Continuous-duty circuit

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Samuel87

Member
Location
Fresno CA
Occupation
Electrician
Hi there.

You are employed in a large industrial plant. A 480-V, 5000-W heater is used to melt lead in a large tank. It has been decided that the heater is not sufficient to raise the temperature of the lead to the desired level. A second 5000-W heater is to be installed on the same circuit.

What will be the circuit current after installation of the second heater, and what is the minimum size circuit breaker that can be used if this is a continuous-duty circuit?

Is this well done?

when do we use the 80%?

I am confused with that

Thanks.

IMG_1D94990A8AC7-1.jpeg
 

Dsg319

Senior Member
Location
West Virginia
Occupation
Wv Master “lectrician”
Hi there.

You are employed in a large industrial plant. A 480-V, 5000-W heater is used to melt lead in a large tank. It has been decided that the heater is not sufficient to raise the temperature of the lead to the desired level. A second 5000-W heater is to be installed on the same circuit.

What will be the circuit current after installation of the second heater, and what is the minimum size circuit breaker that can be used if this is a continuous-duty circuit?

Is this well done?

when do we use the 80%?

I am confused with that

Thanks.

View attachment 2565592
Looks good to be. #10s and 30amp OCPD.

For this application it’s simpler to use the 125% instead of 80% which are basically the same.

Take your load current that is continuous and time by 125%
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
When do we use 80% instead 125%?
They are reciprocals. Which to use depends on the constant (known value) vs the variable (unknown value). Use whichever makes the math accurate and/or easier for you.

To say a circuit must be sized to carry 125% of a continuous load and to say that a continuous load must not exceed 80% of a circuit's capacity are saying the same thing.
 

Dsg319

Senior Member
Location
West Virginia
Occupation
Wv Master “lectrician”
When do we use 80% instead 125%?
As Larry stated they will yield the same result.

20amp load x (continuous load)125%=25amps. 25amp wire and ocpd needed

If I know alls I have is a 25amp ocpd and 25amp wire then.

25amp x .80= 20 ( I can have a continuous load up to 20 amperes)

To keep it simple and easy to think about just say you have a 20amp breaker you can only load it to 80% if it’s a continuous load.which is 16 amperes.

If you have a continuous load of 16 amps to size the circuit you times by 125%

They reverse each other.
 

tthh

Senior Member
Location
Denver
Occupation
Retired Engineer
I find the easiest way to think of it like this...

If you know the continuous current needed, multiply by 125% and that is your minimum wire and breaker requirement.
On the other hand, if you know the circuit wire and breaker capacity, multiply that by 80% to get the maximum continuous current.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
As Larry stated they will yield the same result.

20amp load x (continuous load)125%=25amps. 25amp wire and ocpd needed

If I know alls I have is a 25amp ocpd and 25amp wire then.

25amp x .80= 20 ( I can have a continuous load up to 20 amperes)

To keep it simple and easy to think about just say you have a 20amp breaker you can only load it to 80% if it’s a continuous load.which is 16 amperes.

If you have a continuous load of 16 amps to size the circuit you times by 125%

They reverse each other.
It is just a matter of how you want to push the buttons on your calculator. I am probably going to do 16/.8 because that is two less key strokes than when you do 16 * 1.25
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
This is simple-enough math that you should be able to do it in your head, at least with whole numbers.
 
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