if it shows any burning.

For a resistive load like a heater, amperage decreases if voltage decreases.

For an inductive load like a motor, amperage increases if voltage decreases.
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
For a resistive load like a heater, amperage decreases if voltage decreases.

For an inductive load like a motor, amperage increases if voltage decreases.

Electric motor current increases because of slower rotor speed and less REMF when voltage is decreased. Other inductive loads (without moving parts) will have the voltage and current rise and fall together similar to a purely resistive load.
 

iwire

Moderator
Staff member
Location
Massachusetts
Doesn't Ohm's Law tell us that with decreased voltage, we have increased current? Seems with a voltage drop, the loads would pull more current. I'd like to see the #10 conductor & see if it shows any burning.

Yes and no.

With a restive load like a heater less voltage will allays result in less current.

With a motor load it is less cut and dry. Often with motors (depending on the mechanical load) the motor will draw less current with less voltage.

That is the function of a reduced voltage motor starter, it reduces the max inrush current but it also makes the motor take longer to come up to full speed.
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
Yes and no.

With a restive load like a heater less voltage will allays result in less current.

With a motor load it is less cut and dry. Often with motors (depending on the mechanical load) the motor will draw less current with less voltage.

That is the function of a reduced voltage motor starter, it reduces the max inrush current but it also makes the motor take longer to come up to full speed.

I would like to know where the myth about inductive loads acting the opposite of resistive loads with varying voltage came from.

DC motors will draw less current with less voltage. At least the ones I checked (automotive related) always did.

AC motors will have current based upon the difference of the 'rotating' magnetic fields in the stator and the rotation of the rotor (and it's magnetic fields) which is called rotor slip and the results of the interaction of the two.

In essence, part of an AC motor acts like a generator and provides REMF back to the system resulting in a lower current when the rotor slip is low. If the rotor slows down for any reason (increased load OR decreased voltage), slip increases, less REMF is produced and the current goes up in the conductors.

In the last year of my apprenticeship we had an excellent video about how the above worked. It showed current and voltage flow and amount as it frolicked around the rotor and stator. Being able to 'see' how they intertwined made it much easier to understand. It also makes it easier to understand why we need VFD's to control motor speed instead of just resistors.

A DC motor will also have an REMF effect, but it's not as pronounced. The motor does not have to stay at a certain RPM or rely on a certain voltage to make full use of it. We see the effects of REMF on DC motors as in rush current, not so much with reduced supply voltage. Blower motors for auto heaters are voltage controlled using large resistors to cut the voltage. The blower resistors used to be located under the hood and needed replacement on occasion. Reduced REMF in a heavily loaded DC motor will also occur, but measurements will show that a decrease in voltage is not the cause.
 
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ramsy

Roger Ruhle dba NoFixNoPay
Location
LA basin, CA
Occupation
Service Electrician 2020 NEC
..Other inductive loads (without moving parts) will have the voltage and current rise and fall together similar to a purely resistive load.
Why does NEC 220.18(B) require the current nameplate for inductive-lighting loads, rather than watts?
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
Why does NEC 220.18(B) require the current nameplate for inductive-lighting loads, rather than watts?

Because OCPDs have to be sized taking current into consideration.

I don't see what that has to do with my statement.

What happens to the current in an inductive lighting load when the voltage drops? Remember, the impedance of the inductive portion of the circuit does not change with voltage. It's dependent on the frequency and the value of the inductor.

If you do the math, you will see that in such a circuit, a decrease in voltage will result in a decrease in current.
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
Why does NEC 220.18(B) require the current nameplate for inductive-lighting loads, rather than watts?

Also, nameplate current is an input value, the value the breaker will (should) see. Watts are an output value and the OCPD really doesn't care about that.

For instance. A 1000 watt MH ballast lists 9.5 amps at 120 VAC. 1000 watts at 120 volts is only 8.3 amps. The circuit has to be sized using the input value, not the output value.

My radio transmits 100 watts, peak. Yet it needs nearly twice that coming in to operate at that level. At 12 volts, it needs 20 amps to make a full 100 watt (vs output at 8.3 at 12 just for math purpose) signal.

I have both an ammeter and a voltmeter on the power supply for the radio. The current and amperage drop and rise in unison so long as the load does not change.
 

kwired

Electron manager
Location
NE Nebraska
Why does NEC 220.18(B) require the current nameplate for inductive-lighting loads, rather than watts?

I say for a couple reasons.

1. these types of lighting are usually rated by lamp wattage and not input current.
input current is usually higher than just dividing lamp wattage by input voltage, you often have ballast losses on top of the lamp watts.

2. they are inductive and power factor needs consideration also.
 

kwired

Electron manager
Location
NE Nebraska
I think the statement that for motors the current goes up if the voltage goes down is primarily applicable to AC induction motors and not all motors in general.

An AC motor has a synchronous speed based primarily on number of poles in the winding and input frequency. The motor is constantly trying to achieve synchronous speed, but when you reduce voltage you reduce torque and increase slip, the motor will draw more current to try to overcome this and attempt to attain synchronous speed.

Reduced voltage starting will lower locked rotor current when first energizing, but will not provide enough torque to accelerate load if voltage remains too low and will lead to higher running and accelerating current if it remains too low.

Edit: I probably should have said, reduced voltage starting will lower inrush current upon energizing and not locked rotor current.
 
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ramsy

Roger Ruhle dba NoFixNoPay
Location
LA basin, CA
Occupation
Service Electrician 2020 NEC
You are on the right track, the impedance in the smaller conductors was causing voltage drop..
If the concept of insulation testing after a wire pull is not well understood, perhaps they're not ready for breaker-trip curves affected by wire impedance.
 

ramsy

Roger Ruhle dba NoFixNoPay
Location
LA basin, CA
Occupation
Service Electrician 2020 NEC
I say for a couple reasons.

1. these types of lighting are usually rated by lamp wattage and not input current.
input current is usually higher than just dividing lamp wattage by input voltage, you often have ballast losses on top of the lamp watts.

2. they are inductive and power factor needs consideration also.
Thanks
 

mivey

Senior Member
I would like to know where the myth about inductive loads acting the opposite of resistive loads with varying voltage came from.
I'll tell you how utilities model distribution systems. One way is by using "ZIP" load models that consist of different percentages of:

constant impedance Z: where I changes linearly with V but power varies with the square of the voltage
constant current I: where I does not change with V but power varies linearly with the voltage
constant power PQ or kVA: where I changes inversely with V but power is constant

Resistive type load are modeled as constant impedances. Loads with electric motors, regulated power supplies, and loads beyond tap-changing transformers, etc. are modeled as constant power loads. Real constant current loads are rare and are usually industrial like some welders, electroplating, etc. but the constant current models serve as models for an approximate even mixture of Z & P loads.

Industrial loads are mostly P with some Z (~ 3:1 to 4:1). Residential with a lot of A/C is more P than Z because of the motor load (~ 2:1). Residential without a lot of A/C is more Z than P (~ 2:1 but even up to 3:1 or 4:1 in old areas with little motor load). Commerial runs from 50/50 P/Z or even higher P/Z ratios but usually not nearly as high as the industrials.

So, from a different perspective, not so much a myth as you might think.
 
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